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About the Authors

Bruce A. Averill

Bruce A. Averill grew up in New England. He then received his B.S. with high honors in chemistry at Michigan State University in 1969, and his Ph.D. in inorganic chemistry at MIT in 1973. After three years as an NIH and NSF Postdoctoral Fellow at Brandeis University and the University of Wisconsin, he began his independent academic career at Michigan State University in 1976.

He was promoted in 1982, after which he moved to the University of Virginia, where he was promoted to Professor in 1988. In 1994, Dr. Averill moved to the University of Amsterdam in the Netherlands as Professor of Biochemistry. He then returned to the United States to the University of Toledo in 2001, where he was a Distinguished University Professor. He was then named a Jefferson Science Policy Fellow at the U.S. State Department, where he remained for several years as a senior energy consultant. He is currently the founder and senior partner of Stategic Energy Security Solutions, which creates public/private partnerships to ensure global energy security. Dr. Averill’s academic research interests are centered on the role of metal ions in biology. He is also an expert on cyber-security.

In his European position, Dr. Averill headed a European Union research network comprised of seven research groups from seven different European countries and a staff of approximately fifty research personnel. In addition, he was responsible for the research theme on Biocatalysis within the E. C. Slater Institute of the University of Amsterdam, which consisted of himself as head and a team of 21 professionals, ranging from associate professors to masters students at any given time.

Dr. Averill’s research has attracted a great deal of attention in the scientific community. His published work is frequently cited by other researchers, and he has been invited to give more than 100 presentations at educational and research institutions and at national and international scientific meetings. Among his numerous awards, Dr. Averill has been an Honorary Woodrow Wilson Fellow, an NSF Predoctoral Fellow, an NIH and NSF Postdoctoral Fellow, and an Alfred P. Sloan Foundation Fellow; he has also received an NSF Special Creativity Award.

Over the years, Dr. Averill has published more than 135 articles dealing with chemical, physical, and biological subjects in refereed journals, and he has also published 15 chapters in books and more than 80 abstracts from national and international meetings. In addition, he has co-edited a graduate text on catalysis, and he has taught courses at all levels, including general chemistry, biochemistry, advanced inorganic, and physical methods.

Aside from his research program, Dr. Averill is an enthusiastic sailor and an avid reader. He also enjoys traveling with his family, and at some point in the future he would like to sail around the world in a classic wooden boat.

Patricia Eldredge

Patricia Eldredge was raised in the U.S. diplomatic service, and has traveled and lived around the world. She has degrees from the Ohio State University, the University of Central Florida, the University of Virginia, and the University of North Carolina, Chapel Hill, where she obtained her Ph.D. in inorganic chemistry following several years as an analytical research chemist in industry. In addition, she has advanced offshore sailing qualifications from both the Royal Yachting Association in Britain and the American Sailing Association.

In 1989, Dr. Eldredge was named the Science Policy Fellow for the American Chemical Society. While in Washington, D.C., she examined the impact of changes in federal funding priorities on academic research funding. She was awarded a Postdoctoral Research Fellowship with Oak Ridge Associated Universities, working with the U.S. Department of Energy on heterogeneous catalysis and coal liquefaction. Subsequently, she returned to the University of Virginia as a Research Scientist and a member of the General Faculty.

In 1992, Dr. Eldredge relocated to Europe for several years. While there, she studied advanced Maritime Engineering, Materials, and Oceanography at the University of Southampton in England, arising from her keen interest in naval architecture.

Upon her return to the United States in 2002, she was a Visiting Assistant Professor and a Senior Research Scientist at the University of Toledo. Her research interests included the use of protein scaffolds to synthesize biologically relevant clusters. Dr. Eldredge has published more than a dozen articles dealing with synthetic inorganic chemistry and catalysis, including several seminal studies describing new synthetic approaches to metal-sulfur clusters. She has also been awarded a patent for her work on catalytic coal liquefaction.

Her diverse teaching experience includes courses on chemistry for the life sciences, introductory chemistry, general, organic, and analytical chemistry. When not authoring textbooks, Dr. Eldredge enjoys traveling, offshore sailing, political activism, and caring for her Havanese dogs.

Acknowledgments

The authors would like to thank the following individuals who reviewed the text and whose contributions were invaluable in shaping the product:

Dedication

To Harvey, who opened the door

and to the Virginia Tech community for its resilience and strength. We Remember.

Preface

In this new millenium, as the world faces new and extreme challenges, the importance of acquiring a solid foundation in chemical principles has become increasingly important to understand the challenges that lie ahead. Moreover, as the world becomes more integrated and interdependent, so too do the scientific disciplines. The divisions between fields such as chemistry, physics, biology, environmental sciences, geology, and materials science, among others, have become less clearly defined. The goal of this text is to address the increasing close relationship among various disciplines and to show the relevance of chemistry to contemporary issues in a pedagogically approachable manner.

Because of the enthusiasm of the majority of first-year chemistry students for biologically and medically relevant topics, this text uses an integrated approach that includes explicit discussions of biological and environmental applications of chemistry. Topics relevant to materials science are also introduced to meet the more specific needs of engineering students. To facilitate integration of such material, simple organic structures, nomenclature, and reactions are introduced very early in the text, and both organic and inorganic examples are used wherever possible. This approach emphasizes the distinctions between ionic and covalent bonding, thus enhancing the students’ chance of success in the organic chemistry course that traditionally follows general chemistry.

The overall goal is to produce a text that introduces the students to the relevance and excitement of chemistry. Although much of first-year chemistry is taught as a service course, there is no reason that the intrinsic excitement and potential of chemistry cannot be the focal point of the text and the course. We emphasize the positive aspects of chemistry and its relationship to students’ lives, which requires bringing in applications early and often. Unfortunately, one cannot assume that students in such courses today are highly motivated to study chemistry for its own sake. The explicit discussion of biological, environmental, and materials issues from a chemical perspective is intended to motivate the students and help them appreciate the relevance of chemistry to their lives. Material that has traditionally been relegated to boxes, and thus perhaps perceived as peripheral by the students, has been incorporated into the text to serve as a learning tool.

To begin the discussion of chemistry rapidly, the traditional first chapter introducing units, significant figures, conversion factors, dimensional analysis, and so on, has been reorganized. The material has been placed in the chapters where the relevant concepts are first introduced, thus providing three advantages: it eliminates the tedium of the traditional approach, which introduces mathematical operations at the outset, and thus avoids the perception that chemistry is a mathematics course; it avoids the early introduction of operations such as logarithms and exponents, which are typically not encountered again for several chapters and may easily be forgotten when they are needed; and third, it provides a review for those students who have already had relatively sophisticated high school chemistry and math courses, although the sections are designed primarily for students unfamiliar with the topic.

Our specific objectives include the following:

  1. To write the text at a level suitable for science majors, but using a less formal writing style that will appeal to modern students.
  2. To produce a truly integrated text that gives the student who takes only a single year of chemistry an overview of the most important subdisciplines of chemistry, including organic, inorganic, biological, materials, environmental, and nuclear chemistry, thus emphasizing unifying concepts.
  3. To introduce fundamental concepts in the first two-thirds of the chapter, then applications relevant to the health sciences or engineers. This provides a flexible text that can be tailored to the specific needs and interests of the audience.
  4. To ensure the accuracy of the material presented, which is enhanced by the author’s breadth of professional experience and experience as active chemical researchers.
  5. To produce a spare, clean, uncluttered text that is less distracting to the student, where each piece of art serves as a pedagogical device.
  6. To introduce the distinction between ionic and covalent bonding and reactions early in the text, and to continue to build on this foundation in the subsequent discussion, while emphasizing the relationship between structure and reactivity.
  7. To utilize established pedagogical devices to maximize students’ ability to learn directly from the text. These include copious worked examples in the text, problem-solving strategies, and similar unworked exercises with solutions. End-of-chapter problems are designed to ensure that students have grasped major concepts in addition to testing their ability to solve numerical, problems. Problems emphasizing applications are drawn from many disciplines.
  8. To emphasize an intuitive and predictive approach to problem solving that relies on a thorough understanding of key concepts and recognition of important patterns rather than on memorization. Many patterns are indicated throughout the text as notes in the margin.

The text is organized by units that discuss introductory concepts, atomic and molecular structure, the states of matter, kinetics and equilibria, and descriptive inorganic chemistry. The text breaks the traditional chapter on liquids and solids into two to expand the coverage of important and topics such as semiconductors and superconductors, polymers, and engineering materials.

In summary, this text represents a step in the evolution of the general chemistry text toward one that reflects the increasing overlap between chemistry and other disciplines. Most importantly, the text discusses exciting and relevant aspects of biological, environmental, and materials science that are usually relegated to the last few chapters, and it provides a format that allows the instructor to tailor the emphasis to the needs of the class. By the end of , the student will have already been introduced to environmental topics such as acid rain, the ozone layer, and periodic extinctions, and to biological topics such as antibiotics and the caloric content of foods. Nonetheless, the new material is presented in such a way as to minimally perturb the traditional sequence of topics in a first-year course, making the adaptation easier for instructors.

Chapter 1 Introduction to Chemistry

As you begin your study of college chemistry, those of you who do not intend to become professional chemists may well wonder why you need to study chemistry. You will soon discover that a basic understanding of chemistry is useful in a wide range of disciplines and career paths. You will also discover that an understanding of chemistry helps you make informed decisions about many issues that affect you, your community, and your world. A major goal of this text is to demonstrate the importance of chemistry in your daily life and in our collective understanding of both the physical world we occupy and the biological realm of which we are a part. The objectives of this chapter are twofold: (1) to introduce the breadth, the importance, and some of the challenges of modern chemistry and (2) to present some of the fundamental concepts and definitions you will need to understand how chemists think and work.

An atomic corral for electrons. A corral of 48 iron atoms (yellow-orange) on a smooth copper surface (cyan-purple) confines the electrons on the surface of the copper, producing a pattern of “ripples” in the distribution of the electrons. Scientists assembled the 713-picometer-diameter corral by individually positioning iron atoms with the tip of a scanning tunneling microscope. (Note that 1 picometer is equivalent to 1 × 10-12 meters.)

1.1 Chemistry in the Modern World

Learning Objective

  1. To recognize the breadth, depth, and scope of chemistry.

ChemistryThe study of matter and the changes that material substances undergo. is the study of matter and the changes that material substances undergo. Of all the scientific disciplines, it is perhaps the most extensively connected to other fields of study. Geologists who want to locate new mineral or oil deposits use chemical techniques to analyze and identify rock samples. Oceanographers use chemistry to track ocean currents, determine the flux of nutrients into the sea, and measure the rate of exchange of nutrients between ocean layers. Engineers consider the relationships between the structures and the properties of substances when they specify materials for various uses. Physicists take advantage of the properties of substances to detect new subatomic particles. Astronomers use chemical signatures to determine the age and distance of stars and thus answer questions about how stars form and how old the universe is. The entire subject of environmental science depends on chemistry to explain the origin and impacts of phenomena such as air pollution, ozone layer depletion, and global warming.

The disciplines that focus on living organisms and their interactions with the physical world rely heavily on biochemistryThe application of chemistry to the study of biological processes., the application of chemistry to the study of biological processes. A living cell contains a large collection of complex molecules that carry out thousands of chemical reactions, including those that are necessary for the cell to reproduce. Biological phenomena such as vision, taste, smell, and movement result from numerous chemical reactions. Fields such as medicine, pharmacology, nutrition, and toxicology focus specifically on how the chemical substances that enter our bodies interact with the chemical components of the body to maintain our health and well-being. For example, in the specialized area of sports medicine, a knowledge of chemistry is needed to understand why muscles get sore after exercise as well as how prolonged exercise produces the euphoric feeling known as “runner’s high.”

Examples of the practical applications of chemistry are everywhere (). Engineers need to understand the chemical properties of the substances when designing biologically compatible implants for joint replacements or designing roads, bridges, buildings, and nuclear reactors that do not collapse because of weakened structural materials such as steel and cement. Archaeology and paleontology rely on chemical techniques to date bones and artifacts and identify their origins. Although law is not normally considered a field related to chemistry, forensic scientists use chemical methods to analyze blood, fibers, and other evidence as they investigate crimes. In particular, DNA matching—comparing biological samples of genetic material to see whether they could have come from the same person—has been used to solve many high-profile criminal cases as well as clear innocent people who have been wrongly accused or convicted. Forensics is a rapidly growing area of applied chemistry. In addition, the proliferation of chemical and biochemical innovations in industry is producing rapid growth in the area of patent law. Ultimately, the dispersal of information in all the fields in which chemistry plays a part requires experts who are able to explain complex chemical issues to the public through television, print journalism, the Internet, and popular books.

Figure 1.1 Chemistry in Everyday Life

Although most people do not recognize it, chemistry and chemical compounds are crucial ingredients in almost everything we eat, wear, and use.

By this point, it shouldn’t surprise you to learn that chemistry was essential in explaining a pivotal event in the history of Earth: the disappearance of the dinosaurs. Although dinosaurs ruled Earth for more than 150 million years, fossil evidence suggests that they became extinct rather abruptly approximately 66 million years ago. Proposed explanations for their extinction have ranged from an epidemic caused by some deadly microbe or virus to more gradual phenomena such as massive climate changes. In 1978 Luis Alvarez (a Nobel Prize–winning physicist), the geologist Walter Alvarez (Luis’s son), and their coworkers discovered a thin layer of sedimentary rock formed 66 million years ago that contained unusually high concentrations of iridium, a rather rare metal (part (a) in ). This layer was deposited at about the time dinosaurs disappeared from the fossil record. Although iridium is very rare in most rocks, accounting for only 0.0000001% of Earth’s crust, it is much more abundant in comets and asteroids. Because corresponding samples of rocks at sites in Italy and Denmark contained high iridium concentrations, the Alvarezes suggested that the impact of a large asteroid with Earth led to the extinction of the dinosaurs. When chemists analyzed additional samples of 66-million-year-old sediments from sites around the world, all were found to contain high levels of iridium. In addition, small grains of quartz in most of the iridium-containing layers exhibit microscopic cracks characteristic of high-intensity shock waves (part (b) in ). These grains apparently originated from terrestrial rocks at the impact site, which were pulverized on impact and blasted into the upper atmosphere before they settled out all over the world.

Figure 1.2 Evidence for the Asteroid Impact That May Have Caused the Extinction of the Dinosaurs

(a) Luis and Walter Alvarez are standing in front of a rock formation in Italy that shows the thin white layer of iridium-rich clay deposited at the time the dinosaurs became extinct. The concentration of iridium is 30 times higher in this layer than in the rocks immediately above and below it. There are no significant differences between the clay layer and the surrounding rocks in the concentrations of any of the 28 other elements examined. (b) Microphotographs of an unshocked quartz grain (left) and a quartz grain from the iridium-rich layer exhibiting microscopic cracks resulting from shock (right).

Scientists calculate that a collision of Earth with a stony asteroid about 10 kilometers (6 miles) in diameter, traveling at 25 kilometers per second (about 56,000 miles per hour), would almost instantaneously release energy equivalent to the explosion of about 100 million megatons of TNT (trinitrotoluene). This is more energy than that stored in the entire nuclear arsenal of the world. The energy released by such an impact would set fire to vast areas of forest, and the smoke from the fires and the dust created by the impact would block the sunlight for months or years, eventually killing virtually all green plants and most organisms that depend on them. This could explain why about 70% of all species—not just dinosaurs—disappeared at the same time. Scientists also calculate that this impact would form a crater at least 125 kilometers (78 miles) in diameter. Recently, satellite images from a Space Shuttle mission confirmed that a huge asteroid or comet crashed into Earth’s surface across the Yucatan’s northern tip in the Gulf of Mexico 65 million years ago, leaving a partially submerged crater 180 kilometers (112 miles) in diameter (). Thus simple chemical measurements of the abundance of one element in rocks led to a new and dramatic explanation for the extinction of the dinosaurs. Though still controversial, this explanation is supported by additional evidence, much of it chemical.

Figure 1.3 Asteroid Impact

The location of the asteroid impact crater near what is now the tip of the Yucatan Peninsula in Mexico.

This is only one example of how chemistry has been applied to an important scientific problem. Other chemical applications and explanations that we will discuss in this text include how astronomers determine the distance of galaxies and how fish can survive in subfreezing water under polar ice sheets. We will also consider ways in which chemistry affects our daily lives: the addition of iodine to table salt; the development of more effective drugs to treat diseases such as cancer, AIDS (acquired immunodeficiency syndrome), and arthritis; the retooling of industry to use nonchlorine-containing refrigerants, propellants, and other chemicals to preserve Earth’s ozone layer; the use of modern materials in engineering; current efforts to control the problems of acid rain and global warming; and the awareness that our bodies require small amounts of some chemical substances that are toxic when ingested in larger doses. By the time you finish this text, you will be able to discuss these kinds of topics knowledgeably, either as a beginning scientist who intends to spend your career studying such problems or as an informed observer who is able to participate in public debates that will certainly arise as society grapples with scientific issues.

Summary

Chemistry is the study of matter and the changes material substances undergo. It is essential for understanding much of the natural world and central to many other scientific disciplines, including astronomy, geology, paleontology, biology, and medicine.

Key Takeaway

  • An understanding of chemistry is essential for understanding much of the natural world and is central to many other disciplines.

1.2 The Scientific Method

Learning Objective

  1. To identify the components of the scientific method.

Scientists search for answers to questions and solutions to problems by using a procedure called the scientific methodThe procedure that scientists use to search for answers to questions and solutions to problems.. This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles ().

Figure 1.4 The Scientific Method

As depicted in this flowchart, the scientific method consists of making observations, formulating hypotheses, and designing experiments. A scientist may enter the cycle at any point.

Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: the outside air temperature is cooler during the winter season, table salt is a crystalline solid, sulfur crystals are yellow, and dissolving a penny in dilute nitric acid forms a blue solution and a brown gas. Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: the melting point of crystalline sulfur is 115.21 degrees Celsius, and 35.9 grams of table salt—whose chemical name is sodium chloride—dissolve in 100 grams of water at 20 degrees Celsius. For the question of the dinosaurs’ extinction, the initial observation was quantitative: iridium concentrations in sediments dating to 66 million years ago were 20–160 times higher than normal.

After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a hypothesisA tentative explanation for scientific observations that puts the system being studied into a form that can be tested., a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. For example, the observation that we experience alternating periods of light and darkness corresponding to observed movements of the sun, moon, clouds, and shadows is consistent with either of two hypotheses: (1) Earth rotates on its axis every 24 hours, alternately exposing one side to the sun, or (2) the sun revolves around Earth every 24 hours. Suitable experiments can be designed to choose between these two alternatives. For the disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists can collect additional data that either support or refute it.

After a hypothesis has been formed, scientists conduct experiments to test its validity. ExperimentsA systematic observation or measurement, preferably made under controlled conditions—that is, conditions in which the variable of interest is clearly distinguished from any others. are systematic observations or measurements, preferably made under controlled conditions—that is, under conditions in which a single variable changes. For example, in our extinction scenario, iridium concentrations were measured worldwide and compared. A properly designed and executed experiment enables a scientist to determine whether the original hypothesis is valid. Experiments often demonstrate that the hypothesis is incorrect or that it must be modified. More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in a lawA verbal or mathematical description of a phenomenon that allows for general predictions and says what happens, not why it happens., a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply says what happens; it does not address the question of why. One example of a law, the law of definite proportionsA chemical substance always contains the same proportions of elements by mass., which was discovered by the French scientist Joseph Proust (1754–1826), states that a chemical substance always contains the same proportions of elements by mass. Thus sodium chloride (table salt) always contains the same proportion by mass of sodium to chlorine, in this case 39.34% sodium and 60.66% chlorine by mass, and sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass.You will learn in that some solid compounds do not strictly obey the law of definite proportions. (For a review of common units of measurement, see Essential Skills 1 in .) The law of definite proportions should seem obvious—we would expect the composition of sodium chloride to be consistent—but the head of the US Patent Office did not accept it as a fact until the early 20th century.

Whereas a law states only what happens, a theoryA statement that attempts to explain why nature behaves the way it does. attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time unless a major experimental error is discovered. In contrast, a theory, by definition, is incomplete and imperfect, evolving with time to explain new facts as they are discovered. The theory developed to explain the extinction of the dinosaurs, for example, is that Earth occasionally encounters small- to medium-sized asteroids, and these encounters may have unfortunate implications for the continued existence of most species. This theory is by no means proven, but it is consistent with the bulk of evidence amassed to date. summarizes the application of the scientific method in this case.

Figure 1.5 A Summary of How the Scientific Method Was Used in Developing the Asteroid Impact Theory to Explain the Disappearance of the Dinosaurs from Earth

Example 1

Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation.

  1. Ice always floats on liquid water.
  2. Birds evolved from dinosaurs.
  3. Hot air is less dense than cold air, probably because the components of hot air are moving more rapidly.
  4. When 10 g of ice were added to 100 mL of water at 25°C, the temperature of the water decreased to 15.5°C after the ice melted.
  5. The ingredients of Ivory soap were analyzed to see whether it really is 99.44% pure, as advertised.

Given: components of the scientific method

Asked for: statement classification

Strategy:

Refer to the definitions in this section to determine which category best describes each statement.

Solution:

  1. This is a general statement of a relationship between the properties of liquid and solid water, so it is a law.
  2. This is a possible explanation for the origin of birds, so it is a hypothesis.
  3. This is a statement that tries to explain the relationship between the temperature and the density of air based on fundamental principles, so it is a theory.
  4. The temperature is measured before and after a change is made in a system, so these are quantitative observations.
  5. This is an analysis designed to test a hypothesis (in this case, the manufacturer’s claim of purity), so it is an experiment.

Exercise

Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation.

  1. Measured amounts of acid were added to a Rolaids tablet to see whether it really “consumes 47 times its weight in excess stomach acid.”
  2. Heat always flows from hot objects to cooler ones, not in the opposite direction.
  3. The universe was formed by a massive explosion that propelled matter into a vacuum.
  4. Michael Jordan is the greatest pure shooter ever to play professional basketball.
  5. Limestone is relatively insoluble in water but dissolves readily in dilute acid with the evolution of a gas.
  6. Gas mixtures that contain more than 4% hydrogen in air are potentially explosive.

Answer:

  1. experiment
  2. law
  3. theory
  4. hypothesis
  5. qualitative observation
  6. quantitative observation

Because scientists can enter the cycle shown in at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations.

It is important to remember that scientists have a tendency to formulate hypotheses in familiar terms simply because it is difficult to propose something that has never been encountered or imagined before. As a result, scientists sometimes discount or overlook unexpected findings that disagree with the basic assumptions behind the hypothesis or theory being tested. Fortunately, truly important findings are immediately subject to independent verification by scientists in other laboratories, so science is a self-correcting discipline. When the Alvarezes originally suggested that an extraterrestrial impact caused the extinction of the dinosaurs, the response was almost universal skepticism and scorn. In only 20 years, however, the persuasive nature of the evidence overcame the skepticism of many scientists, and their initial hypothesis has now evolved into a theory that has revolutionized paleontology and geology.

In , we begin our discussion of chemistry with a description of matter. This discussion is followed by a summary of some of the pioneering discoveries that led to our present understanding of the structure of the fundamental unit of chemistry: the atom.

Summary

Chemists expand their knowledge by making observations, carrying out experiments, and testing hypotheses to develop laws to summarize their results and theories to explain them. In doing so, they are using the scientific method.

Key Takeaway

  • Chemists expand their knowledge with the scientific method.

Conceptual Problems

  1. What are the three components of the scientific method? Is it necessary for an individual to conduct experiments to follow the scientific method?

  2. Identify each statement as a theory or a law and explain your reasoning.

    1. The ratio of elements in a pure substance is constant.
    2. An object appears black because it absorbs all the visible light that strikes it.
    3. Energy is neither created nor destroyed.
    4. Metals conduct electricity because their electrons are not tightly bound to a particular nucleus and are therefore free to migrate.
  3. Identify each statement as a theory or a law and explain your reasoning.

    1. A pure chemical substance contains the same proportion of elements by mass.
    2. The universe is expanding.
    3. Oppositely charged particles attract each other.
    4. Life exists on other planets.
  4. Classify each statement as a qualitative observation or a quantitative observation.

    1. Mercury and bromine are the only elements that are liquids at room temperature.
    2. An element is both malleable and ductile.
    3. The density of iron is 7.87 g/cm3.
    4. Lead absorbs sound very effectively.
    5. A meteorite contains 20% nickel by mass.
  5. Classify each statement as a quantitative observation or a qualitative observation.

    1. Nickel deficiency in rats is associated with retarded growth.
    2. Boron is a good conductor of electricity at high temperatures.
    3. There are 1.4–2.3 g of zinc in an average 70 kg adult.
    4. Certain osmium compounds found in air in concentrations as low as 10.7 µg/m3 can cause lung cancer.

Answers

    1. law
    2. theory
    3. law
    4. theory
    1. qualitative
    2. qualitative
    3. quantitative
    4. quantitative

1.3 A Description of Matter

Learning Objective

  1. To classify matter.

Chemists study the structures, physical properties, and chemical properties of material substances. These consist of matterAnything that occupies space and has mass., which is anything that occupies space and has mass. Gold and iridium are matter, as are peanuts, people, and postage stamps. Smoke, smog, and laughing gas are matter. Energy, light, and sound, however, are not matter; ideas and emotions are also not matter.

The massA fundamental property that does not depend on an object’s location; it is the quantity of matter an object contains. of an object is the quantity of matter it contains. Do not confuse an object’s mass with its weightA force caused by the gravitational attraction that operates on an object. The weight of an object depends on its location (c.f. mass)., which is a force caused by the gravitational attraction that operates on the object. Mass is a fundamental property of an object that does not depend on its location.In physical terms, the mass of an object is directly proportional to the force required to change its speed or direction. A more detailed discussion of the differences between weight and mass and the units used to measure them is included in Essential Skills 1 (). Weight, on the other hand, depends on the location of an object. An astronaut whose mass is 95 kg weighs about 210 lb on Earth but only about 35 lb on the moon because the gravitational force he or she experiences on the moon is approximately one-sixth the force experienced on Earth. For practical purposes, weight and mass are often used interchangeably in laboratories. Because the force of gravity is considered to be the same everywhere on Earth’s surface, 2.2 lb (a weight) equals 1.0 kg (a mass), regardless of the location of the laboratory on Earth.

Under normal conditions, there are three distinct states of matter: solids, liquids, and gases (). SolidsOne of three distinct states of matter that, under normal conditions, is relatively rigid and has a fixed volume. are relatively rigid and have fixed shapes and volumes. A rock, for example, is a solid. In contrast, liquidsOne of three distinct states of matter that, under normal conditons, has a fixed volume but flows to assume the shape of its container. have fixed volumes but flow to assume the shape of their containers, such as a beverage in a can. GasesOne of three distinct states of matter that, under normal conditions, has neither a fixed shape nor a fixed volume and expands to completely fill its container., such as air in an automobile tire, have neither fixed shapes nor fixed volumes and expand to completely fill their containers. Whereas the volume of gases strongly depends on their temperature and pressureThe amount of force exerted on a given area. (the amount of force exerted on a given area), the volumes of liquids and solids are virtually independent of temperature and pressure. Matter can often change from one physical state to another in a process called a physical changeA change of state that does not affect the chemical composition of a substance.. For example, liquid water can be heated to form a gas called steam, or steam can be cooled to form liquid water. However, such changes of state do not affect the chemical composition of the substance.

Figure 1.6 The Three States of Matter

Solids have a defined shape and volume. Liquids have a fixed volume but flow to assume the shape of their containers. Gases completely fill their containers, regardless of volume.

Pure Substances and Mixtures

A pure chemical substance is any matter that has a fixed chemical composition and characteristic properties. Oxygen, for example, is a pure chemical substance that is a colorless, odorless gas at 25°C. Very few samples of matter consist of pure substances; instead, most are mixturesA combination of two or more pure substances in variable proportions in which the individual substances retain their respective identities., which are combinations of two or more pure substances in variable proportions in which the individual substances retain their identity. Air, tap water, milk, blue cheese, bread, and dirt are all mixtures. If all portions of a material are in the same state, have no visible boundaries, and are uniform throughout, then the material is homogeneousA mixture in which all portions of a material are in the same state, have no visible boundaries, and are uniform throughout.. Examples of homogeneous mixtures are the air we breathe and the tap water we drink. Homogeneous mixtures are also called solutions. Thus air is a solution of nitrogen, oxygen, water vapor, carbon dioxide, and several other gases; tap water is a solution of small amounts of several substances in water. The specific compositions of both of these solutions are not fixed, however, but depend on both source and location; for example, the composition of tap water in Boise, Idaho, is not the same as the composition of tap water in Buffalo, New York. Although most solutions we encounter are liquid, solutions can also be solid. The gray substance still used by some dentists to fill tooth cavities is a complex solid solution that contains 50% mercury and 50% of a powder that contains mostly silver, tin, and copper, with small amounts of zinc and mercury. Solid solutions of two or more metals are commonly called alloys.

If the composition of a material is not completely uniform, then it is heterogeneousA mixture in which a material is not completely uniform throughout. (e.g., chocolate chip cookie dough, blue cheese, and dirt). Mixtures that appear to be homogeneous are often found to be heterogeneous after microscopic examination. Milk, for example, appears to be homogeneous, but when examined under a microscope, it clearly consists of tiny globules of fat and protein dispersed in water (). The components of heterogeneous mixtures can usually be separated by simple means. Solid-liquid mixtures such as sand in water or tea leaves in tea are readily separated by filtration, which consists of passing the mixture through a barrier, such as a strainer, with holes or pores that are smaller than the solid particles. In principle, mixtures of two or more solids, such as sugar and salt, can be separated by microscopic inspection and sorting. More complex operations are usually necessary, though, such as when separating gold nuggets from river gravel by panning. First solid material is filtered from river water; then the solids are separated by inspection. If gold is embedded in rock, it may have to be isolated using chemical methods.

Figure 1.7 A Heterogeneous Mixture

Under a microscope, whole milk is actually a heterogeneous mixture composed of globules of fat and protein dispersed in water.

Homogeneous mixtures (solutions) can be separated into their component substances by physical processes that rely on differences in some physical property, such as differences in their boiling points. Two of these separation methods are distillation and crystallization. DistillationA physical process used to separate homogeneous mixtures (solutions) into their component substances. Distillation makes use of differences in the volatilities of the component substances. makes use of differences in volatility, a measure of how easily a substance is converted to a gas at a given temperature. shows a simple distillation apparatus for separating a mixture of substances, at least one of which is a liquid. The most volatile component boils first and is condensed back to a liquid in the water-cooled condenser, from which it flows into the receiving flask. If a solution of salt and water is distilled, for example, the more volatile component, pure water, collects in the receiving flask, while the salt remains in the distillation flask.

Figure 1.8 The Distillation of a Solution of Table Salt in Water

The solution of salt in water is heated in the distilling flask until it boils. The resulting vapor is enriched in the more volatile component (water), which condenses to a liquid in the cold condenser and is then collected in the receiving flask.

Mixtures of two or more liquids with different boiling points can be separated with a more complex distillation apparatus. One example is the refining of crude petroleum into a range of useful products: aviation fuel, gasoline, kerosene, diesel fuel, and lubricating oil (in the approximate order of decreasing volatility). Another example is the distillation of alcoholic spirits such as brandy or whiskey. This relatively simple procedure caused more than a few headaches for federal authorities in the 1920s during the era of Prohibition, when illegal stills proliferated in remote regions of the United States.

CrystallizationA physical process used to separate homogeneous mixtures (solutions) into their component substances. Crystallization separates mixtures based on differences in their solubilities. separates mixtures based on differences in solubility, a measure of how much solid substance remains dissolved in a given amount of a specified liquid. Most substances are more soluble at higher temperatures, so a mixture of two or more substances can be dissolved at an elevated temperature and then allowed to cool slowly. Alternatively, the liquid, called the solvent, may be allowed to evaporate. In either case, the least soluble of the dissolved substances, the one that is least likely to remain in solution, usually forms crystals first, and these crystals can be removed from the remaining solution by filtration. dramatically illustrates the process of crystallization.

Figure 1.9 The Crystallization of Sodium Acetate from a Concentrated Solution of Sodium Acetate in Water

The addition of a small “seed” crystal (a) causes the compound to form white crystals, which grow and eventually occupy most of the flask (b).

Most mixtures can be separated into pure substances, which may be either elements or compounds. An elementA pure substance that cannot be broken down into a simpler substance by chemical changes., such as gray, metallic sodium, is a substance that cannot be broken down into simpler ones by chemical changes; a compoundA pure substance that contains two or more elements and has chemical and physical properties that are usually different from those of the elements of which it is composed., such as white, crystalline sodium chloride, contains two or more elements and has chemical and physical properties that are usually different from those of the elements of which it is composed. With only a few exceptions, a particular compound has the same elemental composition (the same elements in the same proportions) regardless of its source or history. The chemical composition of a substance is altered in a process called a chemical changeA process in which the chemical composition of one or more substances is altered.. The conversion of two or more elements, such as sodium and chlorine, to a chemical compound, sodium chloride, is an example of a chemical change, often called a chemical reaction. Currently, about 115 elements are known, but millions of chemical compounds have been prepared from these 115 elements. The known elements are listed in the periodic table (see ).

In general, a reverse chemical process breaks down compounds into their elements. For example, water (a compound) can be decomposed into hydrogen and oxygen (both elements) by a process called electrolysis. In electrolysis, electricity provides the energy needed to separate a compound into its constituent elements (). A similar technique is used on a vast scale to obtain pure aluminum, an element, from its ores, which are mixtures of compounds. Because a great deal of energy is required for electrolysis, the cost of electricity is by far the greatest expense incurred in manufacturing pure aluminum. Thus recycling aluminum is both cost-effective and ecologically sound.

Figure 1.10 The Decomposition of Water to Hydrogen and Oxygen by Electrolysis

Water is a chemical compound; hydrogen and oxygen are elements.

The overall organization of matter and the methods used to separate mixtures are summarized in .

Figure 1.11 Relationships between the Types of Matter and the Methods Used to Separate Mixtures

Example 2

Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution).

  1. filtered tea
  2. freshly squeezed orange juice
  3. a compact disc
  4. aluminum oxide, a white powder that contains a 2:3 ratio of aluminum and oxygen atoms
  5. selenium

Given: a chemical substance

Asked for: its classification

Strategy:

A Decide whether a substance is chemically pure. If it is pure, the substance is either an element or a compound. If a substance can be separated into its elements, it is a compound.

B If a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. If its composition is uniform throughout, it is a homogeneous mixture.

Solution:

  1. A Tea is a solution of compounds in water, so it is not chemically pure. It is usually separated from tea leaves by filtration. B Because the composition of the solution is uniform throughout, it is a homogeneous mixture.
  2. A Orange juice contains particles of solid (pulp) as well as liquid; it is not chemically pure. B Because its composition is not uniform throughout, orange juice is a heterogeneous mixture.
  3. A A compact disc is a solid material that contains more than one element, with regions of different compositions visible along its edge. Hence a compact disc is not chemically pure. B The regions of different composition indicate that a compact disc is a heterogeneous mixture.
  4. A Aluminum oxide is a single, chemically pure compound.
  5. A Selenium is one of the known elements.

Exercise

Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution).

  1. white wine
  2. mercury
  3. ranch-style salad dressing
  4. table sugar (sucrose)

Answer:

  1. solution
  2. element
  3. heterogeneous mixture
  4. compound

Properties of Matter

All matter has physical and chemical properties. Physical propertiesA characteristic that scientists can measure without changing the composition of a sample under study. are characteristics that scientists can measure without changing the composition of the sample under study, such as mass, color, and volume (the amount of space occupied by a sample). Chemical propertiesThe characteristic ability of a substance to react to form new substances. describe the characteristic ability of a substance to react to form new substances; they include its flammability and susceptibility to corrosion. All samples of a pure substance have the same chemical and physical properties. For example, pure copper is always a reddish-brown solid (a physical property) and always dissolves in dilute nitric acid to produce a blue solution and a brown gas (a chemical property).

Physical properties can be extensive or intensive. Extensive propertiesA physical property that varies with the amount of a substance. vary with the amount of the substance and include mass, weight, and volume. Intensive propertiesA physical property that does not depend on the amount of the substance and physical state at a given temperature and pressure., in contrast, do not depend on the amount of the substance; they include color, melting point, boiling point, electrical conductivity, and physical state at a given temperature. For example, elemental sulfur is a yellow crystalline solid that does not conduct electricity and has a melting point of 115.2°C, no matter what amount is examined (). Scientists commonly measure intensive properties to determine a substance’s identity, whereas extensive properties convey information about the amount of the substance in a sample.

Figure 1.12 The Difference between Extensive and Intensive Properties of Matter

Because they differ in size, the two samples of sulfur have different extensive properties, such as mass and volume. In contrast, their intensive properties, including color, melting point, and electrical conductivity, are identical.

Although mass and volume are both extensive properties, their ratio is an important intensive property called density (d)An intensive property of matter, density is the mass per unit volume (usually expressed in g/cm3). At a given temperature, the density of a substance is a constant.. Density is defined as mass per unit volume and is usually expressed in grams per cubic centimeter (g/cm3). As mass increases in a given volume, density also increases. For example, lead, with its greater mass, has a far greater density than the same volume of air, just as a brick has a greater density than the same volume of Styrofoam. At a given temperature and pressure, the density of a pure substance is a constant:

Equation 1.1

density =massvolumed=mv

Pure water, for example, has a density of 0.998 g/cm3 at 25°C.

The average densities of some common substances are in . Notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float.” Example 3 shows how density measurements can be used to identify pure substances.

Table 1.1 Densities of Common Substances

Substance Density at 25°C (g/cm3)
blood 1.035
body fat 0.918
whole milk 1.030
corn oil 0.922
mayonnaise 0.910
honey 1.420

Example 3

The densities of some common liquids are in . Imagine you have five bottles containing colorless liquids (labeled AE). You must identify them by measuring the density of each. Using a pipette, a laboratory instrument for accurately measuring and transferring liquids, you carefully measure 25.00 mL of each liquid into five beakers of known mass (1 mL = 1 cm3). You then weigh each sample on a laboratory balance. Use the tabulated data to calculate the density of each sample. Based solely on your results, can you unambiguously identify all five liquids?If necessary, review the use of significant figures in calculations in Essential Skills 1 () prior to working this example.

Masses of samples: A, 17.72 g; B, 19.75 g; C, 24.91 g; D, 19.65 g; E, 27.80 g

Table 1.2 Densities of Liquids in Example 3

Substance Density at 25°C (g/cm3)
water 0.998
ethanol (the alcohol in beverages) 0.789
methanol (wood alcohol) 0.792
ethylene glycol (used in antifreeze) 1.113
diethyl ether (“ether”; once widely used as an anesthetic) 0.708
isopropanol (rubbing alcohol) 0.785

Given: volume and mass

Asked for: density

Strategy:

A Calculate the density of each liquid from the volumes and masses given.

B Check to make sure that your answer makes sense.

C Compare each calculated density with those given in . If the calculated density of a liquid is not significantly different from that of one of the liquids given in the table, then the unknown liquid is most likely the corresponding liquid.

D If none of the reported densities corresponds to the calculated density, then the liquid cannot be unambiguously identified.

Solution:

A Density is mass per unit volume and is usually reported in grams per cubic centimeter (or grams per milliliter because 1 mL = 1 cm3). The masses of the samples are given in grams, and the volume of all the samples is 25.00 mL (= 25.00 cm3). The density of each sample is calculated by dividing the mass by its volume (). The density of sample A, for example, is

17.72 g25.00 cm3=0.7088 g/cm3

Both the volume and the mass are given to four significant figures, so four significant figures are permitted in the result. (See Essential Skills 1, , for a discussion of significant figures.) The densities of the other samples (in grams per cubic centimeter) are as follows: B, 0.7900; C, 0.9964; D, 0.7860; and E, 1.112.

B Except for sample E, the calculated densities are slightly less than 1 g/cm3. This makes sense because the masses (in grams) of samples AD are all slightly less than the volume of the samples, 25.00 mL. In contrast, the mass of sample E is slightly greater than 25 g, so its density must be somewhat greater than 1 g/cm3.

C Comparing these results with the data given in shows that sample A is probably diethyl ether (0.708 g/cm3 and 0.7088 g/cm3 are not substantially different), sample C is probably water (0.998 g/cm3 in the table versus 0.9964 g/cm3 measured), and sample E is probably ethylene glycol (1.113 g/cm3 in the table versus 1.112 g/cm3 measured).

D Samples B and D are more difficult to identify for two reasons: (1) Both have similar densities (0.7900 and 0.7860 g/cm3), so they may or may not be chemically identical. (2) Within experimental error, the measured densities of B and D are indistinguishable from the densities of ethanol (0.789 g/cm3), methanol (0.792 g/cm3), and isopropanol (0.785 g/cm3). Thus some property other than density must be used to identify each sample.

Exercise

Given the volumes and masses of five samples of compounds used in blending gasoline, together with the densities of several chemically pure liquids, identify as many of the samples as possible.

Sample Volume (mL) Mass (g)
A 337 250.0
B 972 678.1
C 243 190.9
D 119 103.2
E 499 438.7
Substance Density (g/cm3)
benzene 0.8787
toluene 0.8669
m-xylene 0.8684
isooctane 0.6979
methyl t-butyl ether 0.7405
t-butyl alcohol 0.7856

Answer: A, methyl t-butyl ether; B, isooctane; C, t-butyl alcohol; D, toluene or m-xylene; E, benzene

Summary

Matter is anything that occupies space and has mass. The three states of matter are solid, liquid, and gas. A physical change involves the conversion of a substance from one state of matter to another, without changing its chemical composition. Most matter consists of mixtures of pure substances, which can be homogeneous (uniform in composition) or heterogeneous (different regions possess different compositions and properties). Pure substances can be either chemical compounds or elements. Compounds can be broken down into elements by chemical reactions, but elements cannot be separated into simpler substances by chemical means. The properties of substances can be classified as either physical or chemical. Scientists can observe physical properties without changing the composition of the substance, whereas chemical properties describe the tendency of a substance to undergo chemical changes (chemical reactions) that change its chemical composition. Physical properties can be intensive or extensive. Intensive properties are the same for all samples; do not depend on sample size; and include, for example, color, physical state, and melting and boiling points. Extensive properties depend on the amount of material and include mass and volume. The ratio of two extensive properties, mass and volume, is an important intensive property called density.

Key Takeaway

  • Matter can be classified according to physical and chemical properties.

Conceptual Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 1 () before proceeding to the Conceptual Problems.

  1. What is the difference between mass and weight? Is the mass of an object on Earth the same as the mass of the same object on Jupiter? Why or why not?

  2. Is it accurate to say that a substance with a mass of 1 kg weighs 2.2 lb? Why or why not?

  3. What factor must be considered when reporting the weight of an object as opposed to its mass?

  4. Construct a table with the headings “Solid,” “Liquid,” and “Gas.” For any given substance, state what you expect for each of the following:

    1. the relative densities of the three phases
    2. the physical shapes of the three phases
    3. the volumes for the same mass of compound
    4. the sensitivity of the volume of each phase to changes in temperature
    5. the sensitivity of the volume to changes in pressure
  5. Classify each substance as homogeneous or heterogeneous and explain your reasoning.

    1. platinum
    2. a carbonated beverage
    3. bronze
    4. wood
    5. natural gas
    6. Styrofoam
  6. Classify each substance as homogeneous or heterogeneous and explain your reasoning.

    1. snowflakes
    2. gasoline
    3. black tea
    4. plastic wrap
    5. blood
    6. water containing ice cubes
  7. Classify each substance as a pure substance or a mixture and explain your reasoning.

    1. seawater
    2. coffee
    3. 14-karat gold
    4. diamond
    5. distilled water
  8. Classify each substance as a pure substance or a mixture.

    1. cardboard
    2. caffeine
    3. tin
    4. a vitamin tablet
    5. helium gas
  9. Classify each substance as an element or a compound.

    1. sugar
    2. silver
    3. rust
    4. rubbing alcohol
    5. copper
  10. Classify each substance as an element or a compound.

    1. water
    2. iron
    3. hydrogen gas
    4. glass
    5. nylon
  11. What techniques could be used to separate each of the following?

    1. sugar and water from an aqueous solution of sugar
    2. a mixture of sugar and sand
    3. a heterogeneous mixture of solids with different solubilities
  12. What techniques could be used to separate each of the following?

    1. solid calcium chloride from a solution of calcium chloride in water
    2. the components of a solution of vinegar in water
    3. particulates from water in a fish tank
  13. Match each separation technique in (a) with the physical/chemical property that each takes advantage of in (b).

    1. crystallization, distillation, filtration
    2. volatility, physical state, solubility
  14. The following figures illustrate the arrangement of atoms in some samples of matter. Which figures are related by a physical change? By a chemical change?

  15. Classify each statement as an extensive property or an intensive property.

    1. Carbon, in the form of diamond, is one of the hardest known materials.
    2. A sample of crystalline silicon, a grayish solid, has a mass of 14.3 g.
    3. Germanium has a density of 5.32 g/cm3.
    4. Gray tin converts to white tin at 13.2°C.
    5. Lead is a bluish-white metal.
  16. Classify each statement as a physical property or a chemical property.

    1. Fluorine etches glass.
    2. Chlorine interacts with moisture in the lungs to produce a respiratory irritant.
    3. Bromine is a reddish-brown liquid.
    4. Iodine has a density of 11.27 g/L at 0°C.

Numerical Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 1 () before proceeding to the Numerical Problems.

  1. If a person weighs 176 lb on Earth, what is his or her mass on Mars, where the force of gravity is 37% of that on Earth?

  2. If a person weighs 135 lb on Earth, what is his or her mass on Jupiter, where the force of gravity is 236% of that on Earth?

  3. Calculate the volume of 10.00 g of each element and then arrange the elements in order of decreasing volume. The numbers in parentheses are densities.

    1. copper (8.92 g/cm3)
    2. calcium (1.54 g/cm3)
    3. titanium (4.51 g/cm3)
    4. iridium (22.85 g/cm3)
  4. Given 15.00 g of each element, calculate the volume of each and then arrange the elements in order of increasing volume. The numbers in parentheses are densities.

    1. gold (19.32 g/cm3)
    2. lead (11.34 g/cm3)
    3. iron (7.87 g/cm3)
    4. sulfur (2.07 g/cm3)
  5. A silver bar has dimensions of 10.00 cm × 4.00 cm × 1.50 cm, and the density of silver is 10.49 g/cm3. What is the mass of the bar?

  6. Platinum has a density of 21.45 g/cm3. What is the mass of a platinum bar measuring 3.00 cm × 1.50 cm × 0.500 cm?

  7. Complete the following table.

    Density (g/cm3) Mass (g) Volume (cm3) Element
    3.14 79.904 Br
    3.51 3.42 C
    39.1 45.5 K
    11.34 207.2 Pb
    107.868 10.28 Ag
    6.51 14.0 Zr
  8. Gold has a density of 19.30 g/cm3. If a person who weighs 85.00 kg (1 kg = 1000 g) were given his or her weight in gold, what volume (in cm3) would the gold occupy? Are we justified in using the SI unit of mass for the person’s weight in this case?

  9. An irregularly shaped piece of magnesium with a mass of 11.81 g was dropped into a graduated cylinder partially filled with water. The magnesium displaced 6.80 mL of water. What is the density of magnesium?

  10. The density of copper is 8.92 g/cm3. If a 10.00 g sample is placed in a graduated cylinder that contains 15.0 mL of water, what is the total volume that would be occupied?

  11. At 20°C, the density of fresh water is 0.9982 kg/m3, and the density of seawater is 1.025 kg/m3. Will a ship float higher in fresh water or in seawater? Explain your reasoning.

Answers

  1. Unlike weight, mass does not depend on location. The mass of the person is therefore the same on Earth and Mars: 176 lb ÷ 2.2 lb/kg = 80 kg.

    1. Cu: 1.12 cm3
    2. Ca: 6.49 cm3
    3. Ti: 2.22 cm3
    4. Ir: 0.4376 cm3

    Volume decreases: Ca > Ti > Cu > Ir

  2. 629 g

  3. 1.74 g/cm3

1.4 A Brief History of Chemistry

Learning Objective

  1. To understand the development of the atomic model.

It was not until the era of the ancient Greeks that we have any record of how people tried to explain the chemical changes they observed and used. At that time, natural objects were thought to consist of only four basic elements: earth, air, fire, and water. Then, in the fourth century BC, two Greek philosophers, Democritus and Leucippus, suggested that matter was not infinitely divisible into smaller particles but instead consisted of fundamental, indivisible particles called atomsThe fundamental, individual particles of which matter is composed.. Unfortunately, these early philosophers did not have the technology to test their hypothesis. They would have been unlikely to do so in any case because the ancient Greeks did not conduct experiments or use the scientific method. They believed that the nature of the universe could be discovered by rational thought alone.

Over the next two millennia, alchemists, who engaged in a form of chemistry and speculative philosophy during the Middle Ages and Renaissance, achieved many advances in chemistry. Their major goal was to convert certain elements into others by a process they called transmutationThe process of converting one element to another. (). In particular, alchemists wanted to find a way to transform cheaper metals into gold. Although most alchemists did not approach chemistry systematically and many appear to have been outright frauds, alchemists in China, the Arab kingdoms, and medieval Europe made major contributions, including the discovery of elements such as quicksilver (mercury) and the preparation of several strong acids.

Figure 1.13 An Alchemist at Work

Alchemy was a form of chemistry that flourished during the Middle Ages and Renaissance. Although some alchemists were frauds, others made major contributions, including the discovery of several elements and the preparation of strong acids.

Modern Chemistry

The 16th and 17th centuries saw the beginnings of what we now recognize as modern chemistry. During this period, great advances were made in metallurgy, the extraction of metals from ores, and the first systematic quantitative experiments were carried out. In 1661, the Englishman Robert Boyle (1627–91) published The Sceptical Chymist, which described the relationship between the pressure and the volume of air. More important, Boyle defined an element as a substance that cannot be broken down into two or more simpler substances by chemical means. This led to the identification of a large number of elements, many of which were metals. Ironically, Boyle himself never thought that metals were elements.

In the 18th century, the English clergyman Joseph Priestley (1733–1804) discovered oxygen gas and found that many carbon-containing materials burn vigorously in an oxygen atmosphere, a process called combustionThe burning of a material in an oxygen atmosphere.. Priestley also discovered that the gas produced by fermenting beer, which we now know to be carbon dioxide, is the same as one of the gaseous products of combustion. Priestley’s studies of this gas did not continue as he would have liked, however. After he fell into a vat of fermenting beer, brewers prohibited him from working in their factories. Although Priestley did not understand its identity, he found that carbon dioxide dissolved in water to produce seltzer water. In essence, he may be considered the founder of the multibillion-dollar carbonated soft drink industry.

Joseph Priestley (1733–1804)

Priestley was a political theorist and a leading Unitarian minister. He was appointed to Warrington Academy in Lancashire, England, where he developed new courses on history, science, and the arts. During visits to London, Priestley met the leading men of science, including Benjamin Franklin, who encouraged Priestley’s interest in electricity. Priestley’s work on gases began while he was living next to a brewery in Leeds, where he noticed “fixed air” bubbling out of vats of fermenting beer and ale. His scientific discoveries included the relationship between electricity and chemical change, 10 new “airs,” and observations that led to the discovery of photosynthesis. Due to his support for the principles of the French Revolution, Priestley’s house, library, and laboratory were destroyed by a mob in 1791. He and his wife emigrated to the United States in 1794 to join their three sons, who had previously emigrated to Pennsylvania. Priestley never returned to England and died in his new home in Pennsylvania.

Despite the pioneering studies of Priestley and others, a clear understanding of combustion remained elusive. In the late 18th century, however, the French scientist Antoine Lavoisier (1743–94) showed that combustion is the reaction of a carbon-containing substance with oxygen to form carbon dioxide and water and that life depends on a similar reaction, which today we call respiration. Lavoisier also wrote the first modern chemistry text and is widely regarded as the father of modern chemistry. His most important contribution was the law of conservation of massIn any chemical reaction, the mass of the substances that react equals the mass of the products that are formed., which states that in any chemical reaction, the mass of the substances that react equals the mass of the products that are formed. That is, in a chemical reaction, mass is neither lost nor destroyed. Unfortunately, Lavoisier invested in a private corporation that collected taxes for the Crown, and royal tax collectors were not popular during the French Revolution. He was executed on the guillotine at age 51, prematurely terminating his contributions to chemistry.

The Atomic Theory of Matter

In 1803, the English schoolteacher John Dalton (1766–1844) expanded Proust’s development of the law of definite proportions () and Lavoisier’s findings on the conservation of mass in chemical reactions to propose that elements consist of indivisible particles that he called atoms (taking the term from Democritus and Leucippus). Dalton’s atomic theory of matter contains four fundamental hypotheses:

  1. All matter is composed of tiny indivisible particles called atoms.
  2. All atoms of an element are identical in mass and chemical properties, whereas atoms of different elements differ in mass and fundamental chemical properties.
  3. A chemical compound is a substance that always contains the same atoms in the same ratio.
  4. In chemical reactions, atoms from one or more compounds or elements redistribute or rearrange in relation to other atoms to form one or more new compounds. Atoms themselves do not undergo a change of identity in chemical reactions.

This last hypothesis suggested that the alchemists’ goal of transmuting other elements to gold was impossible, at least through chemical reactions. We now know that Dalton’s atomic theory is essentially correct, with four minor modifications:

  1. Not all atoms of an element must have precisely the same mass.
  2. Atoms of one element can be transformed into another through nuclear reactions.
  3. The compositions of many solid compounds are somewhat variable.
  4. Under certain circumstances, some atoms can be divided (split into smaller particles).

These modifications illustrate the effectiveness of the scientific method; later experiments and observations were used to refine Dalton’s original theory.

The Law of Multiple Proportions

Despite the clarity of his thinking, Dalton could not use his theory to determine the elemental compositions of chemical compounds because he had no reliable scale of atomic masses; that is, he did not know the relative masses of elements such as carbon and oxygen. For example, he knew that the gas we now call carbon monoxide contained carbon and oxygen in the ratio 1:1.33 by mass, and a second compound, the gas we call carbon dioxide, contained carbon and oxygen in the ratio 1:2.66 by mass. Because 2.66/1.33 = 2.00, the second compound contained twice as many oxygen atoms per carbon atom as did the first. But what was the correct formula for each compound? If the first compound consisted of particles that contain one carbon atom and one oxygen atom, the second must consist of particles that contain one carbon atom and two oxygen atoms. If the first compound had two carbon atoms and one oxygen atom, the second must have two carbon atoms and two oxygen atoms. If the first had one carbon atom and two oxygen atoms, the second would have one carbon atom and four oxygen atoms, and so forth. Dalton had no way to distinguish among these or more complicated alternatives. However, these data led to a general statement that is now known as the law of multiple proportionsWhen two elements form a series of compounds, the ratios of the masses of the second element that are present per gram of the first element can almost always be expressed as the ratios of integers. (The same law holds for the mass ratios of compounds forming a series that contains more than two elements.): when two elements form a series of compounds, the ratios of the masses of the second element that are present per gram of the first element can almost always be expressed as the ratios of integers. (The same law holds for mass ratios of compounds forming a series that contains more than two elements.) Example 4 shows how the law of multiple proportions can be applied to determine the identity of a compound.

Example 4

A chemist is studying a series of simple compounds of carbon and hydrogen. The following table lists the masses of hydrogen that combine with 1 g of carbon to form each compound.

Compound Mass of Hydrogen (g)
A 0.0839
B 0.1678
C 0.2520
D
  1. Determine whether these data follow the law of multiple proportions.
  2. Calculate the mass of hydrogen that would combine with 1 g of carbon to form D, the fourth compound in the series.

Given: mass of hydrogen per gram of carbon for three compounds

Asked for:

  1. ratios of masses of hydrogen to carbon
  2. mass of hydrogen per gram of carbon for fourth compound in series

Strategy:

A Select the lowest mass to use as the denominator and then calculate the ratio of each of the other masses to that mass. Include other ratios if appropriate.

B If the ratios are small whole integers, the data follow the law of multiple proportions.

C Decide whether the ratios form a numerical series. If so, then determine the next member of that series and predict the ratio corresponding to the next compound in the series.

D Use proportions to calculate the mass of hydrogen per gram of carbon in that compound.

Solution:

A Compound A has the lowest mass of hydrogen, so we use it as the denominator. The ratios of the remaining masses of hydrogen, B and C, that combine with 1 g of carbon are as follows:

CA=0.2520 g0.0839 g=3.00=31BA=0.1678 g0.0839 g=2.00=21CB=0.2520 g0.1678 g=1.50232

B The ratios of the masses of hydrogen that combine with 1 g of carbon are indeed composed of small whole integers (3/1, 2/1, 3/2), as predicted by the law of multiple proportions.

C The ratios B/A and C/A form the series 2/1, 3/1, so the next member of the series should be D/A = 4/1.

D Thus, if compound D exists, it would be formed by combining 4 × 0.0839 g = 0.336 g of hydrogen with 1 g of carbon. Such a compound does exist; it is methane, the major constituent of natural gas.

Exercise

Four compounds containing only sulfur and fluorine are known. The following table lists the masses of fluorine that combine with 1 g of sulfur to form each compound.

Compound Mass of Fluorine (g)
A 3.54
B 2.96
C 2.36
D 0.59
  1. Determine the ratios of the masses of fluorine that combine with 1 g of sulfur in these compounds. Are these data consistent with the law of multiple proportions?
  2. Calculate the mass of fluorine that would combine with 1 g of sulfur to form the next two compounds in the series: E and F.

Answer:

  1. A/D = 6.0 or 6/1; B/D ≈ 5.0, or 5/1; C/D = 4.0, or 4/1; yes
  2. Ratios of 3.0 and 2.0 give 1.8 g and 1.2 g of fluorine/gram of sulfur, respectively. (Neither of these compounds is yet known.)

Avogadro’s Hypothesis

In a further attempt to establish the formulas of chemical compounds, the French chemist Joseph Gay-Lussac (1778–1850) carried out a series of experiments using volume measurements. Under conditions of constant temperature and pressure, he carefully measured the volumes of gases that reacted to make a given chemical compound, together with the volumes of the products if they were gases. Gay-Lussac found, for example, that one volume of chlorine gas always reacted with one volume of hydrogen gas to produce two volumes of hydrogen chloride gas. Similarly, one volume of oxygen gas always reacted with two volumes of hydrogen gas to produce two volumes of water vapor (part (a) in ).

Figure 1.14 Gay-Lussac’s Experiments with Chlorine Gas and Hydrogen Gas

(a) One volume of chlorine gas reacted with one volume of hydrogen gas to produce two volumes of hydrogen chloride gas, and one volume of oxygen gas reacted with two volumes of hydrogen gas to produce two volumes of water vapor. (b) A summary of Avogadro’s hypothesis, which interpreted Gay-Lussac’s results in terms of atoms. Note that the simplest way for two molecules of hydrogen chloride to be produced is if hydrogen and chlorine each consist of molecules that contain two atoms of the element.

Gay-Lussac’s results did not by themselves reveal the formulas for hydrogen chloride and water. The Italian chemist Amadeo Avogadro (1776–1856) developed the key insight that led to the exact formulas. He proposed that when gases are measured at the same temperature and pressure, equal volumes of different gases contain equal numbers of gas particles. Avogadro’s hypothesis, which explained Gay-Lussac’s results, is summarized here and in part (b) in :

one volume(or particle) ofhydrogen+one volume(or particle) ofchlorinetwo volumes(or particles) ofhydrogen chloride

If Dalton’s theory of atoms was correct, then each particle of hydrogen or chlorine had to contain at least two atoms of hydrogen or chlorine because two particles of hydrogen chloride were produced. The simplest—but not the only—explanation was that hydrogen and chlorine contained two atoms each (i.e., they were diatomic) and that hydrogen chloride contained one atom each of hydrogen and chlorine. Applying this reasoning to Gay-Lussac’s results with hydrogen and oxygen leads to the conclusion that water contains two hydrogen atoms per oxygen atom. Unfortunately, because no data supported Avogadro’s hypothesis that equal volumes of gases contained equal numbers of particles, his explanations and formulas for simple compounds were not generally accepted for more than 50 years. Dalton and many others continued to believe that water particles contained one hydrogen atom and one oxygen atom, rather than two hydrogen atoms and one oxygen atom. The historical development of the concept of the atom is summarized in .

Figure 1.15 A Summary of the Historical Development of the Concept of the Atom

Summary

The ancient Greeks first proposed that matter consisted of fundamental particles called atoms. Chemistry took its present scientific form in the 18th century, when careful quantitative experiments by Lavoisier, Proust, and Dalton resulted in the law of definite proportions, the law of conservation of mass, and the law of multiple proportions, which laid the groundwork for Dalton’s atomic theory of matter. In particular, Avogadro’s hypothesis provided the first link between the macroscopic properties of a substance (in this case, the volume of a gas) and the number of atoms or molecules present.

Key Takeaway

  • The development of the atomic model relied on the application of the scientific method over several centuries.

Conceptual Problems

  1. Define combustion and discuss the contributions made by Priestley and Lavoisier toward understanding a combustion reaction.

  2. Chemical engineers frequently use the concept of “mass balance” in their calculations, in which the mass of the reactants must equal the mass of the products. What law supports this practice?

  3. Does the law of multiple proportions apply to both mass ratios and atomic ratios? Why or why not?

  4. What are the four hypotheses of the atomic theory of matter?

  5. Much of the energy in France is provided by nuclear reactions. Are such reactions consistent with Dalton’s hypotheses? Why or why not?

  6. Does 1 L of air contain the same number of particles as 1 L of nitrogen gas? Explain your answer.

Numerical Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 1 () before proceeding to the Numerical Problems.

  1. One of the minerals found in soil has an Al:Si:O atomic ratio of 0.2:0.2:0.5. Is this consistent with the law of multiple proportions? Why or why not? Is the ratio of elements consistent with Dalton’s atomic theory of matter?

  2. Nitrogen and oxygen react to form three different compounds that contain 0.571 g, 1.143 g, and 2.285 g of oxygen/gram of nitrogen, respectively. Is this consistent with the law of multiple proportions? Explain your answer.

  3. Three binary compounds of vanadium and oxygen are known. The following table gives the masses of oxygen that combine with 10.00 g of vanadium to form each compound.

    Compound Mass of Oxygen (g)
    A 4.71
    B 6.27
    C
    1. Determine the ratio of the masses of oxygen that combine with 3.14 g of vanadium in compounds A and B.
    2. Predict the mass of oxygen that would combine with 3.14 g of vanadium to form the third compound in the series.
  4. Three compounds containing titanium, magnesium, and oxygen are known. The following table gives the masses of titanium and magnesium that react with 5.00 g of oxygen to form each compound.

    Compound Mass of Titanium (g) Mass of Magnesium (g)
    A 4.99 2.53
    B 3.74 3.80
    C
    1. Determine the ratios of the masses of titanium and magnesium that combine with 5.00 g of oxygen in these compounds.
    2. Predict the masses of titanium and magnesium that would combine with 5.00 g of oxygen to form another possible compound in the series: C.

1.5 The Atom

Learning Objective

  1. To become familiar with the components and structure of the atom.

To date, about 115 different elements have been discovered; by definition, each is chemically unique. To understand why they are unique, you need to understand the structure of the atom (the fundamental, individual particle of an element) and the characteristics of its components.

Atoms consist of electronsA subatomic particle with a negative charge that resides around the nucleus of all atoms., protonsA subatomic particle with a positive charge that resides in the nucleus of all atoms., and neutronsA subatomic particle with no charge that resides in the nucleus of almost all atoms..This is an oversimplification that ignores the other subatomic particles that have been discovered, but it is sufficient for our discussion of chemical principles. Some properties of these subatomic particles are summarized in , which illustrates three important points.

  1. Electrons and protons have electrical charges that are identical in magnitude but opposite in sign. We usually assign relative charges of −1 and +1 to the electron and proton, respectively.
  2. Neutrons have approximately the same mass as protons but no charge. They are electrically neutral.
  3. The mass of a proton or a neutron is about 1836 times greater than the mass of an electron. Protons and neutrons constitute by far the bulk of the mass of atoms.

The discovery of the electron and the proton was crucial to the development of the modern model of the atom and provides an excellent case study in the application of the scientific method. In fact, the elucidation of the atom’s structure is one of the greatest detective stories in the history of science.

Table 1.3 Properties of Subatomic Particles*

Particle Mass (g) Atomic Mass (amu) Electrical Charge (coulombs) Relative Charge
electron 9.109 × 10−28 0.0005486 −1.602 × 10−19 −1
proton 1.673 × 10−24 1.007276 +1.602 × 10−19 +1
neutron 1.675 × 10−24 1.008665 0 0
* For a review of using scientific notation and units of measurement, see Essential Skills 1 ().

The Electron

Figure 1.16 A Gas Discharge Tube Producing Cathode Rays

When a high voltage is applied to a gas contained at low pressure in a gas discharge tube, electricity flows through the gas, and energy is emitted in the form of light.

Long before the end of the 19th century, it was well known that applying a high voltage to a gas contained at low pressure in a sealed tube (called a gas discharge tube) caused electricity to flow through the gas, which then emitted light (). Researchers trying to understand this phenomenon found that an unusual form of energy was also emitted from the cathode, or negatively charged electrode; hence this form of energy was called cathode rays. In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the ultimate form of matter. He demonstrated that cathode rays could be deflected, or bent, by magnetic or electric fields, which indicated that cathode rays consist of charged particles (). More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; we now call these particles electrons. Most important for chemistry, Thomson found that the mass-to-charge ratio of cathode rays was independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.

Figure 1.17 Deflection of Cathode Rays by an Electric Field

As the cathode rays travel toward the right, they are deflected toward the positive electrode (+), demonstrating that they are negatively charged.

Subsequently, the American scientist Robert Millikan (1868–1953) carried out a series of experiments using electrically charged oil droplets, which allowed him to calculate the charge on a single electron. With this information and Thomson’s mass-to-charge ratio, Millikan determined the mass of an electron:

masscharge×charge= mass

It was at this point that two separate lines of investigation began to converge, both aimed at determining how and why matter emits energy.

Radioactivity

The second line of investigation began in 1896, when the French physicist Henri Becquerel (1852–1908) discovered that certain minerals, such as uranium salts, emitted a new form of energy. Becquerel’s work was greatly extended by Marie Curie (1867–1934) and her husband, Pierre (1854–1906); all three shared the Nobel Prize in Physics in 1903. Marie Curie coined the term radioactivityThe spontaneous emission of energy rays (radiation) by matter. (from the Latin radius, meaning “ray”) to describe the emission of energy rays by matter. She found that one particular uranium ore, pitchblende, was substantially more radioactive than most, which suggested that it contained one or more highly radioactive impurities. Starting with several tons of pitchblende, the Curies isolated two new radioactive elements after months of work: polonium, which was named for Marie’s native Poland, and radium, which was named for its intense radioactivity. Pierre Curie carried a vial of radium in his coat pocket to demonstrate its greenish glow, a habit that caused him to become ill from radiation poisoning well before he was run over by a horse-drawn wagon and killed instantly in 1906. Marie Curie, in turn, died of what was almost certainly radiation poisoning.

Radium bromide illuminated by its own radioactive glow. This 1922 photo was taken in the dark in the Curie laboratory.

Building on the Curies’ work, the British physicist Ernest Rutherford (1871–1937) performed decisive experiments that led to the modern view of the structure of the atom. While working in Thomson’s laboratory shortly after Thomson discovered the electron, Rutherford showed that compounds of uranium and other elements emitted at least two distinct types of radiation. One was readily absorbed by matter and seemed to consist of particles that had a positive charge and were massive compared to electrons. Because it was the first kind of radiation to be discovered, Rutherford called these substances α particles. Rutherford also showed that the particles in the second type of radiation, β particles, had the same charge and mass-to-charge ratio as Thomson’s electrons; they are now known to be high-speed electrons. A third type of radiation, γ rays, was discovered somewhat later and found to be similar to a lower-energy form of radiation called x-rays, now used to produce images of bones and teeth.

These three kinds of radiation—α particles, β particles, and γ rays—are readily distinguished by the way they are deflected by an electric field and by the degree to which they penetrate matter. As illustrates, α particles and β particles are deflected in opposite directions; α particles are deflected to a much lesser extent because of their higher mass-to-charge ratio. In contrast, γ rays have no charge, so they are not deflected by electric or magnetic fields. shows that α particles have the least penetrating power and are stopped by a sheet of paper, whereas β particles can pass through thin sheets of metal but are absorbed by lead foil or even thick glass. In contrast, γ-rays can readily penetrate matter; thick blocks of lead or concrete are needed to stop them.

Figure 1.18 Effect of an Electric Field on α Particles, β Particles, and γ Rays

A negative electrode deflects negatively charged β particles, whereas a positive electrode deflects positively charged α particles. Uncharged γ rays are unaffected by an electric field. (Relative deflections are not shown to scale.)

Figure 1.19 Relative Penetrating Power of the Three Types of Radiation

A sheet of paper stops comparatively massive α particles, whereas β particles easily penetrate paper but are stopped by a thin piece of lead foil. Uncharged γ rays penetrate the paper and lead foil; a much thicker piece of lead or concrete is needed to absorb them.

The Atomic Model

Once scientists concluded that all matter contains negatively charged electrons, it became clear that atoms, which are electrically neutral, must also contain positive charges to balance the negative ones. Thomson proposed that the electrons were embedded in a uniform sphere that contained both the positive charge and most of the mass of the atom, much like raisins in plum pudding or chocolate chips in a cookie ().

Figure 1.20 Thomson’s Plum Pudding or Chocolate Chip Cookie Model of the Atom

In this model, the electrons are embedded in a uniform sphere of positive charge.

In a single famous experiment, however, Rutherford showed unambiguously that Thomson’s model of the atom was impossible. Rutherford aimed a stream of α particles at a very thin gold foil target (part (a) in ) and examined how the α particles were scattered by the foil. Gold was chosen because it could be easily hammered into extremely thin sheets with a thickness that minimized the number of atoms in the target. If Thomson’s model of the atom were correct, the positively charged α particles should crash through the uniformly distributed mass of the gold target like cannonballs through the side of a wooden house. They might be moving a little slower when they emerged, but they should pass essentially straight through the target (part (b) in ). To Rutherford’s amazement, a small fraction of the α particles were deflected at large angles, and some were reflected directly back at the source (part (c) in ). According to Rutherford, “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.”

Figure 1.21 A Summary of Rutherford’s Experiments

(a) A representation of the apparatus Rutherford used to detect deflections in a stream of α particles aimed at a thin gold foil target. The particles were produced by a sample of radium. (b) If Thomson’s model of the atom were correct, the α particles should have passed straight through the gold foil. (c) But a small number of α particles were deflected in various directions, including right back at the source. This could be true only if the positive charge were much more massive than the α particle. It suggested that the mass of the gold atom is concentrated in a very small region of space, which he called the nucleus.

Rutherford’s results were not consistent with a model in which the mass and positive charge are distributed uniformly throughout the volume of an atom. Instead, they strongly suggested that both the mass and positive charge are concentrated in a tiny fraction of the volume of an atom, which Rutherford called the nucleusThe central core of an atom where protons and any neutrons reside.. It made sense that a small fraction of the α particles collided with the dense, positively charged nuclei in either a glancing fashion, resulting in large deflections, or almost head-on, causing them to be reflected straight back at the source.

Although Rutherford could not explain why repulsions between the positive charges in nuclei that contained more than one positive charge did not cause the nucleus to disintegrate, he reasoned that repulsions between negatively charged electrons would cause the electrons to be uniformly distributed throughout the atom’s volume.Today we know that strong nuclear forces, which are much stronger than electrostatic interactions, hold the protons and the neutrons together in the nucleus. For this and other insights, Rutherford was awarded the Nobel Prize in Chemistry in 1908. Unfortunately, Rutherford would have preferred to receive the Nobel Prize in Physics because he thought that physics was superior to chemistry. In his opinion, “All science is either physics or stamp collecting.” (The authors of this text do not share Rutherford’s view!)

Subsequently, Rutherford established that the nucleus of the hydrogen atom was a positively charged particle, for which he coined the name proton in 1920. He also suggested that the nuclei of elements other than hydrogen must contain electrically neutral particles with approximately the same mass as the proton. The neutron, however, was not discovered until 1932, when James Chadwick (1891–1974, a student of Rutherford; Nobel Prize in Physics, 1935) discovered it. As a result of Rutherford’s work, it became clear that an α particle contains two protons and neutrons and is therefore simply the nucleus of a helium atom.

The historical development of the different models of the atom’s structure is summarized in . Rutherford’s model of the atom is essentially the same as the modern one, except that we now know that electrons are not uniformly distributed throughout an atom’s volume. Instead, they are distributed according to a set of principles described in . shows how the model of the atom has evolved over time from the indivisible unit of Dalton to the modern view taught today.

Figure 1.22 A Summary of the Historical Development of Models of the Components and Structure of the Atom

The dates in parentheses are the years in which the key experiments were performed.

Figure 1.23 The Evolution of Atomic Theory, as Illustrated by Models of the Oxygen Atom

Bohr’s model and the current model are described in .

Summary

Atoms, the smallest particles of an element that exhibit the properties of that element, consist of negatively charged electrons around a central nucleus composed of more massive positively charged protons and electrically neutral neutrons. Radioactivity is the emission of energetic particles and rays (radiation) by some substances. Three important kinds of radiation are α particles (helium nuclei), β particles (electrons traveling at high speed), and γ rays (similar to x-rays but higher in energy).

Key Takeaway

  • The atom consists of discrete particles that govern its chemical and physical behavior.

Conceptual Problems

  1. Describe the experiment that provided evidence that the proton is positively charged.

  2. What observation led Rutherford to propose the existence of the neutron?

  3. What is the difference between Rutherford’s model of the atom and the model chemists use today?

  4. If cathode rays are not deflected when they pass through a region of space, what does this imply about the presence or absence of a magnetic field perpendicular to the path of the rays in that region?

  5. Describe the outcome that would be expected from Rutherford’s experiment if the charge on α particles had remained the same but the nucleus were negatively charged. If the nucleus were neutral, what would have been the outcome?

  6. Describe the differences between an α particle, a β particle, and a γ ray. Which has the greatest ability to penetrate matter?

Numerical Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 1 () before proceeding to the Numerical Problems.

  1. Using the data in and the periodic table (see ), calculate the percentage of the mass of a silicon atom that is due to

    1. electrons.
    2. protons.
  2. Using the data in and the periodic table (see ), calculate the percentage of the mass of a helium atom that is due to

    1. electrons.
    2. protons.
  3. The radius of an atom is approximately 104 times larger than the radius of its nucleus. If the radius of the nucleus were 1.0 cm, what would be the radius of the atom in centimeters? in miles?

  4. The total charge on an oil drop was found to be 3.84 × 10−18 coulombs. What is the total number of electrons contained in the drop?

1.6 Isotopes and Atomic Masses

Learning Objective

  1. To know the meaning of isotopes and atomic masses.

Rutherford’s nuclear model of the atom helped explain why atoms of different elements exhibit different chemical behavior. The identity of an element is defined by its atomic number (Z)The number of protons in the nucleus of an atom of an element., the number of protons in the nucleus of an atom of the element. The atomic number is therefore different for each element. The known elements are arranged in order of increasing Z in the periodic tableA chart of the chemical elements arranged in rows of increasing atomic number so that the elements in each column (group) have similar chemical properties. (Figure 1.24 "The Periodic Table Showing the Elements in Order of Increasing "; also see Chapter 32 "Appendix H: Periodic Table of Elements"),We will explain the rationale for the peculiar format of the periodic table in Chapter 7 "The Periodic Table and Periodic Trends". in which each element is assigned a unique one-, two-, or three-letter symbol. The names of the elements are listed in the periodic table, along with their symbols, atomic numbers, and atomic masses. The chemistry of each element is determined by its number of protons and electrons. In a neutral atom, the number of electrons equals the number of protons.

Figure 1.24 The Periodic Table Showing the Elements in Order of Increasing Z

As described in Section 1.7 "Introduction to the Periodic Table", the metals are on the bottom left in the periodic table, and the nonmetals are at the top right. The semimetals lie along a diagonal line separating the metals and nonmetals.

In most cases, the symbols for the elements are derived directly from each element’s name, such as C for carbon, U for uranium, Ca for calcium, and Po for polonium. Elements have also been named for their properties [such as radium (Ra) for its radioactivity], for the native country of the scientist(s) who discovered them [polonium (Po) for Poland], for eminent scientists [curium (Cm) for the Curies], for gods and goddesses [selenium (Se) for the Greek goddess of the moon, Selene], and for other poetic or historical reasons. Some of the symbols used for elements that have been known since antiquity are derived from historical names that are no longer in use; only the symbols remain to remind us of their origin. Examples are Fe for iron, from the Latin ferrum; Na for sodium, from the Latin natrium; and W for tungsten, from the German wolfram. Examples are in Table 1.4 "Element Symbols Based on Names No Longer in Use". As you work through this text, you will encounter the names and symbols of the elements repeatedly, and much as you become familiar with characters in a play or a film, their names and symbols will become familiar.

Table 1.4 Element Symbols Based on Names No Longer in Use

Element Symbol Derivation Meaning
antimony Sb stibium Latin for “mark”
copper Cu cuprum from Cyprium, Latin name for the island of Cyprus, the major source of copper ore in the Roman Empire
gold Au aurum Latin for “gold”
iron Fe ferrum Latin for “iron”
lead Pb plumbum Latin for “heavy”
mercury Hg hydrargyrum Latin for “liquid silver”
potassium K kalium from the Arabic al-qili, “alkali”
silver Ag argentum Latin for “silver”
sodium Na natrium Latin for “sodium”
tin Sn stannum Latin for “tin”
tungsten W wolfram German for “wolf stone” because it interfered with the smelting of tin and was thought to devour the tin

Recall from Section 1.5 "The Atom" that the nuclei of most atoms contain neutrons as well as protons. Unlike protons, the number of neutrons is not absolutely fixed for most elements. Atoms that have the same number of protons, and hence the same atomic number, but different numbers of neutrons are called isotopesAtoms that have the same numbers of protons but different numbers of neutrons.. All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A)The number of protons and neutrons in the nucleus of an atom of an element., the sum of the numbers of protons and neutrons.

The element carbon (C) has an atomic number of 6, which means that all neutral carbon atoms contain 6 protons and 6 electrons. In a typical sample of carbon-containing material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. An isotope of any element can be uniquely represented as XZA, where X is the atomic symbol of the element. The isotope of carbon that has 6 neutrons is therefore C612. The subscript indicating the atomic number is actually redundant because the atomic symbol already uniquely specifies Z. Consequently, C612 is more often written as 12C, which is read as “carbon-12.” Nevertheless, the value of Z is commonly included in the notation for nuclear reactions because these reactions involve changes in Z, as described in Chapter 20 "Nuclear Chemistry".

In addition to 12C, a typical sample of carbon contains 1.11% C613 (13C), with 7 neutrons and 6 protons, and a trace of C614 (14C), with 8 neutrons and 6 protons. The nucleus of 14C is not stable, however, but undergoes a slow radioactive decay that is the basis of the carbon-14 dating technique used in archaeology (see Chapter 14 "Chemical Kinetics"). Many elements other than carbon have more than one stable isotope; tin, for example, has 10 isotopes. The properties of some common isotopes are in Table 1.5 "Properties of Selected Isotopes".

Table 1.5 Properties of Selected Isotopes

Element Symbol Atomic Mass (amu) Isotope Mass Number Isotope Masses (amu) Percent Abundances (%)
hydrogen H 1.0079 1 1.007825 99.9855
2 2.014102 0.0115
boron B 10.81 10 10.012937 19.91
11 11.009305 80.09
carbon C 12.011 12 12 (defined) 99.89
13 13.003355 1.11
oxygen O 15.9994 16 15.994915 99.757
17 16.999132 0.0378
18 17.999161 0.205
iron Fe 55.845 54 53.939611 5.82
56 55.934938 91.66
57 56.935394 2.19
58 57.933276 0.33
uranium U 238.03 234 234.040952 0.0054
235 235.043930 0.7204
238 238.050788 99.274

Example 5

An element with three stable isotopes has 82 protons. The separate isotopes contain 124, 125, and 126 neutrons. Identify the element and write symbols for the isotopes.

Given: number of protons and neutrons

Asked for: element and atomic symbol

Strategy:

A Refer to the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements") and use the number of protons to identify the element.

B Calculate the mass number of each isotope by adding together the numbers of protons and neutrons.

C Give the symbol of each isotope with the mass number as the superscript and the number of protons as the subscript, both written to the left of the symbol of the element.

Solution:

A The element with 82 protons (atomic number of 82) is lead: Pb.

B For the first isotope, A = 82 protons + 124 neutrons = 206. Similarly, A = 82 + 125 = 207 and A = 82 + 126 = 208 for the second and third isotopes, respectively. The symbols for these isotopes are P82206b,P82207b, and P82208b, which are usually abbreviated as 206Pb, 207Pb, and 208Pb.

Exercise

Identify the element with 35 protons and write the symbols for its isotopes with 44 and 46 neutrons.

Answer: B3579r and B3581r or, more commonly, 79Br and 81Br.

Although the masses of the electron, the proton, and the neutron are known to a high degree of precision (Table 1.3 "Properties of Subatomic Particles*"), the mass of any given atom is not simply the sum of the masses of its electrons, protons, and neutrons. For example, the ratio of the masses of 1H (hydrogen) and 2H (deuterium) is actually 0.500384, rather than 0.49979 as predicted from the numbers of neutrons and protons present. Although the difference in mass is small, it is extremely important because it is the source of the huge amounts of energy released in nuclear reactions (Chapter 20 "Nuclear Chemistry").

Because atoms are much too small to measure individually and do not have a charge, there is no convenient way to accurately measure absolute atomic masses. Scientists can measure relative atomic masses very accurately, however, using an instrument called a mass spectrometer. The technique is conceptually similar to the one Thomson used to determine the mass-to-charge ratio of the electron. First, electrons are removed from or added to atoms or molecules, thus producing charged particles called ionsA charged particle produced when one or more electrons is removed from or added to an atom or molecule.. When an electric field is applied, the ions are accelerated into a separate chamber where they are deflected from their initial trajectory by a magnetic field, like the electrons in Thomson’s experiment. The extent of the deflection depends on the mass-to-charge ratio of the ion. By measuring the relative deflection of ions that have the same charge, scientists can determine their relative masses (Figure 1.25 "Determining Relative Atomic Masses Using a Mass Spectrometer"). Thus it is not possible to calculate absolute atomic masses accurately by simply adding together the masses of the electrons, the protons, and the neutrons, and absolute atomic masses cannot be measured, but relative masses can be measured very accurately. It is actually rather common in chemistry to encounter a quantity whose magnitude can be measured only relative to some other quantity, rather than absolutely. We will encounter many other examples later in this text. In such cases, chemists usually define a standard by arbitrarily assigning a numerical value to one of the quantities, which allows them to calculate numerical values for the rest.

Figure 1.25 Determining Relative Atomic Masses Using a Mass Spectrometer

Chlorine consists of two isotopes, 35Cl and 37Cl, in approximately a 3:1 ratio. (a) When a sample of elemental chlorine is injected into the mass spectrometer, electrical energy is used to dissociate the Cl2 molecules into chlorine atoms and convert the chlorine atoms to Cl+ ions. The ions are then accelerated into a magnetic field. The extent to which the ions are deflected by the magnetic field depends on their relative mass-to-charge ratios. Note that the lighter 35Cl+ ions are deflected more than the heavier 37Cl+ ions. By measuring the relative deflections of the ions, chemists can determine their mass-to-charge ratios and thus their masses. (b) Each peak in the mass spectrum corresponds to an ion with a particular mass-to-charge ratio. The abundance of the two isotopes can be determined from the heights of the peaks.

The arbitrary standard that has been established for describing atomic mass is the atomic mass unit (amu)One-twelfth of the mass of one atom of C12; 1 amu = 1.66 × 1024g., defined as one-twelfth of the mass of one atom of 12C. Because the masses of all other atoms are calculated relative to the 12C standard, 12C is the only atom listed in Table 1.5 "Properties of Selected Isotopes" whose exact atomic mass is equal to the mass number. Experiments have shown that 1 amu = 1.66 × 10−24 g.

Mass spectrometric experiments give a value of 0.167842 for the ratio of the mass of 2H to the mass of 12C, so the absolute mass of 2H is

mass of H2mass of C12×mass of C12= 0.167842 × 12 amu = 2.104104 amu

The masses of the other elements are determined in a similar way.

The periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements") lists the atomic masses of all the elements. If you compare these values with those given for some of the isotopes in Table 1.5 "Properties of Selected Isotopes", you can see that the atomic masses given in the periodic table never correspond exactly to those of any of the isotopes. Because most elements exist as mixtures of several stable isotopes, the atomic mass of an element is defined as the weighted average of the masses of the isotopes. For example, naturally occurring carbon is largely a mixture of two isotopes: 98.89% 12C (mass = 12 amu by definition) and 1.11% 13C (mass = 13.003355 amu). The percent abundance of 14C is so low that it can be ignored in this calculation. The average atomic mass of carbon is then calculated as

(0.9889 × 12 amu) + (0.0111 × 13.003355 amu) = 12.01 amu

Carbon is predominantly 12C, so its average atomic mass should be close to 12 amu, which is in agreement with our calculation.

The value of 12.01 is shown under the symbol for C in the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements"), although without the abbreviation amu, which is customarily omitted. Thus the tabulated atomic mass of carbon or any other element is the weighted average of the masses of the naturally occurring isotopes.

Example 6

Naturally occurring bromine consists of the two isotopes listed in the following table:

Isotope Exact Mass (amu) Percent Abundance (%)
79Br 78.9183 50.69
81Br 80.9163 49.31

Calculate the atomic mass of bromine.

Given: exact mass and percent abundance

Asked for: atomic mass

Strategy:

A Convert the percent abundances to decimal form to obtain the mass fraction of each isotope.

B Multiply the exact mass of each isotope by its corresponding mass fraction (percent abundance ÷ 100) to obtain its weighted mass.

C Add together the weighted masses to obtain the atomic mass of the element.

D Check to make sure that your answer makes sense.

Solution:

A The atomic mass is the weighted average of the masses of the isotopes. In general, we can write

atomic mass of element = [(mass of isotope 1 in amu) (mass fraction of isotope 1)] + [(mass of isotope 2) (mass fraction of isotope 2)] + …

Bromine has only two isotopes. Converting the percent abundances to mass fractions gives

B79r: 50.69100=0.5069 B81r: 49.31100=0.4931

B Multiplying the exact mass of each isotope by the corresponding mass fraction gives the isotope’s weighted mass:

79Br: 79.9183 amu × 0.5069 = 40.00 amu 81Br: 80.9163 amu × 0.4931 = 39.90 amu

C The sum of the weighted masses is the atomic mass of bromine is

40.00 amu + 39.90 amu = 79.90 amu

D This value is about halfway between the masses of the two isotopes, which is expected because the percent abundance of each is approximately 50%.

Exercise

Magnesium has the three isotopes listed in the following table:

Isotope Exact Mass (amu) Percent Abundance (%)
24Mg 23.98504 78.70
25Mg 24.98584 10.13
26Mg 25.98259 11.17

Use these data to calculate the atomic mass of magnesium.

Answer: 24.31 amu

Summary

Each atom of an element contains the same number of protons, which is the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. The atomic mass of an element is the weighted average of the masses of the naturally occurring isotopes. When one or more electrons are added to or removed from an atom or molecule, a charged particle called an ion is produced, whose charge is indicated by a superscript after the symbol.

Key Takeaway

  • The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, whereas the number of protons and electrons determines its charge.

Conceptual Problems

  1. Complete the following table for the missing elements, symbols, and numbers of electrons.

    Element Symbol Number of Electrons
    molybdenum
    19
    titanium
    B
    53
    Sm
    helium
    14
  2. Complete the following table for the missing elements, symbols, and numbers of electrons.

    Element Symbol Number of Electrons
    lanthanum
    Ir
    aluminum
    80
    sodium
    Si
    9
    Be
  3. Is the mass of an ion the same as the mass of its parent atom? Explain your answer.

  4. What isotopic standard is used for determining the mass of an atom?

  5. Give the symbol XZA for these elements, all of which exist as a single isotope.

    1. beryllium
    2. ruthenium
    3. phosphorus
    4. aluminum
    5. cesium
    6. praseodymium
    7. cobalt
    8. yttrium
    9. arsenic
  6. Give the symbol XZA for these elements, all of which exist as a single isotope.

    1. fluorine
    2. helium
    3. terbium
    4. iodine
    5. gold
    6. scandium
    7. sodium
    8. niobium
    9. manganese
  7. Identify each element, represented by X, that have the given symbols.

    1. X2655
    2. X3374
    3. X1224
    4. X53127
    5. X1840
    6. X63152

Numerical Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9 "Essential Skills 1") before proceeding to the Numerical Problems.

  1. The isotopes 131I and 60Co are commonly used in medicine. Determine the number of neutrons, protons, and electrons in a neutral atom of each.

  2. Determine the number of protons, neutrons, and electrons in a neutral atom of each isotope:

    1. 97Tc
    2. 113In
    3. 63Ni
    4. 55Fe
  3. Both technetium-97 and americium-240 are produced in nuclear reactors. Determine the number of protons, neutrons, and electrons in the neutral atoms of each.

  4. The following isotopes are important in archaeological research. How many protons, neutrons, and electrons does a neutral atom of each contain?

    1. 207Pb
    2. 16O
    3. 40K
    4. 137Cs
    5. 40Ar
  5. Copper, an excellent conductor of heat, has two isotopes: 63Cu and 65Cu. Use the following information to calculate the average atomic mass of copper:

    Isotope Percent Abundance (%) Atomic Mass (amu)
    63Cu 69.09 62.9298
    65Cu 30.92 64.9278
  6. Silicon consists of three isotopes with the following percent abundances:

    Isotope Percent Abundance (%) Atomic Mass (amu)
    28Si 92.18 27.976926
    29Si 4.71 28.976495
    30Si 3.12 29.973770

    Calculate the average atomic mass of silicon.

  7. Complete the following table for neon. The average atomic mass of neon is 20.1797 amu.

    Isotope Percent Abundance (%) Atomic Mass (amu)
    20Ne 90.92 19.99244
    21Ne 0.257 20.99395
    22Ne
  8. Are X2863 and X2962 isotopes of the same element? Explain your answer.

  9. Complete the following table:

    Isotope Number of Protons Number of Neutrons Number of Electrons
    238X 95
    238U
    75 112
  10. Complete the following table:

    Isotope Number of Protons Number of Neutrons Number of Electrons
    57Fe
    40X 20
    36S
  11. Using a mass spectrometer, a scientist determined the percent abundances of the isotopes of sulfur to be 95.27% for 32S, 0.51% for 33S, and 4.22% for 34S. Use the atomic mass of sulfur from the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements") and the following atomic masses to determine whether these data are accurate, assuming that these are the only isotopes of sulfur: 31.972071 amu for 32S, 32.971459 amu for 33S, and 33.967867 amu for 34S.

  12. The percent abundances of two of the three isotopes of oxygen are 99.76% for 16O, and 0.204% for 18O. Use the atomic mass of oxygen given in the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements") and the following data to determine the mass of 17O: 15.994915 amu for 16O and 17.999160 amu for 18O.

  13. Which element has the higher proportion by mass in NaI?

  14. Which element has the higher proportion by mass in KBr?

1.7 Introduction to the Periodic Table

Learning Objective

  1. To become familiar with the organization of the periodic table.

The elements are arranged in a periodic table (; also see ), which is probably the single most important learning aid in chemistry. It summarizes huge amounts of information about the elements in a way that permits you to predict many of their properties and chemical reactions. The elements are arranged in seven horizontal rows, in order of increasing atomic number from left to right and top to bottom. The rows are called periodsA row of elements in the periodic table., and they are numbered from 1 to 7. The elements are stacked in such a way that elements with similar chemical properties form vertical columns, called groupsA vertical column of elements in the periodic table. Elements with similar chemical properties reside in the same group., numbered from 1 to 18 (older periodic tables use a system based on roman numerals). Groups 1, 2, and 13–18 are the main group elementsAny element in groups 1, 2, and 13–18 in the periodic table. These groups contain metals, semimetals, and nonmetals., listed as A in older tables. Groups 3–12 are in the middle of the periodic table and are the transition elementsAny element in groups 3–12 in the periodic table. All of the transition elements are metals., listed as B in older tables. The two rows of 14 elements at the bottom of the periodic table are the lanthanides and the actinides, whose positions in the periodic table are indicated in group 3. A more comprehensive description of the periodic table is found in .

Metals, Nonmetals, and Semimetals

The heavy orange zigzag line running diagonally from the upper left to the lower right through groups 13–16 in divides the elements into metalsAny element to the left of the zigzag line in the periodic table that runs from boron to astatine. All metals except mercury are solids at room temperature and pressure. (in blue, below and to the left of the line) and nonmetalsAny element to the right of the zigzag line in the periodic table that runs from boron to astatine. Nonmetals may be solids, liquids, or gases at room temperature and pressure. (in bronze, above and to the right of the line). As you might expect, elements colored in gold that lie along the diagonal line exhibit properties intermediate between metals and nonmetals; they are called semimetalsAny element that lies adjacent to the zigzag line in the periodic table that runs from boron to astatine. Semimetals (also called metalloids) exhibit properties intermediate between those of metals and nonmetals..

The distinction between metals and nonmetals is one of the most fundamental in chemistry. Metals—such as copper or gold—are good conductors of electricity and heat; they can be pulled into wires because they are ductileThe ability to be pulled into wires. Metals are ductile, whereas nonmetals are usually brittle.; they can be hammered or pressed into thin sheets or foils because they are malleableThe ability to be hammered or pressed into thin sheets or foils. Metals are malleable, whereas nonmetals are usually brittle.; and most have a shiny appearance, so they are lustrousHaving a shiny appearance. Metals are lustrous, whereas nonmetals are not.. The vast majority of the known elements are metals. Of the metals, only mercury is a liquid at room temperature and pressure; all the rest are solids.

Nonmetals, in contrast, are generally poor conductors of heat and electricity and are not lustrous. Nonmetals can be gases (such as chlorine), liquids (such as bromine), or solids (such as iodine) at room temperature and pressure. Most solid nonmetals are brittle, so they break into small pieces when hit with a hammer or pulled into a wire. As expected, semimetals exhibit properties intermediate between metals and nonmetals.

Example 7

Based on its position in the periodic table, do you expect selenium to be a metal, a nonmetal, or a semimetal?

Given: element

Asked for: classification

Strategy:

Find selenium in the periodic table shown in and then classify the element according to its location.

Solution:

The atomic number of selenium is 34, which places it in period 4 and group 16. In , selenium lies above and to the right of the diagonal line marking the boundary between metals and nonmetals, so it should be a nonmetal. Note, however, that because selenium is close to the metal-nonmetal dividing line, it would not be surprising if selenium were similar to a semimetal in some of its properties.

Exercise

Based on its location in the periodic table, do you expect indium to be a nonmetal, a metal, or a semimetal?

Answer: metal

Descriptive Names

As we noted, the periodic table is arranged so that elements with similar chemical behaviors are in the same group. Chemists often make general statements about the properties of the elements in a group using descriptive names with historical origins. For example, the elements of group 1 are known as the alkali metalsAny element in group 1 of the periodic table., group 2 are the alkaline earth metalsAny element in group 2 of the periodic table., group 17 are the halogensDerived from the Greek for “salt forming,” an element in group 17 of the periodic table., and group 18 are the noble gasesAny element in group 18 of the periodic table. All are unreactive monatomic gases at room temperature and pressure..

The Alkali Metals

The alkali metals are lithium, sodium, potassium, rubidium, cesium, and francium. Hydrogen is unique in that it is generally placed in group 1, but it is not a metal.

The compounds of the alkali metals are common in nature and daily life. One example is table salt (sodium chloride); lithium compounds are used in greases, in batteries, and as drugs to treat patients who exhibit manic-depressive, or bipolar, behavior. Although lithium, rubidium, and cesium are relatively rare in nature, and francium is so unstable and highly radioactive that it exists in only trace amounts, sodium and potassium are the seventh and eighth most abundant elements in Earth’s crust, respectively.

The Alkaline Earth Metals

The alkaline earth metals are beryllium, magnesium, calcium, strontium, barium, and radium. Beryllium, strontium, and barium are rather rare, and radium is unstable and highly radioactive. In contrast, calcium and magnesium are the fifth and sixth most abundant elements on Earth, respectively; they are found in huge deposits of limestone and other minerals.

The Halogens

The halogens are fluorine, chlorine, bromine, iodine, and astatine. The name halogen is derived from the Greek for “salt forming,” which reflects that all the halogens react readily with metals to form compounds, such as sodium chloride and calcium chloride (used in some areas as road salt).

Compounds that contain the fluoride ion are added to toothpaste and the water supply to prevent dental cavities. Fluorine is also found in Teflon coatings on kitchen utensils. Although chlorofluorocarbon propellants and refrigerants are believed to lead to the depletion of Earth’s ozone layer and contain both fluorine and chlorine, the latter is responsible for the adverse effect on the ozone layer. Bromine and iodine are less abundant than chlorine, and astatine is so radioactive that it exists in only negligible amounts in nature.

The Noble Gases

The noble gases are helium, neon, argon, krypton, xenon, and radon. Because the noble gases are composed of only single atoms, they are monatomicA species containing a single atom.. At room temperature and pressure, they are unreactive gases. Because of their lack of reactivity, for many years they were called inert gases or rare gases. However, the first chemical compounds containing the noble gases were prepared in 1962. Although the noble gases are relatively minor constituents of the atmosphere, natural gas contains substantial amounts of helium. Because of its low reactivity, argon is often used as an unreactive (inert) atmosphere for welding and in light bulbs. The red light emitted by neon in a gas discharge tube is used in neon lights.

Note the Pattern

The noble gases are unreactive at room temperature and pressure.

Summary

The periodic table is an arrangement of the elements in order of increasing atomic number. Elements that exhibit similar chemistry appear in vertical columns called groups (numbered 1–18 from left to right); the seven horizontal rows are called periods. Some of the groups have widely used common names, including the alkali metals (group 1) and the alkaline earth metals (group 2) on the far left, and the halogens (group 17) and the noble gases (group 18) on the far right. The elements can be broadly divided into metals, nonmetals, and semimetals. Semimetals exhibit properties intermediate between those of metals and nonmetals. Metals are located on the left of the periodic table, and nonmetals are located on the upper right. They are separated by a diagonal band of semimetals. Metals are lustrous, good conductors of electricity, and readily shaped (they are ductile and malleable), whereas solid nonmetals are generally brittle and poor electrical conductors. Other important groupings of elements in the periodic table are the main group elements, the transition metals, the lanthanides, and the actinides.

Key Takeaway

  • The periodic table is used as a predictive tool.

Conceptual Problems

  1. Classify each element in Conceptual Problem 1 () as a metal, a nonmetal, or a semimetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal.

  2. Classify each element in Conceptual Problem 2 () as a metal, a nonmetal, or a semimetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal.

  3. Classify each element as a metal, a semimetal, or a nonmetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal.

    1. iron
    2. tantalum
    3. sulfur
    4. silicon
    5. chlorine
    6. nickel
    7. potassium
    8. radon
    9. zirconium
  4. Which of these sets of elements are all in the same period?

    1. potassium, vanadium, and ruthenium
    2. lithium, carbon, and chlorine
    3. sodium, magnesium, and sulfur
    4. chromium, nickel, and krypton
  5. Which of these sets of elements are all in the same period?

    1. barium, tungsten, and argon
    2. yttrium, zirconium, and selenium
    3. potassium, calcium, and zinc
    4. scandium, bromine, and manganese
  6. Which of these sets of elements are all in the same group?

    1. sodium, rubidium, and barium
    2. nitrogen, phosphorus, and bismuth
    3. copper, silver, and gold
    4. magnesium, strontium, and samarium
  7. Which of these sets of elements are all in the same group?

    1. iron, ruthenium, and osmium
    2. nickel, palladium, and lead
    3. iodine, fluorine, and oxygen
    4. boron, aluminum, and gallium
  8. Indicate whether each element is a transition metal, a halogen, or a noble gas.

    1. manganese
    2. iridium
    3. fluorine
    4. xenon
    5. lithium
    6. carbon
    7. zinc
    8. sodium
    9. tantalum
    10. hafnium
    11. antimony
    12. cadmium
  9. Which of the elements indicated in color in the periodic table shown below is most likely to exist as a monoatomic gas? As a diatomic gas? Which is most likely to be a semimetal? A reactive metal?

  10. Based on their locations in the periodic table, would you expect these elements to be malleable? Why or why not?

    1. phosphorus
    2. chromium
    3. rubidium
    4. copper
    5. aluminum
    6. bismuth
    7. neodymium
  11. Based on their locations in the periodic table, would you expect these elements to be lustrous? Why or why not?

    1. sulfur
    2. vanadium
    3. nickel
    4. arsenic
    5. strontium
    6. cerium
    7. sodium

Answer

  1. Symbol Type
    Fe metal: transition metal
    Ta metal: transition metal
    S nonmetal
    Si semimetal
    Cl nonmetal (halogen)
    Ni metal: transition metal
    K metal: alkali metal
    Rn nonmetal (noble gas)
    Zr metal: transition metal

1.8 Essential Elements for Life

Learning Objective

  1. To understand the importance of elements to nutrition.

Of the approximately 115 elements known, only the 19 highlighted in purple in are absolutely required in the human diet. These elements—called essential elementsAny of the 19 elements that are absolutely required in the human diet for survival. An additional seven elements are thought to be essential for humans.—are restricted to the first four rows of the periodic table (see ), with only two or three exceptions (molybdenum, iodine, and possibly tin in the fifth row). Some other elements are essential for specific organisms. For example, boron is required for the growth of certain plants, bromine is widely distributed in marine organisms, and tungsten is necessary for some microorganisms.

 

Figure 1.26 The Essential Elements in the Periodic Table

Elements that are known to be essential for human life are shown in purple; elements that are suggested to be essential are shown in green. Elements not known to be essential are shown in gray.

What makes an element “essential”? By definition, an essential element is one that is required for life and whose absence results in death. Because of the experimental difficulties involved in producing deficiencies severe enough to cause death, especially for elements that are required in very low concentrations in the diet, a somewhat broader definition is generally used. An element is considered to be essential if a deficiency consistently causes abnormal development or functioning and if dietary supplementation of that element—and only that element—prevents this adverse effect. Scientists determine whether an element is essential by raising rats, chicks, and other animals on a synthetic diet that has been carefully analyzed and supplemented with acceptable levels of all elements except the element of interest (E). Ultraclean environments, in which plastic cages are used and dust from the air is carefully removed, minimize inadvertent contamination. If the animals grow normally on a diet that is as low as possible in E, then either E is not an essential element or the diet is not yet below the minimum required concentration. If the animals do not grow normally on a low-E diet, then their diets are supplemented with E until a level is reached at which the animals grow normally. This level is the minimum required intake of element E.

Classification of the Essential Elements

The approximate elemental composition of a healthy 70.0 kg (154 lb) adult human is listed in . Note that most living matter consists primarily of the so-called bulk elements: oxygen, carbon, hydrogen, nitrogen, and sulfur—the building blocks of the compounds that constitute our organs and muscles. These five elements also constitute the bulk of our diet; tens of grams per day are required for humans. Six other elements—sodium, magnesium, potassium, calcium, chlorine, and phosphorus—are often referred to as macrominerals because they provide essential ions in body fluids and form the major structural components of the body. In addition, phosphorus is a key constituent of both DNA and RNA: the genetic building blocks of living organisms. The six macrominerals are present in the body in somewhat smaller amounts than the bulk elements, so correspondingly lower levels are required in the diet. The remaining essential elements—called trace elements—are present in very small amounts, ranging from a few grams to a few milligrams in an adult human. Finally, measurable levels of some elements are found in humans but are not required for growth or good health. Examples are rubidium and strontium, whose chemistry is similar to that of the elements immediately above them in the periodic table (potassium and calcium, respectively, which are essential elements). Because the body’s mechanisms for extracting potassium and calcium from foods are not 100% selective, small amounts of rubidium and strontium, which have no known biological function, are absorbed.

Table 1.6 Approximate Elemental Composition of a Typical 70 kg Human

Bulk Elements (kg) Macrominerals (g)
oxygen 44 calcium 1700
carbon 12.6 phosphorus 680
hydrogen 6.6 potassium 250
nitrogen 1.8 chlorine 115
sulfur 0.1 sodium 70
magnesium 42
Trace Elements (mg)
iron 5000 lead 35
silicon 3000 barium 21
zinc 1750 molybdenum 14
rubidium 360 boron 14
copper 280 arsenic ~3
strontium 280 cobalt ~3
bromine 140 chromium ~3
tin 140 nickel ~3
manganese 70 selenium ~2
iodine 70 lithium ~2
aluminum 35 vanadium ~2

The Trace Elements

Because it is difficult to detect low levels of some essential elements, the trace elements were relatively slow to be recognized as essential. Iron was the first. In the 17th century, anemia was proved to be caused by an iron deficiency and often was cured by supplementing the diet with extracts of rusty nails. It was not until the 19th century, however, that trace amounts of iodine were found to eliminate goiter (an enlarged thyroid gland). This is why common table salt is “iodized”: a small amount of iodine is added. Copper was shown to be essential for humans in 1928, and manganese, zinc, and cobalt soon after that. Molybdenum was not known to be an essential element until 1953, and the need for chromium, selenium, vanadium, fluorine, and silicon was demonstrated only in the last 50 years. It seems likely that in the future other elements, possibly including tin, will be found to be essential at very low levels.

Many compounds of trace elements, such as arsenic, selenium, and chromium, are toxic and can even cause cancer, yet these elements are identified as essential elements in . In fact, there is some evidence that one bacterium has replaced phosphorus with arsenic, although the finding is controversial. This has opened up the possibility of a “shadow biosphere” on Earth in which life evolved from an as yet undetected common ancestor. How can elements toxic to life be essential? First, the toxicity of an element often depends on its chemical form—for example, only certain compounds of chromium are toxic, whereas others are used in mineral supplements. Second, as shown in , every element has three possible levels of dietary intake: deficient, optimum, and toxic in order of increasing concentration in the diet. Very low intake levels lead to symptoms of deficiency. Over some range of higher intake levels, an organism is able to maintain its tissue concentrations of the element at a level that optimizes biological functions. Finally, at some higher intake level, the normal regulatory mechanisms are overloaded, causing toxic symptoms to appear. Each element has its own characteristic curve. Both the width of the plateau and the specific concentration corresponding to the center of the plateau region differ by as much as several orders of magnitude for different elements. In the adult human, for example, the recommended daily dietary intake is 10–18 mg of iron, 2–3 mg of copper, and less than 0.1 mg of chromium and selenium.

Figure 1.27 Possible Concentrations of an Essential Element in the Diet

The deficient, optimum, and toxic concentrations are different for different elements.

Amplification

How can elements that are present in such minuscule amounts have such large effects on an organism’s health? Our knowledge of the pathways by which each of the known trace elements affects health is far from complete, but certain general features are clear. The trace elements participate in an amplification mechanism; that is, they are essential components of larger biological molecules that are capable of interacting with or regulating the levels of relatively large amounts of other molecules. For example, vitamin B12 contains a single atom of cobalt, which is essential for its biological function. If the molecule whose level is controlled by the trace element can regulate the level of another molecule, and more and more molecules, then the potential exists for extreme amplification of small variations in the level of the trace element. One goal of modern chemical research is to elucidate in detail the roles of the essential elements. In subsequent chapters, we will introduce some results of this research to demonstrate the biological importance of many of the elements and their compounds.

Summary

About 19 of the approximately 115 known elements are essential for humans. An essential element is one whose absence results in abnormal biological function or development that is prevented by dietary supplementation with that element. Living organisms contain relatively large amounts of oxygen, carbon, hydrogen, nitrogen, and sulfur (these five elements are known as the bulk elements), along with sodium, magnesium, potassium, calcium, chlorine, and phosphorus (these six elements are known as macrominerals). The other essential elements are the trace elements, which are present in very small quantities. Dietary intakes of elements range from deficient to optimum to toxic with increasing quantities; the optimum levels differ greatly for the essential elements.

Key Takeaway

  • The absence of some elements can result in abnormal biological function or development.

1.9 Essential Skills 1

Topics

  • Measurement
  • Scientific Notation
  • Significant Figures
  • Accuracy and Precision

This section describes some of the fundamental mathematical skills you will need to complete the questions and problems in this text. For some of you, this discussion will serve as a review, whereas others may be encountering at least some of the ideas and techniques for the first time. We will introduce other mathematical skills in subsequent Essential Skills sections as the need arises. Be sure you are familiar with the topics discussed here before you start the problems.

Measurement

Instruments of Measurement

Graduated glassware is used to deliver variable volumes of liquid.

Volumetric glassware is used to deliver (pipette) or contain (volumetric flask) a single volume accurately when filled to the calibration mark.

A balance is used to measure mass.

A variety of instruments are available for making direct measurements of the macroscopic properties of a chemical substance. For example, we usually measure the volumeThe amount of space occupied by a sample of matter. of a liquid sample with pipettes, burets, graduated cylinders, and volumetric flasks, whereas we usually measure the mass of a solid or liquid substance with a balance. Measurements on an atomic or molecular scale, in contrast, require specialized instrumentation, such as the mass spectrometer described in .

SI Units

All reported measurements must include an appropriate unit of measurement because to say that a substance has “a mass of 10,” for example, does not tell whether the mass was measured in grams, pounds, tons, or some other unit. To establish worldwide standards for the consistent measurement of important physical and chemical properties, an international body called the General Conference on Weights and Measures devised the Système internationale d’unités (or SI)A system of units based on metric units that requires measurements to be expressed in decimal form. There are seven base units in the SI system.. The International System of Units is based on metric units and requires that measurements be expressed in decimal form. lists the seven base units of the SI system; all other SI units of measurement are derived from them.

By attaching prefixes to the base unit, the magnitude of the unit is indicated; each prefix indicates that the base unit is multiplied by a specified power of 10. The prefixes, their symbols, and their numerical significance are given in . To study chemistry, you need to know the information presented in and .

Table 1.7 SI Base Units

Base Quantity Unit Name Abbreviation
mass kilogram kg
length meter m
time second s
temperature kelvin K
electric current ampere A
amount of substance mole mol
luminous intensity candela cd

Table 1.8 Prefixes Used with SI Units

Prefix Symbol Value Power of 10 Meaning
tera T 1,000,000,000,000 1012 trillion
giga G 1,000,000,000 109 billion
mega M 1,000,000 106 million
kilo k 1000 103 thousand
hecto h 100 102 hundred
deca da 10 101 ten
1 100 one
deci d 0.1 10−1 tenth
centi c 0.01 10−2 hundredth
milli m 0.001 10−3 thousandth
micro μ 0.000001 10−6 millionth
nano n 0.000000001 10−9 billionth
pico p 0.000000000001 10−12 trillionth
femto f 0.000000000000001 10−15 quadrillionth

Units of Mass, Volume, and Length

The units of measurement you will encounter most frequently in chemistry are those for mass, volume, and length. The basic SI unit for mass is the kilogram (kg), but in the laboratory, mass is usually expressed in either grams (g) or milligrams (mg): 1000 g = 1 kg, 1000 mg = 1 g, and 1,000,000 mg = 1 kg. Units for volume are derived from the cube of the SI unit for length, which is the meter (m). Thus the basic SI unit for volume is cubic meters (length × width × height = m3). In chemistry, however, volumes are usually reported in cubic centimeters (cm3) and cubic decimeters (dm3) or milliliters (mL) and liters (L), although the liter is not an SI unit of measurement. The relationships between these units are as follows:

1 L = 1000 mL = 1 dm3 1 mL = 1 cm3 1000 cm3 = 1 L

Scientific Notation

Chemists often work with numbers that are exceedingly large or small. For example, entering the mass in grams of a hydrogen atom into a calculator requires a display with at least 24 decimal places. A system called scientific notationA system that expresses numbers in the form N × 10n, where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10) and n is an integer that can be either positive or negative (100 = 1). The purpose of scientific notation is to simplify the manipulation of numbers with large or small magnitudes. avoids much of the tedium and awkwardness of manipulating numbers with large or small magnitudes. In scientific notation, these numbers are expressed in the form

N × 10n

where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10), and n is a positive or negative integer (100 = 1). The number 10 is called the base because it is this number that is raised to the power n. Although a base number may have values other than 10, the base number in scientific notation is always 10.

A simple way to convert numbers to scientific notation is to move the decimal point as many places to the left or right as needed to give a number from 1 to 10 (N). The magnitude of n is then determined as follows:

  • If the decimal point is moved to the left n places, n is positive.
  • If the decimal point is moved to the right n places, n is negative.

Another way to remember this is to recognize that as the number N decreases in magnitude, the exponent increases and vice versa. The application of this rule is illustrated in Skill Builder ES1.

Skill Builder ES1

Convert each number to scientific notation.

  1. 637.8
  2. 0.0479
  3. 7.86
  4. 12,378
  5. 0.00032
  6. 61.06700
  7. 2002.080
  8. 0.01020

Solution

  1. To convert 637.8 to a number from 1 to 10, we move the decimal point two places to the left:

    637.8

    Because the decimal point was moved two places to the left, n = 2. In scientific notation, 637.8 = 6.378 × 102.

  2. To convert 0.0479 to a number from 1 to 10, we move the decimal point two places to the right:

    0.0479

    Because the decimal point was moved two places to the right, n = −2. In scientific notation, 0.0479 = 4.79 × 10−2.

  3. 7.86 × 100: this is usually expressed simply as 7.86. (Recall that 100 = 1.)
  4. 1.2378 × 104; because the decimal point was moved four places to the left, n = 4.
  5. 3.2 × 10−4; because the decimal point was moved four places to the right, n = −4.
  6. 6.106700 × 101: this is usually expressed as 6.1067 × 10.
  7. 2.002080 × 103
  8. 1.020 × 10−2

Addition and Subtraction

Before numbers expressed in scientific notation can be added or subtracted, they must be converted to a form in which all the exponents have the same value. The appropriate operation is then carried out on the values of N. Skill Builder ES2 illustrates how to do this.

Skill Builder ES2

Carry out the appropriate operation on each number and then express the answer in scientific notation.

  1. (1.36 × 102) + (4.73 × 103)
  2. (6.923 × 10−3) − (8.756 × 10−4)

Solution

  1. Both exponents must have the same value, so these numbers are converted to either (1.36 × 102) + (47.3 × 102) or (0.136 × 103) + (4.73 × 103). Choosing either alternative gives the same answer, reported to two decimal places:

    (1.36 × 102) + (47.3 × 102) = (1.36 + 47.3) × 102 = 48.66 × 102 = 4.87 × 103 (0.136 × 103) + (4.73 × 103) = (0.136 + 4.73) × 103 = 4.87 × 103

    In converting 48.66 × 102 to scientific notation, n has become more positive by 1 because the value of N has decreased.

  2. Converting the exponents to the same value gives either (6.923 × 10−3) − (0.8756 × 10−3) or (69.23 × 10−4) − (8.756 × 10−4). Completing the calculations gives the same answer, expressed to three decimal places:

    (6.923 × 10−3) − (0.8756 × 10−3) = (6.923 − 0.8756) × 10−3 = 6.047 × 10−3 (69.23 × 10−4) − (8.756 × 10−4) = (69.23 − 8.756) × 10−4 = 60.474 × 10−4 = 6.047 × 10−3

Multiplication and Division

When multiplying numbers expressed in scientific notation, we multiply the values of N and add together the values of n. Conversely, when dividing, we divide N in the dividend (the number being divided) by N in the divisor (the number by which we are dividing) and then subtract n in the divisor from n in the dividend. In contrast to addition and subtraction, the exponents do not have to be the same in multiplication and division. Examples of problems involving multiplication and division are shown in Skill Builder ES3.

Skill Builder ES3

Perform the appropriate operation on each expression and express your answer in scientific notation.

  1. (6.022 × 1023)(6.42 × 10−2)
  2. 1.67×10249.12×1028
  3. (6.63×1034)(6.0×10)8.52×102

Solution

  1. In multiplication, we add the exponents:

    (6.022 × 1023)(6.42 × 102)= (6.022)(6.42) × 10[23 + (2)]= 38.× 1021= 3.87 × 1022
  2. In division, we subtract the exponents:

    1.67×10249.12×1028=1.679.12×10[24  (28)]=0.183×104=1.83×103
  3. This problem has both multiplication and division:

    (6.63×1034)(6.0×10)(8.52×102)=39.788.52×10[34 + 1  (2)]=4.7×1031

Significant Figures

No measurement is free from error. Error is introduced by (1) the limitations of instruments and measuring devices (such as the size of the divisions on a graduated cylinder) and (2) the imperfection of human senses. Although errors in calculations can be enormous, they do not contribute to uncertainty in measurements. Chemists describe the estimated degree of error in a measurement as the uncertaintyThe estimated degree of error in a measurement. The degree of uncertainty in a measurement can be indicated by reporting all significant figures plus one. of the measurement, and they are careful to report all measured values using only significant figuresNumbers that describe the value without exaggerating the degree to which it is known to be accurate., numbers that describe the value without exaggerating the degree to which it is known to be accurate. Chemists report as significant all numbers known with absolute certainty, plus one more digit that is understood to contain some uncertainty. The uncertainty in the final digit is usually assumed to be ±1, unless otherwise stated.

The following rules have been developed for counting the number of significant figures in a measurement or calculation:

  1. Any nonzero digit is significant.
  2. Any zeros between nonzero digits are significant. The number 2005, for example, has four significant figures.
  3. Any zeros used as a placeholder preceding the first nonzero digit are not significant. So 0.05 has one significant figure because the zeros are used to indicate the placement of the digit 5. In contrast, 0.050 has two significant figures because the last two digits correspond to the number 50; the last zero is not a placeholder. As an additional example, 5.0 has two significant figures because the zero is used not to place the 5 but to indicate 5.0.
  4. When a number does not contain a decimal point, zeros added after a nonzero number may or may not be significant. An example is the number 100, which may be interpreted as having one, two, or three significant figures. (Note: treat all trailing zeros in exercises and problems in this text as significant unless you are specifically told otherwise.)
  5. Integers obtained either by counting objects or from definitions are exact numbersAn integer obtained either by counting objects or from definitions (e.g., 1 in. = 2.54 cm). Exact numbers have infinitely many significant figures., which are considered to have infinitely many significant figures. If we have counted four objects, for example, then the number 4 has an infinite number of significant figures (i.e., it represents 4.000…). Similarly, 1 foot (ft) is defined to contain 12 inches (in), so the number 12 in the following equation has infinitely many significant figures:
1 ft = 12 in

An effective method for determining the number of significant figures is to convert the measured or calculated value to scientific notation because any zero used as a placeholder is eliminated in the conversion. When 0.0800 is expressed in scientific notation as 8.00 × 10−2, it is more readily apparent that the number has three significant figures rather than five; in scientific notation, the number preceding the exponential (i.e., N) determines the number of significant figures. Skill Builder ES4 provides practice with these rules.

Skill Builder ES4

Give the number of significant figures in each. Identify the rule for each.

  1. 5.87
  2. 0.031
  3. 52.90
  4. 00.2001
  5. 500
  6. 6 atoms

Solution

  1. three (rule 1)
  2. two (rule 3); in scientific notation, this number is represented as 3.1 × 10−2, showing that it has two significant figures.
  3. four (rule 3)
  4. four (rule 2); this number is 2.001 × 10−1 in scientific notation, showing that it has four significant figures.
  5. one, two, or three (rule 4)
  6. infinite (rule 5)

Skill Builder ES5

Which measuring apparatus would you use to deliver 9.7 mL of water as accurately as possible? To how many significant figures can you measure that volume of water with the apparatus you selected?

Solution

Use the 10 mL graduated cylinder, which will be accurate to two significant figures.

Mathematical operations are carried out using all the digits given and then rounding the final result to the correct number of significant figures to obtain a reasonable answer. This method avoids compounding inaccuracies by successively rounding intermediate calculations. After you complete a calculation, you may have to round the last significant figure up or down depending on the value of the digit that follows it. If the digit is 5 or greater, then the number is rounded up. For example, when rounded to three significant figures, 5.215 is 5.22, whereas 5.213 is 5.21. Similarly, to three significant figures, 5.005 kg becomes 5.01 kg, whereas 5.004 kg becomes 5.00 kg. The procedures for dealing with significant figures are different for addition and subtraction versus multiplication and division.

When we add or subtract measured values, the value with the fewest significant figures to the right of the decimal point determines the number of significant figures to the right of the decimal point in the answer. Drawing a vertical line to the right of the column corresponding to the smallest number of significant figures is a simple method of determining the proper number of significant figures for the answer:

3240.7+21.2363261.936

The line indicates that the digits 3 and 6 are not significant in the answer. These digits are not significant because the values for the corresponding places in the other measurement are unknown (3240.7??). Consequently, the answer is expressed as 3261.9, with five significant figures. Again, numbers greater than or equal to 5 are rounded up. If our second number in the calculation had been 21.256, then we would have rounded 3261.956 to 3262.0 to complete our calculation.

When we multiply or divide measured values, the answer is limited to the smallest number of significant figures in the calculation; thus, 42.9 × 8.323 = 357.057 = 357. Although the second number in the calculation has four significant figures, we are justified in reporting the answer to only three significant figures because the first number in the calculation has only three significant figures. An exception to this rule occurs when multiplying a number by an integer, as in 12.793 × 12. In this case, the number of significant figures in the answer is determined by the number 12.973, because we are in essence adding 12.973 to itself 12 times. The correct answer is therefore 155.516, an increase of one significant figure, not 155.52.

When you use a calculator, it is important to remember that the number shown in the calculator display often shows more digits than can be reported as significant in your answer. When a measurement reported as 5.0 kg is divided by 3.0 L, for example, the display may show 1.666666667 as the answer. We are justified in reporting the answer to only two significant figures, giving 1.7 kg/L as the answer, with the last digit understood to have some uncertainty.

In calculations involving several steps, slightly different answers can be obtained depending on how rounding is handled, specifically whether rounding is performed on intermediate results or postponed until the last step. Rounding to the correct number of significant figures should always be performed at the end of a series of calculations because rounding of intermediate results can sometimes cause the final answer to be significantly in error.

In practice, chemists generally work with a calculator and carry all digits forward through subsequent calculations. When working on paper, however, we often want to minimize the number of digits we have to write out. Because successive rounding can compound inaccuracies, intermediate roundings need to be handled correctly. When working on paper, always round an intermediate result so as to retain at least one more digit than can be justified and carry this number into the next step in the calculation. The final answer is then rounded to the correct number of significant figures at the very end.

In the worked examples in this text, we will often show the results of intermediate steps in a calculation. In doing so, we will show the results to only the correct number of significant figures allowed for that step, in effect treating each step as a separate calculation. This procedure is intended to reinforce the rules for determining the number of significant figures, but in some cases it may give a final answer that differs in the last digit from that obtained using a calculator, where all digits are carried through to the last step. Skill Builder ES6 provides practice with calculations using significant figures.

Skill Builder ES6

Complete the calculations and report your answers using the correct number of significant figures.

  1. 87.25 mL + 3.0201 mL
  2. 26.843 g + 12.23 g
  3. 6 × 12.011
  4. 2(1.008) g + 15.99 g
  5. 137.3 + 2(35.45)
  6. 118.72 g35.5 g
  7. 47.23 g207.25.92 g
  8. 77.6046.4674.8
  9. 24.862.03.26(0.98)
  10. (15.9994 × 9) + 2.0158

Solution

  1. 90.27 mL
  2. 39.07 g
  3. 72.066 (See rule 5 under “Significant Figures.”)
  4. 2(1.008) g + 15.99 g = 2.016 g + 15.99 g = 18.01 g
  5. 137.3 + 2(35.45) = 137.3 + 70.90 = 208.2
  6. 59.35 g − 35.5 g = 23.9 g
  7. 47.23 g − 35.0 g = 12.2 g
  8. 12.00 − 4.8 = 7.2
  9. 12 − 3.2 = 9
  10. 143.9946 + 2.0158 = 146.0104

Accuracy and Precision

Measurements may be accurateThe measured value is the same as the true value., meaning that the measured value is the same as the true value; they may be preciseMultiple measurements give nearly identical values., meaning that multiple measurements give nearly identical values (i.e., reproducible results); they may be both accurate and precise; or they may be neither accurate nor precise. The goal of scientists is to obtain measured values that are both accurate and precise.

Suppose, for example, that the mass of a sample of gold was measured on one balance and found to be 1.896 g. On a different balance, the same sample was found to have a mass of 1.125 g. Which was correct? Careful and repeated measurements, including measurements on a calibrated third balance, showed the sample to have a mass of 1.895 g. The masses obtained from the three balances are in the following table:

Balance 1 Balance 2 Balance 3
1.896 g 1.125 g 1.893 g
1.895 g 1.158 g 1.895 g
1.894 g 1.067 g 1.895 g

Whereas the measurements obtained from balances 1 and 3 are reproducible (precise) and are close to the accepted value (accurate), those obtained from balance 2 are neither. Even if the measurements obtained from balance 2 had been precise (if, for example, they had been 1.125, 1.124, and 1.125), they still would not have been accurate. We can assess the precision of a set of measurements by calculating the average deviation of the measurements as follows:

  1. Calculate the average value of all the measurements:

    average =sum of measurementsnumber of measurements
  2. Calculate the deviation of each measurement, which is the absolute value of the difference between each measurement and the average value:

    deviation = |measurement − average|

    where | | means absolute value (i.e., convert any negative number to a positive number).

  3. Add all the deviations and divide by the number of measurements to obtain the average deviation:

    average =sum of deviationsnumber of measurements

Then we can express the precision as a percentage by dividing the average deviation by the average value of the measurements and multiplying the result by 100. In the case of balance 2, the average value is

1.125 g + 1.158 + 1.067 g3= 1.117 g

The deviations are 1.125 g − 1.117 g = 0.008 g, 1.158 g − 1.117 g = 0.041 g, and |1.067 g − 1.117 g| = 0.050 g. So the average deviation is

0.008 g + 0.041 g + 0.050 g3= 0.033 g

The precision of this set of measurements is therefore

0.033g1.117g×100 = 3.0%

When a series of measurements is precise but not accurate, the error is usually systematic. Systematic errors can be caused by faulty instrumentation or faulty technique. The difference between accuracy and precision is demonstrated in Skill Builder ES7.

Skill Builder ES7

The following archery targets show marks that represent the results of four sets of measurements. Which target shows

  1. a precise but inaccurate set of measurements?
  2. an accurate but imprecise set of measurements?
  3. a set of measurements that is both precise and accurate?
  4. a set of measurements that is neither precise nor accurate?

Solution

  1. (c)
  2. (a)
  3. (b)
  4. (d)

Skill Builder ES8

  1. A 1-carat diamond has a mass of 200.0 mg. When a jeweler repeatedly weighed a 2-carat diamond, he obtained measurements of 450.0 mg, 459.0 mg, and 463.0 mg. Were the jeweler’s measurements accurate? Were they precise?
  2. A single copper penny was tested three times to determine its composition. The first analysis gave a composition of 93.2% zinc and 2.8% copper, the second gave 92.9% zinc and 3.1% copper, and the third gave 93.5% zinc and 2.5% copper. The actual composition of the penny was 97.6% zinc and 2.4% copper. Were the results accurate? Were they precise?

Solution

  1. The expected mass of a 2-carat diamond is 2 × 200.0 mg = 400.0 mg. The average of the three measurements is 457.3 mg, about 13% greater than the true mass. These measurements are not particularly accurate.

    The deviations of the measurements are 7.3 mg, 1.7 mg, and 5.7 mg, respectively, which give an average deviation of 4.9 mg and a precision of

    4.9mg457.3mg×100 = 1.1%

    These measurements are rather precise.

  2. The average values of the measurements are 93.2% zinc and 2.8% copper versus the true values of 97.6% zinc and 2.4% copper. Thus these measurements are not very accurate, with errors of −4.5% and + 17% for zinc and copper, respectively. (The sum of the measured zinc and copper contents is only 96.0% rather than 100%, which tells us that either there is a significant error in one or both measurements or some other element is present.)

    The deviations of the measurements are 0.0%, 0.3%, and 0.3% for both zinc and copper, which give an average deviation of 0.2% for both metals. We might therefore conclude that the measurements are equally precise, but that is not the case. Recall that precision is the average deviation divided by the average value times 100. Because the average value of the zinc measurements is much greater than the average value of the copper measurements (93.2% versus 2.8%), the copper measurements are much less precise.

    precision (Zn) =0.2%93.2%×100 = 0.2%precision (Cu) =0.2%2.8%×100 = 7%

1.10 End-of-Chapter Material

Application Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9 "Essential Skills 1") before proceeding to the Application Problems. Problems marked with a ♦ involve multiple concepts.

  1. In 1953, James Watson and Francis Crick spent three days analyzing data to develop a model that was consistent with the known facts about the structure of DNA, the chemical substance that is the basis for life. They were awarded the Nobel Prize in Physiology or Medicine for their work. Based on this information, would you classify their proposed model for the structure of DNA as an experiment, a law, a hypothesis, or a theory? Explain your reasoning.

  2. In each scenario, state the observation and the hypothesis.

    1. A recently discovered Neanderthal throat bone has been found to be similar in dimensions and appearance to that of modern humans; therefore, some scientists believe that Neanderthals could talk.
    2. Because DNA profiles from samples of human tissue are widely used in criminal trials, DNA sequences from plant residue on clothing can be used to place a person at the scene of a crime.
  3. Small quantities of gold from far underground are carried to the surface by groundwater, where the gold can be taken up by certain plants and stored in their leaves. By identifying the kinds of plants that grow around existing gold deposits, one should be able to use this information to discover potential new gold deposits.

    1. State the observation.
    2. State the hypothesis.
    3. Devise an experiment to test the hypothesis.
  4. Large amounts of nitrogen are used by the electronics industry to provide a gas blanket over a component during production. This ensures that undesired reactions with oxygen will not occur. Classify each statement as an extensive property or an intensive property of nitrogen.

    1. Nitrogen is a colorless gas.
    2. A volume of 22.4 L of nitrogen gas weighs 28 g at 0°C.
    3. Liquid nitrogen boils at 77.4 K.
    4. Nitrogen gas has a density of 1.25 g/L at 0°C.
  5. Oxygen is the third most abundant element in the universe and makes up about two-thirds of the human body. Classify each statement as an extensive property or an intensive property of oxygen.

    1. Liquid oxygen boils at 90.2 K.
    2. Liquid oxygen is pale blue.
    3. A volume of 22.4 L of oxygen gas weighs 32 g at 0°C.
    4. Oxygen has a density of 1.43 g/L at 0°C.
  6. One of the first high-temperature superconductors was found to contain elements in the ratio 1Y:2Ba:3Cu:6.8O. A material that contains elements in the ratio 1Y:2Ba:3Cu:6O, however, was not a high-temperature superconductor. Do these materials obey the law of multiple proportions? Is the ratio of elements in each compound consistent with Dalton’s law of indivisible atoms?

  7. ♦ There has been increased evidence that human activities are causing changes in Earth’s atmospheric chemistry. Recent research efforts have focused on atmospheric ozone (O3) concentrations. The amount of ozone in the atmosphere is influenced by concentrations of gases that contain only nitrogen and oxygen, among others. The following table gives the masses of nitrogen that combine with 1.00 g of oxygen to form three of these compounds.

    Compound Mass of Nitrogen (g)
    A 0.875
    B 0.438
    C 0.350
    1. Determine the ratios of the masses of nitrogen that combine with 1.00 g of oxygen in these compounds. Are these data consistent with the law of multiple proportions?
    2. Predict the mass of nitrogen that would combine with 1.00 g of oxygen to form another possible compound in the series.
  8. Indium has an average atomic mass of 114.818 amu. One of its two isotopes has an atomic mass of 114.903 amu with a percent abundance of 95.70. What is the mass of the other isotope?

  9. Earth’s core is largely composed of iron, an element that is also a major component of black sands on beaches. Iron has four stable isotopes. Use the data to calculate the average atomic mass of iron.

    Isotope Percent Abundance (%) Atomic Mass (amu)
    54Fe 5.82 53.9396
    56Fe 91.66 55.9349
    57Fe 2.19 56.9354
    58Fe 0.33 57.9333
  10. ♦ Because ores are deposited during different geologic periods, lead ores from different mining regions of the world can contain different ratios of isotopes. Archaeologists use these differences to determine the origin of lead artifacts. For example, the following table lists the percent abundances of three lead isotopes from one artifact recovered from Rio Tinto in Spain.

    Isotope Percent Abundance (%) Atomic mass (amu)
    204Pb 203.973028
    206Pb 24.41 205.974449
    207Pb 20.32 206.97580
    208Pb 50.28 207.976636
    1. If the only other lead isotope in the artifact is 204Pb, what is its percent abundance?
    2. What is the average atomic mass of lead if the only other isotope in the artifact is 204Pb?
    3. An artifact from Laurion, Greece, was found to have a 207Pb:206Pb ratio of 0.8307. From the data given, can you determine whether the lead in the artifact from Rio Tinto came from the same source as the lead in the artifact from Laurion, Greece?
  11. The macrominerals sodium, magnesium, potassium, calcium, chlorine, and phosphorus are widely distributed in biological substances, although their distributions are far from uniform. Classify these elements by both their periods and their groups and then state whether each is a metal, a nonmetal, or a semimetal. If a metal, is the element a transition metal?

  12. The composition of fingernails is sensitive to exposure to certain elements, including sodium, magnesium, aluminum, chlorine, potassium, calcium, selenium, vanadium, chromium, manganese, iron, cobalt, copper, zinc, scandium, arsenic, and antimony. Classify these elements by both their periods and their groups and then determine whether each is a metal, a nonmetal, or a semimetal. Of the metals, which are transition metals? Based on your classifications, predict other elements that could prove to be detectable in fingernails.

  13. Mercury levels in hair have been used to identify individuals who have been exposed to toxic levels of mercury. Is mercury an essential element? a trace element?

  14. Trace elements are usually present at levels of less than 50 mg/kg of body weight. Classify the essential trace elements by their groups and periods in the periodic table. Based on your classifications, would you predict that arsenic, cadmium, and lead are potential essential trace elements?

Chapter 2 Molecules, Ions, and Chemical Formulas

Chapter 1 "Introduction to Chemistry" introduced some of the fundamental concepts of chemistry, with particular attention to the basic properties of atoms and elements. These entities are the building blocks of all substances we encounter, yet most common substances do not consist of only pure elements or individual atoms. Instead, nearly all substances are chemical compounds or mixtures of chemical compounds. Although there are only about 115 elements (of which about 86 occur naturally), millions of chemical compounds are known, with a tremendous range of physical and chemical properties. Consequently, the emphasis of modern chemistry (and this text) is on understanding the relationship between the structures and properties of chemical compounds.

Petroleum refining. Using chemicals, catalysts, heat, and pressure, a petroleum refinery will separate, combine, and rearrange the structure and bonding patterns of the basic carbon-hydrogen molecules found in crude oil. The final products include gasoline, paraffin, diesel fuel, lubricants, and bitumen.

In this chapter, you will learn how to describe the composition of chemical compounds. We introduce you to chemical nomenclature—the language of chemistry—that will enable you to recognize and name the most common kinds of compounds. An understanding of chemical nomenclature not only is essential for your study of chemistry but also has other benefits—for example, it helps you understand the labels on products found in the supermarket and the pharmacy. You will also be better equipped to understand many of the important environmental and medical issues that face society. By the end of this chapter, you will be able to describe what happens chemically when a doctor prepares a cast to stabilize a broken bone, and you will know the composition of common substances such as laundry bleach, the active ingredient in baking powder, and the foul-smelling compound responsible for the odor of spoiled fish. Finally, you will be able to explain the chemical differences among different grades of gasoline.

2.1 Chemical Compounds

Learning Objective

  1. To understand the differences between covalent and ionic bonding.

The atoms in all substances that contain more than one atom are held together by electrostatic interactionsAn interaction between electrically charged particles such as protons and electrons.—interactions between electrically charged particles such as protons and electrons. Electrostatic attractionAn electrostatic interaction between oppositely charged species (positive and negative) that results in a force that causes them to move toward each other. between oppositely charged species (positive and negative) results in a force that causes them to move toward each other, like the attraction between opposite poles of two magnets. In contrast, electrostatic repulsionAn electrostatic interaction between two species that have the same charge (both positive or both negative) that results in a force that causes them to repel each other. between two species with the same charge (either both positive or both negative) results in a force that causes them to repel each other, as do the same poles of two magnets. Atoms form chemical compounds when the attractive electrostatic interactions between them are stronger than the repulsive interactions. Collectively, we refer to the attractive interactions between atoms as chemical bondsAn attractive interaction between atoms that holds them together in compounds..

Chemical bonds are generally divided into two fundamentally different kinds: ionic and covalent. In reality, however, the bonds in most substances are neither purely ionic nor purely covalent, but they are closer to one of these extremes. Although purely ionic and purely covalent bonds represent extreme cases that are seldom encountered in anything but very simple substances, a brief discussion of these two extremes helps us understand why substances that have different kinds of chemical bonds have very different properties. Ionic compoundsA compound consisting of positively charged ions (cations) and negatively charged ions (anions) held together by strong electrostatic forces. consist of positively and negatively charged ions held together by strong electrostatic forces, whereas covalent compoundsA compound that consists of discrete molecules. generally consist of moleculesA group of atoms in which one or more pairs of electrons are shared between bonded atoms., which are groups of atoms in which one or more pairs of electrons are shared between bonded atoms. In a covalent bondThe electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons they share., the atoms are held together by the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons they share. We begin our discussion of structures and formulas by describing covalent compounds. The energetic factors involved in bond formation are described in more quantitative detail in .

Note the Pattern

Ionic compounds consist of ions of opposite charges held together by strong electrostatic forces, whereas pairs of electrons are shared between bonded atoms in covalent compounds.

Covalent Molecules and Compounds

Just as an atom is the simplest unit that has the fundamental chemical properties of an element, a molecule is the simplest unit that has the fundamental chemical properties of a covalent compound. Some pure elements exist as covalent molecules. Hydrogen, nitrogen, oxygen, and the halogens occur naturally as the diatomic (“two atoms”) molecules H2, N2, O2, F2, Cl2, Br2, and I2 (part (a) in ). Similarly, a few pure elements are polyatomicMolecules that contain more than two atoms. (“many atoms”) molecules, such as elemental phosphorus and sulfur, which occur as P4 and S8 (part (b) in ).

Each covalent compound is represented by a molecular formulaA representation of a covalent compound that consists of the atomic symbol for each component element (in a prescribed order) accompanied by a subscript indicating the number of atoms of that element in the molecule. The subscript is written only if the number is greater than 1., which gives the atomic symbol for each component element, in a prescribed order, accompanied by a subscript indicating the number of atoms of that element in the molecule. The subscript is written only if the number of atoms is greater than 1. For example, water, with two hydrogen atoms and one oxygen atom per molecule, is written as H2O. Similarly, carbon dioxide, which contains one carbon atom and two oxygen atoms in each molecule, is written as CO2.

Figure 2.1 Elements That Exist as Covalent Molecules

(a) Several elements naturally exist as diatomic molecules, in which two atoms (E) are joined by one or more covalent bonds to form a molecule with the general formula E2. (b) A few elements naturally exist as polyatomic molecules, which contain more than two atoms. For example, phosphorus exists as P4 tetrahedra—regular polyhedra with four triangular sides—with a phosphorus atom at each vertex. Elemental sulfur consists of a puckered ring of eight sulfur atoms connected by single bonds. Selenium is not shown due to the complexity of its structure.

Covalent compounds that contain predominantly carbon and hydrogen are called organic compoundsA covalent compound that contains predominantly carbon and hydrogen.. The convention for representing the formulas of organic compounds is to write carbon first, followed by hydrogen and then any other elements in alphabetical order (e.g., CH4O is methyl alcohol, a fuel). Compounds that consist primarily of elements other than carbon and hydrogen are called inorganic compoundsAn ionic or covalent compound that consists primarily of elements other than carbon and hydrogen.; they include both covalent and ionic compounds. In inorganic compounds, the component elements are listed beginning with the one farthest to the left in the periodic table (see ), such as we see in CO2 or SF6. Those in the same group are listed beginning with the lower element and working up, as in ClF. By convention, however, when an inorganic compound contains both hydrogen and an element from groups 13–15, the hydrogen is usually listed last in the formula. Examples are ammonia (NH3) and silane (SiH4). Compounds such as water, whose compositions were established long before this convention was adopted, are always written with hydrogen first: Water is always written as H2O, not OH2. The conventions for inorganic acids, such as hydrochloric acid (HCl) and sulfuric acid (H2SO4), are described in .

Note the Pattern

For organic compounds: write C first, then H, and then the other elements in alphabetical order. For molecular inorganic compounds: start with the element at far left in the periodic table; list elements in same group beginning with the lower element and working up.

Example 1

Write the molecular formula of each compound.

  1. The phosphorus-sulfur compound that is responsible for the ignition of so-called strike anywhere matches has 4 phosphorus atoms and 3 sulfur atoms per molecule.
  2. Ethyl alcohol, the alcohol of alcoholic beverages, has 1 oxygen atom, 2 carbon atoms, and 6 hydrogen atoms per molecule.
  3. Freon-11, once widely used in automobile air conditioners and implicated in damage to the ozone layer, has 1 carbon atom, 3 chlorine atoms, and 1 fluorine atom per molecule.

Given: identity of elements present and number of atoms of each

Asked for: molecular formula

Strategy:

A Identify the symbol for each element in the molecule. Then identify the substance as either an organic compound or an inorganic compound.

B If the substance is an organic compound, arrange the elements in order beginning with carbon and hydrogen and then list the other elements alphabetically. If it is an inorganic compound, list the elements beginning with the one farthest left in the periodic table. List elements in the same group starting with the lower element and working up.

C From the information given, add a subscript for each kind of atom to write the molecular formula.

Solution:

  1. A The molecule has 4 phosphorus atoms and 3 sulfur atoms. Because the compound does not contain mostly carbon and hydrogen, it is inorganic. B Phosphorus is in group 15, and sulfur is in group 16. Because phosphorus is to the left of sulfur, it is written first. C Writing the number of each kind of atom as a right-hand subscript gives P4S3 as the molecular formula.
  2. A Ethyl alcohol contains predominantly carbon and hydrogen, so it is an organic compound. B The formula for an organic compound is written with the number of carbon atoms first, the number of hydrogen atoms next, and the other atoms in alphabetical order: CHO. C Adding subscripts gives the molecular formula C2H6O.
  3. A Freon-11 contains carbon, chlorine, and fluorine. It can be viewed as either an inorganic compound or an organic compound (in which fluorine has replaced hydrogen). The formula for Freon-11 can therefore be written using either of the two conventions.

    B According to the convention for inorganic compounds, carbon is written first because it is farther left in the periodic table. Fluorine and chlorine are in the same group, so they are listed beginning with the lower element and working up: CClF. Adding subscripts gives the molecular formula CCl3F.

    C We obtain the same formula for Freon-11 using the convention for organic compounds. The number of carbon atoms is written first, followed by the number of hydrogen atoms (zero) and then the other elements in alphabetical order, also giving CCl3F.

Exercise

Write the molecular formula for each compound.

  1. Nitrous oxide, also called “laughing gas,” has 2 nitrogen atoms and 1 oxygen atom per molecule. Nitrous oxide is used as a mild anesthetic for minor surgery and as the propellant in cans of whipped cream.
  2. Sucrose, also known as cane sugar, has 12 carbon atoms, 11 oxygen atoms, and 22 hydrogen atoms.
  3. Sulfur hexafluoride, a gas used to pressurize “unpressurized” tennis balls and as a coolant in nuclear reactors, has 6 fluorine atoms and 1 sulfur atom per molecule.

Answer:

  1. N2O
  2. C12H22O11
  3. SF6

Representations of Molecular Structures

Molecular formulas give only the elemental composition of molecules. In contrast, structural formulasA representation of a molecule that shows which atoms are bonded to one another and, in some cases, the approximate arrangement of atoms in space. show which atoms are bonded to one another and, in some cases, the approximate arrangement of the atoms in space. Knowing the structural formula of a compound enables chemists to create a three-dimensional model, which provides information about how that compound will behave physically and chemically.

The structural formula for H2 can be drawn as H–H and that for I2 as I–I, where the line indicates a single pair of shared electrons, a single bondA chemical bond formed when two atoms share a single pair of electrons.. Two pairs of electrons are shared in a double bondA chemical bond formed when two atoms share two pairs of electrons., which is indicated by two lines— for example, O2 is O=O. Three electron pairs are shared in a triple bondA chemical bond formed when two atoms share three pairs of electrons., which is indicated by three lines—for example, N2 is N≡N (see ). Carbon is unique in the extent to which it forms single, double, and triple bonds to itself and other elements. The number of bonds formed by an atom in its covalent compounds is not arbitrary. As you will learn in , hydrogen, oxygen, nitrogen, and carbon have a very strong tendency to form substances in which they have one, two, three, and four bonds to other atoms, respectively ().

Figure 2.2 Molecules That Contain Single, Double, and Triple Bonds

Hydrogen (H2) has a single bond between atoms. Oxygen (O2) has a double bond between atoms, indicated by two lines (=). Nitrogen (N2) has a triple bond between atoms, indicated by three lines (≡). Each bond represents an electron pair.

Table 2.1 The Number of Bonds That Selected Atoms Commonly Form to Other Atoms

Atom Number of Bonds
H (group 1) 1
O (group 16) 2
N (group 15) 3
C (group 14) 4

The structural formula for water can be drawn as follows:

Because the latter approximates the experimentally determined shape of the water molecule, it is more informative. Similarly, ammonia (NH3) and methane (CH4) are often written as planar molecules:

As shown in , however, the actual three-dimensional structure of NH3 looks like a pyramid with a triangular base of three hydrogen atoms. The structure of CH4, with four hydrogen atoms arranged around a central carbon atom as shown in , is tetrahedral. That is, the hydrogen atoms are positioned at every other vertex of a cube. Many compounds—carbon compounds, in particular—have four bonded atoms arranged around a central atom to form a tetrahedron.

Figure 2.3 The Three-Dimensional Structures of Water, Ammonia, and Methane

(a) Water is a V-shaped molecule, in which all three atoms lie in a plane. (b) In contrast, ammonia has a pyramidal structure, in which the three hydrogen atoms form the base of the pyramid and the nitrogen atom is at the vertex. (c) The four hydrogen atoms of methane form a tetrahedron; the carbon atom lies in the center.

CH4. Methane has a three-dimensional, tetrahedral structure.

, , and illustrate different ways to represent the structures of molecules. It should be clear that there is no single “best” way to draw the structure of a molecule; the method you use depends on which aspect of the structure you want to emphasize and how much time and effort you want to spend. shows some of the different ways to portray the structure of a slightly more complex molecule: methanol. These representations differ greatly in their information content. For example, the molecular formula for methanol (part (a) in ) gives only the number of each kind of atom; writing methanol as CH4O tells nothing about its structure. In contrast, the structural formula (part (b) in ) indicates how the atoms are connected, but it makes methanol look as if it is planar (which it is not). Both the ball-and-stick model (part (c) in ) and the perspective drawing (part (d) in ) show the three-dimensional structure of the molecule. The latter (also called a wedge-and-dash representation) is the easiest way to sketch the structure of a molecule in three dimensions. It shows which atoms are above and below the plane of the paper by using wedges and dashes, respectively; the central atom is always assumed to be in the plane of the paper. The space-filling model (part (e) in ) illustrates the approximate relative sizes of the atoms in the molecule, but it does not show the bonds between the atoms. Also, in a space-filling model, atoms at the “front” of the molecule may obscure atoms at the “back.”

Figure 2.4 Different Ways of Representing the Structure of a Molecule

(a) The molecular formula for methanol gives only the number of each kind of atom present. (b) The structural formula shows which atoms are connected. (c) The ball-and-stick model shows the atoms as spheres and the bonds as sticks. (d) A perspective drawing (also called a wedge-and-dash representation) attempts to show the three-dimensional structure of the molecule. (e) The space-filling model shows the atoms in the molecule but not the bonds. (f) The condensed structural formula is by far the easiest and most common way to represent a molecule.

Although a structural formula, a ball-and-stick model, a perspective drawing, and a space-filling model provide a significant amount of information about the structure of a molecule, each requires time and effort. Consequently, chemists often use a condensed structural formula (part (f) in ), which omits the lines representing bonds between atoms and simply lists the atoms bonded to a given atom next to it. Multiple groups attached to the same atom are shown in parentheses, followed by a subscript that indicates the number of such groups. For example, the condensed structural formula for methanol is CH3OH, which tells us that the molecule contains a CH3 unit that looks like a fragment of methane (CH4). Methanol can therefore be viewed either as a methane molecule in which one hydrogen atom has been replaced by an –OH group or as a water molecule in which one hydrogen atom has been replaced by a –CH3 fragment. Because of their ease of use and information content, we use condensed structural formulas for molecules throughout this text. Ball-and-stick models are used when needed to illustrate the three-dimensional structure of molecules, and space-filling models are used only when it is necessary to visualize the relative sizes of atoms or molecules to understand an important point.

Example 2

Write the molecular formula for each compound. The condensed structural formula is given.

  1. Sulfur monochloride (also called disulfur dichloride) is a vile-smelling, corrosive yellow liquid used in the production of synthetic rubber. Its condensed structural formula is ClSSCl.
  2. Ethylene glycol is the major ingredient in antifreeze. Its condensed structural formula is HOCH2CH2OH.
  3. Trimethylamine is one of the substances responsible for the smell of spoiled fish. Its condensed structural formula is (CH3)3N.

Given: condensed structural formula

Asked for: molecular formula

Strategy:

A Identify every element in the condensed structural formula and then determine whether the compound is organic or inorganic.

B As appropriate, use either organic or inorganic convention to list the elements. Then add appropriate subscripts to indicate the number of atoms of each element present in the molecular formula.

Solution:

The molecular formula lists the elements in the molecule and the number of atoms of each.

  1. A Each molecule of sulfur monochloride has two sulfur atoms and two chlorine atoms. Because it does not contain mostly carbon and hydrogen, it is an inorganic compound. B Sulfur lies to the left of chlorine in the periodic table, so it is written first in the formula. Adding subscripts gives the molecular formula S2Cl2.
  2. A Counting the atoms in ethylene glycol, we get six hydrogen atoms, two carbon atoms, and two oxygen atoms per molecule. The compound consists mostly of carbon and hydrogen atoms, so it is organic. B As with all organic compounds, C and H are written first in the molecular formula. Adding appropriate subscripts gives the molecular formula C2H6O2.
  3. A The condensed structural formula shows that trimethylamine contains three CH3 units, so we have one nitrogen atom, three carbon atoms, and nine hydrogen atoms per molecule. Because trimethylamine contains mostly carbon and hydrogen, it is an organic compound. B According to the convention for organic compounds, C and H are written first, giving the molecular formula C3H9N.

Exercise

Write the molecular formula for each molecule.

  1. Chloroform, which was one of the first anesthetics and was used in many cough syrups until recently, contains one carbon atom, one hydrogen atom, and three chlorine atoms. Its condensed structural formula is CHCl3.
  2. Hydrazine is used as a propellant in the attitude jets of the space shuttle. Its condensed structural formula is H2NNH2.
  3. Putrescine is a pungent-smelling compound first isolated from extracts of rotting meat. Its condensed structural formula is H2NCH2CH2CH2CH2NH2. This is often written as H2N(CH2)4NH2 to indicate that there are four CH2 fragments linked together.

Answer:

  1. CHCl3
  2. N2H4
  3. C4H12N2

Ionic Compounds

The substances described in the preceding discussion are composed of molecules that are electrically neutral; that is, the number of positively charged protons in the nucleus is equal to the number of negatively charged electrons. In contrast, ions are atoms or assemblies of atoms that have a net electrical charge. Ions that contain fewer electrons than protons have a net positive charge and are called cationsAn ion that has fewer electrons than protons, resulting in a net positive charge.. Conversely, ions that contain more electrons than protons have a net negative charge and are called anionsAn ion that has fewer protons than electrons, resulting in a net negative charge.. Ionic compounds contain both cations and anions in a ratio that results in no net electrical charge.

Note the Pattern

Ionic compounds contain both cations and anions in a ratio that results in zero electrical charge.

In covalent compounds, electrons are shared between bonded atoms and are simultaneously attracted to more than one nucleus. In contrast, ionic compounds contain cations and anions rather than discrete neutral molecules. Ionic compounds are held together by the attractive electrostatic interactions between cations and anions. In an ionic compound, the cations and anions are arranged in space to form an extended three-dimensional array that maximizes the number of attractive electrostatic interactions and minimizes the number of repulsive electrostatic interactions (). As shown in , the electrostatic energy of the interaction between two charged particles is proportional to the product of the charges on the particles and inversely proportional to the distance between them:

Equation 2.1

electrostatic energyQ1Q2r

where Q1 and Q2 are the electrical charges on particles 1 and 2, and r is the distance between them. When Q1 and Q2 are both positive, corresponding to the charges on cations, the cations repel each other and the electrostatic energy is positive. When Q1 and Q2 are both negative, corresponding to the charges on anions, the anions repel each other and the electrostatic energy is again positive. The electrostatic energy is negative only when the charges have opposite signs; that is, positively charged species are attracted to negatively charged species and vice versa. As shown in , the strength of the interaction is proportional to the magnitude of the charges and decreases as the distance between the particles increases. We will return to these energetic factors in , where they are described in greater quantitative detail.

Note the Pattern

If the electrostatic energy is positive, the particles repel each other; if the electrostatic energy is negative, the particles are attracted to each other.

Figure 2.5 Covalent and Ionic Bonding

(a) In molecular hydrogen (H2), two hydrogen atoms share two electrons to form a covalent bond. (b) The ionic compound NaCl forms when electrons from sodium atoms are transferred to chlorine atoms. The resulting Na+ and Cl ions form a three-dimensional solid that is held together by attractive electrostatic interactions.

Figure 2.6 The Effect of Charge and Distance on the Strength of Electrostatic Interactions

As the charge on ions increases or the distance between ions decreases, so does the strength of the attractive (−…+) or repulsive (−…− or +…+) interactions. The strength of these interactions is represented by the thickness of the arrows.

One example of an ionic compound is sodium chloride (NaCl; ), formed from sodium and chlorine. In forming chemical compounds, many elements have a tendency to gain or lose enough electrons to attain the same number of electrons as the noble gas closest to them in the periodic table. When sodium and chlorine come into contact, each sodium atom gives up an electron to become a Na+ ion, with 11 protons in its nucleus but only 10 electrons (like neon), and each chlorine atom gains an electron to become a Cl ion, with 17 protons in its nucleus and 18 electrons (like argon), as shown in part (b) in . Solid sodium chloride contains equal numbers of cations (Na+) and anions (Cl), thus maintaining electrical neutrality. Each Na+ ion is surrounded by 6 Cl ions, and each Cl ion is surrounded by 6 Na+ ions. Because of the large number of attractive Na+Cl interactions, the total attractive electrostatic energy in NaCl is great.

Figure 2.7 Sodium Chloride: an Ionic Solid

The planes of an NaCl crystal reflect the regular three-dimensional arrangement of its Na+ (purple) and Cl (green) ions.

Consistent with a tendency to have the same number of electrons as the nearest noble gas, when forming ions, elements in groups 1, 2, and 3 tend to lose one, two, and three electrons, respectively, to form cations, such as Na+ and Mg2+. They then have the same number of electrons as the nearest noble gas: neon. Similarly, K+, Ca2+, and Sc3+ have 18 electrons each, like the nearest noble gas: argon. In addition, the elements in group 13 lose three electrons to form cations, such as Al3+, again attaining the same number of electrons as the noble gas closest to them in the periodic table. Because the lanthanides and actinides formally belong to group 3, the most common ion formed by these elements is M3+, where M represents the metal. Conversely, elements in groups 17, 16, and 15 often react to gain one, two, and three electrons, respectively, to form ions such as Cl, S2−, and P3−. Ions such as these, which contain only a single atom, are called monatomic ionsAn ion with only a single atom.. You can predict the charges of most monatomic ions derived from the main group elements by simply looking at the periodic table and counting how many columns an element lies from the extreme left or right. For example, you can predict that barium (in group 2) will form Ba2+ to have the same number of electrons as its nearest noble gas, xenon, that oxygen (in group 16) will form O2− to have the same number of electrons as neon, and cesium (in group 1) will form Cs+ to also have the same number of electrons as xenon. Note that this method does not usually work for most of the transition metals, as you will learn in . Some common monatomic ions are in .

Note the Pattern

Elements in groups 1, 2, and 3 tend to form 1+, 2+, and 3+ ions, respectively; elements in groups 15, 16, and 17 tend to form 3−, 2−, and 1− ions, respectively.

Table 2.2 Some Common Monatomic Ions and Their Names

Group 1 Group 2 Group 3 Group 13 Group 15 Group 16 Group 17

Li+

lithium

Be2+

beryllium

N3−

nitride

(azide)

O2−

oxide

F

fluoride

Na+

sodium

Mg2+

magnesium

Al3+

aluminum

P3−

phosphide

S2−

sulfide

Cl

chloride

K+

potassium

Ca2+

calcium

Sc3+

scandium

Ga3+

gallium

As3−

arsenide

Se2−

selenide

Br

bromide

Rb+

rubidium

Sr2+

strontium

Y3+

yttrium

In3+

indium

Te2−

telluride

I

iodide

Cs+

cesium

Ba2+

barium

La3+

lanthanum

Example 3

Predict the charge on the most common monatomic ion formed by each element.

  1. aluminum, used in the quantum logic clock, the world’s most precise clock
  2. selenium, used to make ruby-colored glass
  3. yttrium, used to make high-performance spark plugs

Given: element

Asked for: ionic charge

Strategy:

A Identify the group in the periodic table to which the element belongs. Based on its location in the periodic table, decide whether the element is a metal, which tends to lose electrons; a nonmetal, which tends to gain electrons; or a semimetal, which can do either.

B After locating the noble gas that is closest to the element, determine the number of electrons the element must gain or lose to have the same number of electrons as the nearest noble gas.

Solution:

  1. A Aluminum is a metal in group 13; consequently, it will tend to lose electrons. B The nearest noble gas to aluminum is neon. Aluminum will lose three electrons to form the Al3+ ion, which has the same number of electrons as neon.
  2. A Selenium is a nonmetal in group 16, so it will tend to gain electrons. B The nearest noble gas is krypton, so we predict that selenium will gain two electrons to form the Se2− ion, which has the same number of electrons as krypton.
  3. A Yttrium is in group 3, and elements in this group are metals that tend to lose electrons. B The nearest noble gas to yttrium is krypton, so yttrium is predicted to lose three electrons to form Y3+, which has the same number of electrons as krypton.

Exercise

Predict the charge on the most common monatomic ion formed by each element.

  1. calcium, used to prevent osteoporosis
  2. iodine, required for the synthesis of thyroid hormones
  3. zirconium, widely used in nuclear reactors

Answer:

  1. Ca2+
  2. I
  3. Zr4+

Physical Properties of Ionic and Covalent Compounds

In general, ionic and covalent compounds have different physical properties. Ionic compounds usually form hard crystalline solids that melt at rather high temperatures and are very resistant to evaporation. These properties stem from the characteristic internal structure of an ionic solid, illustrated schematically in part (a) in , which shows the three-dimensional array of alternating positive and negative ions held together by strong electrostatic attractions. In contrast, as shown in part (b) in , most covalent compounds consist of discrete molecules held together by comparatively weak intermolecular forces (the forces between molecules), even though the atoms within each molecule are held together by strong intramolecular covalent bonds (the forces within the molecule). Covalent substances can be gases, liquids, or solids at room temperature and pressure, depending on the strength of the intermolecular interactions. Covalent molecular solids tend to form soft crystals that melt at rather low temperatures and evaporate relatively easily.Some covalent substances, however, are not molecular but consist of infinite three-dimensional arrays of covalently bonded atoms and include some of the hardest materials known, such as diamond. This topic will be addressed in . The covalent bonds that hold the atoms together in the molecules are unaffected when covalent substances melt or evaporate, so a liquid or vapor of discrete, independent molecules is formed. For example, at room temperature, methane, the major constituent of natural gas, is a gas that is composed of discrete CH4 molecules. A comparison of the different physical properties of ionic compounds and covalent molecular substances is given in .

Table 2.3 The Physical Properties of Typical Ionic Compounds and Covalent Molecular Substances

Ionic Compounds Covalent Molecular Substances
hard solids gases, liquids, or soft solids
high melting points low melting points
nonvolatile volatile

Figure 2.8 Interactions in Ionic and Covalent Solids

(a) The positively and negatively charged ions in an ionic solid such as sodium chloride (NaCl) are held together by strong electrostatic interactions. (b) In this representation of the packing of methane (CH4) molecules in solid methane, a prototypical molecular solid, the methane molecules are held together in the solid only by relatively weak intermolecular forces, even though the atoms within each methane molecule are held together by strong covalent bonds.

Summary

The atoms in chemical compounds are held together by attractive electrostatic interactions known as chemical bonds. Ionic compounds contain positively and negatively charged ions in a ratio that results in an overall charge of zero. The ions are held together in a regular spatial arrangement by electrostatic forces. Most covalent compounds consist of molecules, groups of atoms in which one or more pairs of electrons are shared by at least two atoms to form a covalent bond. The atoms in molecules are held together by the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons shared by the nuclei. The molecular formula of a covalent compound gives the types and numbers of atoms present. Compounds that contain predominantly carbon and hydrogen are called organic compounds, whereas compounds that consist primarily of elements other than carbon and hydrogen are inorganic compounds. Diatomic molecules contain two atoms, and polyatomic molecules contain more than two. A structural formula indicates the composition and approximate structure and shape of a molecule. Single bonds, double bonds, and triple bonds are covalent bonds in which one, two, and three pairs of electrons, respectively, are shared between two bonded atoms. Atoms or groups of atoms that possess a net electrical charge are called ions; they can have either a positive charge (cations) or a negative charge (anions). Ions can consist of one atom (monatomic ions) or several (polyatomic ions). The charges on monatomic ions of most main group elements can be predicted from the location of the element in the periodic table. Ionic compounds usually form hard crystalline solids with high melting points. Covalent molecular compounds, in contrast, consist of discrete molecules held together by weak intermolecular forces and can be gases, liquids, or solids at room temperature and pressure.

Key Takeaway

  • There are two fundamentally different kinds of chemical bonds (covalent and ionic) that cause substances to have very different properties.

Conceptual Problems

  1. Ionic and covalent compounds are held together by electrostatic attractions between oppositely charged particles. Describe the differences in the nature of the attractions in ionic and covalent compounds. Which class of compounds contains pairs of electrons shared between bonded atoms?

  2. Which contains fewer electrons than the neutral atom—the corresponding cation or the anion?

  3. What is the difference between an organic compound and an inorganic compound?

  4. What is the advantage of writing a structural formula as a condensed formula?

  5. The majority of elements that exist as diatomic molecules are found in one group of the periodic table. Identify the group.

  6. Discuss the differences between covalent and ionic compounds with regard to

    1. the forces that hold the atoms together.
    2. melting points.
    3. physical states at room temperature and pressure.
  7. Why do covalent compounds generally tend to have lower melting points than ionic compounds?

Answer

  1. Covalent compounds generally melt at lower temperatures than ionic compounds because the intermolecular interactions that hold the molecules together in a molecular solid are weaker than the electrostatic attractions that hold oppositely charged ions together in an ionic solid.

Numerical Problems

  1. The structural formula for chloroform (CHCl3) was shown in Example 2. Based on this information, draw the structural formula of dichloromethane (CH2Cl2).

  2. What is the total number of electrons present in each ion?

    1. F
    2. Rb+
    3. Ce3+
    4. Zr4+
    5. Zn2+
    6. Kr2+
    7. B3+
  3. What is the total number of electrons present in each ion?

    1. Ca2+
    2. Se2−
    3. In3+
    4. Sr2+
    5. As3+
    6. N3−
    7. Tl+
  4. Predict how many electrons are in each ion.

    1. an oxygen ion with a −2 charge
    2. a beryllium ion with a +2 charge
    3. a silver ion with a +1 charge
    4. a selenium ion with a +4 charge
    5. an iron ion with a +2 charge
    6. a chlorine ion with a −1 charge
  5. Predict how many electrons are in each ion.

    1. a copper ion with a +2 charge
    2. a molybdenum ion with a +4 charge
    3. an iodine ion with a −1 charge
    4. a gallium ion with a +3 charge
    5. an ytterbium ion with a +3 charge
    6. a scandium ion with a +3 charge
  6. Predict the charge on the most common monatomic ion formed by each element.

    1. chlorine
    2. phosphorus
    3. scandium
    4. magnesium
    5. arsenic
    6. oxygen
  7. Predict the charge on the most common monatomic ion formed by each element.

    1. sodium
    2. selenium
    3. barium
    4. rubidium
    5. nitrogen
    6. aluminum
  8. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group of the periodic table in which the element is found.

    1. X492+
    2. X11
    3. X 8162–
  9. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group of the periodic table in which the element is found.

    1. X37+
    2. X 919
    3. X13273+

Answers

    1. 27
    2. 38
    3. 54
    4. 28
    5. 67
    6. 18
    1. Li, Li+, 2nd period, group 1
    2. F, F, 2nd period, group 17
    3. Al, Al3+, 3nd period, group 13

2.2 Chemical Formulas

Learning Objective

  1. To describe the composition of a chemical compound.

When chemists synthesize a new compound, they may not yet know its molecular or structural formula. In such cases, they usually begin by determining its empirical formulaA formula for a compound that consists of the atomic symbol for each component element accompanied by a subscript indicating the relative number of atoms of that element in the compound, reduced to the smallest whole numbers., the relative numbers of atoms of the elements in a compound, reduced to the smallest whole numbers. Because the empirical formula is based on experimental measurements of the numbers of atoms in a sample of the compound, it shows only the ratios of the numbers of the elements present. The difference between empirical and molecular formulas can be illustrated with butane, a covalent compound used as the fuel in disposable lighters. The molecular formula for butane is C4H10. The ratio of carbon atoms to hydrogen atoms in butane is 4:10, which can be reduced to 2:5. The empirical formula for butane is therefore C2H5. The formula unitThe absolute grouping of atoms or ions represented by the empirical formula. is the absolute grouping of atoms or ions represented by the empirical formula of a compound, either ionic or covalent. Butane, for example, has the empirical formula C2H5, but it contains two C2H5 formula units, giving a molecular formula of C4H10.

Because ionic compounds do not contain discrete molecules, empirical formulas are used to indicate their compositions. All compounds, whether ionic or covalent, must be electrically neutral. Consequently, the positive and negative charges in a formula unit must exactly cancel each other. If the cation and the anion have charges of equal magnitude, such as Na+ and Cl, then the compound must have a 1:1 ratio of cations to anions, and the empirical formula must be NaCl. If the charges are not the same magnitude, then a cation:anion ratio other than 1:1 is needed to produce a neutral compound. In the case of Mg2+ and Cl, for example, two Cl ions are needed to balance the two positive charges on each Mg2+ ion, giving an empirical formula of MgCl2. Similarly, the formula for the ionic compound that contains Na+ and O2− ions is Na2O.

Note the Pattern

Ionic compounds do not contain discrete molecules, so empirical formulas are used to indicate their compositions.

Binary Ionic Compounds

An ionic compound that contains only two elements, one present as a cation and one as an anion, is called a binary ionic compoundAn ionic compound that contains only two elements, one present as a cation and one as an anion.. One example is MgCl2, a coagulant used in the preparation of tofu from soybeans. For binary ionic compounds, the subscripts in the empirical formula can also be obtained by crossing charges: use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically as follows:

Crossing charges. One method for obtaining subscripts in the empirical formula is by crossing charges.

When crossing charges, you will sometimes find it necessary to reduce the subscripts to their simplest ratio to write the empirical formula. Consider, for example, the compound formed by Mg2+ and O2−. Using the absolute values of the charges on the ions as subscripts gives the formula Mg2O2:

This simplifies to its correct empirical formula MgO. The empirical formula has one Mg2+ ion and one O2− ion.

Example 4

Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair.

  1. Ga3+ and As3−
  2. Eu3+ and O2−
  3. calcium and chlorine

Given: ions or elements

Asked for: empirical formula for binary ionic compound

Strategy:

A If not given, determine the ionic charges based on the location of the elements in the periodic table.

B Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the lowest numbers to write the empirical formula. Check to make sure the empirical formula is electrically neutral.

Solution:

  1. B Using the absolute values of the charges on the ions as the subscripts gives Ga3As3:

    Reducing the subscripts to the smallest whole numbers gives the empirical formula GaAs, which is electrically neutral [+3 + (−3) = 0]. Alternatively, we could recognize that Ga3+ and As3− have charges of equal magnitude but opposite signs. One Ga3+ ion balances the charge on one As3− ion, and a 1:1 compound will have no net charge. Because we write subscripts only if the number is greater than 1, the empirical formula is GaAs. GaAs is gallium arsenide, which is widely used in the electronics industry in transistors and other devices.

  2. B Because Eu3+ has a charge of +3 and O2− has a charge of −2, a 1:1 compound would have a net charge of +1. We must therefore find multiples of the charges that cancel. We cross charges, using the absolute value of the charge on one ion as the subscript for the other ion:

    The subscript for Eu3+ is 2 (from O2−), and the subscript for O2− is 3 (from Eu3+), giving Eu2O3; the subscripts cannot be reduced further. The empirical formula contains a positive charge of 2(+3) = +6 and a negative charge of 3(−2) = −6, for a net charge of 0. The compound Eu2O3 is neutral. Europium oxide is responsible for the red color in television and computer screens.

  3. A Because the charges on the ions are not given, we must first determine the charges expected for the most common ions derived from calcium and chlorine. Calcium lies in group 2, so it should lose two electrons to form Ca2+. Chlorine lies in group 17, so it should gain one electron to form Cl.

    B Two Cl ions are needed to balance the charge on one Ca2+ ion, which leads to the empirical formula CaCl2. We could also cross charges, using the absolute value of the charge on Ca2+ as the subscript for Cl and the absolute value of the charge on Cl as the subscript for Ca:

    The subscripts in CaCl2 cannot be reduced further. The empirical formula is electrically neutral [+2 + 2(−1) = 0]. This compound is calcium chloride, one of the substances used as “salt” to melt ice on roads and sidewalks in winter.

Exercise

Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair.

  1. Li+ and N3−
  2. Al3+ and O2−
  3. lithium and oxygen

Answer:

  1. Li3N
  2. Al2O3
  3. Li2O

Polyatomic Ions

Polyatomic ionsA group of two or more atoms that has a net electrical charge. are groups of atoms that bear a net electrical charge, although the atoms in a polyatomic ion are held together by the same covalent bonds that hold atoms together in molecules. Just as there are many more kinds of molecules than simple elements, there are many more kinds of polyatomic ions than monatomic ions. Two examples of polyatomic cations are the ammonium (NH4+) and the methylammonium (CH3NH3+) ions. Polyatomic anions are much more numerous than polyatomic cations; some common examples are in .

Table 2.4 Common Polyatomic Ions and Their Names

Formula Name of Ion
NH4+ ammonium
CH3NH3+ methylammonium
OH hydroxide
O22− peroxide
CN cyanide
SCN thiocyanate
NO2 nitrite
NO3 nitrate
CO32− carbonate
HCO3 hydrogen carbonate, or bicarbonate
SO32− sulfite
SO42− sulfate
HSO4 hydrogen sulfate, or bisulfate
PO43− phosphate
HPO42− hydrogen phosphate
H2PO4 dihydrogen phosphate
ClO hypochlorite
ClO2 chlorite
ClO3 chlorate
ClO4 perchlorate
MnO4 permanganate
CrO42− chromate
Cr2O72− dichromate
C2O42− oxalate
HCO2 formate
CH3CO2 acetate
C6H5CO2 benzoate

The method we used to predict the empirical formulas for ionic compounds that contain monatomic ions can also be used for compounds that contain polyatomic ions. The overall charge on the cations must balance the overall charge on the anions in the formula unit. Thus K+ and NO3 ions combine in a 1:1 ratio to form KNO3 (potassium nitrate or saltpeter), a major ingredient in black gunpowder. Similarly, Ca2+ and SO42− form CaSO4 (calcium sulfate), which combines with varying amounts of water to form gypsum and plaster of Paris. The polyatomic ions NH4+ and NO3 form NH4NO3 (ammonium nitrate), which is a widely used fertilizer and, in the wrong hands, an explosive. One example of a compound in which the ions have charges of different magnitudes is calcium phosphate, which is composed of Ca2+ and PO43− ions; it is a major component of bones. The compound is electrically neutral because the ions combine in a ratio of three Ca2+ ions [3(+2) = +6] for every two ions [2(−3) = −6], giving an empirical formula of Ca3(PO4)2; the parentheses around PO4 in the empirical formula indicate that it is a polyatomic ion. Writing the formula for calcium phosphate as Ca3P2O8 gives the correct number of each atom in the formula unit, but it obscures the fact that the compound contains readily identifiable PO43− ions.

Example 5

Write the empirical formula for the compound formed from each ion pair.

  1. Na+ and HPO42−
  2. potassium cation and cyanide anion
  3. calcium cation and hypochlorite anion

Given: ions

Asked for: empirical formula for ionic compound

Strategy:

A If it is not given, determine the charge on a monatomic ion from its location in the periodic table. Use to find the charge on a polyatomic ion.

B Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the smallest whole numbers when writing the empirical formula.

Solution:

  1. B Because HPO42− has a charge of −2 and Na+ has a charge of +1, the empirical formula requires two Na+ ions to balance the charge of the polyatomic ion, giving Na2HPO4. The subscripts are reduced to the lowest numbers, so the empirical formula is Na2HPO4. This compound is sodium hydrogen phosphate, which is used to provide texture in processed cheese, puddings, and instant breakfasts.
  2. A The potassium cation is K+, and the cyanide anion is CN. B Because the magnitude of the charge on each ion is the same, the empirical formula is KCN. Potassium cyanide is highly toxic, and at one time it was used as rat poison. This use has been discontinued, however, because too many people were being poisoned accidentally.
  3. A The calcium cation is Ca2+, and the hypochlorite anion is ClO. B Two ClO ions are needed to balance the charge on one Ca2+ ion, giving Ca(ClO)2. The subscripts cannot be reduced further, so the empirical formula is Ca(ClO)2. This is calcium hypochlorite, the “chlorine” used to purify water in swimming pools.

Exercise

Write the empirical formula for the compound formed from each ion pair.

  1. Ca2+ and H2PO4
  2. sodium cation and bicarbonate anion
  3. ammonium cation and sulfate anion

Answer:

  1. Ca(H2PO4)2: calcium dihydrogen phosphate is one of the ingredients in baking powder.
  2. NaHCO3: sodium bicarbonate is found in antacids and baking powder; in pure form, it is sold as baking soda.
  3. (NH4)2SO4: ammonium sulfate is a common source of nitrogen in fertilizers.

Hydrates

Many ionic compounds occur as hydratesA compound that contains specific ratios of loosely bound water molecules, called waters of hydration., compounds that contain specific ratios of loosely bound water molecules, called waters of hydrationThe loosely bound water molecules in hydrate compounds. These waters of hydration can often be removed by simply heating the compound.. Waters of hydration can often be removed simply by heating. For example, calcium dihydrogen phosphate can form a solid that contains one molecule of water per Ca(H2PO4)2 unit and is used as a leavening agent in the food industry to cause baked goods to rise. The empirical formula for the solid is Ca(H2PO4)2·H2O. In contrast, copper sulfate usually forms a blue solid that contains five waters of hydration per formula unit, with the empirical formula CuSO4·5H2O. When heated, all five water molecules are lost, giving a white solid with the empirical formula CuSO4 ().

Figure 2.9 Loss of Water from a Hydrate with Heating

When blue CuSO4·5H2O is heated, two molecules of water are lost at 30°C, two more at 110°C, and the last at 250°C to give white CuSO4.

Compounds that differ only in the numbers of waters of hydration can have very different properties. For example, CaSO4·½H2O is plaster of Paris, which was often used to make sturdy casts for broken arms or legs, whereas CaSO4·2H2O is the less dense, flakier gypsum, a mineral used in drywall panels for home construction. When a cast would set, a mixture of plaster of Paris and water crystallized to give solid CaSO4·2H2O. Similar processes are used in the setting of cement and concrete.

Summary

An empirical formula gives the relative numbers of atoms of the elements in a compound, reduced to the lowest whole numbers. The formula unit is the absolute grouping represented by the empirical formula of a compound, either ionic or covalent. Empirical formulas are particularly useful for describing the composition of ionic compounds, which do not contain readily identifiable molecules. Some ionic compounds occur as hydrates, which contain specific ratios of loosely bound water molecules called waters of hydration.

Key Takeaway

  • The composition of a compound is represented by an empirical or molecular formula, each consisting of at least one formula unit.

Conceptual Problems

  1. What are the differences and similarities between a polyatomic ion and a molecule?

  2. Classify each compound as ionic or covalent.

    1. Zn3(PO4)2
    2. C6H5CO2H
    3. K2Cr2O7
    4. CH3CH2SH
    5. NH4Br
    6. CCl2F2
  3. Classify each compound as ionic or covalent. Which are organic compounds and which are inorganic compounds?

    1. CH3CH2CO2H
    2. CaCl2
    3. Y(NO3)3
    4. H2S
    5. NaC2H3O2
  4. Generally, one cannot determine the molecular formula directly from an empirical formula. What other information is needed?

  5. Give two pieces of information that we obtain from a structural formula that we cannot obtain from an empirical formula.

  6. The formulas of alcohols are often written as ROH rather than as empirical formulas. For example, methanol is generally written as CH3OH rather than CH4O. Explain why the ROH notation is preferred.

  7. The compound dimethyl sulfide has the empirical formula C2H6S and the structural formula CH3SCH3. What information do we obtain from the structural formula that we do not get from the empirical formula? Write the condensed structural formula for the compound.

  8. What is the correct formula for magnesium hydroxide—MgOH2 or Mg(OH)2? Why?

  9. Magnesium cyanide is written as Mg(CN)2, not MgCN2. Why?

  10. Does a given hydrate always contain the same number of waters of hydration?

Answer

  1. The structural formula gives us the connectivity of the atoms in the molecule or ion, as well as a schematic representation of their arrangement in space. Empirical formulas tell us only the ratios of the atoms present. The condensed structural formula of dimethylsulfide is (CH3)2S.

Numerical Problems

  1. Write the formula for each compound.

    1. magnesium sulfate, which has 1 magnesium atom, 4 oxygen atoms, and 1 sulfur atom
    2. ethylene glycol (antifreeze), which has 6 hydrogen atoms, 2 carbon atoms, and 2 oxygen atoms
    3. acetic acid, which has 2 oxygen atoms, 2 carbon atoms, and 4 hydrogen atoms
    4. potassium chlorate, which has 1 chlorine atom, 1 potassium atom, and 3 oxygen atoms
    5. sodium hypochlorite pentahydrate, which has 1 chlorine atom, 1 sodium atom, 6 oxygen atoms, and 10 hydrogen atoms
  2. Write the formula for each compound.

    1. cadmium acetate, which has 1 cadmium atom, 4 oxygen atoms, 4 carbon atoms, and 6 hydrogen atoms
    2. barium cyanide, which has 1 barium atom, 2 carbon atoms, and 2 nitrogen atoms
    3. iron(III) phosphate dihydrate, which has 1 iron atom, 1 phosphorus atom, 6 oxygen atoms, and 4 hydrogen atoms
    4. manganese(II) nitrate hexahydrate, which has 1 manganese atom, 12 hydrogen atoms, 12 oxygen atoms, and 2 nitrogen atoms
    5. silver phosphate, which has 1 phosphorus atom, 3 silver atoms, and 4 oxygen atoms
  3. Complete the following table by filling in the formula for the ionic compound formed by each cation-anion pair.

    Ion K+ Fe3+ NH4+ Ba2+
    Cl KCl
    SO42−
    PO43−
    NO3
    OH
  4. Write the empirical formula for the binary compound formed by the most common monatomic ions formed by each pair of elements.

    1. zinc and sulfur
    2. barium and iodine
    3. magnesium and chlorine
    4. silicon and oxygen
    5. sodium and sulfur
  5. Write the empirical formula for the binary compound formed by the most common monatomic ions formed by each pair of elements.

    1. lithium and nitrogen
    2. cesium and chlorine
    3. germanium and oxygen
    4. rubidium and sulfur
    5. arsenic and sodium
  6. Write the empirical formula for each compound.

    1. Na2S2O4
    2. B2H6
    3. C6H12O6
    4. P4O10
    5. KMnO4
  7. Write the empirical formula for each compound.

    1. Al2Cl6
    2. K2Cr2O7
    3. C2H4
    4. (NH2)2CNH
    5. CH3COOH

Answers

    1. MgSO4
    2. C2H6O2
    3. C2H4O2
    4. KClO3
    5. NaOCl·5H2O
  1. Ion K + Fe 3+ NH 4+ Ba 2+
    Cl KCl FeCl3 NH4Cl BaCl2
    SO 4 2− K2SO4 Fe2(SO4)3 (NH4)2SO4 BaSO4
    PO 4 3− K3PO4 FePO4 (NH4)3PO4 Ba3(PO4)2
    NO 3 KNO3 Fe(NO3)3 NH4NO3 Ba(NO3)2
    OH KOH Fe(OH)3 NH4OH Ba(OH)2
    1. Li3N
    2. CsCl
    3. GeO2
    4. Rb2S
    5. Na3As
    1. AlCl3
    2. K2Cr2O7
    3. CH2
    4. CH5N3
    5. CH2O

2.3 Naming Ionic Compounds

Learning Objective

  1. To name ionic compounds.

The empirical and molecular formulas discussed in the preceding section are precise and highly informative, but they have some disadvantages. First, they are inconvenient for routine verbal communication. For example, saying “C-A-three-P-O-four-two” for Ca3(PO4)2 is much more difficult than saying “calcium phosphate.” In addition, you will see in that many compounds have the same empirical and molecular formulas but different arrangements of atoms, which result in very different chemical and physical properties. In such cases, it is necessary for the compounds to have different names that distinguish among the possible arrangements.

Many compounds, particularly those that have been known for a relatively long time, have more than one name: a common name (sometimes more than one) and a systematic name, which is the name assigned by adhering to specific rules. Like the names of most elements, the common names of chemical compounds generally have historical origins, although they often appear to be unrelated to the compounds of interest. For example, the systematic name for KNO3 is potassium nitrate, but its common name is saltpeter.

In this text, we use a systematic nomenclature to assign meaningful names to the millions of known substances. Unfortunately, some chemicals that are widely used in commerce and industry are still known almost exclusively by their common names; in such cases, you must be familiar with the common name as well as the systematic one. The objective of this and the next two sections is to teach you to write the formula for a simple inorganic compound from its name—and vice versa—and introduce you to some of the more frequently encountered common names.

We begin with binary ionic compounds, which contain only two elements. The procedure for naming such compounds is outlined in and uses the following steps:

Figure 2.10 Naming an Ionic Compound

  1. Place the ions in their proper order: cation and then anion.
  2. Name the cation.

    1. Metals that form only one cation. As noted in , these metals are usually in groups 1–3, 12, and 13. The name of the cation of a metal that forms only one cation is the same as the name of the metal (with the word ion added if the cation is by itself). For example, Na+ is the sodium ion, Ca2+ is the calcium ion, and Al3+ is the aluminum ion.
    2. Metals that form more than one cation. As shown in , many metals can form more than one cation. This behavior is observed for most transition metals, many actinides, and the heaviest elements of groups 13–15. In such cases, the positive charge on the metal is indicated by a roman numeral in parentheses immediately following the name of the metal. Thus Cu+ is copper(I) (read as “copper one”), Fe2+ is iron(II), Fe3+ is iron(III), Sn2+ is tin(II), and Sn4+ is tin(IV).

      An older system of nomenclature for such cations is still widely used, however. The name of the cation with the higher charge is formed from the root of the element’s Latin name with the suffix -ic attached, and the name of the cation with the lower charge has the same root with the suffix -ous. The names of Fe3+, Fe2+, Sn4+, and Sn2+ are therefore ferric, ferrous, stannic, and stannous, respectively. Even though this text uses the systematic names with roman numerals, you should be able to recognize these common names because they are still often used. For example, on the label of your dentist’s fluoride rinse, the compound chemists call tin(II) fluoride is usually listed as stannous fluoride.

      Some examples of metals that form more than one cation are in along with the names of the ions. Note that the simple Hg+ cation does not occur in chemical compounds. Instead, all compounds of mercury(I) contain a dimeric cation, Hg22+, in which the two Hg atoms are bonded together.

      Table 2.5 Common Cations of Metals That Form More Than One Ion

      Cation Systematic Name Common Name Cation Systematic Name Common Name
      Cr2+ chromium(II) chromous Cu2+ copper(II) cupric
      Cr3+ chromium(III) chromic Cu+ copper(I) cuprous
      Mn2+ manganese(II) manganous* Hg2+ mercury(II) mercuric
      Mn3+ manganese(III) manganic* Hg22+ mercury(I) mercurous
      Fe2+ iron(II) ferrous Sn4+ tin(IV) stannic
      Fe3+ iron(III) ferric Sn2+ tin(II) stannous
      Co2+ cobalt(II) cobaltous* Pb4+ lead(IV) plumbic*
      Co3+ cobalt(III) cobaltic* Pb2+ lead(II) plumbous*
      * Not widely used.
      The isolated mercury(I) ion exists only as the gaseous ion.
    3. Polyatomic cations. The names of the common polyatomic cations that are relatively important in ionic compounds (such as, the ammonium ion) are in .
  3. Name the anion.

    1. Monatomic anions. Monatomic anions are named by adding the suffix -ide to the root of the name of the parent element; thus, Cl is chloride, O2− is oxide, P3− is phosphide, N3− is nitride (also called azide), and C4− is carbide. Because the charges on these ions can be predicted from their position in the periodic table, it is not necessary to specify the charge in the name. Examples of monatomic anions are in .
    2. Polyatomic anions. Polyatomic anions typically have common names that you must learn; some examples are in . Polyatomic anions that contain a single metal or nonmetal atom plus one or more oxygen atoms are called oxoanions (or oxyanions). In cases where only two oxoanions are known for an element, the name of the oxoanion with more oxygen atoms ends in -ate, and the name of the oxoanion with fewer oxygen atoms ends in -ite. For example, NO3− is nitrate and NO2− is nitrite.

      The halogens and some of the transition metals form more extensive series of oxoanions with as many as four members. In the names of these oxoanions, the prefix per- is used to identify the oxoanion with the most oxygen (so that ClO4 is perchlorate and ClO3 is chlorate), and the prefix hypo- is used to identify the anion with the fewest oxygen (ClO2 is chlorite and ClO is hypochlorite). The relationship between the names of oxoanions and the number of oxygen atoms present is diagrammed in . Differentiating the oxoanions in such a series is no trivial matter. For example, the hypochlorite ion is the active ingredient in laundry bleach and swimming pool disinfectant, but compounds that contain the perchlorate ion can explode if they come into contact with organic substances.

  4. Write the name of the compound as the name of the cation followed by the name of the anion.

    It is not necessary to indicate the number of cations or anions present per formula unit in the name of an ionic compound because this information is implied by the charges on the ions. You must consider the charge of the ions when writing the formula for an ionic compound from its name, however. Because the charge on the chloride ion is −1 and the charge on the calcium ion is +2, for example, consistent with their positions in the periodic table, simple arithmetic tells you that calcium chloride must contain twice as many chloride ions as calcium ions to maintain electrical neutrality. Thus the formula is CaCl2. Similarly, calcium phosphate must be Ca3(PO4)2 because the cation and the anion have charges of +2 and −3, respectively. The best way to learn how to name ionic compounds is to work through a few examples, referring to , , , and as needed.

Figure 2.11 Metals That Form More Than One Cation and Their Locations in the Periodic Table

With only a few exceptions, these metals are usually transition metals or actinides.

Figure 2.12 The Relationship between the Names of Oxoanions and the Number of Oxygen Atoms Present

Note the Pattern

Cations are always named before anions.

Most transition metals, many actinides, and the heaviest elements of groups 13–15 can form more than one cation.

Example 6

Write the systematic name (and the common name if applicable) for each ionic compound.

  1. LiCl
  2. MgSO4
  3. (NH4)3PO4
  4. Cu2O

Given: empirical formula

Asked for: name

Strategy:

A If only one charge is possible for the cation, give its name, consulting or if necessary. If the cation can have more than one charge (), specify the charge using roman numerals.

B If the anion does not contain oxygen, name it according to step 3a, using and if necessary. For polyatomic anions that contain oxygen, use and the appropriate prefix and suffix listed in step 3b.

C Beginning with the cation, write the name of the compound.

Solution:

  1. A B Lithium is in group 1, so we know that it forms only the Li+ cation, which is the lithium ion. Similarly, chlorine is in group 7, so it forms the Cl anion, which is the chloride ion. C Because we begin with the name of the cation, the name of this compound is lithium chloride, which is used medically as an antidepressant drug.
  2. A B The cation is the magnesium ion, and the anion, which contains oxygen, is sulfate. C Because we list the cation first, the name of this compound is magnesium sulfate. A hydrated form of magnesium sulfate (MgSO4·7H2O) is sold in drugstores as Epsom salts, a harsh but effective laxative.
  3. A B The cation is the ammonium ion (from ), and the anion is phosphate. C The compound is therefore ammonium phosphate, which is widely used as a fertilizer. It is not necessary to specify that the formula unit contains three ammonium ions because three are required to balance the negative charge on phosphate.
  4. A B The cation is a transition metal that often forms more than one cation (). We must therefore specify the positive charge on the cation in the name: copper(I) or, according to the older system, cuprous. The anion is oxide. C The name of this compound is copper(I) oxide or, in the older system, cuprous oxide. Copper(I) oxide is used as a red glaze on ceramics and in antifouling paints to prevent organisms from growing on the bottoms of boats.

Cu2O. The bottom of a boat is protected with a red antifouling paint containing copper(I) oxide, Cu2O.

Exercise

Write the systematic name (and the common name if applicable) for each ionic compound.

  1. CuCl2
  2. MgCO3
  3. FePO4

Answer:

  1. copper(II) chloride (or cupric chloride)
  2. magnesium carbonate
  3. iron(III) phosphate (or ferric phosphate)

Example 7

Write the formula for each compound.

  1. calcium dihydrogen phosphate
  2. aluminum sulfate
  3. chromium(III) oxide

Given: systematic name

Asked for: formula

Strategy:

A Identify the cation and its charge using the location of the element in the periodic table and , , , and . If the cation is derived from a metal that can form cations with different charges, use the appropriate roman numeral or suffix to indicate its charge.

B Identify the anion using and . Beginning with the cation, write the compound’s formula and then determine the number of cations and anions needed to achieve electrical neutrality.

Solution:

  1. A Calcium is in group 2, so it forms only the Ca2+ ion. B Dihydrogen phosphate is the H2PO4 ion (). Two H2PO4 ions are needed to balance the positive charge on Ca2+, to give Ca(H2PO4)2. A hydrate of calcium dihydrogen phosphate, Ca(H2PO4)2·H2O, is the active ingredient in baking powder.
  2. A Aluminum, near the top of group 13 in the periodic table, forms only one cation, Al3+ (). B Sulfate is SO42− (). To balance the electrical charges, we need two Al3+ cations and three SO42− anions, giving Al2(SO4)3. Aluminum sulfate is used to tan leather and purify drinking water.
  3. A Because chromium is a transition metal, it can form cations with different charges. The roman numeral tells us that the positive charge in this case is +3, so the cation is Cr3+. B Oxide is O2−. Thus two cations (Cr3+) and three anions (O2−) are required to give an electrically neutral compound, Cr2O3. This compound is a common green pigment that has many uses, including camouflage coatings.

Cr2O3. Chromium(III) oxide (Cr2O3) is a common pigment in dark green paints, such as camouflage paint.

Exercise

Write the formula for each compound.

  1. barium chloride
  2. sodium carbonate
  3. iron(III) hydroxide

Answer:

  1. BaCl2
  2. Na2CO3
  3. Fe(OH)3

Summary

Ionic compounds are named according to systematic procedures, although common names are widely used. Systematic nomenclature enables us to write the structure of any compound from its name and vice versa. Ionic compounds are named by writing the cation first, followed by the anion. If a metal can form cations with more than one charge, the charge is indicated by roman numerals in parentheses following the name of the metal. Oxoanions are polyatomic anions that contain a single metal or nonmetal atom and one or more oxygen atoms.

Key Takeaway

  • There is a systematic method used to name ionic compounds.

Conceptual Problems

  1. Name each cation.

    1. K+
    2. Al3+
    3. NH4+
    4. Mg2+
    5. Li+
  2. Name each anion.

    1. Br
    2. CO32−
    3. S2−
    4. NO3
    5. HCO2
    6. F
    7. ClO
    8. C2O42−
  3. Name each anion.

    1. PO43−
    2. Cl
    3. SO32−
    4. CH3CO2
    5. HSO4
    6. ClO4
    7. NO2
    8. O2−
  4. Name each anion.

    1. SO42−
    2. CN
    3. Cr2O72−
    4. N3−
    5. OH
    6. I
    7. O22−
  5. Name each compound.

    1. MgBr2
    2. NH4CN
    3. CaO
    4. KClO3
    5. K3PO4
    6. NH4NO2
    7. NaN3
  6. Name each compound.

    1. NaNO3
    2. Cu3(PO4)2
    3. NaOH
    4. Li4C
    5. CaF2
    6. NH4Br
    7. MgCO3
  7. Name each compound.

    1. RbBr
    2. Mn2(SO4)3
    3. NaClO
    4. (NH4)2SO4
    5. NaBr
    6. KIO3
    7. Na2CrO4
  8. Name each compound.

    1. NH4ClO4
    2. SnCl4
    3. Fe(OH)2
    4. Na2O
    5. MgCl2
    6. K2SO4
    7. RaCl2
  9. Name each compound.

    1. KCN
    2. LiOH
    3. CaCl2
    4. NiSO4
    5. NH4ClO2
    6. LiClO4
    7. La(CN)3

Answer

    1. rubidium bromide
    2. manganese(III) sulfate
    3. sodium hypochlorite
    4. ammonium sulfate
    5. sodium bromide
    6. potassium iodate
    7. sodium chromate

Numerical Problems

  1. For each ionic compound, name the cation and the anion and give the charge on each ion.

    1. BeO
    2. Pb(OH)2
    3. BaS
    4. Na2Cr2O7
    5. ZnSO4
    6. KClO
    7. NaH2PO4
  2. For each ionic compound, name the cation and the anion and give the charge on each ion.

    1. Zn(NO3)2
    2. CoS
    3. BeCO3
    4. Na2SO4
    5. K2C2O4
    6. NaCN
    7. FeCl2
  3. Write the formula for each compound.

    1. magnesium carbonate
    2. aluminum sulfate
    3. potassium phosphate
    4. lead(IV) oxide
    5. silicon nitride
    6. sodium hypochlorite
    7. titanium(IV) chloride
    8. disodium ammonium phosphate
  4. Write the formula for each compound.

    1. lead(II) nitrate
    2. ammonium phosphate
    3. silver sulfide
    4. barium sulfate
    5. cesium iodide
    6. sodium bicarbonate
    7. potassium dichromate
    8. sodium hypochlorite
  5. Write the formula for each compound.

    1. zinc cyanide
    2. silver chromate
    3. lead(II) iodide
    4. benzene
    5. copper(II) perchlorate
  6. Write the formula for each compound.

    1. calcium fluoride
    2. sodium nitrate
    3. iron(III) oxide
    4. copper(II) acetate
    5. sodium nitrite
  7. Write the formula for each compound.

    1. sodium hydroxide
    2. calcium cyanide
    3. magnesium phosphate
    4. sodium sulfate
    5. nickel(II) bromide
    6. calcium chlorite
    7. titanium(IV) bromide
  8. Write the formula for each compound.

    1. sodium chlorite
    2. potassium nitrite
    3. sodium nitride (also called sodium azide)
    4. calcium phosphide
    5. tin(II) chloride
    6. calcium hydrogen phosphate
    7. iron(II) chloride dihydrate
  9. Write the formula for each compound.

    1. potassium carbonate
    2. chromium(III) sulfite
    3. cobalt(II) phosphate
    4. magnesium hypochlorite
    5. nickel(II) nitrate hexahydrate

2.4 Naming Covalent Compounds

Learning Objective

  1. To name covalent compounds that contain up to three elements.

As with ionic compounds, the system that chemists have devised for naming covalent compounds enables us to write the molecular formula from the name and vice versa. In this and the following section, we describe the rules for naming simple covalent compounds. We begin with inorganic compounds and then turn to simple organic compounds that contain only carbon and hydrogen.

Binary Inorganic Compounds

Binary covalent compounds—that is, covalent compounds that contain only two elements—are named using a procedure similar to that used to name simple ionic compounds, but prefixes are added as needed to indicate the number of atoms of each kind. The procedure, diagrammed in , uses the following steps:

Figure 2.13 Naming a Covalent Inorganic Compound

  1. Place the elements in their proper order.

    1. The element farthest to the left in the periodic table is usually named first. If both elements are in the same group, the element closer to the bottom of the column is named first.
    2. The second element is named as if it were a monatomic anion in an ionic compound (even though it is not), with the suffix -ide attached to the root of the element name.
  2. Identify the number of each type of atom present.

    1. Prefixes derived from Greek stems are used to indicate the number of each type of atom in the formula unit (). The prefix mono- (“one”) is used only when absolutely necessary to avoid confusion, just as we omit the subscript 1 when writing molecular formulas.

      To demonstrate steps 1 and 2a, we name HCl as hydrogen chloride (because hydrogen is to the left of chlorine in the periodic table) and PCl5 as phosphorus pentachloride. The order of the elements in the name of BrF3, bromine trifluoride, is determined by the fact that bromine lies below fluorine in group 17.

      Table 2.6 Prefixes for Indicating the Number of Atoms in Chemical Names

      Prefix Number
      mono- 1
      di- 2
      tri- 3
      tetra- 4
      penta- 5
      hexa- 6
      hepta- 7
      octa- 8
      nona- 9
      deca- 10
      undeca- 11
      dodeca- 12
    2. If a molecule contains more than one atom of both elements, then prefixes are used for both. Thus N2O3 is dinitrogen trioxide, as shown in .
    3. In some names, the final a or o of the prefix is dropped to avoid awkward pronunciation. Thus OsO4 is osmium tetroxide rather than osmium tetraoxide.
  3. Write the name of the compound.

    1. Binary compounds of the elements with oxygen are generally named as “element oxide,” with prefixes that indicate the number of atoms of each element per formula unit. For example, CO is carbon monoxide. The only exception is binary compounds of oxygen with fluorine, which are named as oxygen fluorides. (The reasons for this convention will become clear in and .)
    2. Certain compounds are always called by the common names that were assigned long ago when names rather than formulas were used. For example, H2O is water (not dihydrogen oxide); NH3 is ammonia; PH3 is phosphine; SiH4 is silane; and B2H6, a dimer of BH3, is diborane. For many compounds, the systematic name and the common name are both used frequently, so you must be familiar with them. For example, the systematic name for NO is nitrogen monoxide, but it is much more commonly called nitric oxide. Similarly, N2O is usually called nitrous oxide rather than dinitrogen monoxide. Notice that the suffixes -ic and -ous are the same ones used for ionic compounds.

Note the Pattern

Start with the element at the far left in the periodic table and work to the right. If two or more elements are in the same group, start with the bottom element and work up.

Example 8

Write the name of each binary covalent compound.

  1. SF6
  2. N2O4
  3. ClO2

Given: molecular formula

Asked for: name of compound

Strategy:

A List the elements in order according to their positions in the periodic table. Identify the number of each type of atom in the chemical formula and then use to determine the prefixes needed.

B If the compound contains oxygen, follow step 3a. If not, decide whether to use the common name or the systematic name.

Solution:

  1. A Because sulfur is to the left of fluorine in the periodic table, sulfur is named first. Because there is only one sulfur atom in the formula, no prefix is needed. B There are, however, six fluorine atoms, so we use the prefix for six: hexa- (). The compound is sulfur hexafluoride.
  2. A Because nitrogen is to the left of oxygen in the periodic table, nitrogen is named first. Because more than one atom of each element is present, prefixes are needed to indicate the number of atoms of each. According to , the prefix for two is di-, and the prefix for four is tetra-. B The compound is dinitrogen tetroxide (omitting the a in tetra- according to step 2c) and is used as a component of some rocket fuels.
  3. A Although oxygen lies to the left of chlorine in the periodic table, it is not named first because ClO2 is an oxide of an element other than fluorine (step 3a). Consequently, chlorine is named first, but a prefix is not necessary because each molecule has only one atom of chlorine. B Because there are two oxygen atoms, the compound is a dioxide. Thus the compound is chlorine dioxide. It is widely used as a substitute for chlorine in municipal water treatment plants because, unlike chlorine, it does not react with organic compounds in water to produce potentially toxic chlorinated compounds.

Exercise

Write the name of each binary covalent compound.

  1. IF7
  2. N2O5
  3. OF2

Answer:

  1. iodine heptafluoride
  2. dinitrogen pentoxide
  3. oxygen difluoride

Example 9

Write the formula for each binary covalent compound.

  1. sulfur trioxide
  2. diiodine pentoxide

Given: name of compound

Asked for: formula

Strategy:

List the elements in the same order as in the formula, use to identify the number of each type of atom present, and then indicate this quantity as a subscript to the right of that element when writing the formula.

Solution:

  1. Sulfur has no prefix, which means that each molecule has only one sulfur atom. The prefix tri- indicates that there are three oxygen atoms. The formula is therefore SO3. Sulfur trioxide is produced industrially in huge amounts as an intermediate in the synthesis of sulfuric acid.
  2. The prefix di- tells you that each molecule has two iodine atoms, and the prefix penta- indicates that there are five oxygen atoms. The formula is thus I2O5, a compound used to remove carbon monoxide from air in respirators.

Exercise

Write the formula for each binary covalent compound.

  1. silicon tetrachloride
  2. disulfur decafluoride

Answer:

  1. SiCl4
  2. S2F10

The structures of some of the compounds in Example 8 and Example 9 are shown in , along with the location of the “central atom” of each compound in the periodic table. It may seem that the compositions and structures of such compounds are entirely random, but this is not true. After you have mastered the material in and , you will be able to predict the compositions and structures of compounds of this type with a high degree of accuracy.

Figure 2.14 The Structures of Some Covalent Inorganic Compounds and the Locations of the “Central Atoms” in the Periodic Table

The compositions and structures of covalent inorganic compounds are not random. As you will learn in and , they can be predicted from the locations of the component atoms in the periodic table.

Hydrocarbons

Approximately one-third of the compounds produced industrially are organic compounds. All living organisms are composed of organic compounds, as is most of the food you consume, the medicines you take, the fibers in the clothes you wear, and the plastics in the materials you use. introduced two organic compounds: methane (CH4) and methanol (CH3OH). These and other organic compounds appear frequently in discussions and examples throughout this text.

The detection of organic compounds is useful in many fields. In one recently developed application, scientists have devised a new method called “material degradomics” to make it possible to monitor the degradation of old books and historical documents. As paper ages, it produces a familiar “old book smell” from the release of organic compounds in gaseous form. The composition of the gas depends on the original type of paper used, a book’s binding, and the applied media. By analyzing these organic gases and isolating the individual components, preservationists are better able to determine the condition of an object and those books and documents most in need of immediate protection.

The simplest class of organic compounds is the hydrocarbonsThe simplest class of organic molecules, consisting of only carbon and hydrogen., which consist entirely of carbon and hydrogen. Petroleum and natural gas are complex, naturally occurring mixtures of many different hydrocarbons that furnish raw materials for the chemical industry. The four major classes of hydrocarbons are the alkanesA saturated hydrocarbon with only carbon–hydrogen and carbon–carbon single bonds., which contain only carbon–hydrogen and carbon–carbon single bonds; the alkenesAn unsaturated hydrocarbon with at least one carbon–carbon double bond., which contain at least one carbon–carbon double bond; the alkynesAn unsaturated hydrocarbon with at least one carbon–carbon triple bond., which contain at least one carbon–carbon triple bond; and the aromatic hydrocarbonsAn unsaturated hydrocarbon consisting of a ring of six carbon atoms with alternating single and double bonds., which usually contain rings of six carbon atoms that can be drawn with alternating single and double bonds. Alkanes are also called saturated hydrocarbons, whereas hydrocarbons that contain multiple bonds (alkenes, alkynes, and aromatics) are unsaturated.

Alkanes

The simplest alkane is methane (CH4), a colorless, odorless gas that is the major component of natural gas. In larger alkanes whose carbon atoms are joined in an unbranched chain (straight-chain alkanes), each carbon atom is bonded to at most two other carbon atoms. The structures of two simple alkanes are shown in , and the names and condensed structural formulas for the first 10 straight-chain alkanes are in . The names of all alkanes end in -ane, and their boiling points increase as the number of carbon atoms increases.

Figure 2.15 Straight-Chain Alkanes with Two and Three Carbon Atoms

Table 2.7 The First 10 Straight-Chain Alkanes

Name Number of Carbon Atoms Molecular Formula Condensed Structural Formula Boiling Point (°C) Uses
methane 1 CH4 CH4 −162 natural gas constituent
ethane 2 C2H6 CH3CH3 −89 natural gas constituent
propane 3 C3H8 CH3CH2CH3 −42 bottled gas
butane 4 C4H10 CH3CH2CH2CH3 or CH3(CH2)2CH3 0 lighters, bottled gas
pentane 5 C5H12 CH3(CH2)3CH3 36 solvent, gasoline
hexane 6 C6H14 CH3(CH2)4CH3 69 solvent, gasoline
heptane 7 C7H16 CH3(CH2)5CH3 98 solvent, gasoline
octane 8 C8H18 CH3(CH2)6CH3 126 gasoline
nonane 9 C9H20 CH3(CH2)7CH3 151 gasoline
decane 10 C10H22 CH3(CH2)8CH3 174 kerosene

Alkanes with four or more carbon atoms can have more than one arrangement of atoms. The carbon atoms can form a single unbranched chain, or the primary chain of carbon atoms can have one or more shorter chains that form branches. For example, butane (C4H10) has two possible structures. Normal butane (usually called n-butane) is CH3CH2CH2CH3, in which the carbon atoms form a single unbranched chain. In contrast, the condensed structural formula for isobutane is (CH3)2CHCH3, in which the primary chain of three carbon atoms has a one-carbon chain branching at the central carbon. Three-dimensional representations of both structures are as follows:

The systematic names for branched hydrocarbons use the lowest possible number to indicate the position of the branch along the longest straight carbon chain in the structure. Thus the systematic name for isobutane is 2-methylpropane, which indicates that a methyl group (a branch consisting of –CH3) is attached to the second carbon of a propane molecule. Similarly, you will learn in that one of the major components of gasoline is commonly called isooctane; its structure is as follows:

As you can see, the compound has a chain of five carbon atoms, so it is a derivative of pentane. There are two methyl group branches at one carbon atom and one methyl group at another. Using the lowest possible numbers for the branches gives 2,2,4-trimethylpentane for the systematic name of this compound.

Alkenes

The simplest alkenes are ethylene, C2H4 or CH2=CH2, and propylene, C3H6 or CH3CH=CH2 (part (a) in ). The names of alkenes that have more than three carbon atoms use the same stems as the names of the alkanes () but end in -ene instead of -ane.

Once again, more than one structure is possible for alkenes with four or more carbon atoms. For example, an alkene with four carbon atoms has three possible structures. One is CH2=CHCH2CH3 (1-butene), which has the double bond between the first and second carbon atoms in the chain. The other two structures have the double bond between the second and third carbon atoms and are forms of CH3CH=CHCH3 (2-butene). All four carbon atoms in 2-butene lie in the same plane, so there are two possible structures (part (a) in ). If the two methyl groups are on the same side of the double bond, the compound is cis-2-butene (from the Latin cis, meaning “on the same side”). If the two methyl groups are on opposite sides of the double bond, the compound is trans-2-butene (from the Latin trans, meaning “across”). These are distinctly different molecules: cis-2-butene melts at −138.9°C, whereas trans-2-butene melts at −105.5°C.

Figure 2.16 Some Simple (a) Alkenes, (b) Alkynes, and (c) Cyclic Hydrocarbons

The positions of the carbon atoms in the chain are indicated by C1 or C2.

Just as a number indicates the positions of branches in an alkane, the number in the name of an alkene specifies the position of the first carbon atom of the double bond. The name is based on the lowest possible number starting from either end of the carbon chain, so CH3CH2CH=CH2 is called 1-butene, not 3-butene. Note that CH2=CHCH2CH3 and CH3CH2CH=CH2 are different ways of writing the same molecule (1-butene) in two different orientations.

The name of a compound does not depend on its orientation. As illustrated for 1-butene, both condensed structural formulas and molecular models show different orientations of the same molecule. Don’t let orientation fool you; you must be able to recognize the same structure no matter what its orientation.

Note the Pattern

The positions of groups or multiple bonds are always indicated by the lowest number possible.

Alkynes

The simplest alkyne is acetylene, C2H2 or HC≡CH (part (b) in ). Because a mixture of acetylene and oxygen burns with a flame that is hot enough (>3000°C) to cut metals such as hardened steel, acetylene is widely used in cutting and welding torches. The names of other alkynes are similar to those of the corresponding alkanes but end in -yne. For example, HC≡CCH3 is propyne, and CH3C≡CCH3 is 2-butyne because the multiple bond begins on the second carbon atom.

Note the Pattern

The number of bonds between carbon atoms in a hydrocarbon is indicated in the suffix:

  • alkane: only carbon–carbon single bonds
  • alkene: at least one carbon–carbon double bond
  • alkyne: at least one carbon–carbon triple bond

Cyclic Hydrocarbons

In a cyclic hydrocarbonA hydrocarbon in which the ends of the carbon chain are connected to form a ring of covalently bonded carbon atoms., the ends of a hydrocarbon chain are connected to form a ring of covalently bonded carbon atoms. Cyclic hydrocarbons are named by attaching the prefix cyclo- to the name of the alkane, the alkene, or the alkyne. The simplest cyclic alkanes are cyclopropane (C3H6) a flammable gas that is also a powerful anesthetic, and cyclobutane (C4H8) (part (c) in ). The most common way to draw the structures of cyclic alkanes is to sketch a polygon with the same number of vertices as there are carbon atoms in the ring; each vertex represents a CH2 unit. The structures of the cycloalkanes that contain three to six carbon atoms are shown schematically in .

Figure 2.17 The Simple Cycloalkanes

Aromatic Hydrocarbons

Alkanes, alkenes, alkynes, and cyclic hydrocarbons are generally called aliphatic hydrocarbonsAlkanes, alkenes, alkynes, and cyclic hydrocarbons (hydrocarbons that are not aromatic).. The name comes from the Greek aleiphar, meaning “oil,” because the first examples were extracted from animal fats. In contrast, the first examples of aromatic hydrocarbons, also called arenes, were obtained by the distillation and degradation of highly scented (thus aromatic) resins from tropical trees.

The simplest aromatic hydrocarbon is benzene (C6H6), which was first obtained from a coal distillate. The word aromatic now refers to benzene and structurally similar compounds. As shown in part (a) in , it is possible to draw the structure of benzene in two different but equivalent ways, depending on which carbon atoms are connected by double bonds or single bonds. Toluene is similar to benzene, except that one hydrogen atom is replaced by a –CH3 group; it has the formula C7H8 (part (b) in ). As you will soon learn, the chemical behavior of aromatic compounds differs from the behavior of aliphatic compounds. Benzene and toluene are found in gasoline, and benzene is the starting material for preparing substances as diverse as aspirin and nylon.

Figure 2.18 Two Aromatic Hydrocarbons: (a) Benzene and (b) Toluene

illustrates two of the molecular structures possible for hydrocarbons that have six carbon atoms. As you can see, compounds with the same molecular formula can have very different structures.

Figure 2.19 Two Hydrocarbons with the Molecular Formula C6H12

Example 10

Write the condensed structural formula for each hydrocarbon.

  1. n-heptane
  2. 2-pentene
  3. 2-butyne
  4. cyclooctene

Given: name of hydrocarbon

Asked for: condensed structural formula

Strategy:

A Use the prefix to determine the number of carbon atoms in the molecule and whether it is cyclic. From the suffix, determine whether multiple bonds are present.

B Identify the position of any multiple bonds from the number(s) in the name and then write the condensed structural formula.

Solution:

  1. A The prefix hept- tells us that this hydrocarbon has seven carbon atoms, and n- indicates that the carbon atoms form a straight chain. The suffix -ane tells that it is an alkane, with no carbon–carbon double or triple bonds. B The condensed structural formula is CH3CH2CH2CH2CH2CH2CH3, which can also be written as CH3(CH2)5CH3.
  2. A The prefix pent- tells us that this hydrocarbon has five carbon atoms, and the suffix -ene indicates that it is an alkene, with a carbon–carbon double bond. B The 2- tells us that the double bond begins on the second carbon of the five-carbon atom chain. The condensed structural formula of the compound is therefore CH3CH=CHCH2CH3.

  3. A The prefix but- tells us that the compound has a chain of four carbon atoms, and the suffix -yne indicates that it has a carbon–carbon triple bond. B The 2- tells us that the triple bond begins on the second carbon of the four-carbon atom chain. So the condensed structural formula for the compound is CH3C≡CCH3.

  4. A The prefix cyclo- tells us that this hydrocarbon has a ring structure, and oct- indicates that it contains eight carbon atoms, which we can draw as

    The suffix -ene tells us that the compound contains a carbon–carbon double bond, but where in the ring do we place the double bond? B Because all eight carbon atoms are identical, it doesn’t matter. We can draw the structure of cyclooctene as

Exercise

Write the condensed structural formula for each hydrocarbon.

  1. n-octane
  2. 2-hexene
  3. 1-heptyne
  4. cyclopentane

Answer:

  1. CH3(CH2)6CH3
  2. CH3CH=CHCH2CH2CH3
  3. HC≡C(CH2)4CH3
  4.  

The general name for a group of atoms derived from an alkane is an alkyl group. The name of an alkyl group is derived from the name of the alkane by adding the suffix -yl. Thus the –CH3 fragment is a methyl group, the –CH2CH3 fragment is an ethyl group, and so forth, where the dash represents a single bond to some other atom or group. Similarly, groups of atoms derived from aromatic hydrocarbons are aryl groups, which sometimes have unexpected names. For example, the –C6H5 fragment is derived from benzene, but it is called a phenyl group. In general formulas and structures, alkyl and aryl groups are often abbreviated as RThe abbreviation used for alkyl groups and aryl groups in general formulas and structures..

Structures of alkyl and aryl groups. The methyl group is an example of an alkyl group, and the phenyl group is an example of an aryl group.

Alcohols

Replacing one or more hydrogen atoms of a hydrocarbon with an –OH group gives an alcoholA class of organic compounds obtained by replacing one or more of the hydrogen atoms of a hydrocarbon with an −OH group., represented as ROH. The simplest alcohol (CH3OH) is called either methanol (its systematic name) or methyl alcohol (its common name) (see ). Methanol is the antifreeze in automobile windshield washer fluids, and it is also used as an efficient fuel for racing cars, most notably in the Indianapolis 500. Ethanol (or ethyl alcohol, CH3CH2OH) is familiar as the alcohol in fermented or distilled beverages, such as beer, wine, and whiskey; it is also used as a gasoline additive (). The simplest alcohol derived from an aromatic hydrocarbon is C6H5OH, phenol (shortened from phenyl alcohol), a potent disinfectant used in some sore throat medications and mouthwashes.

Ethanol, which is easy to obtain from fermentation processes, has successfully been used as an alternative fuel for several decades. Although it is a “green” fuel when derived from plants, it is an imperfect substitute for fossil fuels because it is less efficient than gasoline. Moreover, because ethanol absorbs water from the atmosphere, it can corrode an engine’s seals. Thus other types of processes are being developed that use bacteria to create more complex alcohols, such as octanol, that are more energy efficient and that have a lower tendency to absorb water. As scientists attempt to reduce mankind’s dependence on fossil fuels, the development of these so-called biofuels is a particularly active area of research.

Summary

Covalent inorganic compounds are named by a procedure similar to that used for ionic compounds, using prefixes to indicate the numbers of atoms in the molecular formula. The simplest organic compounds are the hydrocarbons, which contain only carbon and hydrogen. Alkanes contain only carbon–hydrogen and carbon–carbon single bonds, alkenes contain at least one carbon–carbon double bond, and alkynes contain one or more carbon–carbon triple bonds. Hydrocarbons can also be cyclic, with the ends of the chain connected to form a ring. Collectively, alkanes, alkenes, and alkynes are called aliphatic hydrocarbons. Aromatic hydrocarbons, or arenes, are another important class of hydrocarbons that contain rings of carbon atoms related to the structure of benzene (C6H6). A derivative of an alkane or an arene from which one hydrogen atom has been removed is called an alkyl group or an aryl group, respectively. Alcohols are another common class of organic compound, which contain an –OH group covalently bonded to either an alkyl group or an aryl group (often abbreviated R).

Key Takeaway

  • Covalent inorganic compounds are named using a procedure similar to that used for ionic compounds, whereas hydrocarbons use a system based on the number of bonds between carbon atoms.

Conceptual Problems

  1. Benzene (C6H6) is an organic compound, and KCl is an ionic compound. The sum of the masses of the atoms in each empirical formula is approximately the same. How would you expect the two to compare with regard to each of the following? What species are present in benzene vapor?

    1. melting point
    2. type of bonding
    3. rate of evaporation
    4. structure
  2. Can an inorganic compound be classified as a hydrocarbon? Why or why not?

  3. Is the compound NaHCO3 a hydrocarbon? Why or why not?

  4. Name each compound.

    1. NiO
    2. TiO2
    3. N2O
    4. CS2
    5. SO3
    6. NF3
    7. SF6
  5. Name each compound.

    1. HgCl2
    2. IF5
    3. N2O5
    4. Cl2O
    5. HgS
    6. PCl5
  6. For each structural formula, write the condensed formula and the name of the compound.

    1.  

    2.  

    3.  

    4.  

    5.  

  7. For each structural formula, write the condensed formula and the name of the compound.

    1.  

    2.  

    3.  

    4.  

  8. Would you expect PCl3 to be an ionic compound or a covalent compound? Explain your reasoning.

  9. What distinguishes an aromatic hydrocarbon from an aliphatic hydrocarbon?

  10. The following general formulas represent specific classes of hydrocarbons. Refer to and and and identify the classes.

    1. CnH2n+ 2
    2. CnH2n
    3. CnH2n− 2
  11. Using R to represent an alkyl or aryl group, show the general structure of an

    1. alcohol.
    2. phenol.

Answer

    1. ROH (where R is an alkyl group)
    2. ROH (where R is an aryl group)

Numerical Problems

  1. Write the formula for each compound.

    1. dinitrogen monoxide
    2. silicon tetrafluoride
    3. boron trichloride
    4. nitrogen trifluoride
    5. phosphorus tribromide
  2. Write the formula for each compound.

    1. dinitrogen trioxide
    2. iodine pentafluoride
    3. boron tribromide
    4. oxygen difluoride
    5. arsenic trichloride
  3. Write the formula for each compound.

    1. thallium(I) selenide
    2. neptunium(IV) oxide
    3. iron(II) sulfide
    4. copper(I) cyanide
    5. nitrogen trichloride
  4. Name each compound.

    1. RuO4
    2. PbO2
    3. MoF6
    4. Hg2(NO3)2·2H2O
    5. WCl4
  5. Name each compound.

    1. NbO2
    2. MoS2
    3. P4S10
    4. Cu2O
    5. ReF5
  6. Draw the structure of each compound.

    1. propyne
    2. ethanol
    3. n-hexane
    4. cyclopropane
    5. benzene
  7. Draw the structure of each compound.

    1. 1-butene
    2. 2-pentyne
    3. cycloheptane
    4. toluene
    5. phenol

Answers

    1. N2O
    2. SiF4
    3. BCl3
    4. NF3
    5. PBr3
    1. Tl2Se
    2. NpO2
    3. FeS
    4. CuCN
    5. NCl3
    1. niobium (IV) oxide
    2. molybdenum (IV) sulfide
    3. tetraphosphorus decasulfide
    4. copper(I) oxide
    5. rhenium(V) fluoride
    1.  

    2.  

    3.  

    4.  

    5.  

2.5 Acids and Bases

Learning Objective

  1. To identify and name some common acids and bases.

For our purposes at this point in the text, we can define an acidA substance with at least one hydrogen atom that can dissociate to form an anion and an H+ ion (a proton) in aqueous solution, thereby foming an acidic solution. as a substance with at least one hydrogen atom that can dissociate to form an anion and an H+ ion (a proton) in aqueous solution, thereby forming an acidic solution. We can define basesA substance that produces one or more hydroxide ions (OH) and a cation when dissolved in aqueous solution, thereby forming a basic solution. as compounds that produce hydroxide ions (OH) and a cation when dissolved in water, thus forming a basic solution. Solutions that are neither basic nor acidic are neutral. We will discuss the chemistry of acids and bases in more detail in , , and , but in this section we describe the nomenclature of common acids and identify some important bases so that you can recognize them in future discussions. Pure acids and bases and their concentrated aqueous solutions are commonly encountered in the laboratory. They are usually highly corrosive, so they must be handled with care.

Acids

The names of acids differentiate between (1) acids in which the H+ ion is attached to an oxygen atom of a polyatomic anion (these are called oxoacidsAn acid in which the dissociable H+ ion is attached to an oxygen atom of a polyatomic anion., or occasionally oxyacids) and (2) acids in which the H+ ion is attached to some other element. In the latter case, the name of the acid begins with hydro- and ends in -ic, with the root of the name of the other element or ion in between. Recall that the name of the anion derived from this kind of acid always ends in -ide. Thus hydrogen chloride (HCl) gas dissolves in water to form hydrochloric acid (which contains H+ and Cl ions), hydrogen cyanide (HCN) gas forms hydrocyanic acid (which contains H+ and CN ions), and so forth (). Examples of this kind of acid are commonly encountered and very important. For instance, your stomach contains a dilute solution of hydrochloric acid to help digest food. When the mechanisms that prevent the stomach from digesting itself malfunction, the acid destroys the lining of the stomach and an ulcer forms.

Note the Pattern

Acids are distinguished by whether the H+ ion is attached to an oxygen atom of a polyatomic anion or some other element.

Table 2.8 Some Common Acids That Do Not Contain Oxygen

Formula Name in Aqueous Solution Name of Gaseous Species
HF hydrofluoric acid hydrogen fluoride
HCl hydrochloric acid hydrogen chloride
HBr hydrobromic acid hydrogen bromide
HI hydroiodic acid hydrogen iodide
HCN hydrocyanic acid hydrogen cyanide
H2S hydrosulfuric acid hydrogen sulfide

If an acid contains one or more H+ ions attached to oxygen, it is a derivative of one of the common oxoanions, such as sulfate (SO42−) or nitrate (NO3). These acids contain as many H+ ions as are necessary to balance the negative charge on the anion, resulting in a neutral species such as H2SO4 and HNO3.

The names of acids are derived from the names of anions according to the following rules:

  1. If the name of the anion ends in -ate, then the name of the acid ends in -ic. For example, because NO3 is the nitrate ion, HNO3 is nitric acid. Similarly, ClO4 is the perchlorate ion, so HClO4 is perchloric acid. Two important acids are sulfuric acid (H2SO4) from the sulfate ion (SO42−) and phosphoric acid (H3PO4) from the phosphate ion (PO43−). These two names use a slight variant of the root of the anion name: sulfate becomes sulfuric and phosphate becomes phosphoric.
  2. If the name of the anion ends in -ite, then the name of the acid ends in -ous. For example, OCl is the hypochlorite ion, and HOCl is hypochlorous acid; NO2 is the nitrite ion, and HNO2 is nitrous acid; and SO32− is the sulfite ion, and H2SO3 is sulfurous acid. The same roots are used whether the acid name ends in -ic or -ous; thus, sulfite becomes sulfurous.

The relationship between the names of the oxoacids and the parent oxoanions is illustrated in , and some common oxoacids are in .

Figure 2.20 The Relationship between the Names of the Oxoacids and the Names of the Parent Oxoanions

Table 2.9 Some Common Oxoacids

Formula Name
HNO2 nitrous acid
HNO3 nitric acid
H2SO3 sulfurous acid
H2SO4 sulfuric acid
H3PO4 phosphoric acid
H2CO3 carbonic acid
HClO hypochlorous acid
HClO2 chlorous acid
HClO3 chloric acid
HClO4 perchloric acid

Example 11

Name and give the formula for each acid.

  1. the acid formed by adding a proton to the hypobromite ion (OBr)
  2. the acid formed by adding two protons to the selenate ion (SeO42−)

Given: anion

Asked for: parent acid

Strategy:

Refer to and to find the name of the acid. If the acid is not listed, use the guidelines given previously.

Solution:

Neither species is listed in or , so we must use the information given previously to derive the name of the acid from the name of the polyatomic anion.

  1. The anion name, hypobromite, ends in -ite, so the name of the parent acid ends in -ous. The acid is therefore hypobromous acid (HOBr).
  2. Selenate ends in -ate, so the name of the parent acid ends in -ic. The acid is therefore selenic acid (H2SeO4).

Exercise

Name and give the formula for each acid.

  1. the acid formed by adding a proton to the perbromate ion (BrO4)
  2. the acid formed by adding three protons to the arsenite ion (AsO33−)

Answer:

  1. perbromic acid; HBrO4
  2. arsenous acid; H3AsO3

Many organic compounds contain the carbonyl groupA carbon atom double-bonded to an oxygen atom. It is a characteristic feature of many organic compounds, including carboxylic acids., in which there is a carbon–oxygen double bond. In carboxylic acidsAn organic compound that contains an −OH group covalently bonded to the carbon atom of a carbonyl group. The general formula of a carboxylic acid is RCO2H. In water a carboxylic acid dissociates to produce an acidic solution., an –OH group is covalently bonded to the carbon atom of the carbonyl group. Their general formula is RCO2H, sometimes written as RCOOH:

where R can be an alkyl group, an aryl group, or a hydrogen atom. The simplest example, HCO2H, is formic acid, so called because it is found in the secretions of stinging ants (from the Latin formica, meaning “ant”). Another example is acetic acid (CH3CO2H), which is found in vinegar. Like many acids, carboxylic acids tend to have sharp odors. For example, butyric acid (CH3CH2CH2CO2H), is responsible for the smell of rancid butter, and the characteristic odor of sour milk and vomit is due to lactic acid [CH3CH(OH)CO2H]. Some common carboxylic acids are shown in .

Figure 2.21 Some Common Carboxylic Acids

Although carboxylic acids are covalent compounds, when they dissolve in water, they dissociate to produce H+ ions (just like any other acid) and RCO2 ions. Note that only the hydrogen attached to the oxygen atom of the CO2group dissociates to form an H+ion. In contrast, the hydrogen atom attached to the oxygen atom of an alcohol does not dissociate to form an H+ ion when an alcohol is dissolved in water. The reasons for the difference in behavior between carboxylic acids and alcohols will be discussed in .

Note the Pattern

Only the hydrogen attached to the oxygen atom of the CO2 group dissociates to form an H+ ion.

Bases

We will present more comprehensive definitions of bases in later chapters, but virtually every base you encounter in the meantime will be an ionic compound, such as sodium hydroxide (NaOH) and barium hydroxide [Ba(OH)2], that contain the hydroxide ion and a metal cation. These have the general formula M(OH)n. It is important to recognize that alcohols, with the general formula ROH, are covalent compounds, not ionic compounds; consequently, they do not dissociate in water to form a basic solution (containing OH ions). When a base reacts with any of the acids we have discussed, it accepts a proton (H+). For example, the hydroxide ion (OH) accepts a proton to form H2O. Thus bases are also referred to as proton acceptors.

Concentrated aqueous solutions of ammonia (NH3) contain significant amounts of the hydroxide ion, even though the dissolved substance is not primarily ammonium hydroxide (NH4OH) as is often stated on the label. Thus aqueous ammonia solution is also a common base. Replacing a hydrogen atom of NH3 with an alkyl group results in an amineAn organic compound that has the general formula RNH2, where R is an alkyl group. Amines, like ammonia, are bases. (RNH2), which is also a base. Amines have pungent odors—for example, methylamine (CH3NH2) is one of the compounds responsible for the foul odor associated with spoiled fish. The physiological importance of amines is suggested in the word vitamin, which is derived from the phrase vital amines. The word was coined to describe dietary substances that were effective at preventing scurvy, rickets, and other diseases because these substances were assumed to be amines. Subsequently, some vitamins have indeed been confirmed to be amines.

Note the Pattern

Metal hydroxides (MOH) yield OH ions and are bases, alcohols (ROH) do not yield OH or H+ ions and are neutral, and carboxylic acids (RCO2H) yield H+ ions and are acids.

Summary

Common acids and the polyatomic anions derived from them have their own names and rules for nomenclature. The nomenclature of acids differentiates between oxoacids, in which the H+ ion is attached to an oxygen atom of a polyatomic ion, and acids in which the H+ ion is attached to another element. Carboxylic acids are an important class of organic acids. Ammonia is an important base, as are its organic derivatives, the amines.

Key Takeaway

  • Common acids and polyatomic anions derived from them have their own names and rules for nomenclature.

Conceptual Problems

  1. Name each acid.

    1. HCl
    2. HBrO3
    3. HNO3
    4. H2SO4
    5. HIO3
  2. Name each acid.

    1. HBr
    2. H2SO3
    3. HClO3
    4. HCN
    5. H3PO4
  3. Name the aqueous acid that corresponds to each gaseous species.

    1. hydrogen bromide
    2. hydrogen cyanide
    3. hydrogen iodide
  4. For each structural formula, write the condensed formula and the name of the compound.

    1.  

    2.  

  5. For each structural formula, write the condensed formula and the name of the compound.

    1.  

    2.  

  6. When each compound is added to water, is the resulting solution acidic, neutral, or basic?

    1. CH3CH2OH
    2. Mg(OH)2
    3. C6H5CO2H
    4. LiOH
    5. C3H7CO2H
    6. H2SO4
  7. Draw the structure of the simplest example of each type of compound.

    1. alkane
    2. alkene
    3. alkyne
    4. aromatic hydrocarbon
    5. alcohol
    6. carboxylic acid
    7. amine
    8. cycloalkane
  8. Identify the class of organic compound represented by each compound.

    1.  

    2. CH3CH2OH
    3. HC≡CH
    4.  

    5. C3H7NH2
    6. CH3CH=CHCH2CH3
    7.  

    8.  

  9. Identify the class of organic compound represented by each compound.

    1.  

    2.  

    3.  

    4.  

    5.  

    6. CH3C≡CH
    7.  

    8.  

Numerical Problems

  1. Write the formula for each compound.

    1. hypochlorous acid
    2. perbromic acid
    3. hydrobromic acid
    4. sulfurous acid
    5. sodium perbromate
  2. Write the formula for each compound.

    1. hydroiodic acid
    2. hydrogen sulfide
    3. phosphorous acid
    4. perchloric acid
    5. calcium hypobromite
  3. Name each compound.

    1. HBr
    2. H2SO3
    3. HCN
    4. HClO4
    5. NaHSO4
  4. Name each compound.

    1. H2SO4
    2. HNO2
    3. K2HPO4
    4. H3PO3
    5. Ca(H2PO4)2·H2O

2.6 Industrially Important Chemicals

Learning Objective

  1. To appreciate the scope of the chemical industry and its contributions to modern society.

It isn’t easy to comprehend the scale on which the chemical industry must operate to supply the huge amounts of chemicals required in modern industrial societies. lists the names and formulas of the chemical industry’s “top 25” for 2002—the 25 chemicals produced in the largest quantity in the United States that year—along with the amounts produced, in billions of pounds. To put these numbers in perspective, consider that the 88.80 billion pounds of sulfuric acid produced in the United States in 2002 has a volume of 21.90 million cubic meters (2.19 × 107 m3), enough to fill the Pentagon, probably the largest office building in the world, about 22 times.

Figure 2.22 Top 25 Chemicals Produced in the United States in 2002*

According to , 11 of the top 15 compounds produced in the United States are inorganic, and the total mass of inorganic chemicals produced is almost twice the mass of organic chemicals. Yet the diversity of organic compounds used in industry is such that over half of the top 25 compounds (13 out of 25) are organic.

Why are such huge quantities of chemical compounds produced annually? They are used both directly as components of compounds and materials that we encounter on an almost daily basis and indirectly in the production of those compounds and materials. The single largest use of industrial chemicals is in the production of foods: 7 of the top 15 chemicals are either fertilizers (ammonia, urea, and ammonium nitrate) or used primarily in the production of fertilizers (sulfuric acid, nitric acid, nitrogen, and phosphoric acid). Many of the organic chemicals on the list are used primarily as ingredients in the plastics and related materials that are so prevalent in contemporary society. Ethylene and propylene, for example, are used to produce polyethylene and polypropylene, which are made into plastic milk bottles, sandwich bags, indoor-outdoor carpets, and other common items. Vinyl chloride, in the form of polyvinylchloride, is used in everything from pipes to floor tiles to trash bags. Though not listed in , butadiene and carbon black are used in the manufacture of synthetic rubber for tires, and phenol and formaldehyde are ingredients in plywood, fiberglass, and many hard plastic items.

We do not have the space in this text to consider the applications of all these compounds in any detail, but we will return to many of them after we have developed the concepts necessary to understand their underlying chemistry. Instead, we conclude this chapter with a brief discussion of petroleum refining as it relates to gasoline and octane ratings and a look at the production and use of the topmost industrial chemical, sulfuric acid.

Petroleum

The petroleum that is pumped out of the ground at locations around the world is a complex mixture of several thousand organic compounds, including straight-chain alkanes, cycloalkanes, alkenes, and aromatic hydrocarbons with four to several hundred carbon atoms. The identities and relative abundances of the components vary depending on the source. So Texas crude oil is somewhat different from Saudi Arabian crude oil. In fact, the analysis of petroleum from different deposits can produce a “fingerprint” of each, which is useful in tracking down the sources of spilled crude oil. For example, Texas crude oil is “sweet,” meaning that it contains a small amount of sulfur-containing molecules, whereas Saudi Arabian crude oil is “sour,” meaning that it contains a relatively large amount of sulfur-containing molecules.

Gasoline

Petroleum is converted to useful products such as gasoline in three steps: distillation, cracking, and reforming. Recall from that distillation separates compounds on the basis of their relative volatility, which is usually inversely proportional to their boiling points. Part (a) in shows a cutaway drawing of a column used in the petroleum industry for separating the components of crude oil. The petroleum is heated to approximately 400°C (750°F), at which temperature it has become a mixture of liquid and vapor. This mixture, called the feedstock, is introduced into the refining tower. The most volatile components (those with the lowest boiling points) condense at the top of the column where it is cooler, while the less volatile components condense nearer the bottom. Some materials are so nonvolatile that they collect at the bottom without evaporating at all. Thus the composition of the liquid condensing at each level is different. These different fractions, each of which usually consists of a mixture of compounds with similar numbers of carbon atoms, are drawn off separately. Part (b) in shows the typical fractions collected at refineries, the number of carbon atoms they contain, their boiling points, and their ultimate uses. These products range from gases used in natural and bottled gas to liquids used in fuels and lubricants to gummy solids used as tar on roads and roofs.

Figure 2.23 The Distillation of Petroleum

(a) This is a diagram of a distillation column used for separating petroleum fractions. (b) Petroleum fractions condense at different temperatures, depending on the number of carbon atoms in the molecules, and are drawn off from the column. The most volatile components (those with the lowest boiling points) condense at the top of the column, and the least volatile (those with the highest boiling points) condense at the bottom.

The economics of petroleum refining are complex. For example, the market demand for kerosene and lubricants is much lower than the demand for gasoline, yet all three fractions are obtained from the distillation column in comparable amounts. Furthermore, most gasolines and jet fuels are blends with very carefully controlled compositions that cannot vary as their original feedstocks did. To make petroleum refining more profitable, the less volatile, lower-value fractions must be converted to more volatile, higher-value mixtures that have carefully controlled formulas. The first process used to accomplish this transformation is crackingA process in petroleum refining in which the larger and heavier hydrocarbons in kerosene and higher-boiling-point fractions are heated to high temperatures, causing the carbon–carbon bonds to break (“crack”), thus producing a more volatile mixture., in which the larger and heavier hydrocarbons in the kerosene and higher-boiling-point fractions are heated to temperatures as high as 900°C. High-temperature reactions cause the carbon–carbon bonds to break, which converts the compounds to lighter molecules similar to those in the gasoline fraction. Thus in cracking, a straight-chain alkane with a number of carbon atoms corresponding to the kerosene fraction is converted to a mixture of hydrocarbons with a number of carbon atoms corresponding to the lighter gasoline fraction. The second process used to increase the amount of valuable products is called reformingThe second process used in petroleum refining, which is the chemical conversion of straight-chain alkanes to either branched-chain alkanes or mixtures of aromatic hydrocarbons.; it is the chemical conversion of straight-chain alkanes to either branched-chain alkanes or mixtures of aromatic hydrocarbons. Using metals such as platinum brings about the necessary chemical reactions. The mixtures of products obtained from cracking and reforming are separated by fractional distillation.

Octane Ratings

The quality of a fuel is indicated by its octane ratingA measure of a fuel’s ability to burn in a combustion engine without knocking or pinging (indications of premature combustion). The higher the octane rating, the higher quality the fuel., which is a measure of its ability to burn in a combustion engine without knocking or pinging. Knocking and pinging signal premature combustion (), which can be caused either by an engine malfunction or by a fuel that burns too fast. In either case, the gasoline-air mixture detonates at the wrong point in the engine cycle, which reduces the power output and can damage valves, pistons, bearings, and other engine components. The various gasoline formulations are designed to provide the mix of hydrocarbons least likely to cause knocking or pinging in a given type of engine performing at a particular level.

Figure 2.24 The Burning of Gasoline in an Internal Combustion Engine

(a) Normally, fuel is ignited by the spark plug, and combustion spreads uniformly outward. (b) Gasoline with an octane rating that is too low for the engine can ignite prematurely, resulting in uneven burning that causes knocking and pinging.

The octane scale was established in 1927 using a standard test engine and two pure compounds: n-heptane and isooctane (2,2,4-trimethylpentane). n-Heptane, which causes a great deal of knocking on combustion, was assigned an octane rating of 0, whereas isooctane, a very smooth-burning fuel, was assigned an octane rating of 100. Chemists assign octane ratings to different blends of gasoline by burning a sample of each in a test engine and comparing the observed knocking with the amount of knocking caused by specific mixtures of n-heptane and isooctane. For example, the octane rating of a blend of 89% isooctane and 11% n-heptane is simply the average of the octane ratings of the components weighted by the relative amounts of each in the blend. Converting percentages to decimals, we obtain the octane rating of the mixture:

0.89(100) + 0.11(0) = 89

A gasoline that performs at the same level as a blend of 89% isooctane and 11% n-heptane is assigned an octane rating of 89; this represents an intermediate grade of gasoline. Regular gasoline typically has an octane rating of 87; premium has a rating of 93 or higher.

As shown in , many compounds that are now available have octane ratings greater than 100, which means they are better fuels than pure isooctane. In addition, antiknock agents, also called octane enhancers, have been developed. One of the most widely used for many years was tetraethyllead [(C2H5)4Pb], which at approximately 3 g/gal gives a 10–15-point increase in octane rating. Since 1975, however, lead compounds have been phased out as gasoline additives because they are highly toxic. Other enhancers, such as methyl t-butyl ether (MTBE), have been developed to take their place. They combine a high octane rating with minimal corrosion to engine and fuel system parts. Unfortunately, when gasoline containing MTBE leaks from underground storage tanks, the result has been contamination of the groundwater in some locations, resulting in limitations or outright bans on the use of MTBE in certain areas. As a result, the use of alternative octane enhancers such as ethanol, which can be obtained from renewable resources such as corn, sugar cane, and, eventually, corn stalks and grasses, is increasing.

Figure 2.25 The Octane Ratings of Some Hydrocarbons and Common Additives

Example 12

You have a crude (i.e., unprocessed or straight-run) petroleum distillate consisting of 10% n-heptane, 10% n-hexane, and 80% n-pentane by mass, with an octane rating of 52. What percentage of MTBE by mass would you need to increase the octane rating of the distillate to that of regular-grade gasoline (a rating of 87), assuming that the octane rating is directly proportional to the amounts of the compounds present? Use the information presented in .

Given: composition of petroleum distillate, initial octane rating, and final octane rating

Asked for: percentage of MTBE by mass in final mixture

Strategy:

A Define the unknown as the percentage of MTBE in the final mixture. Then subtract this unknown from 100% to obtain the percentage of petroleum distillate.

B Multiply the percentage of MTBE and the percentage of petroleum distillate by their respective octane ratings; add these values to obtain the overall octane rating of the new mixture.

C Solve for the unknown to obtain the percentage of MTBE needed.

Solution:

A The question asks what percentage of MTBE will give an overall octane rating of 87 when mixed with the straight-run fraction. From , the octane rating of MTBE is 116. Let x be the percentage of MTBE, and let 100 − x be the percentage of petroleum distillate.

B Multiplying the percentage of each component by its respective octane rating and setting the sum equal to the desired octane rating of the mixture (87) times 100 gives

final octane rating of mixture =87(100)=52(100x)+116x=520052x+116x=5200+64x

C Solving the equation gives x = 55%. Thus the final mixture must contain 55% MTBE by mass.

To obtain a composition of 55% MTBE by mass, you would have to add more than an equal mass of MTBE (actually 0.55/0.45, or 1.2 times) to the straight-run fraction. This is 1.2 tons of MTBE per ton of straight-run gasoline, which would be prohibitively expensive. Thus there are sound economic reasons for reforming the kerosene fractions to produce toluene and other aromatic compounds, which have high octane ratings and are much cheaper than MTBE.

Exercise

As shown in , toluene is one of the fuels suitable for use in automobile engines. How much toluene would have to be added to a blend of the petroleum fraction in this example containing 15% MTBE by mass to increase the octane rating to that of premium gasoline (93)?

Answer: The final blend is 56% toluene by mass, which requires a ratio of 56/44, or 1.3 tons of toluene per ton of blend.

Sulfuric Acid

Sulfuric acid is one of the oldest chemical compounds known. It was probably first prepared by alchemists who burned sulfate salts such as FeSO4·7H2O, called green vitriol from its color and glassy appearance (from the Latin vitrum, meaning “glass”). Because pure sulfuric acid was found to be useful for dyeing textiles, enterprising individuals looked for ways to improve its production. By the mid-18th century, sulfuric acid was being produced in multiton quantities by the lead-chamber process, which was invented by John Roebuck in 1746. In this process, sulfur was burned in a large room lined with lead, and the resulting fumes were absorbed in water.

Production

The production of sulfuric acid today is likely to start with elemental sulfur obtained through an ingenious technique called the Frasch process, which takes advantage of the low melting point of elemental sulfur (115.2°C). Large deposits of elemental sulfur are found in porous limestone rocks in the same geological formations that often contain petroleum. In the Frasch process, water at high temperature (160°C) and high pressure is pumped underground to melt the sulfur, and compressed air is used to force the liquid sulfur-water mixture to the surface (). The material that emerges from the ground is more than 99% pure sulfur. After it solidifies, it is pulverized and shipped in railroad cars to the plants that produce sulfuric acid, as shown here.

Transporting sulfur. A train carries elemental sulfur through the White Canyon of the Thompson River in British Columbia, Canada.

Figure 2.26 Extraction of Elemental Sulfur from Underground Deposits

In the Frasch process for extracting sulfur, very hot water at high pressure is injected into the sulfur-containing rock layer to melt the sulfur. The resulting mixture of liquid sulfur and hot water is forced up to the surface by compressed air.

An increasing number of sulfuric acid manufacturers have begun to use sulfur dioxide (SO2) as a starting material instead of elemental sulfur. Sulfur dioxide is recovered from the burning of oil and gas, which contain small amounts of sulfur compounds. When not recovered, SO2 is released into the atmosphere, where it is converted to an environmentally hazardous form that leads to acid rain ().

If sulfur is the starting material, the first step in the production of sulfuric acid is the combustion of sulfur with oxygen to produce SO2. Next, SO2 is converted to SO3 by the contact process, in which SO2 and O2 react in the presence of V2O5 to achieve about 97% conversion to SO3. The SO3 can then be treated with a small amount of water to produce sulfuric acid. Usually, however, the SO3 is absorbed in concentrated sulfuric acid to produce oleum, a more potent form called fuming sulfuric acid. Because of its high SO3 content (approximately 99% by mass), oleum is cheaper to ship than concentrated sulfuric acid. At the point of use, the oleum is diluted with water to give concentrated sulfuric acid (very carefully because dilution generates enormous amounts of heat). Because SO2 is a pollutant, the small amounts of unconverted SO2 are recovered and recycled to minimize the amount released into the air.

Uses

Two-thirds of the sulfuric acid produced in the United States is used to make fertilizers, most of which contain nitrogen, phosphorus, and potassium (in a form called potash). In earlier days, phosphate-containing rocks were simply ground up and spread on fields as fertilizer, but the extreme insolubility of many salts that contain the phosphate ion (PO43−) limits the availability of phosphorus from these sources. Sulfuric acid serves as a source of protons (H+ ions) that react with phosphate minerals to produce more soluble salts containing HPO42− or H2PO4 as the anion, which are much more readily taken up by plants. In this context, sulfuric acid is used in two principal ways: (1) the phosphate rocks are treated with concentrated sulfuric acid to produce “superphosphate,” a mixture of 32% CaHPO4 and Ca(H2PO4)2·H2O, 50% CaSO4·2H2O, approximately 3% absorbed phosphoric acid, and other nutrients; and (2) sulfuric acid is used to produce phosphoric acid (H3PO4), which can then be used to convert phosphate rocks to “triple superphosphate,” which is largely Ca(H2PO4)2·H2O.

Sulfuric acid is also used to produce potash, one of the other major ingredients in fertilizers. The name potash originally referred to potassium carbonate (obtained by boiling wood ashes with water in iron pots), but today it also refers to compounds such as potassium hydroxide (KOH) and potassium oxide (K2O). The usual source of potassium in fertilizers is actually potassium sulfate (K2SO4), which is produced by several routes, including the reaction of concentrated sulfuric acid with solid potassium chloride (KCl), which is obtained as the pure salt from mineral deposits.

Summary

Many chemical compounds are prepared industrially in huge quantities and used to produce foods, fuels, plastics, and other such materials. Petroleum refining takes a complex mixture of naturally occurring hydrocarbons as a feedstock and, through a series of steps involving distillation, cracking, and reforming, converts them to mixtures of simpler organic compounds with desirable properties. A major use of petroleum is in the production of motor fuels such as gasoline. The performance of such fuels in engines is described by their octane rating, which depends on the identity of the compounds present and their relative abundance in the blend.

Sulfuric acid is the compound produced in the largest quantity in the industrial world. Much of the sulfur used in the production of sulfuric acid is obtained via the Frasch process, in which very hot water forces liquid sulfur out of the ground in nearly pure form. Sulfuric acid is produced by the reaction of sulfur dioxide with oxygen in the presence of vanadium(V) oxide (the contact process), followed by absorption of the sulfur trioxide in concentrated sulfuric acid to produce oleum. Most sulfuric acid is used to prepare fertilizers.

Key Takeaway

  • Many chemical compounds are prepared industrially in huge quantities to prepare the materials we need and use in our daily lives.

Conceptual Problems

  1. Describe the processes used for converting crude oil to transportation fuels.

  2. If your automobile engine is knocking, is the octane rating of your gasoline too low or too high? Explain your answer.

  3. Tetraethyllead is no longer used as a fuel additive to prevent knocking. Instead, fuel is now marketed as “unleaded.” Why is tetraethyllead no longer used?

  4. If you were to try to extract sulfur from an underground source, what process would you use? Describe briefly the essential features of this process.

  5. Why are phosphate-containing minerals used in fertilizers treated with sulfuric acid?

Answer

  1. Phosphate salts contain the highly-charged PO43− ion, salts of which are often insoluble. Protonation of the PO43− ion by strong acids such as H2SO4 leads to the formation of the HPO42− and H2PO4 ions. Because of their decreased negative charge, salts containing these anions are usually much more soluble, allowing the anions to be readily taken up by plants when they are applied as fertilizer.

Numerical Problem

  1. In Example 12, the crude petroleum had an overall octane rating of 52. What is the composition of a solution of MTBE and n-heptane that has this octane rating?

2.7 End-of-Chapter Material

Application Problems

    Problems marked with a ♦ involve multiple concepts.

  1. Carbon tetrachloride (CCl4) was used as a dry cleaning solvent until it was found to cause liver cancer. Based on the structure of chloroform given in , draw the structure of carbon tetrachloride.

  2. Ammonium nitrate and ammonium sulfate are used in fertilizers as a source of nitrogen. The ammonium cation is tetrahedral. Refer to to draw the structure of the ammonium ion.

  3. The white light in fireworks displays is produced by burning magnesium in air, which contains oxygen. What compound is formed?

  4. Sodium hydrogen sulfite, which is used for bleaching and swelling leather and to preserve flavor in almost all commercial wines, is made from sulfur dioxide. What are the formulas for these two sulfur-containing compounds?

  5. Carbonic acid is used in carbonated drinks. When combined with lithium hydroxide, it produces lithium carbonate, a compound used to increase the brightness of pottery glazes and as a primary treatment for depression and bipolar disorder. Write the formula for both of these carbon-containing compounds.

  6. Vinegar is a dilute solution of acetic acid, an organic acid, in water. What grouping of atoms would you expect to find in the structural formula for acetic acid?

  7. ♦ Sodamide, or sodium amide, is prepared from sodium metal and gaseous ammonia. Sodamide contains the amide ion (NH2), which reacts with water to form the hydroxide anion by removing an H+ ion from water. Sodium amide is also used to prepare sodium cyanide.

    1. Write the formula for each of these sodium-containing compounds.
    2. What are the products of the reaction of sodamide with water?
  8. A mixture of isooctane, n-pentane, and n-heptane is known to have an octane rating of 87. Use the data in to calculate how much isooctane and n-heptane are present if the mixture is known to contain 30% n-pentane.

  9. A crude petroleum distillate consists of 60% n-pentane, 25% methanol, and the remainder n-hexane by mass ().

    1. What is the octane rating?
    2. How much MTBE would have to be added to increase the octane rating to 93?
  10. Premium gasoline sold in much of the central United States has an octane rating of 93 and contains 10% ethanol. What is the octane rating of the gasoline fraction before ethanol is added? (See .)

Answers

  1. MgO, magnesium oxide

  2. Carbonic acid is H2CO3; lithium carbonate is Li2CO3.

    1. Sodamide is NaNH2, and sodium cyanide is NaCN.
    2. Sodium hydroxide (NaOH) and ammonia (NH3).
    1. 68
    2. 52 g of MTBE must be added to 48 g of the crude distillate.

Chapter 3 Chemical Reactions

Chapter 2 "Molecules, Ions, and Chemical Formulas" introduced you to a wide variety of chemical compounds, many of which have interesting applications. For example, nitrous oxide, a mild anesthetic, is also used as the propellant in cans of whipped cream, while copper(I) oxide is used as both a red glaze for ceramics and in antifouling bottom paints for boats. In addition to the physical properties of substances, chemists are also interested in their chemical reactionsA process in which a substance is converted to one or more other substances with different compositions and properties., processes in which a substance is converted to one or more other substances with different compositions and properties. Our very existence depends on chemical reactions, such as those between oxygen in the air we breathe and nutrient molecules in the foods we eat. Other reactions cook those foods, heat our homes, and provide the energy to run our cars. Many of the materials and pharmaceuticals that we take for granted today, such as silicon nitride for the sharp edge of cutting tools and antibiotics such as amoxicillin, were unknown only a few years ago. Their development required that chemists understand how substances combine in certain ratios and under specific conditions to produce a new substance with particular properties.

Sodium. The fourth most abundant alkali metal on Earth, sodium is a highly reactive element that is never found free in nature. When heated to 250°C, it bursts into flames if exposed to air.

We begin this chapter by describing the relationship between the mass of a sample of a substance and its composition. We then develop methods for determining the quantities of compounds produced or consumed in chemical reactions, and we describe some fundamental types of chemical reactions. By applying the concepts and skills introduced in this chapter, you will be able to explain what happens to the sugar in a candy bar you eat, what reaction occurs in a battery when you start your car, what may be causing the “ozone hole” over Antarctica, and how we might prevent the hole’s growth.

3.1 The Mole and Molar Masses

Learning Objective

  1. To calculate the molecular mass of a covalent compound and the formula mass of an ionic compound and to calculate the number of atoms, molecules, or formula units in a sample of a substance.

As you learned in Chapter 1 "Introduction to Chemistry", the mass number is the sum of the numbers of protons and neutrons present in the nucleus of an atom. The mass number is an integer that is approximately equal to the numerical value of the atomic mass. Although the mass number is unitless, it is assigned units called atomic mass units (amu). Because a molecule or a polyatomic ion is an assembly of atoms whose identities are given in its molecular or ionic formula, we can calculate the average atomic mass of any molecule or polyatomic ion from its composition by adding together the masses of the constituent atoms. The average mass of a monatomic ion is the same as the average mass of an atom of the element because the mass of electrons is so small that it is insignificant in most calculations.

Molecular and Formula Masses

The molecular massThe sum of the average masses of the atoms in one molecule of a substance, each multiplied by its subscript. of a substance is the sum of the average masses of the atoms in one molecule of a substance. It is calculated by adding together the atomic masses of the elements in the substance, each multiplied by its subscript (written or implied) in the molecular formula. Because the units of atomic mass are atomic mass units, the units of molecular mass are also atomic mass units. The procedure for calculating molecular masses is illustrated in Example 1.

Example 1

Calculate the molecular mass of ethanol, whose condensed structural formula is CH3CH2OH. Among its many uses, ethanol is a fuel for internal combustion engines.

Given: molecule

Asked for: molecular mass

Strategy:

A Determine the number of atoms of each element in the molecule.

B Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.

C Add together the masses to give the molecular mass.

Solution:

A The molecular formula of ethanol may be written in three different ways: CH3CH2OH (which illustrates the presence of an ethyl group, CH3CH2−, and an −OH group), C2H5OH, and C2H6O; all show that ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom.

B Taking the atomic masses from the periodic table, we obtain

× atomic mass of carbon = 2 atoms(12.011 amuatom)=24.022 amu× atomic mass of hydrogen = 6 atoms(1.0079 amuatom)=6.0474 amu× atomic mass of oxygen = 1 atom(15.9994 amuatom)=15.9994 amu

C Adding together the masses gives the molecular mass:

24.022 amu + 6.0474 amu + 15.9994 amu = 46.069 amu

Alternatively, we could have used unit conversions to reach the result in one step, as described in Essential Skills 2 (Section 3.7 "Essential Skills 2"):

[atoms C(12.011 amuatom C)]+[atoms H(1.0079 amuatom H)]+[atoms O(15.9994 amuatom O)]=46.069 amu

The same calculation can also be done in a tabular format, which is especially helpful for more complex molecules:

2C(2 atoms)(12.011 amu/atom)=24.022 amu6H(6 atoms)(1.0079 amu/atom)=6.0474 amu+1O(1 atom)(15.9994 amu/atom)=15.9994 amuC2H6Omolecular mass of ethanol=46.069 amu

Exercise

Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, whose condensed structural formula is CCl3F. Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows:

Answer: 137.368 amu

Unlike molecules, which have covalent bonds, ionic compounds do not have a readily identifiable molecular unit. So for ionic compounds we use the formula mass (also called the empirical formula massAnother name for formula mass.) of the compound rather than the molecular mass. The formula massThe sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript. is the sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript (written or implied). It is directly analogous to the molecular mass of a covalent compound. Once again, the units are atomic mass units.

Note the Pattern

Atomic mass, molecular mass, and formula mass all have the same units: atomic mass units.

Example 2

Calculate the formula mass of Ca3(PO4)2, commonly called calcium phosphate. This compound is the principal source of calcium found in bovine milk.

Given: ionic compound

Asked for: formula mass

Strategy:

A Determine the number of atoms of each element in the empirical formula.

B Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.

C Add together the masses to give the formula mass.

Solution:

A The empirical formula—Ca3(PO4)2—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca2+ ions and two PO43− ions. The formula mass of this molecular unit is calculated by adding together the atomic masses of three calcium atoms, two phosphorus atoms, and eight oxygen atoms.

B Taking atomic masses from the periodic table, we obtain

× atomic mass of calcium = 3 atoms(40.078 amuatom)=120.234 amu× atomic mass of phosphorus = 2 atoms(30.973761 amuatom)=61.947522 amu× atomic mass of oxygen = 8 atoms(15.9994 amuatom)=127.9952 amu

C Adding together the masses gives the formula mass of Ca3(PO4)2:

120.234 amu + 61.947522 amu + 127.9952 amu = 310.177 amu

We could also find the formula mass of Ca3(PO4)2 in one step by using unit conversions or a tabular format:

[atoms Ca(40.078 amuatom Ca)]+[atoms P(30.973761 amuatom P)]+[atoms O(15.9994 amuatom O)]=310.177 amu 3Ca(3 atoms)(40.078 amu/atom)=120.234 amu2P(2 atoms)(30.973761 amu/atom)=61.947522 amu+8O(8 atoms)(15.9994 amu/atom)=127.9952 amuCa3P2O8formula mass of Ca3(PO4)2=310.177 amu

Exercise

Calculate the formula mass of Si3N4, commonly called silicon nitride. It is an extremely hard and inert material that is used to make cutting tools for machining hard metal alloys.

Answer: 140.29 amu

The Mole

In Chapter 1 "Introduction to Chemistry", we described Dalton’s theory that each chemical compound has a particular combination of atoms and that the ratios of the numbers of atoms of the elements present are usually small whole numbers. We also described the law of multiple proportions, which states that the ratios of the masses of elements that form a series of compounds are small whole numbers. The problem for Dalton and other early chemists was to discover the quantitative relationship between the number of atoms in a chemical substance and its mass. Because the masses of individual atoms are so minuscule (on the order of 10−23 g/atom), chemists do not measure the mass of individual atoms or molecules. In the laboratory, for example, the masses of compounds and elements used by chemists typically range from milligrams to grams, while in industry, chemicals are bought and sold in kilograms and tons. To analyze the transformations that occur between individual atoms or molecules in a chemical reactionA process in which a substance is converted to one or more other substances with different compositions and properties., it is therefore absolutely essential for chemists to know how many atoms or molecules are contained in a measurable quantity in the laboratory—a given mass of sample. The unit that provides this link is the mole (mol)The quantity of a substance that contains the same number of units (e.g., atoms or molecules) as the number of carbon atoms in exactly 12 g of isotopically pure carbon-12., from the Latin moles, meaning “pile” or “heap” (not from the small subterranean animal!).

Many familiar items are sold in numerical quantities that have unusual names. For example, cans of soda come in a six-pack, eggs are sold by the dozen (12), and pencils often come in a gross (12 dozen, or 144). Sheets of printer paper are packaged in reams of 500, a seemingly large number. Atoms are so small, however, that even 500 atoms are too small to see or measure by most common techniques. Any readily measurable mass of an element or compound contains an extraordinarily large number of atoms, molecules, or ions, so an extraordinarily large numerical unit is needed to count them. The mole is used for this purpose.

A mole is defined as the amount of a substance that contains the number of carbon atoms in exactly 12 g of isotopically pure carbon-12. According to the most recent experimental measurements, this mass of carbon-12 contains 6.022142 × 1023 atoms, but for most purposes 6.022 × 1023 provides an adequate number of significant figures. Just as 1 mol of atoms contains 6.022 × 1023 atoms, 1 mol of eggs contains 6.022 × 1023 eggs. The number in a mole is called Avogadro’s numberThe number of units (e.g., atoms, molecules, or formula units) in 1 mol: 6.022×1023., after the 19th-century Italian scientist who first proposed a relationship between the volumes of gases and the numbers of particles they contain.

It is not obvious why eggs come in dozens rather than 10s or 14s, or why a ream of paper contains 500 sheets rather than 400 or 600. The definition of a mole—that is, the decision to base it on 12 g of carbon-12—is also arbitrary. The important point is that 1 mol of carbon—or of anything else, whether atoms, compact discs, or houses—always has the same number of objects: 6.022 × 1023.

Note the Pattern

One mole always has the same number of objects: 6.022 × 1023.

To appreciate the magnitude of Avogadro’s number, consider a mole of pennies. Stacked vertically, a mole of pennies would be 4.5 × 1017 mi high, or almost six times the diameter of the Milky Way galaxy. If a mole of pennies were distributed equally among the entire population on Earth, each person would get more than one trillion dollars. Clearly, the mole is so large that it is useful only for measuring very small objects, such as atoms.

The concept of the mole allows us to count a specific number of individual atoms and molecules by weighing measurable quantities of elements and compounds. To obtain 1 mol of carbon-12 atoms, we would weigh out 12 g of isotopically pure carbon-12. Because each element has a different atomic mass, however, a mole of each element has a different mass, even though it contains the same number of atoms (6.022 × 1023). This is analogous to the fact that a dozen extra large eggs weighs more than a dozen small eggs, or that the total weight of 50 adult humans is greater than the total weight of 50 children. Because of the way in which the mole is defined, for every element the number of grams in a mole is the same as the number of atomic mass units in the atomic mass of the element. For example, the mass of 1 mol of magnesium (atomic mass = 24.305 amu) is 24.305 g. Because the atomic mass of magnesium (24.305 amu) is slightly more than twice that of a carbon-12 atom (12 amu), the mass of 1 mol of magnesium atoms (24.305 g) is slightly more than twice that of 1 mol of carbon-12 (12 g). Similarly, the mass of 1 mol of helium (atomic mass = 4.002602 amu) is 4.002602 g, which is about one-third that of 1 mol of carbon-12. Using the concept of the mole, we can now restate Dalton’s theory: 1 mol of a compound is formed by combining elements in amounts whose mole ratios are small whole numbers. For example, 1 mol of water (H2O) has 2 mol of hydrogen atoms and 1 mol of oxygen atoms.

Molar Mass

The molar massThe mass in grams of 1 mol of a substance. of a substance is defined as the mass in grams of 1 mol of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element; for a covalent molecular compound, it is the mass of 1 mol of molecules of that compound; for an ionic compound, it is the mass of 1 mol of formula units. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 1023 atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively.

Note the Pattern

The molar mass of any substance is its atomic mass, molecular mass, or formula mass in grams per mole.

The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 1023 carbon atoms—is therefore 12.011 g/mol:

Substance (formula) Atomic, Molecular, or Formula Mass (amu) Molar Mass (g/mol)
carbon (C) 12.011 (atomic mass) 12.011
ethanol (C2H5OH) 46.069 (molecular mass) 46.069
calcium phosphate [Ca3(PO4)2] 310.177 (formula mass) 310.177

The molar mass of naturally occurring carbon is different from that of carbon-12 and is not an integer because carbon occurs as a mixture of carbon-12, carbon-13, and carbon-14. One mole of carbon still has 6.022 × 1023 carbon atoms, but 98.89% of those atoms are carbon-12, 1.11% are carbon-13, and a trace (about 1 atom in 1012) are carbon-14. (For more information, see Section 1.6 "Isotopes and Atomic Masses".) Similarly, the molar mass of uranium is 238.03 g/mol, and the molar mass of iodine is 126.90 g/mol. When we deal with elements such as iodine and sulfur, which occur as a diatomic molecule (I2) and a polyatomic molecule (S8), respectively, molar mass usually refers to the mass of 1 mol of atoms of the element—in this case I and S, not to the mass of 1 mol of molecules of the element (I2 and S8).

The molar mass of ethanol is the mass of ethanol (C2H5OH) that contains 6.022 × 1023 ethanol molecules. As you calculated in Example 1, the molecular mass of ethanol is 46.069 amu. Because 1 mol of ethanol contains 2 mol of carbon atoms (2 × 12.011 g), 6 mol of hydrogen atoms (6 × 1.0079 g), and 1 mol of oxygen atoms (1 × 15.9994 g), its molar mass is 46.069 g/mol. Similarly, the formula mass of calcium phosphate [Ca3(PO4)2] is 310.177 amu, so its molar mass is 310.177 g/mol. This is the mass of calcium phosphate that contains 6.022 × 1023 formula units. Figure 3.1 "Samples of 1 Mol of Some Common Substances" shows samples that contain precisely one molar mass of several common substances.

Figure 3.1 Samples of 1 Mol of Some Common Substances

The mole is the basis of quantitative chemistry. It provides chemists with a way to convert easily between the mass of a substance and the number of individual atoms, molecules, or formula units of that substance. Conversely, it enables chemists to calculate the mass of a substance needed to obtain a desired number of atoms, molecules, or formula units. For example, to convert moles of a substance to mass, we use the relationship

Equation 3.1

(moles)(molar mass) → mass

or, more specifically,

moles(gramsmole)= grams

Conversely, to convert the mass of a substance to moles, we use

Equation 3.2

(massmolar mass)moles(gramsgrams/mole)=grams(molegrams)= moles

Be sure to pay attention to the units when converting between mass and moles.

Figure 3.2 "A Flowchart for Converting between Mass; the Number of Moles; and the Number of Atoms, Molecules, or Formula Units" is a flowchart for converting between mass; the number of moles; and the number of atoms, molecules, or formula units. The use of these conversions is illustrated in Example 3 and Example 4.

Figure 3.2 A Flowchart for Converting between Mass; the Number of Moles; and the Number of Atoms, Molecules, or Formula Units

Example 3

For 35.00 g of ethylene glycol (HOCH2CH2OH), which is used in inks for ballpoint pens, calculate the number of

  1. moles.
  2. molecules.

Given: mass and molecular formula

Asked for: number of moles and number of molecules

Strategy:

A Use the molecular formula of the compound to calculate its molecular mass in grams per mole.

B Convert from mass to moles by dividing the mass given by the compound’s molar mass.

C Convert from moles to molecules by multiplying the number of moles by Avogadro’s number.

Solution:

  1. A The molecular mass of ethylene glycol can be calculated from its molecular formula using the method illustrated in Example 1:

    2C(2 atoms)(12.011 amu/atom)=24.022 amu6H(6 atoms)(1.0079 amu/atom)=6.0474 amu+2O(2 atoms)(15.9994 amu/atom)=31.9988 amuC2H6O2molecular mass of ethylene glycol=62.068 amu

    The molar mass of ethylene glycol is 62.068 g/mol.

    B The number of moles of ethylene glycol present in 35.00 g can be calculated by dividing the mass (in grams) by the molar mass (in grams per mole):

    mass of ethylene glycol (g)molar mass (g/mol)= moles ethylene glycol (mol)

    So

    35.00 g ethylene glycol(1 mol ethylene glycol62.068 g ethylene glycol)=0.5639 mol ethylene glycol

    It is always a good idea to estimate the answer before you do the actual calculation. In this case, the mass given (35.00 g) is less than the molar mass, so the answer should be less than 1 mol. The calculated answer (0.5639 mol) is indeed less than 1 mol, so we have probably not made a major error in the calculations.

  2. C To calculate the number of molecules in the sample, we multiply the number of moles by Avogadro’s number:

    molecules of ethylene glycol = 0.5639 mol(6.022×1023 moleculesmol)= 3.396×1023 molecules

    Because we are dealing with slightly more than 0.5 mol of ethylene glycol, we expect the number of molecules present to be slightly more than one-half of Avogadro’s number, or slightly more than 3 × 1023 molecules, which is indeed the case.

Exercise

For 75.0 g of CCl3F (Freon-11), calculate the number of

  1. moles.
  2. molecules.

Answer:

  1. 0.546 mol
  2. 3.29 × 1023 molecules

Example 4

Calculate the mass of 1.75 mol of each compound.

  1. S2Cl2 (common name: sulfur monochloride; systematic name: disulfur dichloride)
  2. Ca(ClO)2 (calcium hypochlorite)

Given: number of moles and molecular or empirical formula

Asked for: mass

Strategy:

A Calculate the molecular mass of the compound in grams from its molecular formula (if covalent) or empirical formula (if ionic).

B Convert from moles to mass by multiplying the moles of the compound given by its molar mass.

Solution:

We begin by calculating the molecular mass of S2Cl2 and the formula mass of Ca(ClO)2.

  1. A The molar mass of S2Cl2 is obtained from its molecular mass as follows:

    2S(2 atoms)(32.065 amu/atom)=64.130 amu+2Cl(2 atoms)(35.453 amu/atom)=70.906 amuS2Cl2molecular mass of S2Cl2=135.036 amu

    The molar mass of S2Cl2 is 135.036 g/mol.

    B The mass of 1.75 mol of S2Cl2 is calculated as follows:

    moles S2Cl2[molar mass(gmol)]mass of S2Cl2(g)1.75 mol S2Cl2(135.036 g S2Cl2mol S2Cl2)=236 g S2Cl2
  2. A The formula mass of Ca(ClO)2 is obtained as follows:

    1Ca(1 atom)(40.078 amu/atom)=40.078 amu2Cl(2 atoms)(35.453 amu/atom)=70.906 amu+2O(2 atoms)(15.9994 amu/atom)=31.9988 amuCa(ClO)2formula mass of Ca(ClO)2=142.983 amu

    The molar mass of Ca(ClO)2 142.983 g/mol.

    B The mass of 1.75 mol of Ca(ClO)2 is calculated as follows:

    moles Ca(ClO)2[molar mass Ca(ClO)21 mol Ca(ClO)2]=mass Ca(ClO)21.75 mol Ca(ClO)2[142.983 g Ca(ClO)2mol Ca(ClO)2]=250 g Ca(ClO)2

    Because 1.75 mol is less than 2 mol, the final quantity in grams in both cases should be less than twice the molar mass, which it is.

Exercise

Calculate the mass of 0.0122 mol of each compound.

  1. Si3N4 (silicon nitride), used as bearings and rollers
  2. (CH3)3N (trimethylamine), a corrosion inhibitor

Answer:

  1. 1.71 g
  2. 0.721 g

Summary

The molecular mass and the formula mass of a compound are obtained by adding together the atomic masses of the atoms present in the molecular formula or empirical formula, respectively; the units of both are atomic mass units (amu). The mole is a unit used to measure the number of atoms, molecules, or (in the case of ionic compounds) formula units in a given mass of a substance. The mole is defined as the amount of substance that contains the number of carbon atoms in exactly 12 g of carbon-12 and consists of Avogadro’s number (6.022 × 1023) of atoms of carbon-12. The molar mass of a substance is defined as the mass of 1 mol of that substance, expressed in grams per mole, and is equal to the mass of 6.022 × 1023 atoms, molecules, or formula units of that substance.

Key Takeaway

  • To analyze chemical transformations, it is essential to use a standardized unit of measure called the mole.

Conceptual Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Conceptual Problems.

  1. Describe the relationship between an atomic mass unit and a gram.

  2. Is it correct to say that ethanol has a formula mass of 46? Why or why not?

  3. If 2 mol of sodium react completely with 1 mol of chlorine to produce sodium chloride, does this mean that 2 g of sodium reacts completely with 1 g of chlorine to give the same product? Explain your answer.

  4. Construct a flowchart to show how you would calculate the number of moles of silicon in a 37.0 g sample of orthoclase (KAlSi3O8), a mineral used in the manufacture of porcelain.

  5. Construct a flowchart to show how you would calculate the number of moles of nitrogen in a 22.4 g sample of nitroglycerin that contains 18.5% nitrogen by mass.

Answer

  1.  

    A = %N by mass, expressed as a decimal =1molar  mass  of  nitrogen  in  g g  nitroglycerin×AgN×Bmol N

Numerical Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Numerical Problems.

  1. Derive an expression that relates the number of molecules in a sample of a substance to its mass and molecular mass.

  2. Calculate the molecular mass or formula mass of each compound.

    1. KCl (potassium chloride)
    2. NaCN (sodium cyanide)
    3. H2S (hydrogen sulfide)
    4. NaN3 (sodium azide)
    5. H2CO3 (carbonic acid)
    6. K2O (potassium oxide)
    7. Al(NO3)3 (aluminum nitrate)
    8. Cu(ClO4)2 [copper(II) perchlorate]
  3. Calculate the molecular mass or formula mass of each compound.

    1. V2O4 (vanadium(IV) oxide)
    2. CaSiO3 (calcium silicate)
    3. BiOCl (bismuth oxychloride)
    4. CH3COOH (acetic acid)
    5. Ag2SO4 (silver sulfate)
    6. Na2CO3 (sodium carbonate)
    7. (CH3)2CHOH (isopropyl alcohol)
  4. Calculate the molar mass of each compound.

    1.  

    2.  

    3.  

    4.  

    5.  

  5. Calculate the molar mass of each compound.

    1.  

    2.  

    3.  

    4.  

  6. For each compound, write the condensed formula, name the compound, and give its molar mass.

    1.  

    2.  

  7. For each compound, write the condensed formula, name the compound, and give its molar mass.

    1.  

    2.  

  8. Calculate the number of moles in 5.00 × 102 g of each substance. How many molecules or formula units are present in each sample?

    1. CaO (lime)
    2. CaCO3 (chalk)
    3. C12H22O11 [sucrose (cane sugar)]
    4. NaOCl (bleach)
    5. CO2 (dry ice)
  9. Calculate the mass in grams of each sample.

    1. 0.520 mol of N2O4
    2. 1.63 mol of C6H4Br2
    3. 4.62 mol of (NH4)2SO3
  10. Give the number of molecules or formula units in each sample.

    1. 1.30 × 10−2 mol of SCl2
    2. 1.03 mol of N2O5
    3. 0.265 mol of Ag2Cr2O7
  11. Give the number of moles in each sample.

    1. 9.58 × 1026 molecules of Cl2
    2. 3.62 × 1027 formula units of KCl
    3. 6.94 × 1028 formula units of Fe(OH)2
  12. Solutions of iodine are used as antiseptics and disinfectants. How many iodine atoms correspond to 11.0 g of molecular iodine (I2)?

  13. What is the total number of atoms in each sample?

    1. 0.431 mol of Li
    2. 2.783 mol of methanol (CH3OH)
    3. 0.0361 mol of CoCO3
    4. 1.002 mol of SeBr2O
  14. What is the total number of atoms in each sample?

    1. 0.980 mol of Na
    2. 2.35 mol of O2
    3. 1.83 mol of Ag2S
    4. 1.23 mol of propane (C3H8)
  15. What is the total number of atoms in each sample?

    1. 2.48 g of HBr
    2. 4.77 g of CS2
    3. 1.89 g of NaOH
    4. 1.46 g of SrC2O4
  16. Decide whether each statement is true or false and explain your reasoning.

    1. There are more molecules in 0.5 mol of Cl2 than in 0.5 mol of H2.
    2. One mole of H2 has 6.022 × 1023 hydrogen atoms.
    3. The molecular mass of H2O is 18.0 amu.
    4. The formula mass of benzene is 78 amu.
  17. Complete the following table.

    Substance Mass (g) Number of Moles Number of Molecules or Formula Units Number of Atoms or Ions
    MgCl2 37.62
    AgNO3 2.84
    BH4Cl 8.93 × 1025
    K2S 7.69 × 1026
    H2SO4 1.29
    C6H14 11.84
    HClO3 2.45 × 1026
  18. Give the formula mass or the molecular mass of each substance.

    1. PbClF
    2. Cu2P2O7
    3. BiONO3
    4. Tl2SeO4
  19. Give the formula mass or the molecular mass of each substance.

    1. MoCl5
    2. B2O3
    3. UO2CO3
    4. NH4UO2AsO4

3.2 Determining Empirical and Molecular Formulas

Learning Objectives

  1. To determine the empirical formula of a compound from its composition by mass.
  2. To derive the molecular formula of a compound from its empirical formula.

When a new chemical compound, such as a potential new pharmaceutical, is synthesized in the laboratory or isolated from a natural source, chemists determine its elemental composition, its empirical formula, and its structure to understand its properties. In this section, we focus on how to determine the empirical formula of a compound and then use it to determine the molecular formula if the molar mass of the compound is known.

Calculating Mass Percentages

The law of definite proportions states that a chemical compound always contains the same proportion of elements by mass; that is, the percent compositionThe percentage of each element present in a pure substance. With few exceptions, the percent composition of a chemical compound is constant (see law of definite proportions).—the percentage of each element present in a pure substance—is constant (although we now know there are exceptions to this law). For example, sucrose (cane sugar) is 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. This means that 100.00 g of sucrose always contains 42.11 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen. First we will use the molecular formula of sucrose (C12H22O11) to calculate the mass percentage of the component elements; then we will show how mass percentages can be used to determine an empirical formula.

According to its molecular formula, each molecule of sucrose contains 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. A mole of sucrose molecules therefore contains 12 mol of carbon atoms, 22 mol of hydrogen atoms, and 11 mol of oxygen atoms. We can use this information to calculate the mass of each element in 1 mol of sucrose, which will give us the molar mass of sucrose. We can then use these masses to calculate the percent composition of sucrose. To three decimal places, the calculations are the following:

Equation 3.3

mass of C/mol of sucrose = 12 mol C×12.011 g Cmol C=144.132 g Cmass of H/mol of sucrose = 22 mol H×1.008 g Hmol H=22.176 g Hmass of O/mol of sucrose = 11 mol O×15.999 g Omol O=175.989 g O

Thus 1 mol of sucrose has a mass of 342.297 g; note that more than half of the mass (175.989 g) is oxygen, and almost half of the mass (144.132 g) is carbon.

The mass percentage of each element in sucrose is the mass of the element present in 1 mol of sucrose divided by the molar mass of sucrose, multiplied by 100 to give a percentage. The result is shown to two decimal places:

mass % C in sucrose =mass of C/mol sucrosemolar mass of sucrose×100=144.132 g C342.297 g/mol×100=42.12%mass % H in sucrose =mass of H/mol sucrosemolar mass of sucrose×100=22.176 g H342.297 g/mol×100=6.48%mass % O in sucrose =mass of O/mol sucrosemolar mass of sucrose×100=175.989 g O342.297 g/mol×100=51.41%

You can check your work by verifying that the sum of the percentages of all the elements in the compound is 100%:

42.12% + 6.48% + 51.41% = 100.01%

If the sum is not 100%, you have made an error in your calculations. (Rounding to the correct number of decimal places can, however, cause the total to be slightly different from 100%.) Thus 100.00 g of sucrose contains 42.12 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen; to two decimal places, the percent composition of sucrose is indeed 42.12% carbon, 6.48% hydrogen, and 51.41% oxygen.

We could also calculate the mass percentages using atomic masses and molecular masses, with atomic mass units. Because the answer we are seeking is a ratio, expressed as a percentage, the units of mass cancel whether they are grams (using molar masses) or atomic mass units (using atomic and molecular masses).

Example 5

Aspartame is the artificial sweetener sold as NutraSweet and Equal. Its molecular formula is C14H18N2O5.

  1. Calculate the mass percentage of each element in aspartame.
  2. Calculate the mass of carbon in a 1.00 g packet of Equal, assuming it is pure aspartame.

Given: molecular formula and mass of sample

Asked for: mass percentage of all elements and mass of one element in sample

Strategy:

A Use atomic masses from the periodic table to calculate the molar mass of aspartame.

B Divide the mass of each element by the molar mass of aspartame; then multiply by 100 to obtain percentages.

C To find the mass of an element contained in a given mass of aspartame, multiply the mass of aspartame by the mass percentage of that element, expressed as a decimal.

Solution:

  1. A We calculate the mass of each element in 1 mol of aspartame and the molar mass of aspartame, here to three decimal places:

    14C(14 mol C)(12.011 g/mol C)= 168.154 g18H(18 mol H)(1.008 g/mol H)=18.114 g2N(2 mol N)(14.007 g/mol N)= 28.014 g+5O (5 mol O)(15.999 g/mol O)=79.995 gC14H18N2O5molar mass of aspartame=294.277 g/mol

    Thus more than half the mass of 1 mol of aspartame (294.277 g) is carbon (168.154 g).

    B To calculate the mass percentage of each element, we divide the mass of each element in the compound by the molar mass of aspartame and then multiply by 100 to obtain percentages, here reported to two decimal places:

    mass % C =168.154 g C294.277 g aspartame×100 = 57.14% Cmass % H =18.114 g H294.277 g aspartame×100 = 6.16% Hmass % N =28.014 g N294.277 g aspartame×100 = 9.52% Nmass % O =79.995 g O294.277 g aspartame×100 = 27.18% O

    As a check, we can add the percentages together:

    57.14% + 6.16% + 9.52% + 27.18% = 100.00%

    If you obtain a total that differs from 100% by more than about ±1%, there must be an error somewhere in the calculation.

  2. C The mass of carbon in 1.00 g of aspartame is calculated as follows:

    mass of C = 1.00 g aspartame×57.14 g C100 g aspartame=0.571 g C

Exercise

Calculate the mass percentage of each element in aluminum oxide (Al2O3). Then calculate the mass of aluminum in a 3.62 g sample of pure aluminum oxide.

Answer: 52.93% aluminum; 47.08% oxygen; 1.92 g Al

Determining the Empirical Formula of Penicillin

Just as we can use the empirical formula of a substance to determine its percent composition, we can use the percent composition of a sample to determine its empirical formula, which can then be used to determine its molecular formula. Such a procedure was actually used to determine the empirical and molecular formulas of the first antibiotic to be discovered: penicillin.

Antibiotics are chemical compounds that selectively kill microorganisms, many of which cause diseases. Although we may take antibiotics for granted today, penicillin was discovered only about 80 years ago. The subsequent development of a wide array of other antibiotics for treating many common diseases has contributed greatly to the substantial increase in life expectancy over the past 50 years. The discovery of penicillin is a historical detective story in which the use of mass percentages to determine empirical formulas played a key role.

In 1928, Alexander Fleming, a young microbiologist at the University of London, was working with a common bacterium that causes boils and other infections such as blood poisoning. For laboratory study, bacteria are commonly grown on the surface of a nutrient-containing gel in small, flat culture dishes. One day Fleming noticed that one of his cultures was contaminated by a bluish-green mold similar to the mold found on spoiled bread or fruit. Such accidents are rather common, and most laboratory workers would have simply thrown the cultures away. Fleming noticed, however, that the bacteria were growing everywhere on the gel except near the contaminating mold (part (a) in ), and he hypothesized that the mold must be producing a substance that either killed the bacteria or prevented their growth. To test this hypothesis, he grew the mold in a liquid and then filtered the liquid and added it to various bacteria cultures. The liquid killed not only the bacteria Fleming had originally been studying but also a wide range of other disease-causing bacteria. Because the mold was a member of the Penicillium family (named for their pencil-shaped branches under the microscope) (part (b) in ), Fleming called the active ingredient in the broth penicillin.

Figure 3.3 Penicillium

(a) Penicillium mold is growing in a culture dish; the photo shows its effect on bacterial growth. (b) In this photomicrograph of Penicillium, its rod- and pencil-shaped branches are visible. The name comes from the Latin penicillus, meaning “paintbrush.”

Although Fleming was unable to isolate penicillin in pure form, the medical importance of his discovery stimulated researchers in other laboratories. Finally, in 1940, two chemists at Oxford University, Howard Florey (1898–1968) and Ernst Chain (1906–1979), were able to isolate an active product, which they called penicillin G. Within three years, penicillin G was in widespread use for treating pneumonia, gangrene, gonorrhea, and other diseases, and its use greatly increased the survival rate of wounded soldiers in World War II. As a result of their work, Fleming, Florey, and Chain shared the Nobel Prize in Medicine in 1945.

As soon as they had succeeded in isolating pure penicillin G, Florey and Chain subjected the compound to a procedure called combustion analysis (described later in this section) to determine what elements were present and in what quantities. The results of such analyses are usually reported as mass percentages. They discovered that a typical sample of penicillin G contains 53.9% carbon, 4.8% hydrogen, 7.9% nitrogen, 9.0% sulfur, and 6.5% sodium by mass. The sum of these numbers is only 82.1%, rather than 100.0%, which implies that there must be one or more additional elements. A reasonable candidate is oxygen, which is a common component of compounds that contain carbon and hydrogen;Do not assume that the “missing” mass is always due to oxygen. It could be any other element. for technical reasons, however, it is difficult to analyze for oxygen directly. If we assume that all the missing mass is due to oxygen, then penicillin G contains (100.0% − 82.1%) = 17.9% oxygen. From these mass percentages, the empirical formula and eventually the molecular formula of the compound can be determined.

To determine the empirical formula from the mass percentages of the elements in a compound such as penicillin G, we need to convert the mass percentages to relative numbers of atoms. For convenience, we assume that we are dealing with a 100.0 g sample of the compound, even though the sizes of samples used for analyses are generally much smaller, usually in milligrams. This assumption simplifies the arithmetic because a 53.9% mass percentage of carbon corresponds to 53.9 g of carbon in a 100.0 g sample of penicillin G; likewise, 4.8% hydrogen corresponds to 4.8 g of hydrogen in 100.0 g of penicillin G; and so forth for the other elements. We can then divide each mass by the molar mass of the element to determine how many moles of each element are present in the 100.0 g sample:

Equation 3.4

mass (g)molar mass (g/mol)= (g)(molg)= mol53.9 g C(1 mol C12.011 g C)= 4.49 mol C4.8 g H(1 mol H1.008 g H)= 4.8 mol H7.9 g N(1 mol N14.007 g N)= 0.56 mol N9.0 g S(1 mol S32.065 g S)= 0.28 mol S6.5 g Na(1 mol Na22.990 g Na)= 0.28 mol Na17.9 g O(1 mol O15.999 g O)= 1.12 mol O

Thus 100.0 g of penicillin G contains 4.49 mol of carbon, 4.8 mol of hydrogen, 0.56 mol of nitrogen, 0.28 mol of sulfur, 0.28 mol of sodium, and 1.12 mol of oxygen (assuming that all the missing mass was oxygen). The number of significant figures in the numbers of moles of elements varies between two and three because some of the analytical data were reported to only two significant figures.

These results tell us the ratios of the moles of the various elements in the sample (4.49 mol of carbon to 4.8 mol of hydrogen to 0.56 mol of nitrogen, and so forth), but they are not the whole-number ratios we need for the empirical formula—the empirical formula expresses the relative numbers of atoms in the smallest whole numbers possible. To obtain whole numbers, we divide the numbers of moles of all the elements in the sample by the number of moles of the element present in the lowest relative amount, which in this example is sulfur or sodium. The results will be the subscripts of the elements in the empirical formula. To two significant figures, the results are

Equation 3.5

C: 4.490.28= 16H: 4.80.28= 17N: 0.560.28= 2.0S: 0.280.28= 1.0Na: 0.280.28= 1.0O: 1.120.28= 4.0

The empirical formula of penicillin G is therefore C16H17N2NaO4S. Other experiments have shown that penicillin G is actually an ionic compound that contains Na+ cations and [C16H17N2O4S] anions in a 1:1 ratio. The complex structure of penicillin G () was not determined until 1948.

Figure 3.4 Structural Formula and Ball-and-Stick Model of the Anion of Penicillin G

In some cases, one or more of the subscripts in a formula calculated using this procedure may not be integers. Does this mean that the compound of interest contains a nonintegral number of atoms? No; rounding errors in the calculations as well as experimental errors in the data can result in nonintegral ratios. When this happens, you must exercise some judgment in interpreting the results, as illustrated in Example 6. In particular, ratios of 1.50, 1.33, or 1.25 suggest that you should multiply all subscripts in the formula by 2, 3, or 4, respectively. Only if the ratio is within 5% of an integral value should you consider rounding to the nearest integer.

Example 6

Calculate the empirical formula of the ionic compound calcium phosphate, a major component of fertilizer and a polishing agent in toothpastes. Elemental analysis indicates that it contains 38.77% calcium, 19.97% phosphorus, and 41.27% oxygen.

Given: percent composition

Asked for: empirical formula

Strategy:

A Assume a 100 g sample and calculate the number of moles of each element in that sample.

B Obtain the relative numbers of atoms of each element in the compound by dividing the number of moles of each element in the 100 g sample by the number of moles of the element present in the smallest amount.

C If the ratios are not integers, multiply all subscripts by the same number to give integral values.

D Because this is an ionic compound, identify the anion and cation and write the formula so that the charges balance.

Solution:

A A 100 g sample of calcium phosphate contains 38.77 g of calcium, 19.97 g of phosphorus, and 41.27 g of oxygen. Dividing the mass of each element in the 100 g sample by its molar mass gives the number of moles of each element in the sample:

moles Ca = 38.77 g Ca×1 mol Ca40.078 g Ca= 0.9674 mol Camoles P = 19.97 g P×1 mol P30.9738 g P= 0.6447 mol Pmoles O = 41.27 g O×1 mol O15.9994 g O= 2.5800 mol O

B To obtain the relative numbers of atoms of each element in the compound, divide the number of moles of each element in the 100-g sample by the number of moles of the element in the smallest amount, in this case phosphorus:

P: 0.6447 mol P0.6447 mol P= 1.000Ca: 0.96740.6447= 1.501O: 2.58000.6447= 4.002

C We could write the empirical formula of calcium phosphate as Ca1.501P1.000O4.002, but the empirical formula should show the ratios of the elements as small whole numbers. To convert the result to integral form, multiply all the subscripts by 2 to get Ca3.002P2.000O8.004. The deviation from integral atomic ratios is small and can be attributed to minor experimental errors; therefore, the empirical formula is Ca3P2O8.

D The calcium ion (Ca2+) is a cation, so to maintain electrical neutrality, phosphorus and oxygen must form a polyatomic anion. We know from that phosphorus and oxygen form the phosphate ion (PO43−; see ). Because there are two phosphorus atoms in the empirical formula, two phosphate ions must be present. So we write the formula of calcium phosphate as Ca3(PO4)2.

Exercise

Calculate the empirical formula of ammonium nitrate, an ionic compound that contains 35.00% nitrogen, 5.04% hydrogen, and 59.96% oxygen by mass; refer to if necessary. Although ammonium nitrate is widely used as a fertilizer, it can be dangerously explosive. For example, it was a major component of the explosive used in the 1995 Oklahoma City bombing.

Answer: N2H4O3 is NH4+NO3, written as NH4NO3

Combustion Analysis

One of the most common ways to determine the elemental composition of an unknown hydrocarbon is an analytical procedure called combustion analysis. A small, carefully weighed sample of an unknown compound that may contain carbon, hydrogen, nitrogen, and/or sulfur is burned in an oxygen atmosphere,Other elements, such as metals, can be determined by other methods. and the quantities of the resulting gaseous products (CO2, H2O, N2, and SO2, respectively) are determined by one of several possible methods. One procedure used in combustion analysis is outlined schematically in , and a typical combustion analysis is illustrated in Example 7.

Figure 3.5 Steps for Obtaining an Empirical Formula from Combustion Analysis

Example 7

Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO2 and 11.30 mg of H2O. Determine the empirical formula of naphthalene.

Given: mass of sample and mass of combustion products

Asked for: empirical formula

Strategy:

A Use the masses and molar masses of the combustion products, CO2 and H2O, to calculate the masses of carbon and hydrogen present in the original sample of naphthalene.

B Use those masses and the molar masses of the elements to calculate the empirical formula of naphthalene.

Solution:

A Upon combustion, 1 mol of CO2 is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H2O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO2 and H2O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams:

mass of C = 69.00 mg CO2×g1000 mg×mol CO244.010 g CO2×mol Cmol CO2×12.011 gmol C = 1.883×102 g Cmass of H = 11.30 mg H2O×g1000 mg×mol H2O18.015 g H2O×mol Hmol H2O×1.0079 gmol H= 1.264×103 g H

B To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount:

moles C = 1.883×102 g C×1 mol C12.011 g C= 1.568×103 mol Cmoles H = 1.264×103 g H×1 mol H1.0079 g H= 1.254×103 mol H

Dividing each number by the number of moles of the element present in the smaller amount gives

H: 1.254×1031.254×103= 1.000C: 1.568×1031.254×103= 1.250

Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C1.25H1.0. Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C5H4 as the empirical formula of naphthalene. In fact, the molecular formula of naphthalene is C10H8, which is consistent with our results.

Exercise

  1. Xylene, an organic compound that is a major component of many gasoline blends, contains carbon and hydrogen only. Complete combustion of a 17.12 mg sample of xylene in oxygen yielded 56.77 mg of CO2 and 14.53 mg of H2O. Determine the empirical formula of xylene.
  2. The empirical formula of benzene is CH (its molecular formula is C6H6). If 10.00 mg of benzene is subjected to combustion analysis, what mass of CO2 and H2O will be produced?

Answer:

  1. The empirical formula is C4H5. (The molecular formula of xylene is actually C8H10.)
  2. 33.81 mg of CO2; 6.92 mg of H2O

From Empirical Formula to Molecular Formula

The empirical formula gives only the relative numbers of atoms in a substance in the smallest possible ratio. For a covalent substance, we are usually more interested in the molecular formula, which gives the actual number of atoms of each kind present per molecule. Without additional information, however, it is impossible to know whether the formula of penicillin G, for example, is C16H17N2NaO4S or an integral multiple, such as C32H34N4Na2O8S2, C48H51N6Na3O12S3, or (C16H17N2NaO4S)n, where n is an integer. (The actual structure of penicillin G is shown in .)

Consider glucose, the sugar that circulates in our blood to provide fuel for our bodies and especially for our brains. Results from combustion analysis of glucose report that glucose contains 39.68% carbon and 6.58% hydrogen. Because combustion occurs in the presence of oxygen, it is impossible to directly determine the percentage of oxygen in a compound by using combustion analysis; other more complex methods are necessary. If we assume that the remaining percentage is due to oxygen, then glucose would contain 53.79% oxygen. A 100.0 g sample of glucose would therefore contain 39.68 g of carbon, 6.58 g of hydrogen, and 53.79 g of oxygen. To calculate the number of moles of each element in the 100.0 g sample, we divide the mass of each element by its molar mass:

Equation 3.6

moles C = 39.68 g C×1 mol C12.011 g C= 3.304 mol Cmoles H = 6.58 g H×1 mol H1.0079 g H= 6.53 mol Hmoles O = 53.79 g O×1 mol O15.9994 g O= 3.362 mol O

Once again, we find the subscripts of the elements in the empirical formula by dividing the number of moles of each element by the number of moles of the element present in the smallest amount:

C: 3.3043.304= 1.000H: 6.533.304= 1.98O: 3.3623.304= 1.018

The oxygen:carbon ratio is 1.018, or approximately 1, and the hydrogen:carbon ratio is approximately 2. The empirical formula of glucose is therefore CH2O, but what is its molecular formula?

Many known compounds have the empirical formula CH2O, including formaldehyde, which is used to preserve biological specimens and has properties that are very different from the sugar circulating in our blood. At this point, we cannot know whether glucose is CH2O, C2H4O2, or any other (CH2O)n. We can, however, use the experimentally determined molar mass of glucose (180 g/mol) to resolve this dilemma.

First, we calculate the formula mass, the molar mass of the formula unit, which is the sum of the atomic masses of the elements in the empirical formula multiplied by their respective subscripts. For glucose,

Equation 3.7

formula mass of CH2=[mol C(12.011 gmol C)]+[mol H(1.0079 gmol H)]+[mol O(15.9994 gmol O)]= 30.026 g

This is much smaller than the observed molar mass of 180 g/mol.

Second, we determine the number of formula units per mole. For glucose, we can calculate the number of (CH2O) units—that is, the n in (CH2O)n—by dividing the molar mass of glucose by the formula mass of CH2O:

Equation 3.8

n=180 g30.026 g/CH2O=5.996 CH2O formula units

Each glucose contains six CH2O formula units, which gives a molecular formula for glucose of (CH2O)6, which is more commonly written as C6H12O6. The molecular structures of formaldehyde and glucose, both of which have the empirical formula CH2O, are shown in .

Figure 3.6 Structural Formulas and Ball-and-Stick Models of (a) Formaldehyde and (b) Glucose

Example 8

Calculate the molecular formula of caffeine, a compound found in coffee, tea, and cola drinks that has a marked stimulatory effect on mammals. The chemical analysis of caffeine shows that it contains 49.18% carbon, 5.39% hydrogen, 28.65% nitrogen, and 16.68% oxygen by mass, and its experimentally determined molar mass is 196 g/mol.

Given: percent composition and molar mass

Asked for: molecular formula

Strategy:

A Assume 100 g of caffeine. From the percentages given, use the procedure given in Example 6 to calculate the empirical formula of caffeine.

B Calculate the formula mass and then divide the experimentally determined molar mass by the formula mass. This gives the number of formula units present.

C Multiply each subscript in the empirical formula by the number of formula units to give the molecular formula.

Solution:

A We begin by dividing the mass of each element in 100.0 g of caffeine (49.18 g of carbon, 5.39 g of hydrogen, 28.65 g of nitrogen, 16.68 g of oxygen) by its molar mass. This gives the number of moles of each element in 100 g of caffeine.

moles C = 49.18 g C×1 mol C12.011 g C= 4.095 mol Cmoles H = 5.39 g H×1 mol H1.0079 g H= 5.35 mol Hmoles N = 28.65 g N×1 mol N14.0067 g N= 2.045 mol Nmoles O = 16.68 g O×1 mol O15.9994 g O= 1.043 mol O

To obtain the relative numbers of atoms of each element present, divide the number of moles of each element by the number of moles of the element present in the least amount:

O: 1.0431.043= 1.000C: 4.0951.043= 3.926H: 5.351.043= 5.13N: 2.0451.043= 1.960

These results are fairly typical of actual experimental data. None of the atomic ratios is exactly integral but all are within 5% of integral values. Just as in Example 6, it is reasonable to assume that such small deviations from integral values are due to minor experimental errors, so round to the nearest integer. The empirical formula of caffeine is thus C4H5N2O.

B The molecular formula of caffeine could be C4H5N2O, but it could also be any integral multiple of this. To determine the actual molecular formula, we must divide the experimentally determined molar mass by the formula mass. The formula mass is calculated as follows:

4C(4 atoms C)(12.011 g/atom C)=48.044 g5H(5 atoms H)(1.0079 g/atom H)=5.0395 g2N(2 atoms N)(14.0067 g/atom N)=28.0134 g+1O(1 atom O)(15.9994 g/atom O)=15.9994 gC4H5N2Oformula mass of caffeine=97.096 g

Dividing the measured molar mass of caffeine (196 g/mol) by the calculated formula mass gives

196 g/mol97.096 g/C4H5N2O=2.022 C4H5N2O empirical formula units

C There are two C4H5N2O formula units in caffeine, so the molecular formula must be (C4H5N2O)2 = C8H10N4O2. The structure of caffeine is as follows:

Exercise

Calculate the molecular formula of Freon-114, which has 13.85% carbon, 41.89% chlorine, and 44.06% fluorine. The experimentally measured molar mass of this compound is 171 g/mol. Like Freon-11, Freon-114 is a commonly used refrigerant that has been implicated in the destruction of the ozone layer.

Answer: C2Cl2F4

Summary

The empirical formula of a substance can be calculated from the experimentally determined percent composition, the percentage of each element present in a pure substance by mass. In many cases, these percentages can be determined by combustion analysis. If the molar mass of the compound is known, the molecular formula can be determined from the empirical formula.

Key Takeaway

  • The empirical formula of a substance can be calculated from its percent composition, and the molecular formula can be determined from the empirical formula and the compound’s molar mass.

Conceptual Problems

  1. What is the relationship between an empirical formula and a molecular formula?

  2. Construct a flowchart showing how you would determine the empirical formula of a compound from its percent composition.

Numerical Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 2 () before proceeding to the Numerical Problems.

  1. What is the mass percentage of water in each hydrate?

    1. H3AsO4·0·5H2O
    2. NH4NiCl3·6H2O
    3. Al(NO3)3·9H2O
  2. What is the mass percentage of water in each hydrate?

    1. CaSO4·2H2O
    2. Fe(NO3)3·9H2O
    3. (NH4)3ZrOH(CO3)3·2H2O
  3. Which of the following has the greatest mass percentage of oxygen—KMnO4, K2Cr2O7, or Fe2O3?

  4. Which of the following has the greatest mass percentage of oxygen—ThOCl2, MgCO3, or NO2Cl?

  5. Calculate the percent composition of the element shown in bold in each compound.

    1. SbBr3
    2. As2I4
    3. AlPO4
    4. C6H10O
  6. Calculate the percent composition of the element shown in bold in each compound.

    1. HBrO3
    2. CsReO4
    3. C3H8O
    4. FeSO4
  7. A sample of a chromium compound has a molar mass of 151.99 g/mol. Elemental analysis of the compound shows that it contains 68.43% chromium and 31.57% oxygen. What is the identity of the compound?

  8. The percentages of iron and oxygen in the three most common binary compounds of iron and oxygen are given in the following table. Write the empirical formulas of these three compounds.

    Compound % Iron % Oxygen Empirical Formula
    1 69.9 30.1
    2 77.7 22.3
    3 72.4 27.6
  9. What is the mass percentage of water in each hydrate?

    1. LiCl·H2O
    2. MgSO4·7H2O
    3. Sr(NO3)2·4H2O
  10. What is the mass percentage of water in each hydrate?

    1. CaHPO4·2H2O
    2. FeCl2·4H2O
    3. Mg(NO3)2·4H2O
  11. Two hydrates were weighed, heated to drive off the waters of hydration, and then cooled. The residues were then reweighed. Based on the following results, what are the formulas of the hydrates?

    Compound Initial Mass (g) Mass after Cooling (g)
    NiSO4·xH2O 2.08 1.22
    CoCl2·xH2O 1.62 0.88
  12. Which contains the greatest mass percentage of sulfur—FeS2, Na2S2O4, or Na2S?

  13. Given equal masses of each, which contains the greatest mass percentage of sulfur—NaHSO4 or K2SO4?

  14. Calculate the mass percentage of oxygen in each polyatomic ion.

    1. bicarbonate
    2. chromate
    3. acetate
    4. sulfite
  15. Calculate the mass percentage of oxygen in each polyatomic ion.

    1. oxalate
    2. nitrite
    3. dihydrogen phosphate
    4. thiocyanate
  16. The empirical formula of garnet, a gemstone, is Fe3Al2Si3O12. An analysis of a sample of garnet gave a value of 13.8% for the mass percentage of silicon. Is this consistent with the empirical formula?

  17. A compound has the empirical formula C2H4O, and its formula mass is 88 g. What is its molecular formula?

  18. Mirex is an insecticide that contains 22.01% carbon and 77.99% chlorine. It has a molecular mass of 545.59 g. What is its empirical formula? What is its molecular formula?

  19. How many moles of CO2 and H2O will be produced by combustion analysis of 0.010 mol of styrene?

  20. How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0080 mol of aniline?

  21. How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0074 mol of aspartame?

  22. How many moles of CO2, H2O, N2, and SO2 will be produced by combustion analysis of 0.0060 mol of penicillin G?

  23. Combustion of a 34.8 mg sample of benzaldehyde, which contains only carbon, hydrogen, and oxygen, produced 101 mg of CO2 and 17.7 mg of H2O.

    1. What was the mass of carbon and hydrogen in the sample?
    2. Assuming that the original sample contained only carbon, hydrogen, and oxygen, what was the mass of oxygen in the sample?
    3. What was the mass percentage of oxygen in the sample?
    4. What is the empirical formula of benzaldehyde?
    5. The molar mass of benzaldehyde is 106.12 g/mol. What is its molecular formula?
  24. Salicylic acid is used to make aspirin. It contains only carbon, oxygen, and hydrogen. Combustion of a 43.5 mg sample of this compound produced 97.1 mg of CO2 and 17.0 mg of H2O.

    1. What is the mass of oxygen in the sample?
    2. What is the mass percentage of oxygen in the sample?
    3. What is the empirical formula of salicylic acid?
    4. The molar mass of salicylic acid is 138.12 g/mol. What is its molecular formula?
  25. Given equal masses of the following acids, which contains the greatest amount of hydrogen that can dissociate to form H+—nitric acid, hydroiodic acid, hydrocyanic acid, or chloric acid?

  26. Calculate the formula mass or the molecular mass of each compound.

    1. heptanoic acid (a seven-carbon carboxylic acid)
    2. 2-propanol (a three-carbon alcohol)
    3. KMnO4
    4. tetraethyllead
    5. sulfurous acid
    6. ethylbenzene (an eight-carbon aromatic hydrocarbon)
  27. Calculate the formula mass or the molecular mass of each compound.

    1. MoCl5
    2. B2O3
    3. bromobenzene
    4. cyclohexene
    5. phosphoric acid
    6. ethylamine
  28. Given equal masses of butane, cyclobutane, and propene, which contains the greatest mass of carbon?

  29. Given equal masses of urea [(NH2)2CO] and ammonium sulfate, which contains the most nitrogen for use as a fertilizer?

Answers

  1. To two decimal places, the percentages are:

    1. 5.97%
    2. 37.12%
    3. 43.22%
  2. % oxygen: KMnO4, 40.50%; K2Cr2O7, 38.07%; Fe2O3, 30.06%

  3. To two decimal places, the percentages are:

    1. 66.32% Br
    2. 22.79% As
    3. 25.40% P
    4. 73.43% C
  4. Cr2O3.

  5. To two decimal places, the percentages are:

    1. 29.82%
    2. 51.16%
    3. 25.40%
  6. NiSO4 · 6H2O and CoCl2 · 6H2O

  7. NaHSO4

    1. 72.71%
    2. 69.55%
    3. 65.99%
    4. 0%
  8. C4H8O2

    1. 27.6 mg C and 1.98 mg H
    2. 5.2 mg O
    3. 15%
    4. C7H6O
    5. C7H6O
  9. hydrocyanic acid, HCN

  10. To two decimal places, the values are:

    1. 273.23 amu
    2. 69.62 amu
    3. 157.01 amu
    4. 82.14 amu
    5. 98.00 amu
    6. 45.08 amu
  11. Urea

3.3 Chemical Equations

Learning Objectives

  1. To describe a chemical reaction.
  2. To calculate the quantities of compounds produced or consumed in a chemical reaction.

As shown in Figure 3.7 "An Ammonium Dichromate Volcano: Change during a Chemical Reaction", applying a small amount of heat to a pile of orange ammonium dichromate powder results in a vigorous reaction known as the ammonium dichromate volcano. Heat, light, and gas are produced as a large pile of fluffy green chromium(III) oxide forms. We can describe this reaction with a chemical equationAn expression that gives the identities and quantities of the substances in a chemical reaction. Chemical formulas are used to indicate the reactants on the left and the products on the right. An arrow points from reactants to products., an expression that gives the identities and quantities of the substances in a chemical reaction. Chemical formulas and other symbols are used to indicate the starting material(s), or reactant(s)The starting material(s) in a chemical reaction., which by convention are written on the left side of the equation, and the final compound(s), or product(s)The final compound(s) produced in a chemical reaction., which are written on the right. An arrow points from the reactant to the products:

Figure 3.7 An Ammonium Dichromate Volcano: Change during a Chemical Reaction

The starting material (left) is solid ammonium dichromate. A chemical reaction (right) transforms it to solid chromium(III) oxide, depicted showing a portion of its chained structure, nitrogen gas, and water vapor. (In addition, energy in the form of heat and light is released.) During the reaction, the distribution of atoms changes, but the number of atoms of each element does not change. Because the numbers of each type of atom are the same in the reactants and the products, the chemical equation is balanced.

Equation 3.9

(NH4)2Cr2O7reactantCr2O3+N2+4H2Oproducts

The arrow is read as “yields” or “reacts to form.” So Equation 3.9 tells us that ammonium dichromate (the reactant) yields chromium(III) oxide, nitrogen, and water (the products).

The equation for this reaction is even more informative when written as

Equation 3.10

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g)

Equation 3.10 is identical to Equation 3.9 except for the addition of abbreviations in parentheses to indicate the physical state of each species. The abbreviations are (s) for solid, (l) for liquid, (g) for gas, and (aq) for an aqueous solution, a solution of the substance in water.

Consistent with the law of conservation of mass, the numbers of each type of atom are the same on both sides of Equation 3.9 and Equation 3.10. (For more information on the law of conservation of mass, see Section 1.4 "A Brief History of Chemistry".) As illustrated in Figure 3.7 "An Ammonium Dichromate Volcano: Change during a Chemical Reaction", each side has two chromium atoms, seven oxygen atoms, two nitrogen atoms, and eight hydrogen atoms. In a balanced chemical equation, both the numbers of each type of atom and the total charge are the same on both sides. Equation 3.9 and Equation 3.10 are balanced chemical equations. What is different on each side of the equation is how the atoms are arranged to make molecules or ions. A chemical reaction represents a change in the distribution of atoms but not in the number of atoms. In this reaction, and in most chemical reactions, bonds are broken in the reactants (here, Cr–O and N–H bonds), and new bonds are formed to create the products (here, O–H and N≡N bonds). If the numbers of each type of atom are different on the two sides of a chemical equation, then the equation is unbalanced, and it cannot correctly describe what happens during the reaction. To proceed, the equation must first be balanced.

Note the Pattern

A chemical reaction changes only the distribution of atoms, not the number of atoms.

Interpreting Chemical Equations

In addition to providing qualitative information about the identities and physical states of the reactants and products, a balanced chemical equation provides quantitative information. Specifically, it tells the relative amounts of reactants and products consumed or produced in a reaction. The number of atoms, molecules, or formula units of a reactant or a product in a balanced chemical equation is the coefficientA number greater than 1 preceding a formula in a balanced chemical equation and indicating the number of atoms, molecules, or formula units of a reactant or a product. of that species (e.g., the 4 preceding H2O in Equation 3.9). When no coefficient is written in front of a species, the coefficient is assumed to be 1. As illustrated in Figure 3.8 "The Relationships among Moles, Masses, and Formula Units of Compounds in the Balanced Chemical Reaction for the Ammonium Dichromate Volcano", the coefficients allow us to interpret Equation 3.9 in any of the following ways:

  • Two NH4+ ions and one Cr2O72− ion yield 1 formula unit of Cr2O3, 1 N2 molecule, and 4 H2O molecules.
  • One mole of (NH4)2Cr2O7 yields 1 mol of Cr2O3, 1 mol of N2, and 4 mol of H2O.
  • A mass of 252 g of (NH4)2Cr2O7 yields 152 g of Cr2O3, 28 g of N2, and 72 g of H2O.
  • A total of 6.022 × 1023 formula units of (NH4)2Cr2O7 yields 6.022 × 1023 formula units of Cr2O3, 6.022 × 1023 molecules of N2, and 24.09 × 1023 molecules of H2O.

Figure 3.8 The Relationships among Moles, Masses, and Formula Units of Compounds in the Balanced Chemical Reaction for the Ammonium Dichromate Volcano

These are all chemically equivalent ways of stating the information given in the balanced chemical equation, using the concepts of the mole, molar or formula mass, and Avogadro’s number. The ratio of the number of moles of one substance to the number of moles of another is called the mole ratioThe ratio of the number of moles of one substance to the number of moles of another, as depicted by a balanced chemical equation.. For example, the mole ratio of H2O to N2 in Equation 3.9 is 4:1. The total mass of reactants equals the total mass of products, as predicted by Dalton’s law of conservation of mass: 252 g of (NH4)2Cr2O7 yields 152 + 28 + 72 = 252 g of products. The chemical equation does not, however, show the rate of the reaction (rapidly, slowly, or not at all) or whether energy in the form of heat or light is given off. We will consider these issues in more detail in later chapters.

An important chemical reaction was analyzed by Antoine Lavoisier, an 18th-century French chemist, who was interested in the chemistry of living organisms as well as simple chemical systems. In a classic series of experiments, he measured the carbon dioxide and heat produced by a guinea pig during respiration, in which organic compounds are used as fuel to produce energy, carbon dioxide, and water. Lavoisier found that the ratio of heat produced to carbon dioxide exhaled was similar to the ratio observed for the reaction of charcoal with oxygen in the air to produce carbon dioxide—a process chemists call combustion. Based on these experiments, he proposed that “Respiration is a combustion, slow it is true, but otherwise perfectly similar to that of charcoal.” Lavoisier was correct, although the organic compounds consumed in respiration are substantially different from those found in charcoal. One of the most important fuels in the human body is glucose (C6H12O6), which is virtually the only fuel used in the brain. Thus combustion and respiration are examples of chemical reactions.

Example 9

The balanced chemical equation for the combustion of glucose in the laboratory (or in the brain) is as follows:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

Construct a table showing how to interpret the information in this equation in terms of

  1. a single molecule of glucose.
  2. moles of reactants and products.
  3. grams of reactants and products represented by 1 mol of glucose.
  4. numbers of molecules of reactants and products represented by 1 mol of glucose.

Given: balanced chemical equation

Asked for: molecule, mole, and mass relationships

Strategy:

A Use the coefficients from the balanced chemical equation to determine both the molecular and mole ratios.

B Use the molar masses of the reactants and products to convert from moles to grams.

C Use Avogadro’s number to convert from moles to the number of molecules.

Solution:

This equation is balanced as written: each side has 6 carbon atoms, 18 oxygen atoms, and 12 hydrogen atoms. We can therefore use the coefficients directly to obtain the desired information.

  1. A One molecule of glucose reacts with 6 molecules of O2 to yield 6 molecules of CO2 and 6 molecules of H2O.
  2. One mole of glucose reacts with 6 mol of O2 to yield 6 mol of CO2 and 6 mol of H2O.
  3. B To interpret the equation in terms of masses of reactants and products, we need their molar masses and the mole ratios from part b. The molar masses in grams per mole are as follows: glucose, 180.16; O2, 31.9988; CO2, 44.010; and H2O, 18.015.

    mass of reactants = mass of productsg glucose + g O2=g CO2+g H2Omol glucose(180.16 gmol glucose)+mol O2(31.9988 gmol O2)=mol CO2(44.010 gmol CO2)+mol H2O(18.015 gmol H2O)372.15 g = 372.15 g
  4. C One mole of glucose contains Avogadro’s number (6.022 × 1023) of glucose molecules. Thus 6.022 × 1023 glucose molecules react with (6 × 6.022 × 1023) = 3.613 × 1024 oxygen molecules to yield (6 × 6.022 × 1023) = 3.613 × 1024 molecules each of CO2 and H2O.

    In tabular form:

    C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)
    a. 1 molecule   6 molecules   6 molecules   6 molecules
    b. 1 mol 6 mol 6 mol 6 mol
    c. 180.16 g 191.9928 g 264.06 g 108.09 g
    d. 6.022 × 1023 molecules 3.613 × 1024 molecules 3.613 × 1024 molecules 3.613 × 1024 molecules

Exercise

Ammonium nitrate is a common fertilizer, but under the wrong conditions it can be hazardous. In 1947, a ship loaded with ammonium nitrate caught fire during unloading and exploded, destroying the town of Texas City, Texas. The explosion resulted from the following reaction:

2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)

Construct a table showing how to interpret the information in the equation in terms of

  1. individual molecules and ions.
  2. moles of reactants and products.
  3. grams of reactants and products given 2 mol of ammonium nitrate.
  4. numbers of molecules or formula units of reactants and products given 2 mol of ammonium nitrate.

Answer:

2NH4NO3(s) 2N2(g) + 4H2O(g) + O2(g)
a. 2NH4+ ions and 2NO3 ions   2 molecules   4 molecules   1 molecule
b. 2 mol 2 mol 4 mol 1 mol
c. 160.0864 g 56.0268 g 72.0608 g 31.9988 g
d. 1.204 × 1024 formula units 1.204 × 1024 molecules 2.409 × 1024 molecules 6.022 × 1023 molecules

Ammonium nitrate can be hazardous. This aerial photograph of Texas City, Texas, shows the devastation caused by the explosion of a shipload of ammonium nitrate on April 16, 1947.

Balancing Simple Chemical Equations

When a chemist encounters a new reaction, it does not usually come with a label that shows the balanced chemical equation. Instead, the chemist must identify the reactants and products and then write them in the form of a chemical equation that may or may not be balanced as first written. Consider, for example, the combustion of n-heptane (C7H16), an important component of gasoline:

Equation 3.11

C7H16(l) + O2(g) → CO2(g) + H2O(g)

The complete combustion of any hydrocarbon with sufficient oxygen always yields carbon dioxide and water (Figure 3.9 "An Example of a Combustion Reaction").

Figure 3.9 An Example of a Combustion Reaction

The wax in a candle is a high-molecular-mass hydrocarbon, which produces gaseous carbon dioxide and water vapor in a combustion reaction. When the candle is allowed to burn inside a flask, drops of water, one of the products of combustion, condense at the top of the inner surface of the flask.

Equation 3.11 is not balanced: the numbers of each type of atom on the reactant side of the equation (7 carbon atoms, 16 hydrogen atoms, and 2 oxygen atoms) is not the same as the numbers of each type of atom on the product side (1 carbon atom, 2 hydrogen atoms, and 3 oxygen atoms). Consequently, we must adjust the coefficients of the reactants and products to give the same numbers of atoms of each type on both sides of the equation. Because the identities of the reactants and products are fixed, we cannot balance the equation by changing the subscripts of the reactants or the products. To do so would change the chemical identity of the species being described, as illustrated in Figure 3.10 "Balancing Equations".

Figure 3.10 Balancing Equations

You cannot change subscripts in a chemical formula to balance a chemical equation; you can change only the coefficients. Changing subscripts changes the ratios of atoms in the molecule and the resulting chemical properties. For example, water (H2O) and hydrogen peroxide (H2O2) are chemically distinct substances. H2O2 decomposes to H2O and O2 gas when it comes in contact with the metal platinum, whereas no such reaction occurs between water and platinum.

The simplest and most generally useful method for balancing chemical equations is “inspection,” better known as trial and error. We present an efficient approach to balancing a chemical equation using this method.

Steps in Balancing a Chemical Equation

  1. Identify the most complex substance.
  2. Beginning with that substance, choose an element that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element on both sides.
  3. Balance polyatomic ions (if present) as a unit.
  4. Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients.
  5. Count the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced.

To demonstrate this approach, let’s use the combustion of n-heptane (Equation 3.11) as an example.

  1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is C7H16. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.
  2. Adjust the coefficients. Try to adjust the coefficients of the molecules on the other side of the equation to obtain the same numbers of atoms on both sides. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side:

    Equation 3.12

    C7H16 + O2 → 7CO2 + H2O
  3. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction.
  4. Balance the remaining atoms. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side:

    Equation 3.13

    C7H16 + O2 → 7CO2 + 8H2O

    The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side:

    Equation 3.14

    C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(g)
  5. Check your work. The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a chemical equation is balanced.

The assumption that the final balanced chemical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start. Consider, for example, a similar reaction, the combustion of isooctane (C8H18). Because the combustion of any hydrocarbon with oxygen produces carbon dioxide and water, the unbalanced chemical equation is as follows:

Equation 3.15

C8H18(l) + O2(g) → CO2(g) + H2O(g)
  1. Identify the most complex substance. Begin the balancing process by assuming that the final balanced chemical equation contains a single molecule of isooctane.
  2. Adjust the coefficients. The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO2 molecules in the products:

    Equation 3.16

    C8H18 + O2 → 8CO2 + H2O
  3. Balance polyatomic ions as a unit. This step does not apply to this equation.
  4. Balance the remaining atoms. Eighteen hydrogen atoms in isooctane means that there must be 9 H2O molecules in the products:

    Equation 3.17

    C8H18 + O2 → 8CO2 + 9H2O

    The carbon and hydrogen atoms are now balanced, but we have 25 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the least complex substance, O2, but because there are 2 oxygen atoms per O2 molecule, we must use a fractional coefficient (25/2) to balance the oxygen atoms:

    Equation 3.18

    C8H18 + 25/2O2 → 8CO2 + 9H2O

    Equation 3.18 is now balanced, but we usually write equations with whole-number coefficients. We can eliminate the fractional coefficient by multiplying all coefficients on both sides of the chemical equation by 2:

    Equation 3.19

    2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g)
  5. Check your work. The balanced chemical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side.

Balancing equations requires some practice on your part as well as some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly.

Example 10

The reaction of the mineral hydroxyapatite [Ca5(PO4)3(OH)] with phosphoric acid and water gives Ca(H2PO4)2·H2O (calcium dihydrogen phosphate monohydrate). Write and balance the equation for this reaction.

Given: reactants and product

Asked for: balanced chemical equation

Strategy:

A Identify the product and the reactants and then write the unbalanced chemical equation.

B Follow the steps for balancing a chemical equation.

Solution:

A We must first identify the product and reactants and write an equation for the reaction. The formulas for hydroxyapatite and calcium dihydrogen phosphate monohydrate are given in the problem. Recall from Chapter 2 "Molecules, Ions, and Chemical Formulas" that phosphoric acid is H3PO4. The initial (unbalanced) equation is as follows:

Ca5(PO4)3(OH)(s) + H3PO4(aq) + H2O(l) → Ca(H2PO4)2·H2O(s)
  1. B Identify the most complex substance. We start by assuming that only one molecule or formula unit of the most complex substance, Ca5(PO4)3(OH), appears in the balanced chemical equation.
  2. Adjust the coefficients. Because calcium is present in only one reactant and one product, we begin with it. One formula unit of Ca5(PO4)3(OH) contains 5 calcium atoms, so we need 5 Ca(H2PO4)2·H2O on the right side:

    Ca5(PO4)3(OH) + H3PO4 + H2O → 5Ca(H2PO4)2·H2O
  3. Balance polyatomic ions as a unit. It is usually easier to balance an equation if we recognize that certain combinations of atoms occur on both sides. In this equation, the polyatomic phosphate ion (PO43−), shows up in three places.In H3PO4, the phosphate ion is combined with three H+ ions to make phosphoric acid (H3PO4), whereas in Ca(H2PO4)2·H2O it is combined with two H+ ions to give the dihydrogen phosphate ion. Thus it is easier to balance PO4 as a unit rather than counting individual phosphorus and oxygen atoms. There are 10 PO4 units on the right side but only 4 on the left. The simplest way to balance the PO4 units is to place a coefficient of 7 in front of H3PO4:

    Ca5(PO4)3(OH) + 7H3PO4 + H2O → 5Ca(H2PO4)2·H2O

    Although OH is also a polyatomic ion, it does not appear on both sides of the equation. So oxygen and hydrogen must be balanced separately.

  4. Balance the remaining atoms. We now have 30 hydrogen atoms on the right side but only 24 on the left. We can balance the hydrogen atoms using the least complex substance, H2O, by placing a coefficient of 4 in front of H2O on the left side, giving a total of 4 H2O molecules:

    Ca5(PO4)3(OH)(s) + 7H3PO4(aq) + 4H2O(l) → 5Ca(H2PO4)2·H2O(s)

    The equation is now balanced. Even though we have not explicitly balanced the oxygen atoms, there are 45 oxygen atoms on each side.

  5. Check your work. Both sides of the equation contain 5 calcium atoms, 7 phosphorus atoms, 30 hydrogen atoms, and 45 oxygen atoms.

Exercise

Fermentation is a biochemical process that enables yeast cells to live in the absence of oxygen. Humans have exploited it for centuries to produce wine and beer and make bread rise. In fermentation, sugars such as glucose are converted to ethanol and carbon dioxide. Write a balanced chemical reaction for the fermentation of glucose.

Commercial use of fermentation. (a) Microbrewery vats are used to prepare beer. (b) The fermentation of glucose by yeast cells is the reaction that makes beer production possible.

Answer: C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)

Summary

In a chemical reaction, one or more substances are transformed to new substances. A chemical reaction is described by a chemical equation, an expression that gives the identities and quantities of the substances involved in a reaction. A chemical equation shows the starting compound(s)—the reactants—on the left and the final compound(s)—the products—on the right, separated by an arrow. In a balanced chemical equation, the numbers of atoms of each element and the total charge are the same on both sides of the equation. The number of atoms, molecules, or formula units of a reactant or product in a balanced chemical equation is the coefficient of that species. The mole ratio of two substances in a chemical reaction is the ratio of their coefficients in the balanced chemical equation.

Key Takeaway

  • A chemical reaction is described by a chemical equation that gives the identities and quantities of the reactants and the products.

Conceptual Problems

  1. How does a balanced chemical equation agree with the law of definite proportions?

  2. What is the difference between S8 and 8S? Use this example to explain why subscripts in a formula must not be changed.

  3. What factors determine whether a chemical equation is balanced?

  4. What information can be obtained from a balanced chemical equation? Does a balanced chemical equation give information about the rate of a reaction?

Numerical Problems

  1. Balance each chemical equation.

    1. KI(aq) + Br2(l) → KBr(aq) + I2(s)
    2. MnO2(s) + HCl(aq) → MnCl2(aq) + Cl2(g) + H2O(l)
    3. Na2O(s) + H2O(l) → NaOH(aq)
    4. Cu(s) + AgNO3(aq) → Cu(NO3)2(aq) + Ag(s)
    5. SO2(g) + H2O(l) → H2SO3(aq)
    6. S2Cl2(l) + NH3(l) → S4N4(s) + S8(s) + NH4Cl(s)
  2. Balance each chemical equation.

    1. Be(s) + O2(g) → BeO(s)
    2. N2O3(g) + H2O(l) → HNO2(aq)
    3. Na(s) + H2O(l) → NaOH(aq) + H2(g)
    4. CaO(s) + HCl(aq) → CaCl2(aq) + H2O(l)
    5. CH3NH2(g) + O2(g) → H2O(g) + CO2(g) + N2(g)
    6. Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
  3. Balance each chemical equation.

    1. N2O5(g) → NO2(g) + O2(g)
    2. NaNO3(s) → NaNO2(s) + O2(g)
    3. Al(s) + NH4NO3(s) → N2(g) + H2O(l) + Al2O3(s)
    4. C3H5N3O9(l) → CO2(g) + N2(g) + H2O(g) + O2(g)
    5. reaction of butane with excess oxygen
    6. IO2F(s) + BrF3(l) → IF5(l) + Br2(l) + O2(g)
  4. Balance each chemical equation.

    1. H2S(g) + O2(g) → H2O(l) + S8(s)
    2. KCl(aq) + HNO3(aq) + O2(g) → KNO3(aq) + Cl2(g) + H2O(l)
    3. NH3(g) + O2(g) → NO(g) + H2O(g)
    4. CH4(g) + O2(g) → CO(g) + H2(g)
    5. NaF(aq) + Th(NO3)4(aq) → NaNO3(aq) + ThF4(s)
    6. Ca5(PO4)3F(s) + H2SO4(aq) + H2O(l) → H3PO4(aq) + CaSO4·2H2O(s) + HF(aq)
  5. Balance each chemical equation.

    1. NaCl(aq) + H2SO4(aq) → Na2SO4(aq) + HCl(g)
    2. K(s) + H2O(l) → KOH(aq) + H2(g)
    3. reaction of octane with excess oxygen
    4. S8(s) + Cl2(g) → S2Cl2(l)
    5. CH3OH(l) + I2(s) + P4(s) → CH3I(l) + H3PO4(aq) + H2O(l)
    6. (CH3)3Al(s) + H2O(l) → CH4(g) + Al(OH)3(s)
  6. Write a balanced chemical equation for each reaction.

    1. Aluminum reacts with bromine.
    2. Sodium reacts with chlorine.
    3. Aluminum hydroxide and acetic acid react to produce aluminum acetate and water.
    4. Ammonia and oxygen react to produce nitrogen monoxide and water.
    5. Nitrogen and hydrogen react at elevated temperature and pressure to produce ammonia.
    6. An aqueous solution of barium chloride reacts with a solution of sodium sulfate.
  7. Write a balanced chemical equation for each reaction.

    1. Magnesium burns in oxygen.
    2. Carbon dioxide and sodium oxide react to produce sodium carbonate.
    3. Aluminum reacts with hydrochloric acid.
    4. An aqueous solution of silver nitrate reacts with a solution of potassium chloride.
    5. Methane burns in oxygen.
    6. Sodium nitrate and sulfuric acid react to produce sodium sulfate and nitric acid.

3.4 Mass Relationships in Chemical Equations

Learning Objective

  1. To calculate the quantities of compounds produced or consumed in a chemical reaction.

A balanced chemical equation gives the identity of the reactants and the products as well as the accurate number of molecules or moles of each that are consumed or produced. StoichiometryA collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation. is a collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation. A stoichiometric quantityThe amount of product or reactant specified by the coefficients in a balanced chemical equation. is the amount of product or reactant specified by the coefficients in a balanced chemical equation. In , for example, you learned how to express the stoichiometry of the reaction for the ammonium dichromate volcano in terms of the atoms, ions, or molecules involved and the numbers of moles, grams, and formula units of each (recognizing, for instance, that 1 mol of ammonium dichromate produces 4 mol of water). This section describes how to use the stoichiometry of a reaction to answer questions like the following: How much oxygen is needed to ensure complete combustion of a given amount of isooctane? (This information is crucial to the design of nonpolluting and efficient automobile engines.) How many grams of pure gold can be obtained from a ton of low-grade gold ore? (The answer determines whether the ore deposit is worth mining.) If an industrial plant must produce a certain number of tons of sulfuric acid per week, how much elemental sulfur must arrive by rail each week?

All these questions can be answered using the concepts of the mole and molar and formula masses, along with the coefficients in the appropriate balanced chemical equation.

Stoichiometry Problems

When we carry out a reaction in either an industrial setting or a laboratory, it is easier to work with masses of substances than with the numbers of molecules or moles. The general method for converting from the mass of any reactant or product to the mass of any other reactant or product using a balanced chemical equation is outlined in and described in the following text.

Steps in Converting between Masses of Reactant and Product

  1. Convert the mass of one substance (substance A) to the corresponding number of moles using its molar mass.
  2. From the balanced chemical equation, obtain the number of moles of another substance (B) from the number of moles of substance A using the appropriate mole ratio (the ratio of their coefficients).
  3. Convert the number of moles of substance B to mass using its molar mass. It is important to remember that some species are in excess by virtue of the reaction conditions. For example, if a substance reacts with the oxygen in air, then oxygen is in obvious (but unstated) excess.

Converting amounts of substances to moles—and vice versa—is the key to all stoichiometry problems, whether the amounts are given in units of mass (grams or kilograms), weight (pounds or tons), or volume (liters or gallons).

Figure 3.11 A Flowchart for Stoichiometric Calculations Involving Pure Substances

The molar masses of the reactants and the products are used as conversion factors so that you can calculate the mass of product from the mass of reactant and vice versa.

To illustrate this procedure, let’s return to the combustion of glucose. We saw earlier that glucose reacts with oxygen to produce carbon dioxide and water:

Equation 3.20

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

Just before a chemistry exam, suppose a friend reminds you that glucose is the major fuel used by the human brain. You therefore decide to eat a candy bar to make sure that your brain doesn’t run out of energy during the exam (even though there is no direct evidence that consumption of candy bars improves performance on chemistry exams). If a typical 2 oz candy bar contains the equivalent of 45.3 g of glucose and the glucose is completely converted to carbon dioxide during the exam, how many grams of carbon dioxide will you produce and exhale into the exam room?

The initial step in solving a problem of this type must be to write the balanced chemical equation for the reaction. Inspection of shows that it is balanced as written, so we can proceed to the strategy outlined in , adapting it as follows:

  1. Use the molar mass of glucose (to one decimal place, 180.2 g/mol) to determine the number of moles of glucose in the candy bar:

    moles glucose = 45.3 g glucose×1 mol glucose180.2 g glucose= 0.251 mol glucose
  2. According to the balanced chemical equation, 6 mol of CO2 is produced per mole of glucose; the mole ratio of CO2 to glucose is therefore 6:1. The number of moles of CO2 produced is thus

    moles CO2= mol glucose×6 mol CO21 mol glucose= 0.251 mol glucose×6 mol CO2mol glucose= 1.51 mol CO2
  3. Use the molar mass of CO2 (44.010 g/mol) to calculate the mass of CO2 corresponding to 1.51 mol of CO2:

    mass of CO2= 1.51 mol CO2×44.010 g CO2mol CO2= 66.5 g CO2

We can summarize these operations as follows:

45.3 g glucose×1 mol glucose180.2 g glucosestep 1×6 mol CO21 mol glucosestep 2×44.010 g CO21 mol CO2step 3= 66.4 g CO2

Discrepancies between the two values are attributed to rounding errors resulting from using stepwise calculations in steps 1–3. (For more information about rounding and significant digits, see Essential Skills 1 in , .) In , you will discover that this amount of gaseous carbon dioxide occupies an enormous volume—more than 33 L. We could use similar methods to calculate the amount of oxygen consumed or the amount of water produced.

We just used the balanced chemical equation to calculate the mass of product that is formed from a certain amount of reactant. We can also use the balanced chemical equation to determine the masses of reactants that are necessary to form a certain amount of product or, as shown in Example 11, the mass of one reactant that is required to consume a given mass of another reactant.

Example 11

The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb).

The US space shuttle Discovery during liftoff. The large cylinder in the middle contains the oxygen and hydrogen that fueled the shuttle’s main engine.

Given: reactants, products, and mass of one reactant

Asked for: mass of other reactant

Strategy:

A Write the balanced chemical equation for the reaction.

B Convert mass of oxygen to moles. From the mole ratio in the balanced chemical equation, determine the number of moles of hydrogen required. Then convert the moles of hydrogen to the equivalent mass in tons.

Solution:

We use the same general strategy for solving stoichiometric calculations as in the preceding example. Because the amount of oxygen is given in tons rather than grams, however, we also need to convert tons to units of mass in grams. Another conversion is needed at the end to report the final answer in tons.

A We first use the information given to write a balanced chemical equation. Because we know the identity of both the reactants and the product, we can write the reaction as follows:

H2(g) + O2(g) → H2O(g)

This equation is not balanced because there are two oxygen atoms on the left side and only one on the right. Assigning a coefficient of 2 to both H2O and H2 gives the balanced chemical equation:

2H2(g) + O2(g) → 2H2O(g)

Thus 2 mol of H2 react with 1 mol of O2 to produce 2 mol of H2O.

  1. B To convert tons of oxygen to units of mass in grams, we multiply by the appropriate conversion factors:

    mass of O2= 1.00 tn×2000 lbtn×453.6 glb=9.07×105 g O2

    Using the molar mass of O2 (32.00 g/mol, to four significant figures), we can calculate the number of moles of O2 contained in this mass of O2:

    mol O2=9.07×105 g O2×1 mol O232.00 g O2= 2.83×104 mol O2
  2. Now use the coefficients in the balanced chemical equation to obtain the number of moles of H2 needed to react with this number of moles of O2:

    mol H2=mol O2×2 mol H21 mol O2           = 2.83×104 mol O2×2 mol H2mol O2=5.66×104 mol H2
  3. The molar mass of H2 (2.016 g/mol) allows us to calculate the corresponding mass of H2:

    mass of H2= 5.66×104 mol H2×2.016 g H2mol H2=1.14×105 g H2

    Finally, convert the mass of H2 to the desired units (tons) by using the appropriate conversion factors:

    tons H2= 1.14×105 g H2×1lb453.6 g×1 tn2000 lb=0.126 tn H2

    The space shuttle had to be designed to carry 0.126 tn of H2 for each 1.00 tn of O2. Even though 2 mol of H2 are needed to react with each mole of O2, the molar mass of H2 is so much smaller than that of O2 that only a relatively small mass of H2 is needed compared to the mass of O2.

Exercise

Alchemists produced elemental mercury by roasting the mercury-containing ore cinnabar (HgS) in air:

HgS(s) + O2(g) → Hg(l) + SO2(g)

The volatility and toxicity of mercury make this a hazardous procedure, which likely shortened the life span of many alchemists. Given 100 g of cinnabar, how much elemental mercury can be produced from this reaction?

Answer: 86.2 g

Limiting Reactants

In all the examples discussed thus far, the reactants were assumed to be present in stoichiometric quantities. Consequently, none of the reactants was left over at the end of the reaction. This is often desirable, as in the case of a space shuttle, where excess oxygen or hydrogen was not only extra freight to be hauled into orbit but also an explosion hazard. More often, however, reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. As a result, one or more of them will not be used up completely but will be left over when the reaction is completed. In this situation, the amount of product that can be obtained is limited by the amount of only one of the reactants. The reactant that restricts the amount of product obtained is called the limiting reactantThe reactant that restricts the amount of product obtained in a chemical reaction.. The reactant that remains after a reaction has gone to completion is in excess.

To be certain you understand these concepts, let’s first consider a nonchemical example. Assume you have invited some friends for dinner and want to bake brownies for dessert. You find two boxes of brownie mix in your pantry and see that each package requires two eggs. The balanced equation for brownie preparation is thus

Equation 3.21

1 box mix + 2 eggs → 1 batch brownies

If you have a dozen eggs, which ingredient will determine the number of batches of brownies that you can prepare? Because each box of brownie mix requires two eggs and you have two boxes, you need four eggs. Twelve eggs is eight more eggs than you need. Although the ratio of eggs to boxes in is 2:1, the ratio in your possession is 6:1. Hence the eggs are the ingredient (reactant) present in excess, and the brownie mix is the limiting reactant (). Even if you had a refrigerator full of eggs, you could make only two batches of brownies.

Figure 3.12 The Concept of a Limiting Reactant in the Preparation of Brownies

Let’s now turn to a chemical example of a limiting reactant: the production of pure titanium. This metal is fairly light (45% lighter than steel and only 60% heavier than aluminum) and has great mechanical strength (as strong as steel and twice as strong as aluminum). Because it is also highly resistant to corrosion and can withstand extreme temperatures, titanium has many applications in the aerospace industry. Titanium is also used in medical implants and portable computer housings because it is light and resistant to corrosion. Although titanium is the ninth most common element in Earth’s crust, it is relatively difficult to extract from its ores. In the first step of the extraction process, titanium-containing oxide minerals react with solid carbon and chlorine gas to form titanium tetrachloride (TiCl4) and carbon dioxide. Titanium tetrachloride is then converted to metallic titanium by reaction with magnesium metal at high temperature:

Equation 3.22

TiCl4(g) + 2Mg(l) → Ti(s) + 2MgCl2(l)

Because titanium ores, carbon, and chlorine are all rather inexpensive, the high price of titanium (about $100 per kilogram) is largely due to the high cost of magnesium metal. Under these circumstances, magnesium metal is the limiting reactant in the production of metallic titanium.

Medical use of titanium. Here is an example of its successful use in joint replacement implants.

Suppose you have 1.00 kg of titanium tetrachloride and 200 g of magnesium metal. How much titanium metal can you produce according to ? Solving this type of problem requires that you carry out the following steps:

  1. Determine the number of moles of each reactant.
  2. Compare the mole ratio of the reactants with the ratio in the balanced chemical equation to determine which reactant is limiting.
  3. Calculate the number of moles of product that can be obtained from the limiting reactant.
  4. Convert the number of moles of product to mass of product.

 

  1. To determine the number of moles of reactants present, you must calculate or look up their molar masses: 189.679 g/mol for titanium tetrachloride and 24.305 g/mol for magnesium. The number of moles of each is calculated as follows:

    moles TiCl4=mass TiCl4molar mass TiCl4= 1000 g TiCl4×1 mol TiCl4189.679 g TiCl4= 5.272 mol TiCl4moles Mg =mass Mgmolar mass Mg= 200 g Mg×1 mol Mg24.305 g Mg= 8.23 mol Mg
  2. You have more moles of magnesium than of titanium tetrachloride, but the ratio is only

    mol Mgmol TiCl4=8.23 mol5.272 mol= 1.56

    Because the ratio of the coefficients in the balanced chemical equation is

    2 mol Mg1 mol TiCl4= 2

    you do not have enough magnesium to react with all the titanium tetrachloride. If this point is not clear from the mole ratio, you should calculate the number of moles of one reactant that is required for complete reaction of the other reactant. For example, you have 8.23 mol of Mg, so you need (8.23 ÷ 2) = 4.12 mol of TiCl4 for complete reaction. Because you have 5.272 mol of TiCl4, titanium tetrachloride is present in excess. Conversely, 5.272 mol of TiCl4 requires 2 × 5.272 = 10.54 mol of Mg, but you have only 8.23 mol. So magnesium is the limiting reactant.

  3. Because magnesium is the limiting reactant, the number of moles of magnesium determines the number of moles of titanium that can be formed:

    moles Ti = 8.23 mol Mg=1 mol Timol Mg= 4.12 mol Ti

    Thus only 4.12 mol of Ti can be formed.

  4. To calculate the mass of titanium metal that you can obtain, multiply the number of moles of titanium by the molar mass of titanium (47.867 g/mol):

    moles Ti = mass Ti × molar mass Ti = 4.12 mol Ti×47.867 g Timol Ti= 197 g Ti

Here is a simple and reliable way to identify the limiting reactant in any problem of this sort:

  1. Calculate the number of moles of each reactant present: 5.272 mol of TiCl4 and 8.23 mol of Mg.
  2. Divide the actual number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation:

    TiCl45.272 mol (actual)1 mol (stoich)= 5.272Mg: 8.23 mol (actual)2 mol (stoich)= 4.12
  3. The reactant with the smallest mole ratio is limiting. Magnesium, with a calculated stoichiometric mole ratio of 4.12, is the limiting reactant.

As you learned in , density is the mass per unit volume of a substance. If we are given the density of a substance, we can use it in stoichiometric calculations involving liquid reactants and/or products, as Example 12 demonstrates.

Example 12

Ethyl acetate (CH3CO2C2H5) is the solvent in many fingernail polish removers and is used to decaffeinate coffee beans and tea leaves. It is prepared by reacting ethanol (C2H5OH) with acetic acid (CH3CO2H); the other product is water. A small amount of sulfuric acid is used to accelerate the reaction, but the sulfuric acid is not consumed and does not appear in the balanced chemical equation. Given 10.0 mL each of acetic acid and ethanol, how many grams of ethyl acetate can be prepared from this reaction? The densities of acetic acid and ethanol are 1.0492 g/mL and 0.7893 g/mL, respectively.

Given: reactants, products, and volumes and densities of reactants

Asked for: mass of product

Strategy:

A Balance the chemical equation for the reaction.

B Use the given densities to convert from volume to mass. Then use each molar mass to convert from mass to moles.

C Using mole ratios, determine which substance is the limiting reactant. After identifying the limiting reactant, use mole ratios based on the number of moles of limiting reactant to determine the number of moles of product.

D Convert from moles of product to mass of product.

Solution:

A We always begin by writing the balanced chemical equation for the reaction:

C2H5OH(l) + CH3CO2H(aq) → CH3CO2C2H5(aq) + H2O(l)

B We need to calculate the number of moles of ethanol and acetic acid that are present in 10.0 mL of each. Recall from that the density of a substance is the mass divided by the volume:

density =massvolume

Rearranging this expression gives mass = (density)(volume). We can replace mass by the product of the density and the volume to calculate the number of moles of each substance in 10.0 mL (remember, 1 mL = 1 cm3):

moles C2H5OH =mass C2H5OHmolar mass C2H5OH=volume C2H5OH×density C2H5OHmolar mass C2H5OH= 10.0 mL C2H5OH×0.7893 g C2H5OHmL C2H5OH×1 mol C2H5OH46.07 g C2H5OH= 0.171 mol C2H5OHmoles CH3CO2=mass CH3CO2Hmolar mass CH3CO2H=volume CH3CO2H×density CH3CO2Hmolar mass CH3CO2H= 10.0 mL CH3CO2H×1.0492 g CH3CO2HmL CH3CO2H×1 mol CH3CO2H60.05 g CH3CO2H= 0.175 mol CH3CO2H

C The number of moles of acetic acid exceeds the number of moles of ethanol. Because the reactants both have coefficients of 1 in the balanced chemical equation, the mole ratio is 1:1. We have 0.171 mol of ethanol and 0.175 mol of acetic acid, so ethanol is the limiting reactant and acetic acid is in excess. The coefficient in the balanced chemical equation for the product (ethyl acetate) is also 1, so the mole ratio of ethanol and ethyl acetate is also 1:1. This means that given 0.171 mol of ethanol, the amount of ethyl acetate produced must also be 0.171 mol:

moles ethyl acetate = mol ethanol×1 mol ethyl acetate1 mol ethanol= 0.171 mol C2H5OH×1 mol CH3CO2C2H5mol C2H5OH= 0.171 mol CH3CO2C2H5 

D The final step is to determine the mass of ethyl acetate that can be formed, which we do by multiplying the number of moles by the molar mass:

mass of ethyl acetate = mol ethyl acetate × molar mass ethyl acetate= 0.171 mol CH3CO2C2H5×88.11 g CH3CO2C2H5mol CH3CO2C2H5= 15.1 g CH3CO2C2H5 

Thus 15.1 g of ethyl acetate can be prepared in this reaction. If necessary, you could use the density of ethyl acetate (0.9003 g/cm3) to determine the volume of ethyl acetate that could be produced:

volume of ethyl acetate = 15.1 g CH3CO2C2H5×1 mL CH3CO2C2H50.9003 g CH3CO2C2H5= 16.8 mL CH3CO2C2H5 

Exercise

Under appropriate conditions, the reaction of elemental phosphorus and elemental sulfur produces the compound P4S10. How much P4S10 can be prepared starting with 10.0 g of P4 and 30.0 g of S8?

Answer: 35.9 g

Percent Yield

You have learned that when reactants are not present in stoichiometric quantities, the limiting reactant determines the maximum amount of product that can be formed from the reactants. The amount of product calculated in this way is the theoretical yieldThe maximum amount of product that can be formed from the reactants in a chemical reaction, which theoretically is the amount of product that would be obtained if the reaction occurred perfectly and the method of purifying the product were 100% efficient., the amount you would obtain if the reaction occurred perfectly and your method of purifying the product were 100% efficient.

In reality, you almost always obtain less product than is theoretically possible because of mechanical losses (such as spilling), separation procedures that are not 100% efficient, competing reactions that form undesired products, and reactions that simply do not go all the way to completion, thus resulting in a mixture of products and reactants. This last possibility is a common occurrence and is the subject of . So the actual yieldThe measured mass of products actually obtained from a reaction. The actual yield is nearly always less than the theoretical yield., the measured mass of products obtained from a reaction, is almost always less than the theoretical yield (often much less). The percent yieldThe ratio of the actual yield of a reaction to the theoretical yield multiplied by 100 to give a percentage. of a reaction is the ratio of the actual yield to the theoretical yield, multiplied by 100 to give a percentage:

Equation 3.23

percent yield =actual yield (g)theoretical yield (g)×100

The method used to calculate the percent yield of a reaction is illustrated in Example 13.

Example 13

Procaine is a key component of Novocain, an injectable local anesthetic used in dental work and minor surgery. Procaine can be prepared in the presence of H2SO4 (indicated above the arrow) by the reaction

C7H7NO2p-aminobenzoic acid+C6H15NO2-diethylaminoethanolH2SO4C13H20N2O2procaine+H2O

If we carried out this reaction using 10.0 g of p-aminobenzoic acid and 10.0 g of 2-diethylaminoethanol, and we isolated 15.7 g of procaine, what was the percent yield?

The preparation of procaine. A reaction of p-aminobenzoic acid with 2-diethylaminoethanol yields procaine and water.

Given: masses of reactants and product

Asked for: percent yield

Strategy:

A Write the balanced chemical equation.

B Convert from mass of reactants and product to moles using molar masses and then use mole ratios to determine which is the limiting reactant. Based on the number of moles of the limiting reactant, use mole ratios to determine the theoretical yield.

C Calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100.

Solution:

A From the formulas given for the reactants and the products, we see that the chemical equation is balanced as written. According to the equation, 1 mol of each reactant combines to give 1 mol of product plus 1 mol of water.

B To determine which reactant is limiting, we need to know their molar masses, which are calculated from their structural formulas: p-aminobenzoic acid (C7H7NO2), 137.14 g/mol; 2-diethylaminoethanol (C6H15NO), 117.19 g/mol. Thus the reaction used the following numbers of moles of reactants:

moles p-aminobenzoic acid = 10.0 g×1 mol137.14 g= 0.0729 mol p-aminobenzoic acidmoles 2-diethylaminoethanol = 10.0 g×1 mol117.19 g= 0.0853 mol 2-diethylaminoethanol

The reaction requires a 1:1 mole ratio of the two reactants, so p-aminobenzoic acid is the limiting reactant. Based on the coefficients in the balanced chemical equation, 1 mol of p-aminobenzoic acid yields 1 mol of procaine. We can therefore obtain only a maximum of 0.0729 mol of procaine. To calculate the corresponding mass of procaine, we use its structural formula (C13H20N2O2) to calculate its molar mass, which is 236.31 g/mol.

theoretical yield of procaine = 0.0729 mol×236.31 gmol= 17.2 g

C The actual yield was only 15.7 g of procaine, so the percent yield was

percent yield =15.7 g17.2 g×100=91.3%

(If the product were pure and dry, this yield would indicate that we have very good lab technique!)

Exercise

Lead was one of the earliest metals to be isolated in pure form. It occurs as concentrated deposits of a distinctive ore called galena (PbS), which is easily converted to lead oxide (PbO) in 100% yield by roasting in air via the following reaction:

2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g)

The resulting PbO is then converted to the pure metal by reaction with charcoal. Because lead has such a low melting point (327°C), it runs out of the ore-charcoal mixture as a liquid that is easily collected. The reaction for the conversion of lead oxide to pure lead is as follows:

PbO(s) + C(s) → Pb(l) + CO(g)

If 93.3 kg of PbO is heated with excess charcoal and 77.3 kg of pure lead is obtained, what is the percent yield?

Crystalline galena (a) and a sample of lead (b). Pure lead is soft enough to be shaped easily with a hammer, unlike the brittle mineral galena, the main ore of lead.

Answer: 89.2%

Percent yield can range from 0% to 100%.In the laboratory, a student will occasionally obtain a yield that appears to be greater than 100%. This usually happens when the product is impure or is wet with a solvent such as water. If this is not the case, then the student must have made an error in weighing either the reactants or the products. The law of conservation of mass applies even to undergraduate chemistry laboratory experiments! A 100% yield means that everything worked perfectly, and you obtained all the product that could have been produced. Anyone who has tried to do something as simple as fill a salt shaker or add oil to a car’s engine without spilling knows how unlikely a 100% yield is. At the other extreme, a yield of 0% means that no product was obtained. A percent yield of 80%–90% is usually considered good to excellent; a yield of 50% is only fair. In part because of the problems and costs of waste disposal, industrial production facilities face considerable pressures to optimize the yields of products and make them as close to 100% as possible.

Summary

The stoichiometry of a reaction describes the relative amounts of reactants and products in a balanced chemical equation. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s). If a quantity of a reactant remains unconsumed after complete reaction has occurred, it is in excess. The reactant that is consumed first and limits the amount of product(s) that can be obtained is the limiting reactant. To identify the limiting reactant, calculate the number of moles of each reactant present and compare this ratio to the mole ratio of the reactants in the balanced chemical equation. The maximum amount of product(s) that can be obtained in a reaction from a given amount of reactant(s) is the theoretical yield of the reaction. The actual yield is the amount of product(s) actually obtained in the reaction; it cannot exceed the theoretical yield. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, expressed as a percentage.

Key Takeaway

  • The stoichiometry of a balanced chemical equation identifies the maximum amount of product that can be obtained.

Conceptual Problems

  1. Engineers use conservation of mass, called a “mass balance,” to determine the amount of product that can be obtained from a chemical reaction. Mass balance assumes that the total mass of reactants is equal to the total mass of products. Is this a chemically valid practice? Explain your answer.

  2. Given the equation 2H2(g) + O2(g) → 2H2O(g), is it correct to say that 10 g of hydrogen will react with 10 g of oxygen to produce 20 g of water vapor?

  3. What does it mean to say that a reaction is stoichiometric?

  4. When sulfur is burned in air to produce sulfur dioxide, what is the limiting reactant? Explain your answer.

  5. Is it possible for the percent yield to be greater than the theoretical yield? Justify your answer.

Numerical Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 2 () before proceeding to the Numerical Problems.

  1. What is the formula mass of each species?

    1. ammonium chloride
    2. sodium cyanide
    3. magnesium hydroxide
    4. calcium phosphate
    5. lithium carbonate
    6. hydrogen sulfite ion
  2. What is the molecular or formula mass of each compound?

    1. potassium permanganate
    2. sodium sulfate
    3. hydrogen cyanide
    4. potassium thiocyanate
    5. ammonium oxalate
    6. lithium acetate
  3. How many moles are in each of the following?

    1. 10.76 g of Si
    2. 8.6 g of Pb
    3. 2.49 g of Mg
    4. 0.94 g of La
    5. 2.68 g of chlorine gas
    6. 0.089 g of As
  4. How many moles are in each of the following?

    1. 8.6 g of CO2
    2. 2.7 g of CaO
    3. 0.89 g of KCl
    4. 4.3 g of SrBr2
    5. 2.5 g of NaOH
    6. 1.87 g of Ca(OH)2
  5. Convert the following to moles and millimoles.

    1. 1.68 g of Ba(OH)2
    2. 0.792 g of H3PO4
    3. 3.21 g of K2S
    4. 0.8692 g of Cu(NO3)2
    5. 10.648 g of Ba3(PO4)2
    6. 5.79 g of (NH4)2SO4
    7. 1.32 g of Pb(C2H3O2)2
    8. 4.29 g of CaCl2·6H2O
  6. Convert the following to moles and millimoles.

    1. 0.089 g of silver nitrate
    2. 1.62 g of aluminum chloride
    3. 2.37 g of calcium carbonate
    4. 1.004 g of iron(II) sulfide
    5. 2.12 g of dinitrogen pentoxide
    6. 2.68 g of lead(II) nitrate
    7. 3.02 g of ammonium phosphate
    8. 5.852 g of sulfuric acid
    9. 4.735 g of potassium dichromate
  7. What is the mass of each substance in grams and milligrams?

    1. 5.68 mol of Ag
    2. 2.49 mol of Sn
    3. 0.0873 mol of Os
    4. 1.74 mol of Si
    5. 0.379 mol of H2
    6. 1.009 mol of Zr
  8. What is the mass of each substance in grams and milligrams?

    1. 2.080 mol of CH3OH
    2. 0.288 mol of P4
    3. 3.89 mol of ZnCl2
    4. 1.800 mol of Fe(CO)5
    5. 0.798 mol of S8
    6. 4.01 mol of NaOH
  9. What is the mass of each compound in kilograms?

    1. 6.38 mol of P4O10
    2. 2.26 mol of Ba(OH)2
    3. 4.35 mol of K3PO4
    4. 2.03 mol of Ni(ClO3)2
    5. 1.47 mol of NH4NO3
    6. 0.445 mol of Co(NO3)3
  10. How many atoms are contained in each?

    1. 2.32 mol of Bi
    2. 0.066 mol of V
    3. 0.267 mol of Ru
    4. 4.87 mol of C
    5. 2.74 g of I2
    6. 1.96 g of Cs
    7. 7.78 g of O2
  11. Convert each number of atoms to milligrams.

    1. 5.89 × 1022 Pt atoms
    2. 2.899 × 1021 Hg atoms
    3. 4.826 × 1022 atoms of chlorine
  12. Write a balanced chemical equation for each reaction and then determine which reactant is in excess.

    1. 2.46 g barium(s) plus 3.89 g bromine(l) in water to give barium bromide
    2. 1.44 g bromine(l) plus 2.42 g potassium iodide(s) in water to give potassium bromide and iodine
    3. 1.852 g of Zn metal plus 3.62 g of sulfuric acid in water to give zinc sulfate and hydrogen gas
    4. 0.147 g of iron metal reacts with 0.924 g of silver acetate in water to give iron(II) acetate and silver metal
    5. 3.142 g of ammonium phosphate reacts with 1.648 g of barium hydroxide in water to give ammonium hydroxide and barium phosphate
  13. Under the proper conditions, ammonia and oxygen will react to form dinitrogen monoxide (nitrous oxide, also called laughing gas) and water. Write a balanced chemical equation for this reaction. Determine which reactant is in excess for each combination of reactants.

    1. 24.6 g of ammonia and 21.4 g of oxygen
    2. 3.8 mol of ammonia and 84.2 g of oxygen
    3. 3.6 × 1024 molecules of ammonia and 318 g of oxygen
    4. 2.1 mol of ammonia and 36.4 g of oxygen
  14. When a piece of zinc metal is placed in aqueous hydrochloric acid, zinc chloride is produced, and hydrogen gas is evolved. Write a balanced chemical equation for this reaction. Determine which reactant is in excess for each combination of reactants.

    1. 12.5 g of HCl and 7.3 g of Zn
    2. 6.2 mol of HCl and 100 g of Zn
    3. 2.1 × 1023 molecules of Zn and 26.0 g of HCl
    4. 3.1 mol of Zn and 97.4 g of HCl
  15. Determine the mass of each reactant needed to give the indicated amount of product. Be sure that the chemical equations are balanced.

    1. NaI(aq) + Cl2(g) → NaCl(aq) + I2(s); 1.0 mol of NaCl
    2. NaCl(aq) + H2SO4(aq) → HCl(g) + Na2SO4(aq); 0.50 mol of HCl
    3. NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq); 1.5 mol of HNO3
  16. Determine the mass of each reactant needed to give the indicated amount of product. Be sure that the chemical equations are balanced.

    1. AgNO3(aq) + CaCl2(s) → AgCl(s) + Ca(NO3)2(aq); 1.25 mol of AgCl
    2. Pb(s) + PbO2(s) + H2SO4(aq) → PbSO4(s) + H2O(l); 3.8 g of PbSO4
    3. H3PO4(aq) + MgCO3(s) → Mg3(PO4)2(s) + CO2(g) + H2O(l); 6.41 g of Mg3(PO4)2
  17. Determine the percent yield of each reaction. Be sure that the chemical equations are balanced. Assume that any reactants for which amounts are not given are in excess. (The symbol Δ indicates that the reactants are heated.)

    1. KClO3(s)ΔKCl(s)+O2(g);2.14 g of KClO3 produces 0.67 g of O2
    2. Cu(s) + H2SO4(aq) → CuSO4(aq) + SO2(g) + H2O(l); 4.00 g of copper gives 1.2 g of sulfur dioxide
    3. AgC2H3O2(aq) + Na3PO4(aq) → Ag3PO4(s) + NaC2H3O2(aq); 5.298 g of silver acetate produces 1.583 g of silver phosphate
  18. Each step of a four-step reaction has a yield of 95%. What is the percent yield for the overall reaction?

  19. A three-step reaction yields of 87% for the first step, 94% for the second, and 55% for the third. What is the percent yield of the overall reaction?

  20. Give a general expression relating the theoretical yield (in grams) of product that can be obtained from x grams of B, assuming neither A nor B is limiting.

    A + 3B → 2C
  21. Under certain conditions, the reaction of hydrogen with carbon monoxide can produce methanol.

    1. Write a balanced chemical equation for this reaction.
    2. Calculate the percent yield if exactly 200 g of methanol is produced from exactly 300 g of carbon monoxide.
  22. Chlorine dioxide is a bleaching agent used in the paper industry. It can be prepared by the following reaction:

    NaClO2(s) + Cl2(g) → ClO2(aq) + NaCl(aq)
    1. What mass of chlorine is needed for the complete reaction of 30.5 g of NaClO2?
    2. Give a general equation for the conversion of x grams of sodium chlorite to chlorine dioxide.
  23. The reaction of propane gas (CH3CH2CH3) with chlorine gas (Cl2) produces two monochloride products: CH3CH2CH2Cl and CH3CHClCH3. The first is obtained in a 43% yield and the second in a 57% yield.

    1. If you use 2.78 g of propane gas, how much chlorine gas would you need for the reaction to go to completion?
    2. How many grams of each product could theoretically be obtained from the reaction starting with 2.78 g of propane?
    3. Use the actual percent yield to calculate how many grams of each product would actually be obtained.
  24. Protactinium (Pa), a highly toxic metal, is one of the rarest and most expensive elements. The following reaction is one method for preparing protactinium metal under relatively extreme conditions:

    2PaI5(s)Δ2Pa(s) + 5I2(s)
    1. Given 15.8 mg of reactant, how many milligrams of protactinium could be synthesized?
    2. If 3.4 mg of Pa was obtained, what was the percent yield of this reaction?
    3. If you obtained 3.4 mg of Pa and the percent yield was 78.6%, how many grams of PaI5 were used in the preparation?
  25. Aniline (C6H5NH2) can be produced from chlorobenzene (C6H5Cl) via the following reaction:

    C6H5Cl(l) + 2NH3(g) → C6H5NH2(l) + NH4Cl(s)

    Assume that 20.0 g of chlorobenzene at 92% purity is mixed with 8.30 g of ammonia.

    1. Which is the limiting reactant?
    2. Which reactant is present in excess?
    3. What is the theoretical yield of ammonium chloride in grams?
    4. If 4.78 g of NH4Cl was recovered, what was the percent yield?
    5. Derive a general expression for the theoretical yield of ammonium chloride in terms of grams of chlorobenzene reactant, if ammonia is present in excess.
  26. A stoichiometric quantity of chlorine gas is added to an aqueous solution of NaBr to produce an aqueous solution of sodium chloride and liquid bromine. Write the chemical equation for this reaction. Then assume an 89% yield and calculate the mass of chlorine given the following:

    1. 9.36 × 1024 formula units of NaCl
    2. 8.5 × 104 mol of Br2
    3. 3.7 × 108 g of NaCl

Answers

    1. 53.941 amu
    2. 49.0072 amu
    3. 58.3197 amu
    4. 310.177 amu
    5. 73.891 amu
    6. 81.071 amu
    1. 0.3831 mol Si
    2. 4.2 × 10−2 mol Pb
    3. 0.102 mol Mg
    4. 6.8 × 10−3 mol La
    5. 3.78 × 10−2 mol Cl2
    6. 1.2 × 10−3 mol As
    1. 9.80 × 10−3 mol or 9.80 mmole Ba(OH)2
    2. 8.08 × 10−3 mol or 8.08 mmole H3PO4
    3. 2.91 × 10−2 mol or 29.1 mmole K2S
    4. 4.634 × 10−3 mol or 4.634 mmole Cu(NO3)2
    5. 1.769 × 10−2 mol 17.69 mmole Ba3(PO4)2
    6. 4.38 × 10−2 mol or 43.8 mmole (NH4)2SO4
    7. 4.06 × 10−3 mol or 4.06 mmole Pb(C2H3O2)2
    8. 1.96 × 10−2 mol or 19.6 mmole CaCl2· 6H2O
    1. 613 g or 6.13 × 105 mg Ag
    2. 296 g or 2.96 × 105 mg Sn
    3. 16.6 g or 1.66 × 104 mg Os
    4. 48.9 g or 4.89 × 104 mg Si
    5. 0.764 g or 764 mg H2
    6. 92.05 g or 9.205 × 104 mg Zr
    1. 1.81 kg P4O10
    2. 0.387 kg Ba(OH)2
    3. 0.923 kg K3PO4
    4. 0.458 kg Ni(ClO3)2
    5. 0.118 kg (NH4)NO3
    6. 0.109 kg Co(NO3)3
    1. 1.91 × 104 mg Pt
    2. 965.6 mg Hg
    3. 2841 mg Cl
  1. The balanced chemical equation for this reaction is

    2NH3 + 2O2 → N2O + 3H2O
    1. NH3
    2. NH3
    3. O2
    4. NH3
    1. 150 g NaI and 35 g Cl2
    2. 29 g NaCl and 25 g H2SO4
    3. 140 g NO2 and 27 g H2O
    1. 80%
    2. 30%
    3. 35.7%
  2. 45%.

    1. CO + 2H2 → CH3OH
    2. 58.28%
    1. 2.24 g Cl2
    2. 4.95 g
    3. 2.13 g CH3CH2CH2Cl plus 2.82 g CH3CHClCH3
    1. chlorobenzene
    2. ammonia
    3. 8.74 g ammonium chloride.
    4. 55%
    5.  

      Theoretical yield (NH4Cl) = mass of chlorobenzene (g) × 0.92 × 53.49 g/mol112.55 g/mol

3.5 Classifying Chemical Reactions

Learning Objectives

  1. To identify fundamental types of chemical reactions.
  2. To predict the types of reactions substances will undergo.

The chemical reactions we have described are only a tiny sampling of the infinite number of chemical reactions possible. How do chemists cope with this overwhelming diversity? How do they predict which compounds will react with one another and what products will be formed? The key to success is to find useful ways to categorize reactions. Familiarity with a few basic types of reactions will help you to predict the products that form when certain kinds of compounds or elements come in contact.

Most chemical reactions can be classified into one or more of five basic types: acid–base reactionsA reaction of the general form acid + base → salt., exchange reactionsA chemical reaction that has the general form AB + C → AC + B or AB + CD → AD + CB., condensation reactionsA chemical reaction that has the general form A + B → AB. Condensation reactions are the reverse of cleavage reactions. Some, but not all, condensation reactions are also oxidation–reduction reactions. (and the reverse, cleavage reactionsA chemical reaction that has the general form AB → A + B. Cleavage reactions are the reverse of condensation reactions.), and oxidation–reduction reactionsA chemical reaction that exhibits a change in the oxidation states of one or more elements in the reactants that has the general form oxidant + reductant → reduced oxidant + oxidized reductant.. The general forms of these five kinds of reactions are summarized in , along with examples of each. It is important to note, however, that many reactions can be assigned to more than one classification, as you will see in our discussion. The classification scheme is only for convenience; the same reaction can be classified in different ways, depending on which of its characteristics is most important. Oxidation–reduction reactions, in which there is a net transfer of electrons from one atom to another, and condensation reactions are discussed in this section. Acid–base reactions and one kind of exchange reaction—the formation of an insoluble salt such as barium sulfate when solutions of two soluble salts are mixed together—will be discussed in .

Table 3.1 Basic Types of Chemical Reactions

Name of Reaction General Form Example(s)
oxidation–reduction (redox) oxidant + reductant → reduced oxidant + oxidized reductant C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(g)
acid–base acid + base → salt NH3(aq) + HNO3(aq) → NH4+(aq) + NO3(aq)
exchange AB + C → AC + B CH3Cl + OH → CH3OH + Cl
AB + CD → AD + CB BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
condensation A + B → AB CO2(g) + H2O(l) → H2CO3(aq)
HBr + H2C=CH2 → CH3CH2Br*
cleavage AB → A + B CaCO3(s) → CaO(s) + CO2(g)
CH3CH2Cl → H2C=CH2 + HCl**
* In more advanced chemisty courses you will learn that this reaction is also called an addition reaction.
** In more advanced chemistry courses you will learn that this reaction is also called an elimination reaction.

Oxidation–Reduction Reactions

The term oxidationThe loss of one or more electrons in a chemical reaction. The substance that loses electrons is said to be oxidized. was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron is exposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metal develops a continuous, coherent, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative charge and form oxide ions (O2−). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss of electrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons. For every oxidation, there must be an associated reduction.

Note the Pattern

Any oxidation must be accompanied by a reduction and vice versa.

Originally, the term reductionThe gain of one or more electrons in a chemical reaction. The substance that gains electrons is said to be reduced. referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, its mass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water). The reaction is as follows:

Equation 3.24

Cu2O(s) + H2(g) → 2Cu(s) + H2O(g)

Oxidation and reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants, which follows the mnemonic oxidation is loss reduction is gain, or oil rig. The oxidation stateThe charge that each atom in a compound would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. of each atom in a compound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form, such as O2 or H2, are assigned an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide is

Equation 3.25

4Al(s) + 3O2(g) → 2Al2O3(s)

Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidation state of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with an oxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al2O3, electrons are transferred as follows (the superscript 0 emphasizes the oxidation state of the elements):

Equation 3.26

4Al0 + 3O20 → 4Al3+ + 6O2−

and are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gained to preserve electrical neutrality. In , for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen:

Equation 3.27

electrons lost = 4 Al atoms×3 e lostAl atom=12 e lostelectrons gained = 6 O atoms×2 e gainedO atom=12 e gained

The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained.

An additional example of a redox reaction, the reaction of sodium metal with oxygen in air, is illustrated in .

Note the Pattern

In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained.

Assigning Oxidation States

Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements are identical to the charges on the monatomic ions. In , you learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodium chloride (NaCl; ), magnesium oxide (MgO), and calcium chloride (CaCl2). In covalent compounds, in contrast, atoms share electrons. Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions.

Figure 3.13 The Reaction of a Neutral Sodium Atom with a Neutral Chlorine Atom

The result is the transfer of one electron from sodium to chlorine, forming the ionic compound NaCl.

A set of rules for assigning oxidation states to atoms in chemical compounds follows. As we discuss atomic and molecular structure in , , , and , the principles underlying these rules will be described more fully.

Rules for Assigning Oxidation StatesNonintegral oxidation states are encountered occasionally. They are usually due to the presence of two or more atoms of the same element with different oxidation states.

  1. The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero.
  2. The oxidation state of a monatomic ion is the same as its charge—for example, Na+ = +1, Cl = −1.
  3. The oxidation state of fluorine in chemical compounds is always −1. Other halogens usually have oxidation states of −1 as well, except when combined with oxygen or other halogens.
  4. Hydrogen is assigned an oxidation state of +1 in its compounds with nonmetals and −1 in its compounds with metals.
  5. Oxygen is normally assigned an oxidation state of −2 in compounds, with two exceptions: in compounds that contain oxygen–fluorine or oxygen–oxygen bonds, the oxidation state of oxygen is determined by the oxidation states of the other elements present.
  6. The sum of the oxidation states of all the atoms in a neutral molecule or ion must equal the charge on the molecule or ion.

In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant, just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1. The net charge is zero, as it must be for any compound.

Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in . Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) as opposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H ion, whereas HCl forms H+ and Cl ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rule also prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF2 but −½ in KO2. Note that an oxidation state of −½ for O in KO2 is perfectly acceptable.

The reduction of copper(I) oxide shown in demonstrates how to apply these rules. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which applies to H2 and Cu. From rule 4, hydrogen in H2O has an oxidation state of +1, and from rule 5, oxygen in both Cu2O and H2O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu2O must have a charge of +1: 2(+1) + (−2) = 0. So the oxidation states are as follows:

Equation 3.28

Cu2+1O–2(s)+H20(g)2Cu0(s)+H2+1O–2(g)

Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). So this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge:

Equation 3.29

electrons lost = 2 H atoms×1 e lostH atom=2 e lostelectrons gained = 2 Cu atoms×1 e gainedCu atom=2 e gained

Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidation state of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way of assigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo.

Example 14

Assign oxidation states to all atoms in each compound.

  1. sulfur hexafluoride (SF6)
  2. methanol (CH3OH)
  3. ammonium sulfate [(NH4)2SO4]
  4. magnetite (Fe3O4)
  5. ethanoic (acetic) acid (CH3CO2H)

Given: molecular or empirical formula

Asked for: oxidation states

Strategy:

Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1.

Solution:

  1. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfur hexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero in a neutral molecule (here SF6), the oxidation state of sulfur must be +6:

    [(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0
  2. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has no net charge, carbon must have an oxidation state of −2:

    [(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0
  3. Note that (NH4)2SO4 is an ionic compound that consists of both a polyatomic cation (NH4+) and a polyatomic anion (SO42−) (see ). We assign oxidation states to the atoms in each polyatomic ion separately. For NH4+, hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3:

    [(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH4+ ion

    For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6:

    [(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion
  4. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron:

    [(4 O atoms)(2)]+[(3 Fe atoms)(+83)]= 0

    Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. In fact, Fe3O4 can be viewed as having two Fe3+ ions and one Fe2+ ion per formula unit, giving a net positive charge of +8 per formula unit. Fe3O4 is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestone because it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the “lodestar.”

  5. Initially, we assign oxidation states to the components of CH3CO2H in the same way as any other compound. Hydrogen and oxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of

    [(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0

    So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atoms present. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. To determine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (−CH3) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation state of +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of the carboxylic acid group (−CO2H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, we assign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of

    [(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3

    To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individual atoms in acetic acid are thus

    C3H3+1C+3O22H+1

    Thus the sum of the oxidation states of the two carbon atoms is indeed zero.

Exercise

Assign oxidation states to all atoms in each compound.

  1. barium fluoride (BaF2)
  2. formaldehyde (CH2O)
  3. potassium dichromate (K2Cr2O7)
  4. cesium oxide (CsO2)
  5. ethanol (CH3CH2OH)

Answer:

  1. Ba, +2; F, −1
  2. C, 0; H, +1; O, −2
  3. K, +1; Cr, +6; O, −2
  4. Cs, +1; O, −½
  5. C, −3; H, +1; C, −1; H, +1; O, −2; H, +1

Oxidants and Reductants

Compounds that are capable of accepting electrons, such as O2 or F2, are called oxidants (or oxidizing agents)A compound that is capable of accepting electrons; thus it is reduced. because they can oxidize other compounds. In the process of accepting electrons, an oxidant is reduced. Compounds that are capable of donating electrons, such as sodium metal or cyclohexane (C6H12), are called reductants (or reducing agents)A compound that is capable of donating electrons; thus it is oxidized. because they can cause the reduction of another compound. In the process of donating electrons, a reductant is oxidized. These relationships are summarized in :

Equation 3.30

oxidant + reductant → oxidation−reduction
O2(g)gains e(is reduced)+4Na(s)loses e(is oxidized)2Na2O(s)redox reaction

Some oxidants have a greater ability than others to remove electrons from other compounds. Oxidants can range from very powerful, capable of oxidizing most compounds with which they come in contact, to rather weak. Both F2 and Cl2 are powerful oxidants: for example, F2 will oxidize H2O in a vigorous, potentially explosive reaction. In contrast, S8 is a rather weak oxidant, and O2 falls somewhere in between. Conversely, reductants vary in their tendency to donate electrons to other compounds. Reductants can also range from very powerful, capable of giving up electrons to almost anything, to weak. The alkali metals are powerful reductants, so they must be kept away from atmospheric oxygen to avoid a potentially hazardous redox reaction.

A combustion reactionAn oxidation–reduction reaction in which the oxidant is O2., first introduced in , is an oxidation–reduction reaction in which the oxidant is O2. One example of a combustion reaction is the burning of a candle, shown in . Consider, for example, the combustion of cyclohexane, a typical hydrocarbon, in excess oxygen. The balanced chemical equation for the reaction, with the oxidation state shown for each atom, is as follows:

Equation 3.31

C6–2H12+1+9O206C+4O2–2+6H2+1O–2

If we compare the oxidation state of each element in the products and the reactants, we see that hydrogen is the only element whose oxidation state does not change; it remains +1. Carbon, however, has an oxidation state of −2 in cyclohexane and +4 in CO2; that is, each carbon atom changes its oxidation state by six electrons during the reaction. Oxygen has an oxidation state of 0 in the reactants, but it gains electrons to have an oxidation state of −2 in CO2 and H2O. Because carbon has been oxidized, cyclohexane is the reductant; because oxygen has been reduced, it is the oxidant. All combustion reactions are therefore oxidation–reduction reactions.

Condensation Reactions

The reaction of bromine with ethylene to give 1,2-dibromoethane, which is used in agriculture to kill nematodes in soil, is as follows:

Equation 3.32

C2H4(g) + Br2(g) → BrCH2CH2Br(g)

According to , this is a condensation reaction because it has the general form A + B → AB. This reaction, however, can also be viewed as an oxidation–reduction reaction, in which electrons are transferred from carbon (−2 → −1) to bromine (0 → −1). Another example of a condensation reaction is the one used for the industrial synthesis of ammonia:

Equation 3.33

3H2(g) + N2(g) → 2NH3(g)

Although this reaction also has the general form of a condensation reaction, hydrogen has been oxidized (0 → +1) and nitrogen has been reduced (0 → −3), so it can also be classified as an oxidation–reduction reaction.

Not all condensation reactions are redox reactions. The reaction of an amine with a carboxylic acid, for example, is a variant of a condensation reaction (A + B → A′B′ + C): two large fragments condense to form a single molecule, and a much smaller molecule, such as H2O, is eliminated. In this reaction, the −OH from the carboxylic acid group and −H from the amine group are eliminated as H2O, and the reaction forms an amide bond (also called a peptide bond) that links the two fragments. Amide bonds are the essential structural unit linking the building blocks of proteins and many polymers together, as described in . Nylon, for example, is produced from a condensation reaction ().

Amide bonds. The reaction of an amine with a carboxylic acid proceeds by eliminating water and forms a new C–N (amide) bond.

Figure 3.14 The Production of Nylon

Example 15

The following reactions have important industrial applications. Using , classify each reaction as an oxidation–reduction reaction, an acid–base reaction, an exchange reaction, a condensation reaction, or a cleavage reaction. For each redox reaction, identify the oxidant and reductant and specify which atoms are oxidized or reduced. (Don’t forget that some reactions can be placed into more than one category.)

  1. C2H4(g) + Cl2(g) → ClCH2CH2Cl(g)
  2. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
  3. CaCO3(s) → CaO(s) + CO2(g)
  4. Ca5(PO4)3(OH)(s) + 7H3PO4(aq) + 4H2O(l) → 5Ca(H2PO4)2·H2O(s)
  5. Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)

Given: balanced chemical equation

Asked for: classification of chemical reaction

Strategy:

A Determine the general form of the equation by referring to and then classify the reaction.

B For redox reactions, assign oxidation states to each atom present in the reactants and the products. If the oxidation state of one or more atoms changes, then the reaction is a redox reaction. If not, the reaction must be one of the other types of reaction listed in .

Solution:

  1. A This reaction is used to prepare 1,2-dichloroethane, one of the top 25 industrial chemicals in . It has the general form A + B → AB, which is typical of a condensation reaction. B Because reactions may fit into more than one category, we need to look at the oxidation states of the atoms:

    C2–2H4+1+Cl20Cl–1C–1H2+1C–1H2+1Cl–1

    The oxidation states show that chlorine is reduced from 0 to −1 and carbon is oxidized from −2 to −1, so this is a redox reaction as well as a condensation reaction. Ethylene is the reductant, and chlorine is the oxidant.

  2. A This reaction is used to prepare silver chloride for making photographic film. The chemical equation has the general form AB + CD → AD + CB, so it is classified as an exchange reaction. B The oxidation states of the atoms are as follows

    Ag+1N+5O3–2+Na+1Cl–1Ag+1Cl–1+Na+1N+5O3–2

    There is no change in the oxidation states, so this is not a redox reaction.

    AgCl(s) precipitates when solutions of AgNO3(aq) and NaCl(aq) are mixed. NaNO3 (aq) is in solution as Na+ and NO3 ions.

  3. A This reaction is used to prepare lime (CaO) from limestone (CaCO3) and has the general form AB → A + B. The chemical equation’s general form indicates that it can be classified as a cleavage reaction, the reverse of a condensation reaction. B The oxidation states of the atoms are as follows:

    Ca+2C+4O3–2Ca+2O–2+C+4O2–2

    Because the oxidation states of all the atoms are the same in the products and the reactant, this is not a redox reaction.

  4. A This reaction is used to prepare “super triple phosphate” in fertilizer. One of the reactants is phosphoric acid, which transfers a proton (H+) to the phosphate and hydroxide ions of hydroxyapatite [Ca5(PO4)3(OH)] to form H2PO4 and H2O, respectively. This is an acid–base reaction, in which H3PO4 is the acid (H+ donor) and Ca5(PO4)3(OH) is the base (H+ acceptor).

    B To determine whether it is also a redox reaction, we assign oxidation states to the atoms:

    Ca5+2(P+5O4–2)3(O–2H+1)+7H3+1P+5O4–2+4H+12O–25Ca+2(H+12P+5O4–2)2·H2+1O–2

    Because there is no change in oxidation state, this is not a redox reaction.

  5. A This reaction occurs in a conventional car battery every time the engine is started. An acid (H2SO4) is present and transfers protons to oxygen in PbO2 to form water during the reaction. The reaction can therefore be described as an acid–base reaction.

    B The oxidation states are as follows:

    Pb0+Pb+4O2–2+2H2+1S+6O4–22Pb+2S+6O4–2+2H2+1O–2

    The oxidation state of lead changes from 0 in Pb and +4 in PbO2 (both reactants) to +2 in PbSO4. This is also a redox reaction, in which elemental lead is the reductant, and PbO2 is the oxidant. Which description is correct? Both.

Schematic drawing of a 12-volt car battery. The locations of the reactants (lead metal in a spongy form with large surface area) and PbO2 are shown. The product (PbSO4) forms as a white solid between the plates.

Exercise

Using , classify each reaction as an oxidation–reduction reaction, an acid–base reaction, an exchange reaction, a condensation reaction, or a cleavage reaction. For each redox reaction, identify the oxidant and the reductant and specify which atoms are oxidized or reduced.

  1. Al(s) + OH(aq) + 3H2O(l) → 3/2H2(g) + [Al(OH)4](aq)
  2. TiCl4(l) + 2Mg(l) → Ti(s) + 2MgCl2(l)
  3. MgCl2(aq) + Na2CO3(aq) → MgCO3(s) + 2NaCl(aq)
  4. CO(g) + Cl2(g) → Cl2CO(l)
  5. H2SO4(l) + 2NH3(g) → (NH4)2SO4(s)

Answer:

  1. Redox reaction; reductant is Al, oxidant is H2O; Al is oxidized, H is reduced. This is the reaction that occurs when Drano is used to clear a clogged drain.
  2. Redox reaction; reductant is Mg, oxidant is TiCl4; Mg is oxidized, Ti is reduced.
  3. Exchange reaction. This reaction is responsible for the scale that develops in coffee makers in areas that have hard water.
  4. Both a condensation reaction and a redox reaction; reductant is CO, oxidant is Cl2; C is oxidized, Cl is reduced. The product of this reaction is phosgene, a highly toxic gas used as a chemical weapon in World War I. Phosgene is now used to prepare polyurethanes, which are used in foams for bedding and furniture and in a variety of coatings.
  5. Acid–base reaction.

Catalysts

Many chemical reactions, including some of those discussed previously, occur more rapidly in the presence of a catalystA substance that increases the rate of a chemical reaction without undergoing a net chemical change itself., which is a substance that participates in a reaction and causes it to occur more rapidly but can be recovered unchanged at the end of a reaction and reused. Because catalysts are not involved in the stoichiometry of a reaction, they are usually shown above the arrow in a net chemical equation. Chemical processes in industry rely heavily on the use of catalysts, which are usually added to a reaction mixture in trace amounts, and most biological reactions do not take place without a biological catalyst or enzymeCatalysts that occur naturally in living organisms and catalyze biological reactions.. Examples of catalyzed reactions in industry are the use of platinum in petroleum cracking and reforming, the reaction of SO2 and O2 in the presence of V2O5 to produce SO3 in the industrial synthesis of sulfuric acid, and the use of sulfuric acid in the synthesis of compounds such as ethyl acetate and procaine. Not only do catalysts greatly increase the rates of reactions, but in some cases such as in petroleum refining, they also control which products are formed. The acceleration of a reaction by a catalyst is called catalysisThe acceleration of a chemical reaction by a catalyst..

A heterogeneous catalyst. This large circular gauze, woven from rhodium-platinum wire, is a heterogeneous catalyst in the commercial production of nitric acid by the oxidation of ammonia.

Catalysts may be classified as either homogeneous or heterogeneous. A homogeneous catalystA catalyst that is uniformly dispersed throughout the reactant mixture to form a solution. is uniformly dispersed throughout the reactant mixture to form a solution. Sulfuric acid, for example, is a homogeneous catalyst used in the synthesis of esters such as procaine (Example 13). An ester has a structure similar to that of a carboxylic acid, in which the hydrogen atom attached to oxygen has been replaced by an R group. They are responsible for the fragrances of many fruits, flowers, and perfumes. Other examples of homogeneous catalysts are the enzymes that allow our bodies to function. In contrast, a heterogeneous catalystA catalyst that is in a different physical state than the reactants. is in a different physical state than the reactants. For economic reasons, most industrial processes use heterogeneous catalysts in the form of solids that are added to solutions of the reactants. Because such catalysts often contain expensive precious metals such as platinum or palladium, it makes sense to formulate them as solids that can be easily separated from the liquid or gaseous reactant-product mixture and recovered. Examples of heterogeneous catalysts are the iron oxides used in the industrial synthesis of ammonia and the catalytic converters found in virtually all modern automobiles, which contain precious metals like palladium and rhodium. Catalysis will be discussed in more detail in when we discuss reaction rates, but you will encounter the term frequently throughout the text.

Summary

Chemical reactions may be classified as an acid–base reaction, an exchange reaction, a condensation reaction and its reverse, a cleavage reaction, and an oxidation–reduction (or redox) reaction. To keep track of electrons in chemical reactions, oxidation states are assigned to atoms in compounds. The oxidation state is the charge an atom would have if all its bonding electrons were transferred completely to the atom that has the greater attraction for electrons. In an oxidation–reduction reaction, one atom must lose electrons and another must gain electrons. Oxidation is the loss of electrons, and an element whose oxidation state increases is said to be oxidized. Reduction is the gain of electrons, and an element whose oxidation state decreases is said to be reduced. Oxidants are compounds that are capable of accepting electrons from other compounds, so they are reduced during an oxidation–reduction reaction. In contrast, reductants are compounds that are capable of donating electrons to other compounds, so they are oxidized during an oxidation–reduction reaction. A combustion reaction is a redox reaction in which the oxidant is O2(g). An amide bond is formed from the condensation reaction between a carboxylic acid and an amine; it is the essential structural unit of proteins and many polymers. A catalyst is a substance that increases the rate of a chemical reaction without undergoing a net chemical change itself. A biological catalyst is called an enzyme. Catalysis is an acceleration in the rate of a reaction caused by the presence of a substance that does not appear in the chemical equation. A homogeneous catalyst is uniformly dispersed in a solution of the reactants, whereas a heterogeneous catalyst is present as a different phase, usually a solid.

Key Takeaway

  • Chemical reactions may be classified as acid–base, exchange, condensation, cleavage, and oxidation–reduction (redox).

Conceptual Problems

  1. What is a combustion reaction? How can it be distinguished from an exchange reaction?

  2. What two products are formed in the combustion of an organic compound containing only carbon, hydrogen, and oxygen? Is it possible to form only these two products from a reaction that is not a combustion reaction? Explain your answer.

  3. What factors determine whether a reaction can be classified as a redox reaction?

  4. Name three characteristics of a balanced redox reaction.

  5. Does an oxidant accept electrons or donate them?

  6. Does the oxidation state of a reductant become more positive or more negative during a redox reaction?

  7. Nitrogen, hydrogen, and ammonia are known to have existed on primordial earth, yet mixtures of nitrogen and hydrogen do not usually react to give ammonia. What natural phenomenon would have enough energy to initiate a reaction between these two primordial gases?

  8. Catalysts are not added to reactions in stoichiometric quantities. Why?

  9. State whether each of the following uses a homogeneous catalyst or a heterogeneous catalyst.

    1. Platinum metal is used in the catalytic converter of an automobile.
    2. Nitrogen is biologically converted to ammonia by an enzyme.
    3. Carbon monoxide and hydrogen combine to form methane and water with a nickel catalyst.
    4. A dissolved rhodium compound is used as a catalyst for the conversion of an alkene to an alkane.
  10. State whether each of the following uses a homogeneous catalyst or a heterogeneous catalyst.

    1. Pellets of ZSM-5, an aluminum- and silicon-containing mineral, are used to catalyze the conversion of methanol to gasoline.
    2. The conversion of glucose to a carboxylic acid occurs with catalysis by the enzyme glucose oxidase.
    3. Metallic rhodium is used to the conversion of carbon monoxide and water to carbon dioxide and hydrogen.
  11. Complete the following table to describe some key differences between homogeneous and heterogeneous catalysis.

    Homogeneous Heterogeneous
    number of phases
    ease of separation from product
    ease of recovery of catalyst
  12. To increase the rate of a reaction, a scientist decided to use a catalyst. Unexpectedly, the scientist discovered that the catalyst decreased the yield of the desired product, rather than increasing it. What might have happened?

Answer

  1. Homogeneous Heterogeneous
    number of phases single phase at least two phases
    ease of separation from product difficult easy
    ease of recovery of catalyst difficult easy

Numerical Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 2 () before proceeding to the Numerical Problems.

  1. Classify each chemical reaction according to the types listed in .

    1. 12FeCl2(s) + 3O2(g) → 8FeCl3(s) + 2Fe2O3(s)
    2. CaCl2(aq) + K2SO4(aq) → CaSO4(s) + 2KCl(aq)
    3. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
    4. Br2(l) + C2H4(g) → BrCH2CH2Br(l)
  2. Classify each chemical reaction according to the types listed in .

    1. 4FeO(s) + O2(g) → 2Fe2O3(s)
    2. Ca3(PO4)2(s) + 3H2SO4(aq) → 3CaSO4(s) + 2H3PO4(aq)
    3. HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)
    4. ethane(g) + oxygen(g) → carbon dioxide(g) + water(g)
  3. Assign oxidation states to the atoms in each compound or ion.

    1. (NH4)2S
    2. the phosphate ion
    3. [AlF6]3−
    4. CuS
    5. HCO3
    6. NH4+
    7. H2SO4
    8. formic acid
    9. n-butanol
  4. Assign oxidation states to the atoms in each compound or ion.

    1. ClO2
    2. HO2
    3. sodium bicarbonate
    4. MnO2
    5. PCl5
    6. [Mg(H2O)6]2+
    7. N2O4
    8. butanoic acid
    9. methanol
  5. Balance this chemical equation:

    NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + CO2(g) + H2O(l)

    What type of reaction is this? Justify your answer.

  6. Assign oxidation states to the atoms in each compound.

    1. iron(III) nitrate
    2. Al2O3
    3. potassium sulfate
    4. Cr2O3
    5. sodium perchlorate
    6. Cu2S
    7. hydrazine (N2H4)
    8. NO2
    9. n-pentanol
  7. Assign oxidation states to the atoms in each compound.

    1. calcium carbonate
    2. NaCl
    3. CO2
    4. potassium dichromate
    5. KMnO4
    6. ferric oxide
    7. Cu(OH)2
    8. Na2SO4
    9. n-hexanol
  8. For each redox reaction, determine the identities of the oxidant, the reductant, the species oxidized, and the species reduced.

    1. H2(g) + I2(s) → 2HI(g)
    2. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
    3. 2F2(g) + 2NaOH(aq) → OF2(g) + 2NaF(aq) + H2O(l)
  9. For each redox reaction, determine the identities of the oxidant, the reductant, the species oxidized, and the species reduced.

    1. 2Na(s) + Cl2(g) → 2NaCl(s)
    2. SiCl4(l) + 2Mg(s) → 2MgCl2(s) + Si(s)
    3. 2H2O2(aq) → 2H2O(l) + O2(g)
  10. Balance each chemical equation. Then identify the oxidant, the reductant, the species oxidized, and the species reduced. (Δ indicates that the reaction requires heating.)

    1. H2O(g) + CO(g) → CO2(g) + H2(g)
    2. the reaction of aluminum oxide, carbon, and chlorine gas at 900ºC to produce aluminum chloride and carbon monoxide
    3. HgO(s)ΔHg(l)+O2(g)
  11. Balance each chemical equation. Then identify the oxidant, the reductant, the species oxidized, and the species reduced. (Δ indicates that the reaction requires heating.)

    1. the reaction of water and carbon at 800ºC to produce hydrogen and carbon monoxide
    2. Mn(s) + S8(s) + CaO(s) → CaS(s) + MnO(s)
    3. the reaction of ethylene and oxygen at elevated temperature in the presence of a silver catalyst to produce ethylene oxide

    4. ZnS(s) + H2SO4(aq) + O2(g) → ZnSO4(aq) + S8(s) + H2O(l)
  12. Silver is tarnished by hydrogen sulfide, an atmospheric contaminant, to form a thin layer of dark silver sulfide (Ag2S) along with hydrogen gas.

    1. Write a balanced chemical equation for this reaction.
    2. Which species has been oxidized and which has been reduced?
    3. Assuming 2.2 g of Ag has been converted to silver sulfide, construct a table showing the reaction in terms of the number of atoms in the reactants and products, the moles of reactants and products, the grams of reactants and products, and the molecules of reactants and products.
  13. The following reaction is used in the paper and pulp industry:

    Na2SO4(aq) + C(s) + NaOH(aq) → Na2CO3(aq) + Na2S(aq) + H2O(l)
    1. Balance the chemical equation.
    2. Identify the oxidant and the reductant.
    3. How much carbon is needed to convert 2.8 kg of sodium sulfate to sodium sulfide?
    4. If the yield of the reaction were only 78%, how many kilograms of sodium carbonate would be produced from 2.80 kg of sodium sulfate?
    5. If 240 g of carbon and 2.80 kg of sodium sulfate were used in the reaction, what would be the limiting reactant (assuming an excess of sodium hydroxide)?
  14. The reaction of A2 (blue) with B2 (yellow) is shown below. The initial reaction mixture is shown on the left and the mixture after the reaction has gone to completion is shown on the right.

    1. Write a balanced chemical equation for the reaction.
    2. Which is the limiting reactant in the initial reaction mixture?
    3. How many moles of the product AB4 could you obtain from a mixture of 0.020 mol A2 and 0.060 mol B2?
  15. The reaction of X4 (orange) with Y2 (black) is shown below. The initial reaction mixture is shown on the left and the mixture after the reaction has gone to completion is shown on the right.

    1. Write a balanced chemical equation for the reaction.
    2. Which is the limiting reactant in the initial reaction mixture?
    3. How many moles of the product XY3 could you obtain from a mixture of 0.100 mol X4 and 0.300 mol Y2?
  16. Methyl butyrate, an artificial apple flavor used in the food industry, is produced by the reaction of butanoic acid with methanol in the presence of an acid catalyst (H+):

    CH3CH2CH2COOH(l)+CH3OH(l) H+ CH3CH2CH2CO2CH3(l)+H2O(l)
    1. Given 7.8 g of butanoic acid, how many grams of methyl butyrate would be synthesized, assuming 100% yield?
    2. The reaction produced 5.5 g of methyl butyrate. What was the percent yield?
    3. Is the catalyst used in this reaction heterogeneous or homogeneous?
  17. In the presence of a platinum catalyst, hydrogen and bromine react at elevated temperatures (300°C) to form hydrogen bromide (heat is indicated by Δ):

    H2(g)+Br2(l) ΔPt 2HBr(g)

    Given the following, calculate the mass of hydrogen bromide produced:

    1. 8.23 × 1022 molecules of H2
    2. 6.1 × 103 mol of H2
    3. 1.3 × 105 g of H2
    4. Is the catalyst used in this reaction heterogeneous or homogeneous?

Answers

    1. redox reaction
    2. exchange
    3. acid–base
    4. condensation
    1. S, −2; N, −3; H, +1
    2. P, +5; O, −2
    3. F, −1; Al, +3
    4. S, −2; Cu, +2
    5. H, +1; O, −2; C, +4
    6. H, +1; N, −3
    7. H, +1; O, −2; S, +6
    8. H, +1, O, −2; C, +2
    9. butanol:

      O,−2; H, +1

      From left to right: C, −3–2–2–1

  1. 2NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l) acid–base reaction

    1. Ca, +2; O, −2; C, +4
    2. Na, +1; Cl, −1
    3. O, −2; C, +4
    4. K, +1; O, −2; Cr, +6
    5. K, +1; O, −2; Mn, +7
    6. O, −2; Fe, +3
    7. O, −2; H, +1; Cu, +2
    8. O, −2; S, +6
    9.  

      Hexanol

      O,−2; H, +1

      From left to right: C: −3, −2, −2, −2, −2, −1

    1. Na is the reductant and is oxidized. Cl2 is the oxidant and is reduced.
    2. Mg is the reductant and is oxidized. Si is the oxidant and is reduced.
    3. H2O2 is both the oxidant and reductant. One molecule is oxidized, and one molecule is reduced.
    1. H2O(g) + C(s) Δ H2(g) + CO(g)

      C is the reductant and is oxidized. H2O is the oxidant and is reduced.

    2. 8Mn(s) + S8(s) + 8CaO(s) → 8CaS(s) + 8MnO(s)

      Mn is the reductant and is oxidized. The S8 is the oxidant and is reduced.

    3. 2C2H4(g) + O2(g) Δ 2C2H4O(g)

      Ethylene is the reductant and is oxidized. O2 is the oxidant and is reduced.

    4. 8ZnS(s) + 8H2SO4(aq) + 4O2(g) → 8ZnSO4(aq) + S8(s) + 8H2O(l)

      Sulfide in ZnS is the reductant and is oxidized. O2 is the oxidant and is reduced.

    1. Na2SO4 + 2C + 4NaOH → 2Na2CO3 + Na2S + 2H2O
    2. The sulfate ion is the oxidant, and the reductant is carbon.
    3. 470 g
    4. 3300 g
    5. carbon
    1. 22.1 g
    2. 9.9 × 105 g
    3. 1.0 × 107 g
    4. heterogeneous

3.6 Chemical Reactions in the Atmosphere

Learning Objective

  1. To become familiar with the various reactions implicated in the destruction of Earth’s ozone layer.

described different classes of chemical reactions. Of the many different chemical reactions that occur in Earth’s atmosphere, some are important and controversial because they affect our quality of life and health. The atmospheric reactions presented in this section provide examples of the various classes of reactions introduced in this chapter that are implicated in the destruction of Earth’s protective ozone layerA concentration of ozone in the stratosphere (about 1015 ozone molecules per liter) that acts as a protective screen, absorbing ultraviolet light that would otherwise reach the surface of the earth, where it would harm plants and animals..

Each year since the mid-1970s, scientists have noted a disappearance of approximately 70% of the ozone (O3) layer above Antarctica during the Antarctic spring, creating what is commonly known as the “ozone hole.” OzoneAn unstable form of oxygen that consists of three oxygen atoms bonded together (O3). A layer of ozone in the stratosphere helps protect the plants and animals on earth from harmful ultraviolet radiation. Ozone is responsible for the pungent smell we associate with lightning discharges and electric motors. It is also toxic. is an unstable form of oxygen that consists of three oxygen atoms bonded together. In September 2009, the Antarctic ozone hole reached 24.1 million km2 (9.3 million mi2), about the size of North America. The largest area ever recorded was in the year 2000, when the hole measured 29.9 million km2 and for the first time extended over a populated area—the city of Punta Arenas, Chile (population 154,000; ). A less extensive zone of depletion has been detected over the Arctic as well. Years of study from the ground, from the air, and from satellites in space have shown that chlorine from industrial chemicals used in spray cans, foam packaging, and refrigeration materials is largely responsible for the catalytic depletion of ozone through a series of condensation, cleavage, and oxidation–reduction reactions.

Figure 3.15 Satellite Photos of Earth Reveal the Sizes of the Antarctic Ozone Hole over Time

Dark blue colors correspond to the thinnest ozone; light blue, green, yellow, orange, and red indicate progressively thicker ozone. In September 2000, the Antarctic ozone hole briefly approached a record 30 million km2.

Earth’s Atmosphere and the Ozone Layer

Earth’s atmosphere at sea level is an approximately 80:20 solution of nitrogen and oxygen gases, with small amounts of carbon dioxide, water vapor, and the noble gases, and trace amounts of a variety of other compounds (). A key feature of the atmosphere is that its composition, temperature, and pressure vary dramatically with altitude. Consequently, scientists have divided the atmosphere into distinct layers, which interact differently with the continuous flux of solar radiation from the top and the land and ocean masses at the bottom. Some of the characteristic features of the layers of the atmosphere are illustrated in .

Table 3.2 The Composition of Earth’s Atmosphere at Sea Level*

Gas Formula Volume (%)
nitrogen N2 78.084
oxygen O2 20.948
argon Ar 0.934
carbon dioxide CO2 0.0314
neon Ne 0.00182
helium He 0.000524
krypton Kr 0.000114
methane CH4 0.0002
hydrogen H2 0.00005
nitrous oxide N2O 0.00005
xenon Xe 0.0000087
* In addition, air contains as much as 7% water vapor (H2O), 0.0001% sulfur dioxide (SO2), 0.00007% ozone (O3), 0.000002% carbon monoxide (CO), and 0.000002% nitrogen dioxide (NO2).
Carbon dioxide levels are highly variable; the typical range is 0.01–0.1%.

Figure 3.16 Variation of Temperature with Altitude in Earth’s Atmosphere

Note the important chemical species present in each layer. The yellow line indicates the temperature at various altitudes.

The troposphereThe lowest layer of the atmosphere, the troposphere extends from earth’s surface to an altitude of about 11–13 km (7–8 miles). The temperature of the troposphere decreases steadily with increasing altitude. is the lowest layer of the atmosphere, extending from Earth’s surface to an altitude of about 11–13 km (7–8 mi). Above the troposphere lies the stratosphere, which extends from 13 km (8 mi) to about 44 km (27 mi). As shown in , the temperature of the troposphere decreases steadily with increasing altitude. Because “hot air rises,” this temperature gradient leads to continuous mixing of the upper and lower regions within the layer. The thermally induced turbulence in the troposphere produces fluctuations in temperature and precipitation that we collectively refer to as “weather.” In contrast, mixing between the layers of the atmosphere occurs relatively slowly, so each layer has distinctive chemistry. We focus our attention on the stratosphere, which contains the highest concentration of ozone.

The sun’s radiation is the major source of energy that initiates chemical reactions in the atmosphere. The sun emits many kinds of radiation, including visible lightRadiation that the human eye can detect., which is radiation that the human eye can detect, and ultraviolet lightHigh-energy radiation that cannot be detected by the human eye but can cause a wide variety of chemical reactions that are harmful to organisms., which is higher energy radiation that cannot be detected by the human eye. This higher energy ultraviolet light can cause a wide variety of chemical reactions that are harmful to organisms. For example, ultraviolet light is used to sterilize items, and, as anyone who has ever suffered a severe sunburn knows, it can produce extensive tissue damage.

Light in the higher energy ultraviolet range is almost totally absorbed by oxygen molecules in the upper layers of the atmosphere, causing the O2 molecules to dissociate into two oxygen atoms in a cleavage reaction:

Equation 3.34

O2(g) light 2O(g)

In , light is written above the arrow to indicate that light is required for the reaction to occur. The oxygen atoms produced in can undergo a condensation reaction with O2 molecules to form ozone:

Equation 3.35

O(g) + O2(g) → O3(g)

Ozone is responsible for the pungent smell we associate with lightning discharges and electric motors. It is also toxic and a significant air pollutant, particularly in cities.

In the stratosphere, the ozone produced via has a major beneficial effect. Ozone absorbs the less-energetic range of ultraviolet light, undergoing a cleavage reaction in the process to give O2 and O:

Equation 3.36

O3(g)lightO2(g) + O(g)

The formation of ozone () and its decomposition () are normally in balance, resulting in essentially constant levels of about 1015 ozone molecules per liter in the stratosphere. This so-called ozone layer acts as a protective screen that absorbs ultraviolet light that would otherwise reach Earth’s surface.

In 1974, F. Sherwood Rowland and Mario Molina published a paper claiming that commonly used chlorofluorocarbon (CFC) compounds were causing major damage to the ozone layer (). CFCs had been used as refrigerants and propellants in aerosol cans for many years, releasing millions of tons of CFC molecules into the atmosphere. Because CFCs are volatile compounds that do not readily undergo chemical reactions, they persist in the atmosphere long enough to be carried to the top of the troposphere, where they eventually enter the stratosphere. There they are exposed to intense ultraviolet light and undergo a cleavage reaction to produce a chlorine atom, which is shown for Freon-11:

Equation 3.37

CCl3F(g)lightCCl2F(g) + Cl(g)

The resulting chlorine atoms act as a homogeneous catalyst in two redox reactions ( and ):

Equation 3.38

Cl(g) + O3(g) → ClO(g) + O2(g)

Equation 3.39

ClO(g) + O(g) → Cl(g) + O2(g)

Adding the two reactions in and gives

Equation 3.40

Cl(g) + O3(g) + ClO(g) + O(g) → ClO(g) + Cl(g) + 2O2(g)

Because chlorine and ClO (chlorine monoxide) appear on both sides of the equation, they can be canceled to give the following net reaction:

Equation 3.41

O3(g) + O(g) → 2O2(g)

In the presence of chlorine atoms, one O3 molecule and one oxygen atom react to give two O2 molecules. Although chlorine is necessary for the overall reaction to occur, it does not appear in the net equation. The chlorine atoms are a catalyst that increases the rate at which ozone is converted to oxygen.

Table 3.3 Common CFCs and Related Compounds

Name Molecular Formula Industrial Name
trichlorofluoromethane CCl3F CFC-11 (Freon-11)
dichlorodifluoromethane CCl2F2 CFC-12 (Freon-12)
chlorotrifluoromethane CClF3 CFC-13 (Freon-13)
bromotrifluoromethane CBrF3 Halon-1301*
bromochlorodifluoromethane CBrClF2 Halon-1211
*Halons, compounds similar to CFCs that contain at least one bromine atom, are used as fire extinguishers in specific applications (e.g., the engine rooms of ships).

Because the stratosphere is relatively isolated from the layers of the atmosphere above and below it, once chlorine-containing species enter the stratosphere, they remain there for long periods of time. Each chlorine atom produced from a CFC molecule can lead to the destruction of large numbers of ozone molecules, thereby decreasing the concentration of ozone in the stratosphere. Eventually, however, the chlorine atom reacts with a water molecule to form hydrochloric acid, which is carried back into the troposphere and then washed out of the atmosphere in rainfall.

The Ozone Hole

Massive ozone depletions were first observed in 1975 over the Antarctic and more recently over the Arctic. Although the reactions in and appear to account for most of the ozone destruction observed at low to middle latitudes, requires intense sunlight to generate chlorine atoms, and sunlight is in very short supply during the polar winters. At high latitudes (near the poles), therefore, a different set of reactions must be responsible for the depletion.

Recent research has shown that, in the absence of oxygen atoms, chlorine monoxide can react with stratospheric nitrogen dioxide in a redox reaction to form chlorine nitrate (ClONO2). When chlorine nitrate is in the presence of trace amounts of HCl or adsorbed on ice particles in stratospheric clouds, additional redox reactions can occur in which chlorine nitrate produces Cl2 or HOCl (hypochlorous acid):

Equation 3.42

HCl(g) + ClONO2(g) → Cl2(g) + HNO3(g)

Equation 3.43

H2O(g) + ClONO2(g) → HOCl(g) + HNO3(g)

Both Cl2 and HOCl undergo cleavage reactions by even weak sunlight to give reactive chlorine atoms. When the sun finally rises after the long polar night, relatively large amounts of Cl2 and HOCl are present and rapidly generate high levels of chlorine atoms. The reactions shown in and then cause ozone levels to fall dramatically.

Stratospheric ozone levels decreased about 2.5% from 1978 to 1988, which coincided with a fivefold increase in the widespread use of CFCs since the 1950s. If the trend were allowed to continue, the results could be catastrophic. Fortunately, many countries have banned the use of CFCs in aerosols. In 1987, representatives from 43 nations signed the Montreal Protocol, committing themselves to reducing CFC emissions by 50% by the year 2000. Later, representatives from a large number of countries, alarmed by data showing the rapid depletion of stratospheric chlorine, agreed to phase out CFCs completely by the early 21st century; the United States banned their use in 1995. The projected effects of these agreements on atmospheric chlorine levels are shown in . Because of the very slow rate at which CFCs are removed from the stratosphere, however, stratospheric chlorine levels will not fall to the level at which the Antarctic ozone hole was first observed until about 2050. The scientific community recognized Molina and Rowland’s work in 1995, when they shared the Nobel Prize in Chemistry.

Figure 3.17 Projected Effects of International Agreements on Atmospheric Chlorine Levels

The graph plots atmospheric chlorine content in chlorine atoms per 109 molecules of O2 plus N2 from 1960 to 1990 (actual data) and 1990 to 2080 (estimated for various schemes for regulating CFC emissions).

Manufacturing companies are now under great political and economic pressure to find alternatives to the CFCs used in the air-conditioning units of cars, houses, and commercial buildings. One approach is to use hydrochlorofluorocarbons (HCFCs), hydrocarbons in which only some of the hydrogen atoms are replaced by chlorine or fluorine, and hydrofluorocarbons (HFCs), which do not contain chlorine (). The C–H bonds in HCFCs and HFCs act as “handles” that permit additional chemical reactions to occur. Consequently, these substances are degraded more rapidly, and most are washed out of the atmosphere before they can reach the stratosphere.

Table 3.4 Selected HCFCs and HFCs

Name Molecular Formula Industrial Name
chlorodifluoromethane CHClF2 HCFC-22 (freon-22)
1-chloro-1,1-difluoroethane CH3CClF2 HCFC-141b
2,2-dichloro-1,1,1-trifluoroethane CHCl2CF3 HCFC-123
1,1,1,2-tetrafluoroethane CH2FCF3 HFC-134a

HFCs are used as a replacement for CFCs. The molecular structure of HFC-134a is shown in this ball-and-stick model.

Nonetheless, the small fraction of HCFCs that reaches the stratosphere will deplete ozone levels just as CFCs do, so they are not the final answer. Indeed, the 1990 London amendment to the Montreal Protocol specifies that HCFCs must be phased out by 2040. Finding a suitable replacement for refrigerants is just one of the challenges facing chemists in the 21st century.

Example 16

Nitric oxide (NO) may also be an important factor in the destruction of the ozone layer. One source of this compound is the combustion of hydrocarbons in jet engines. The fact that high-flying supersonic aircraft inject NO directly into the stratosphere was a major argument against the development of commercial supersonic transports. Do you agree with this decision? Why or why not?

Given: identity of compound

Asked for: assessment of likely role in ozone depletion

Strategy:

Predict what reactions are likely to occur between NO and ozone and then determine whether the reactions are likely to deplete ozone from the atmosphere.

Solution:

Both NO and NO2 are known oxides of nitrogen. Thus NO is likely to react with ozone according to the chemical equation

NO(g) + O3(g) → NO2(g) + O2(g)

resulting in ozone depletion. If NO2(g) also reacts with atomic oxygen according to the equation

NO2(g) + O(g) → NO(g) + O2(g)

then we would have a potential catalytic cycle for ozone destruction similar to that caused by chlorine atoms. Based on these reactions, the development of commercial supersonic transports is not recommended until the environmental impact has undergone additional testing. (Although these reactions have been observed, they do not appear to be a major factor in ozone destruction.)

Exercise

An industrial manufacturer proposed that halons such as CF3Br could be used as replacements for CFC propellants. Do you think that this is a reasonable suggestion or is there a potential problem with such a use?

Answer: Because the compound CF3Br contains carbon, fluorine, and a bromine atom that is chemically similar to chlorine, it is likely that it would also be a catalyst for ozone destruction. There is therefore a potential problem with its use.

Summary

Earth’s atmosphere consists of discrete layers that do not mix readily with one another. The sun emits radiation with a wide range of energies, including visible light, which can be detected by the human eye, and ultraviolet light, which is more energetic than visible light and cannot be detected by the human eye. In the stratosphere, ultraviolet light reacts with O2 molecules to form atomic oxygen. Atomic oxygen then reacts with an O2 molecule to produce ozone (O3). As a result of this reaction, the stratosphere contains an appreciable concentration of ozone molecules that constitutes the ozone layer. The absorption of ultraviolet light in the stratosphere protects Earth’s surface from the sun’s harmful effects. Volatile organic compounds that contain chlorine and fluorine, which are known as chlorofluorocarbons (CFCs), are capable of reaching the stratosphere, where they can react with ultraviolet light to generate chlorine atoms and other chlorine-containing species that catalyze the conversion of ozone to O2, thereby decreasing the amount of O3 in the stratosphere. Replacing chlorofluorocarbons with hydrochlorofluorocarbons (HCFCs) or hydrofluorocarbons (HFCs) is one strategy that has been developed to minimize further damage to Earth’s ozone layer.

Key Takeaway

  • The composition of Earth’s atmosphere is vulnerable to degradation through reactions with common industrial chemicals.

Conceptual Problems

  1. Carbon monoxide is a toxic gas that can be produced from the combustion of wood in wood-burning stoves when excess oxygen is not present. Write a balanced chemical equation showing how carbon monoxide is produced from carbon and suggest what might be done to prevent it from being a reaction product.

  2. Explain why stratospheric ozone depletion has developed over the coldest part of Earth (the poles) and reaches a maximum at the beginning of the polar spring.

  3. What type of reactions produce species that are believed to be responsible for catalytic depletion of ozone in the atmosphere?

Numerical Problem

    Please be sure you are familiar with the topics discussed in Essential Skills 2 () before proceeding to the Numerical Problems.

  1. Sulfur dioxide and hydrogen sulfide are important atmospheric contaminants that have resulted in the deterioration of ancient objects. Sulfur dioxide combines with water to produce sulfurous acid, which then reacts with atmospheric oxygen to produce sulfuric acid. Sulfuric acid is known to attack many metals that were used by ancient cultures. Give the formulas for these four sulfur-containing species. What is the percentage of sulfur in each compound? What is the percentage of oxygen in each?

3.7 Essential Skills 2

Topics

  • Proportions
  • Percentages
  • Unit Conversions

In Essential Skills 1 in , , we introduced you to some of the fundamental mathematical operations you need to successfully manipulate mathematical equations in chemistry. Before proceeding to the problems in , you should become familiar with the additional skills described in this section on proportions, percentages, and unit conversions.

Proportions

We can solve many problems in general chemistry by using ratios, or proportions. For example, if the ratio of some quantity A to some quantity B is known, and the relationship between these quantities is known to be constant, then any change in A (from A1 to A2) produces a proportional change in B (from B1 to B2) and vice versa. The relationship between A1, B1, A2, and B2 can be written as follows:

A1B1=A2B2= constant

To solve this equation for A2, we multiply both sides of the equality by B2, thus canceling B2 from the denominator:

(B2)A1B1=(B2)A2B2B2A1B1=A2

Similarly, we can solve for B2 by multiplying both sides of the equality by 1/A2, thus canceling A2 from the numerator:

(1A2)A1B1=(1A2)A2B2A1A2B1=1B2

If the values of A1, A2, and B1 are known, then we can solve the left side of the equation and invert the answer to obtain B2:

1B2= numerical valueB2=1numerical value

If the value of A1, A2, or B1 is unknown, however, we can solve for B2 by inverting both sides of the equality:

B2=A2B1A1

When you manipulate equations, remember that any operation carried out on one side of the equality must be carried out on the other.

Skill Builder ES1 illustrates how to find the value of an unknown by using proportions.

Skill Builder ES1

If 38.4 g of element A are needed to combine with 17.8 g of element B, then how many grams of element A are needed to combine with 52.3 g of element B?

Solution

We set up the proportions as follows:

A1= 38.4 gB1= 17.8 gA2= ?B2= 52.3 gA1B1=A2B238.4 g17.8 g=A252.3 g

Multiplying both sides of the equation by 52.3 g gives

(38.4 g)(52.3 g)17.8 g=A2(52.3 g)(52.3 g)                      A2= 113 g

Notice that grams cancel to leave us with an answer that is in the correct units. Always check to make sure that your answer has the correct units.

Skill Builder ES2

Solve to find the indicated variable.

  1. 16.4 g41.2 g=x18.3 g
  2. 2.65 m4.02 m=3.28 my
  3. 3.27×103 gx=5.0×101 g3.2 g
  4. Solve for V1: P1P2=V2V1
  5. Solve for T1: P1V1T1=P2V2T2

Solution

  1. Multiply both sides of the equality by 18.3 g to remove this measurement from the denominator:

    (18.3 g)16.4 g41.2 g=(18.3 g)x18.3 g              7.28 g =x
  2. Multiply both sides of the equality by 1/3.28 m, solve the left side of the equation, and then invert to solve for y:

    (13.28 m)2.65 m4.02 m=(13.28 m)3.28 my=1y        y=(4.02)(3.28)2.65=4.98 m
  3. Multiply both sides of the equality by 1/3.27 × 10−3 g, solve the right side of the equation, and then invert to find x:

    (13.27×103 g)3.27×103 gx=(13.27×103 g)5.0×101 g3.2 g=1x                                             x=(3.2 g)(3.27×103 g)5.0×101 g=2.1×102 g
  4. Multiply both sides of the equality by 1/V2, and then invert both sides to obtain V1:

    (1V2)P1P2=(1V2)V2V1   P2V2P1=V1
  5. Multiply both sides of the equality by 1/P1V1 and then invert both sides to obtain T1:

    (1P1V1)P1V1T1=(1P1V1)P2V2T2             1T1=P2V2T2P1V1              T1=T2P1V1P2V2

Percentages

Because many measurements are reported as percentages, many chemical calculations require an understanding of how to manipulate such values. You may, for example, need to calculate the mass percentage of a substance, as described in , or determine the percentage of product obtained from a particular reaction mixture.

You can convert a percentage to decimal form by dividing the percentage by 100:

52.8% = 52.8100 = 0.528

Conversely, you can convert a decimal to a percentage by multiplying the decimal by 100:

0.356 × 100 = 35.6%

Suppose, for example, you want to determine the mass of substance A, one component of a sample with a mass of 27 mg, and you are told that the sample consists of 82% A. You begin by converting the percentage to decimal form:

82% = 82100 = 0.82

The mass of A can then be calculated from the mass of the sample:

0.82 × 27 mg = 22 mg

Skill Builder ES3 provides practice in converting and using percentages.

Skill Builder ES3

Convert each number to a percentage or a decimal.

  1. 29.4%
  2. 0.390
  3. 101%
  4. 1.023

Solution

  1. 29.4100 = 0.294
  2. 0.390 × 100 = 39.0%
  3. 101100 = 1.01
  4. 1.023 × 100 = 102.3%

Skill Builder ES4

Use percentages to answer the following questions, being sure to use the correct number of significant figures (see Essential Skills 1 in , ). Express your answer in scientific notation where appropriate.

  1. What is the mass of hydrogen in 52.83 g of a compound that is 11.2% hydrogen?
  2. What is the percentage of carbon in 28.4 g of a compound that contains 13.79 g of that element?
  3. A compound that is 4.08% oxygen contains 194 mg of that element. What is the mass of the compound?

Solution

  1. 52.83 g×11.2100= 52.83 g × 0.112 = 5.92 g 
  2. 13.79 g carbon28.4 g×100=48.6% carbon
  3. This problem can be solved by using a proportion:

    4.08% oxygen100% compound=194 mgx mgx= 4.75×103 mg (or 4.75 g)

Unit Conversions

As you learned in Essential Skills 1 in , , all measurements must be expressed in the correct units to have any meaning. This sometimes requires converting between different units (). Conversions are carried out using conversion factors, which are are ratios constructed from the relationships between different units or measurements. The relationship between milligrams and grams, for example, can be expressed as either 1 g/1000 mg or 1000 mg/1 g. When making unit conversions, use arithmetic steps accompanied by unit cancellation.

Suppose you have measured a mass in milligrams but need to report the measurement in kilograms. In problems that involve SI units, you can use the definitions of the prefixes given in to get the necessary conversion factors. For example, you can convert milligrams to grams and then convert grams to kilograms:

milligrams → grams → kilograms 1000 mg → 1 g 1000 g → 1 kilogram

If you have measured 928 mg of a substance, you can convert that value to kilograms as follows:

928 mg×1 g1000 mg= 0.928 g0.928 g×1 kg1000 g= 0.000928 kg = 9.28×104 kg

In each arithmetic step, the units cancel as if they were algebraic variables, leaving us with an answer in kilograms. In the conversion to grams, we begin with milligrams in the numerator. Milligrams must therefore appear in the denominator of the conversion factor to produce an answer in grams. The individual steps may be connected as follows:

928 mg×g1000 mg×1 kg1000 g=928 kg106= 928×106 kg = 9.28×104 kg

Skill Builder ES5 provides practice converting between units.

Skill Builder ES5

Use the information in to convert each measurement. Be sure that your answers contain the correct number of significant figures and are expressed in scientific notation where appropriate.

  1. 59.2 cm to decimeters
  2. 3.7 × 105 mg to kilograms
  3. 270 mL to cubic decimeters
  4. 2.04 × 103 g to tons
  5. 9.024 × 1010 s to years

Solution

  1. 59.2 cm×m100 cm×10 dmm= 5.92 dm
  2. 3.7×105 mg×g1000 mg×1 kg1000 g= 3.7×101 kg
  3. 270 mL×L1000 mL×1 dm3L= 270×103 dm3= 2.70×101 dm3
  4. 2.04×103 g×lb453.6 g×1 tn2000 lb= 0.00225 tn = 2.25×103 tn
  5. 9.024×1010 s×min60 s×h60 min×d24 h×1 yr365 d= 2.86 × 103 yr

3.8 End-of-Chapter Material

Application Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 2 () before proceeding to the Application Problems. Problems marked with a ♦ involve multiple concepts.

  1. Hydrogen sulfide is a noxious and toxic gas produced from decaying organic matter that contains sulfur. A lethal concentration in rats corresponds to an inhaled dose of 715 molecules per million molecules of air. How many molecules does this correspond to per mole of air? How many moles of hydrogen sulfide does this correspond to per mole of air?

  2. Bromine, sometimes produced from brines (salt lakes) and ocean water, can be used for bleaching fibers and silks. How many moles of bromine atoms are found in 8.0 g of molecular bromine (Br2)?

  3. Paris yellow is a lead compound that is used as a pigment; it contains 16.09% chromium, 19.80% oxygen, and 64.11% lead. What is the empirical formula of Paris yellow?

  4. A particular chromium compound used for dyeing and waterproofing fabrics has the elemental composition 18.36% chromium, 13.81% potassium, 45.19% oxygen, and 22.64% sulfur. What is the empirical formula of this compound?

  5. Compounds with aluminum and silicon are commonly found in the clay fractions of soils derived from volcanic ash. One of these compounds is vermiculite, which is formed in reactions caused by exposure to weather. Vermiculite has the following formula: Ca0.7[Si6.6Al1.4]Al4O20(OH)4. (The content of calcium, silicon, and aluminum are not shown as integers because the relative amounts of these elements vary from sample to sample.) What is the mass percent of each element in this sample of vermiculite?

  6. ♦ Pheromones are chemical signals secreted by a member of one species to evoke a response in another member of the same species. One honeybee pheromone is an organic compound known as an alarm pheromone, which smells like bananas. It induces an aggressive attack by other honeybees, causing swarms of angry bees to attack the same aggressor. The composition of this alarm pheromone is 64.58% carbon, 10.84% hydrogen, and 24.58% oxygen by mass, and its molecular mass is 130.2 amu.

    1. Calculate the empirical formula of this pheromone.
    2. Determine its molecular formula.
    3. Assuming a honeybee secretes 1.00 × 10−11 g of pure pheromone, how many molecules of pheromone are secreted?
  7. Amoxicillin is a prescription drug used to treat a wide variety of bacterial infections, including infections of the middle ear and the upper and lower respiratory tracts. It destroys the cell walls of bacteria, which causes them to die. The elemental composition of amoxicillin is 52.59% carbon, 5.24% hydrogen, 11.50% nitrogen, 21.89% oxygen, and 8.77% sulfur by mass. What is its empirical formula?

  8. Monosodium glutamate (MSG; molar mass = 169 g/mol), is used as a flavor enhancer in food preparation. It is known to cause headaches and chest pains in some individuals, the so-called Chinese food syndrome. Its composition was found to be 35.51% carbon, 4.77% hydrogen, 8.28% nitrogen, and 13.59% sodium by mass. If the “missing” mass is oxygen, what is the empirical formula of MSG?

  9. Ritalin is a mild central nervous system stimulant that is prescribed to treat attention deficit disorders and narcolepsy (an uncontrollable desire to sleep). Its chemical name is methylphenidate hydrochloride, and its empirical formula is C14H20ClNO2. If you sent a sample of this compound to a commercial laboratory for elemental analysis, what results would you expect for the mass percentages of carbon, hydrogen, and nitrogen?

  10. Fructose, a sugar found in fruit, contains only carbon, oxygen, and hydrogen. It is used in ice cream to prevent a sandy texture. Complete combustion of 32.4 mg of fructose in oxygen produced 47.6 mg of CO2 and 19.4 mg of H2O. What is the empirical formula of fructose?

  11. Coniine, the primary toxin in hemlock, contains only carbon, nitrogen, and hydrogen. When ingested, it causes paralysis and eventual death. Complete combustion of 28.7 mg of coniine produced 79.4 mg of CO2 and 34.4 mg of H2O. What is the empirical formula of the coniine?

  12. Copper and tin alloys (bronzes) with a high arsenic content were presumably used by Bronze Age metallurgists because bronze produced from arsenic-rich ores had superior casting and working properties. The compositions of some representative bronzes of this type are as follows:

    Origin % Composition
    Cu As
    Dead Sea 87.0 12.0
    Central America 90.7 3.8

    If ancient metallurgists had used the mineral As2S3 as their source of arsenic, how much As2S3 would have been required to process 100 g of cuprite (Cu2O) bronzes with these compositions?

  13. ♦ The phrase mad as a hatter refers to mental disorders caused by exposure to mercury(II) nitrate in the felt hat manufacturing trade during the 18th and 19th centuries. An even greater danger to humans, however, arises from alkyl derivatives of mercury.

    1. Give the empirical formula of mercury(II) nitrate.
    2. One alkyl derivative, dimethylmercury, is a highly toxic compound that can cause mercury poisoning in humans. How many molecules are contained in a 5.0 g sample of dimethylmercury?
    3. What is the percentage of mercury in the sample?
  14. Magnesium carbonate, aluminum hydroxide, and sodium bicarbonate are commonly used as antacids. Give the empirical formulas and determine the molar masses of these compounds. Based on their formulas, suggest another compound that might be an effective antacid.

  15. ♦ Nickel(II) acetate, lead(II) phosphate, zinc nitrate, and beryllium oxide have all been reported to induce cancers in experimental animals.

    1. Give the empirical formulas for these compounds.
    2. Calculate their formula masses.
    3. Based on the location of cadmium in the periodic table, would you predict that cadmium chloride might also induce cancer?
  16. ♦ Methane, the major component of natural gas, is found in the atmospheres of Jupiter, Saturn, Uranus, and Neptune.

    1. What is the structure of methane?
    2. Calculate the molecular mass of methane.
    3. Calculate the mass percentage of both elements present in methane.
  17. Sodium saccharin, which is approximately 500 times sweeter than sucrose, is frequently used as a sugar substitute. What are the percentages of carbon, oxygen, and sulfur in this artificial sweetener?

  18. Lactic acid, found in sour milk, dill pickles, and sauerkraut, has the functional groups of both an alcohol and a carboxylic acid. The empirical formula for this compound is CH2O, and its molar mass is 90 g/mol. If this compound were sent to a laboratory for elemental analysis, what results would you expect for carbon, hydrogen, and oxygen content?

  19. The compound 2-nonenal is a cockroach repellant that is found in cucumbers, watermelon, and carrots. Determine its molecular mass.

  20. You have obtained a 720 mg sample of what you believe to be pure fructose, although it is possible that the sample has been contaminated with formaldehyde. Fructose and formaldehyde both have the empirical formula CH2O. Could you use the results from combustion analysis to determine whether your sample is pure?

  21. ♦ The booster rockets in the space shuttles used a mixture of aluminum metal and ammonium perchlorate for fuel. Upon ignition, this mixture can react according to the chemical equation

    Al(s) + NH4ClO4(s) → Al2O3(s) + AlCl3(g) + NO(g) + H2O(g)

    Balance the equation and construct a table showing how to interpret this information in terms of the following:

    1. numbers of individual atoms, molecules, and ions
    2. moles of reactants and products
    3. grams of reactants and products
    4. numbers of molecules of reactants and products given 1 mol of aluminum metal
  22. ♦ One of the byproducts of the manufacturing of soap is glycerol. In 1847, it was discovered that the reaction of glycerol with nitric acid produced nitroglycerin according to the following unbalanced chemical equation:

    Nitroglycerine is both an explosive liquid and a blood vessel dilator that is used to treat a heart condition known as angina.

    1. Balance the chemical equation and determine how many grams of nitroglycerine would be produced from 15.00 g of glycerol.
    2. If 9.00 g of nitric acid had been used in the reaction, which would be the limiting reactant?
    3. What is the theoretical yield in grams of nitroglycerin?
    4. If 9.3 g of nitroglycerin was produced from 9.0 g of nitric acid, what would be the percent yield?
    5. Given the data in part d, how would you rate the success of this reaction according to the criteria mentioned in this chapter?
    6. Derive a general expression for the theoretical yield of nitroglycerin in terms of x grams of glycerol.
  23. ♦ A significant weathering reaction in geochemistry is hydration–dehydration. An example is the transformation of hematite (Fe2O3) to ferrihydrite (Fe10O15·9H2O) as the relative humidity of the soil approaches 100%:

    Fe2O3(s) + H2O(l) → Fe10O15·9H2O(s)

    This reaction occurs during advanced stages of the weathering process.

    1. Balance the chemical equation.
    2. Is this a redox reaction? Explain your answer.
    3. If 1 tn of hematite rock weathered in this manner, how many kilograms of ferrihydrite would be formed?
  24. ♦ Hydrazine (N2H4) is used not only as a rocket fuel but also in industry to remove toxic chromates from waste water according to the following chemical equation:

    4CrO42−(aq) + 3N2H4(l) + 4H2O(l) → 4Cr(OH)3(s) + 3N2(g) + 8OH(aq)

    Identify the species that is oxidized and the species that is reduced. What mass of water is needed for the complete reaction of 15.0 kg of hydrazine? Write a general equation for the mass of chromium(III) hydroxide [Cr(OH)3] produced from x grams of hydrazine.

  25. Corrosion is a term for the deterioration of metals through chemical reaction with their environment. A particularly difficult problem for the archaeological chemist is the formation of CuCl, an unstable compound that is formed by the corrosion of copper and its alloys. Although copper and bronze objects can survive burial for centuries without significant deterioration, exposure to air can cause cuprous chloride to react with atmospheric oxygen to form Cu2O and cupric chloride. The cupric chloride then reacts with the free metal to produce cuprous chloride. Continued reaction of oxygen and water with cuprous chloride causes “bronze disease,” which consists of spots of a pale green, powdery deposit of [CuCl2·3Cu(OH)2·H2O] on the surface of the object that continues to grow. Using this series of reactions described, complete and balance the following equations, which together result in bronze disease:

    Equation 1: ___ + O2 → ___ + ___

    Equation 2: ___ + Cu → ___

    Equation 3: ___ + O2 + H2O → CuCl2·3Cu(OH)2·H2Obronze disease+CuCl2

    1. Which species are the oxidants and the reductants in each equation?
    2. If 8.0% by mass of a 350.0 kg copper statue consisted of CuCl, and the statue succumbed to bronze disease, how many pounds of the powdery green hydrate would be formed?
    3. What factors could affect the rate of deterioration of a recently excavated bronze artifact?
  26. ♦ Iron submerged in seawater will react with dissolved oxygen, but when an iron object, such as a ship, sinks into the seabed where there is little or no free oxygen, the iron remains fresh until it is brought to the surface. Even in the seabed, however, iron can react with salt water according to the following unbalanced chemical equation:

    Fe(s) + NaCl(aq) + H2O(l) → FeCl2(s) + NaOH(aq) + H2(g)

    The ferrous chloride and water then form hydrated ferrous chloride according to the following equation:

    FeCl2(s) + 2H2O(l) → FeCl2·2H2O(s)

    When the submerged iron object is removed from the seabed, the ferrous chloride dihydrate reacts with atmospheric moisture to form a solution that seeps outward, producing a characteristic “sweat” that may continue to emerge for many years. Oxygen from the air oxidizes the solution to ferric resulting in the formation of what is commonly referred to as rust (ferric oxide):

    FeCl2(aq) + O2(g) → FeCl3(aq) + Fe2O3(s)

    The rust layer will continue to grow until arrested.

    1. Balance each chemical equation.
    2. Given a 10.0 tn ship of which 2.60% is now rust, how many kilograms of iron were converted to FeCl2, assuming that the ship was pure iron?
    3. What mass of rust in grams would result?
    4. What is the overall change in the oxidation state of iron for this process?
    5. In the first equation given, what species has been reduced? What species has been oxidized?
  27. ♦ The glass industry uses lead oxide in the production of fine crystal glass, such as crystal goblets. Lead oxide can be formed by the following reaction:

    PbS(s) + O2(g) → PbO(s) + SO2(g)

    Balance the equation and determine what has been oxidized and what has been reduced. How many grams of sulfur dioxide would be produced from 4.0 × 103 g of lead sulfide? Discuss some potential environmental hazards that stem from this reaction.

  28. ♦ The Deacon process is one way to recover Cl2 on-site in industrial plants where the chlorination of hydrocarbons produces HCl. The reaction uses oxygen to oxidize HCl to chlorine, as shown.

    HCl(g) + O2(g) → Cl2(g) + H2O(g)

    The reaction is frequently carried out in the presence of NO as a catalyst.

    1. Balance the chemical equation.
    2. Which compound is the oxidant, and which is the reductant?
    3. If 26 kg of HCl was produced during a chlorination reaction, how many kilograms of water would result from the Deacon process?
  29. In 1834, Eilhardt Mitscherlich of the University of Berlin synthesized benzene (C6H6) by heating benzoic acid (C6H5COOH) with calcium oxide according to this balanced chemical equation:

    C6H5COOH(s) + CaO(s)ΔC6H6(l)+CaCO3(s)

    (Heating is indicated by the symbol Δ.) How much benzene would you expect from the reaction of 16.9 g of benzoic acid and 18.4 g of calcium oxide? Which is the limiting reactant? How many grams of benzene would you expect to obtain from this reaction, assuming a 73% yield?

  30. Aspirin (C9H8O4) is synthesized by the reaction of salicylic acid (C7H6O3) with acetic anhydride (C4H6O3) according to the following equation:

    C7H6O3(s) + C4H6O3(l) → C9H8O4(s) + H2O(l)

    Balance the equation and find the limiting reactant given 10.0 g of acetic anhydride and 8.0 g of salicylic acid. How many grams of aspirin would you expect from this reaction, assuming an 83% yield?

  31. ♦ Hydrofluoric acid etches glass because it dissolves silicon dioxide, as represented in the following chemical equation:

    SiO2(s) + HF(aq) → SiF62−(aq) + H+(aq) + H2O(l)
    1. Balance the equation.
    2. How many grams of silicon dioxide will react with 5.6 g of HF?
    3. How many grams of HF are needed to remove 80% of the silicon dioxide from a 4.0 kg piece of glass? (Assume that the glass is pure silicon dioxide.)
  32. ♦ Lead sulfide and hydrogen peroxide react to form lead sulfate and water. This reaction is used to clean oil paintings that have blackened due to the reaction of the lead-based paints with atmospheric hydrogen sulfide.

    1. Write the balanced chemical equation for the oxidation of lead sulfide by hydrogen peroxide.
    2. What mass of hydrogen peroxide would be needed to remove 3.4 g of lead sulfide?
    3. If the painting had originally been covered with 5.4 g of lead sulfide and you had 3.0 g of hydrogen peroxide, what percent of the lead sulfide could be removed?
  33. ♦ It has been suggested that diacetylene (C4H2, HC≡C–C≡CH) may be the ozone of the outer planets. As the largest hydrocarbon yet identified in planetary atmospheres, diacetylene shields planetary surfaces from ultraviolet radiation and is itself reactive when exposed to light. One reaction of diacetylene is an important route for the formation of higher hydrocarbons, as shown in the following chemical equations:

    C4H2(g) + C4H2(g) → C8H3(g) + H(g) C8H3(g) + C4H2(g) → C10H3(g) + C2H2(g)

    Consider the second reaction.

    1. Given 18.4 mol of C8H3 and 1000 g of C4H2, which is the limiting reactant?
    2. Given 2.8 × 1024 molecules of C8H3 and 250 g of C4H2, which is the limiting reactant?
    3. Given 385 g of C8H3 and 200 g of C4H2, which is in excess? How many grams of excess reactant would remain?
    4. Suggest why this reaction might be of interest to scientists.
  34. ♦ Glucose (C6H12O6) can be converted to ethanol and carbon dioxide using certain enzymes. As alcohol concentrations are increased, however, catalytic activity is inhibited, and alcohol production ceases.

    1. Write a balanced chemical equation for the conversion of glucose to ethanol and carbon dioxide.
    2. Given 12.6 g of glucose, how many grams of ethanol would be produced, assuming complete conversion?
    3. If 4.3 g of ethanol had been produced, what would be the percent yield for this reaction?
    4. Is a heterogeneous catalyst or a homogeneous catalyst used in this reaction?
    5. You have been asked to find a way to increase the rate of this reaction given stoichiometric quantities of each reactant. How would you do this?
  35. Early spacecraft developed by the National Aeronautics and Space Administration for its manned missions used capsules that had a pure oxygen atmosphere. This practice was stopped when a spark from an electrical short in the wiring inside the capsule of the Apollo 1 spacecraft ignited its contents. The resulting explosion and fire killed the three astronauts on board within minutes. What chemical steps could have been taken to prevent this disaster?

Answers

  1. 4.31 × 1020 molecules, 7.15 × 10−4

  2. PbCrO4

  3. To two decimal places, the percentages are: H: 0.54%; O: 51.39%; Al: 19.50%; Si: 24.81%; Ca: 3.75%

  4. C16H19O5N3S

    1. Hg(NO3)2
    2. 1.3 × 1022 molecules
    3. 86.96% mercury by mass.
    1. Ni(O2CCH3)2; Pb3(PO4)2; Zn(NO3)2; BeO
    2. To four significant figures, the values are: Ni(O2CCH3)2, 176.8 amu; Pb3(PO4)2, 811.5 amu; Zn(NO3)2, 189.4 amu; BeO, 25.01 amu.
    3. Yes.
  5. C, 40.98%; O, 23.39%; S, 15.63%

  6. 140.22 amu

  7.  

    3Al(s) + 3NH4ClO4(s) → Al2O3(s) + AlCl3(g) + 3NO(g) + 6H2O(g)

    3Al 3NH4ClO4 Al2O3
    a. 3 atoms 30 atoms, 6 ions 5 atoms
    b. 3 mol 3 mol 1 mol
    c. 81 g 352 g 102 g
    d. 6 × 1023 6 × 1023 2 × 1023
    AlCl3 3NO 6H2O
    a. 4 atoms, 1 molecule 6 atoms, 3 molecules 18 atoms, 6 molecules
    b. 1 mol 3 mol 6 mol
    c. 133 g 90 g 108 g
    d. 2 × 1023 6 × 1023 1.2 × 1022
    1. 5Fe2O3 + 9H2O → Fe10O15 · 9H2O
    2. No.
    3. 1090 kg
  8.  

    Equation 1: 8CuCl + O2 → 2Cu2O + 4CuCl2

    Equation 2: CuCl2 + Cu → 2CuCl Equation 3: 12CuCl + 3O2 + 8H2O → 2[CuCl2 3Cu(OH)2 H2O] + 4CuCl2
    1. Equation 1: Oxygen is the oxidant, and CuCl is the reductant. Equation 2: Copper is the reductant, and copper(II) chloride is the oxidant. Equation 3: Copper(I) chloride is the reductant, and oxygen is the oxidant.
    2. 46 pounds
    3. temperature, humidity, and wind (to bring more O2 into contact with the statue)
  9. 2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g) Sulfur in PbS has been oxidized, and oxygen has been reduced. 1.1 × 103 g SO2 is produced. Lead is a toxic metal. Sulfur dioxide reacts with water to make acid rain.

  10. 10.8 g benzene; limiting reactant is benzoic acid; 7.9 g of benzene

    1. SiO2 + 6HF → SiF62− + 2H+ + 2H2O
    2. 2.8 g
    3. 6400 g HF
    1. C8H3
    2. C8H3
    3. C4H2; 6.0 g
    4. Complex molecules are essential for life. Reactions that help block UV may have implications regarding life on other planets.
  11. The disaster occurred because organic compounds are highly flammable in a pure oxygen atmosphere. Using a mixture of 20% O2 and an inert gas such as N2 or He would have prevented the disaster.

Chapter 4 Reactions in Aqueous Solution

In Chapter 3 "Chemical Reactions", we described chemical reactions in general and introduced some techniques that are used to characterize them quantitatively. For the sake of simplicity, we discussed situations in which the reactants and the products of a given reaction were the only chemical species present. In reality, however, virtually every chemical reaction that takes place within and around us, such as the oxidation of foods to generate energy or the treatment of an upset stomach with an antacid tablet, occur in solution. In fact, many reactions must be carried out in solution and do not take place at all if the solid reactants are simply mixed.

The reaction of mercury(II) acetate with sodium iodide. When colorless aqueous solutions of each reactant are mixed, they produce a red precipitate, mercury(II) iodide, which is the result of an exchange reaction.

As you learned in Chapter 1 "Introduction to Chemistry", a solutionA homogeneous mixture of two or more substances in which solutes are dispersed uniformly throughout the solvent. is a homogeneous mixture in which substances present in lesser amounts, called solutesThe substance or substances present in lesser amounts in a solution., are dispersed uniformly throughout the substance in the greater amount, the solventThe substance present in the greater amount in a solution.. An aqueous solutionA solution in which water is the solvent. is a solution in which the solvent is water, whereas in a nonaqueous solution, the solvent is a substance other than water. Familiar examples of nonaqueous solvents are ethyl acetate, used in nail polish removers, and turpentine, used to clean paint brushes. In this chapter, we focus on reactions that occur in aqueous solution.

There are many reasons for carrying out reactions in solution. For a chemical reaction to occur, individual atoms, molecules, or ions must collide, and collisions between two solids, which are not dispersed at the atomic, molecular, or ionic level, do not occur at a significant rate. In addition, when the amount of a substance required for a reaction is so small that it cannot be weighed accurately, using a solution of that substance, in which the solute is dispersed in a much larger mass of solvent, enables chemists to measure its quantity with great precision. Chemists can also more effectively control the amount of heat consumed or produced in a reaction when the reaction occurs in solution, and sometimes the nature of the reaction itself can be controlled by the choice of solvent.

This chapter introduces techniques for preparing and analyzing aqueous solutions, for balancing equations that describe reactions in solution, and for solving problems using solution stoichiometry. By the time you complete this chapter, you will know enough about aqueous solutions to explain what causes acid rain, why acid rain is harmful, and how a Breathalyzer measures alcohol levels. You will also understand the chemistry of photographic development, be able to explain why rhubarb leaves are toxic, and learn about a possible chemical reason for the decline and fall of the Roman Empire.

4.1 Aqueous Solutions

Learning Objective

  1. To understand how and why solutions form.

The solvent in aqueous solutions is water, which makes up about 70% of the mass of the human body and is essential for life. Many of the chemical reactions that keep us alive depend on the interaction of water molecules with dissolved compounds. Moreover, as we will discuss in , the presence of large amounts of water on Earth’s surface helps maintain its surface temperature in a range suitable for life. In this section, we describe some of the interactions of water with various substances and introduce you to the characteristics of aqueous solutions.

Polar Substances

As shown in , the individual water molecule consists of two hydrogen atoms bonded to an oxygen atom in a bent (V-shaped) structure. As is typical of group 16 elements, the oxygen atom in each O–H covalent bond attracts electrons more strongly than the hydrogen atom does. (For more information on periodic table groups and covalent bonding, see and .) Consequently, the oxygen and hydrogen nuclei do not equally share electrons. Instead, hydrogen atoms are electron poor compared with a neutral hydrogen atom and have a partial positive charge, which is indicated by δ+. The oxygen atom, in contrast, is more electron rich than a neutral oxygen atom, so it has a partial negative charge. This charge must be twice as large as the partial positive charge on each hydrogen for the molecule to have a net charge of zero. Thus its charge is indicated by 2δ. This unequal distribution of charge creates a polar bondA chemical bond in which there is an unequal distribution of charge between the bonding atoms., in which one portion of the molecule carries a partial negative charge, while the other portion carries a partial positive charge (). Because of the arrangement of polar bonds in a water molecule, water is described as a polar substance.

Figure 4.1 The Polar Nature of Water

Each water molecule consists of two hydrogen atoms bonded to an oxygen atom in a bent (V-shaped) structure. Because the oxygen atom attracts electrons more strongly than the hydrogen atoms do, the oxygen atom is partially negatively charged (2δ; blue) and the hydrogen atoms are partially positively charged (δ+; red). For the molecule to have a net charge of zero, the partial negative charge on oxygen must be twice as large as the partial positive charge on each hydrogen.

Because of the asymmetric charge distribution in the water molecule, adjacent water molecules are held together by attractive electrostatic (δ+…δ) interactions between the partially negatively charged oxygen atom of one molecule and the partially positively charged hydrogen atoms of adjacent molecules (). Energy is needed to overcome these electrostatic attractions. In fact, without them, water would evaporate at a much lower temperature, and neither Earth’s oceans nor we would exist!

Figure 4.2 The Structure of Liquid Water

Two views of a water molecule are shown: (a) a ball-and-stick structure and (b) a space-filling model. Water molecules are held together by electrostatic attractions (dotted lines) between the partially negatively charged oxygen atom of one molecule and the partially positively charged hydrogen atoms on adjacent molecules. As a result, the water molecules in liquid water form transient networks with structures similar to that shown. Because the interactions between water molecules are continually breaking and reforming, liquid water does not have a single fixed structure.

As you learned in , ionic compounds such as sodium chloride (NaCl) are also held together by electrostatic interactions—in this case, between oppositely charged ions in the highly ordered solid, where each ion is surrounded by ions of the opposite charge in a fixed arrangement. In contrast to an ionic solid, the structure of liquid water is not completely ordered because the interactions between molecules in a liquid are constantly breaking and reforming.

The unequal charge distribution in polar liquids such as water makes them good solvents for ionic compounds. When an ionic solid dissolves in water, the ions dissociate. That is, the partially negatively charged oxygen atoms of the H2O molecules surround the cations (Na+ in the case of NaCl), and the partially positively charged hydrogen atoms in H2O surround the anions (Cl; ). Individual cations and anions that are each surrounded by their own shell of water molecules are called hydrated ionsIndividual cations and anions that are each surrounded by their own shell of water molecules.. We can describe the dissolution of NaCl in water as

Equation 4.1

NaCl(s)H2O(l)Na+(aq)+Cl(aq)

where (aq) indicates that Na+ and Cl are hydrated ions.

Figure 4.3 The Dissolution of Sodium Chloride in Water

An ionic solid such as sodium chloride dissolves in water because of the electrostatic attraction between the cations (Na+) and the partially negatively charged oxygen atoms of water molecules, and between the anions (Cl) and the partially positively charged hydrogen atoms of water.

Note the Pattern

Polar liquids are good solvents for ionic compounds.

Electrolytes

When electricity, in the form of an electrical potential, is applied to a solution, ions in solution migrate toward the oppositely charged rod or plate to complete an electrical circuit, whereas neutral molecules in solution do not (). Thus solutions that contain ions conduct electricity, while solutions that contain only neutral molecules do not. Electrical current will flow through the circuit shown in and the bulb will glow only if ions are present. The lower the concentration of ions in solution, the weaker the current and the dimmer the glow. Pure water, for example, contains only very low concentrations of ions, so it is a poor electrical conductor.

Note the Pattern

Solutions that contain ions conduct electricity.

Figure 4.4 The Effect of Ions on the Electrical Conductivity of Water

An electrical current will flow and light the bulb only if the solution contains ions. (a) Pure water or an aqueous solution of a nonelectrolyte allows almost no current to flow, and the bulb does not light. (b) A weak electrolyte produces a few ions, allowing some current to flow and the bulb to glow dimly. (c) A strong electrolyte produces many ions, allowing more current to flow and the bulb to shine brightly.

An electrolyteAny compound that can form ions when dissolved in water (c.f. nonelectrolytes). Electrolytes may be strong or weak. is any compound that can form ions when it dissolves in water. When strong electrolytesAn electrolyte that dissociates completely into ions when dissolved in water, thus producing an aqueous solution that conducts electricity very well. dissolve, the constituent ions dissociate completely due to strong electrostatic interactions with the solvent, producing aqueous solutions that conduct electricity very well (). Examples include ionic compounds such as barium chloride (BaCl2) and sodium hydroxide (NaOH), which are both strong electrolytes and dissociate as follows:

Equation 4.2

BaCl2(s)H2O(l)Ba2+(aq)+2Cl(aq)

Equation 4.3

NaOH(s)H2O(l)Na+(aq)+OH(aq)

The single arrows from reactant to products in and indicate that dissociation is complete.

When weak electrolytesA compound that produces relatively few ions when dissolved in water, thus producing an aqueous solution that conducts electricity poorly. dissolve, they produce relatively few ions in solution. This does not mean that the compounds do not dissolve readily in water; many weak electrolytes contain polar bonds and are therefore very soluble in a polar solvent such as water. They do not completely dissociate to form ions, however, because of their weaker electrostatic interactions with the solvent. Because very few of the dissolved particles are ions, aqueous solutions of weak electrolytes do not conduct electricity as well as solutions of strong electrolytes. One such compound is acetic acid (CH3CO2H), which contains the –CO2H unit. Although it is soluble in water, it is a weak acid and therefore also a weak electrolyte. Similarly, ammonia (NH3) is a weak base and therefore a weak electrolyte. The behavior of weak acids and weak bases will be described in more detail when we discuss acid–base reactions in .

NonelectrolytesA substance that dissolves in water to form neutral molecules and has essentially no effect on electrical conductivity. that dissolve in water do so as neutral molecules and thus have essentially no effect on conductivity. Examples of nonelectrolytes that are very soluble in water but that are essentially nonconductive are ethanol, ethylene glycol, glucose, and sucrose, all of which contain the –OH group that is characteristic of alcohols. In , we will discuss why alcohols and carboxylic acids behave differently in aqueous solution; for now, however, you can simply look for the presence of the –OH and –CO2H groups when trying to predict whether a substance is a strong electrolyte, a weak electrolyte, or a nonelectrolyte. In addition to alcohols, two other classes of organic compounds that are nonelectrolytes are aldehydesA class of organic compounds that has the general form RCHO, in which the carbon atom of the carbonyl group is bonded to a hydrogen atom and an R group. The R group may be either another hydrogen atom or an alkyl group (c.f. ketone). and ketonesA class of organic compounds with the general form RC(O)R’, in which the carbon atom of the carbonyl group is bonded to two alkyl groups (c.f. aldehyde). The alkyl groups may be the same or different., whose general structures are shown here. The distinctions between soluble and insoluble substances and between strong, weak, and nonelectrolytes are illustrated in .

Note the Pattern

Ionic substances and carboxylic acids are electrolytes; alcohols, aldehydes, and ketones are nonelectrolytes.

General structure of an aldehyde and a ketone. Notice that both contain the C=O group.

Figure 4.5 The Difference between Soluble and Insoluble Compounds (a) and Strong, Weak, and Nonelectrolytes (b)

When a soluble compound dissolves, its constituent atoms, molecules, or ions disperse throughout the solvent. In contrast, the constituents of an insoluble compound remain associated with one another in the solid. A soluble compound is a strong electrolyte if it dissociates completely into ions, a weak electrolyte if it dissociates only slightly into ions, and a nonelectrolyte if it dissolves to produce only neutral molecules.

Example 1

Predict whether each compound is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in water.

  1. formaldehyde

  2. cesium chloride

Given: compound

Asked for: relative ability to form ions in water

Strategy:

A Classify the compound as ionic or covalent.

B If the compound is ionic and dissolves, it is a strong electrolyte that will dissociate in water completely to produce a solution that conducts electricity well. If the compound is covalent and organic, determine whether it contains the carboxylic acid group. If the compound contains this group, it is a weak electrolyte. If not, it is a nonelectrolyte.

Solution:

  1. A Formaldehyde is an organic compound, so it is covalent. B It contains an aldehyde group, not a carboxylic acid group, so it should be a nonelectrolyte.
  2. A Cesium chloride (CsCl) is an ionic compound that consists of Cs+ and Cl ions. B Like virtually all other ionic compounds that are soluble in water, cesium chloride will dissociate completely into Cs+(aq) and Cl(aq) ions. Hence it should be a strong electrolyte.

Exercise

Predict whether each compound is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in water.

  1. (CH3)2CHOH (2-propanol)

  2. ammonium sulfate

Answer:

  1. nonelectrolyte
  2. strong electrolyte

Summary

Most chemical reactions are carried out in solutions, which are homogeneous mixtures of two or more substances. In a solution, a solute (the substance present in the lesser amount) is dispersed in a solvent (the substance present in the greater amount). Aqueous solutions contain water as the solvent, whereas nonaqueous solutions have solvents other than water.

Polar substances, such as water, contain asymmetric arrangements of polar bonds, in which electrons are shared unequally between bonded atoms. Polar substances and ionic compounds tend to be most soluble in water because they interact favorably with its structure. In aqueous solution, dissolved ions become hydrated; that is, a shell of water molecules surrounds them.

Substances that dissolve in water can be categorized according to whether the resulting aqueous solutions conduct electricity. Strong electrolytes dissociate completely into ions to produce solutions that conduct electricity well. Weak electrolytes produce a relatively small number of ions, resulting in solutions that conduct electricity poorly. Nonelectrolytes dissolve as uncharged molecules and have no effect on the electrical conductivity of water.

Key Takeaway

  • Aqueous solutions can be classified as polar or nonpolar depending on how well they conduct electricity.

Conceptual Problems

  1. What are the advantages to carrying out a reaction in solution rather than simply mixing the pure reactants?

  2. What types of compounds dissolve in polar solvents?

  3. Describe the charge distribution in liquid water. How does this distribution affect its physical properties?

  4. Must a molecule have an asymmetric charge distribution to be polar? Explain your answer.

  5. Why are many ionic substances soluble in water?

  6. Explain the phrase like dissolves like.

  7. What kinds of covalent compounds are soluble in water?

  8. Why do most aromatic hydrocarbons have only limited solubility in water? Would you expect their solubility to be higher, lower, or the same in ethanol compared with water? Why?

  9. Predict whether each compound will dissolve in water and explain why.

    1. toluene
    2. acetic acid
    3. sodium acetate
    4. butanol
    5. pentanoic acid
  10. Predict whether each compound will dissolve in water and explain why.

    1. ammonium chloride
    2. 2-propanol
    3. heptane
    4. potassium dichromate
    5. 2-octanol
  11. Given water and toluene, predict which is the better solvent for each compound and explain your reasoning.

    1. sodium cyanide
    2. benzene
    3. acetic acid
    4. sodium ethoxide (CH3CH2ONa)
  12. Of water and toluene, predict which is the better solvent for each compound and explain your reasoning.

    1. t-butanol
    2. calcium chloride
    3. sucrose
    4. cyclohexene
  13. Compound A is divided into three equal samples. The first sample does not dissolve in water, the second sample dissolves only slightly in ethanol, and the third sample dissolves completely in toluene. What does this suggest about the polarity of A?

  14. You are given a mixture of three solid compounds—A, B, and C—and are told that A is a polar compound, B is slightly polar, and C is nonpolar. Suggest a method for separating these three compounds.

  15. A laboratory technician is given a sample that contains only sodium chloride, sucrose, and cyclodecanone (a ketone). You must tell the technician how to separate these three compounds from the mixture. What would you suggest?

  16. Many over-the-counter drugs are sold as ethanol/water solutions rather than as purely aqueous solutions. Give a plausible reason for this practice.

  17. What distinguishes a weak electrolyte from a strong electrolyte?

  18. Which organic groups result in aqueous solutions that conduct electricity?

  19. It is considered highly dangerous to splash barefoot in puddles during a lightning storm. Why?

  20. Which solution(s) would you expect to conduct electricity well? Explain your reasoning.

    1. an aqueous solution of sodium chloride
    2. a solution of ethanol in water
    3. a solution of calcium chloride in water
    4. a solution of sucrose in water
  21. Which solution(s) would you expect to conduct electricity well? Explain your reasoning.

    1. an aqueous solution of acetic acid
    2. an aqueous solution of potassium hydroxide
    3. a solution of ethylene glycol in water
    4. a solution of ammonium chloride in water
  22. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in an aqueous solution? Explain your reasoning.

    1. potassium hydroxide
    2. ammonia
    3. calcium chloride
    4. butanoic acid
  23. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in an aqueous solution? Explain your reasoning.

    1. magnesium hydroxide
    2. butanol
    3. ammonium bromide
    4. pentanoic acid
  24. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in aqueous solution? Explain your reasoning.

    1. H2SO4
    2. diethylamine
    3. 2-propanol
    4. ammonium chloride
    5. propanoic acid

Answers

  1. Ionic compounds such as NaCl are held together by electrostatic interactions between oppositely charged ions in the highly ordered solid. When an ionic compound dissolves in water, the partially negatively charged oxygen atoms of the H2O molecules surround the cations, and the partially positively charged hydrogen atoms in H2O surround the anions. The favorable electrostatic interactions between water and the ions compensate for the loss of the electrostatic interactions between ions in the solid.

    1. Because toluene is an aromatic hydrocarbon that lacks polar groups, it is unlikely to form a homogenous solution in water.
    2. Acetic acid contains a carboxylic acid group attached to a small alkyl group (a methyl group). Consequently, the polar characteristics of the carboxylic acid group will be dominant, and acetic acid will form a homogenous solution with water.
    3. Because most sodium salts are soluble, sodium acetate should form a homogenous solution with water.
    4. Like all alcohols, butanol contains an −OH group that can interact well with water. The alkyl group is rather large, consisting of a 4-carbon chain. In this case, the nonpolar character of the alkyl group is likely to be as important as the polar character of the –OH, decreasing the likelihood that butanol will form a homogeneous solution with water.
    5. Like acetic acid, pentanoic acid is a carboxylic acid. Unlike acetic acid, however, the alkyl group is rather large, consisting of a 4-carbon chain as in butanol. As with butanol, the nonpolar character of the alkyl group is likely to be as important as the polar character of the carboxylic acid group, making it unlikely that pentanoic acid will form a homogeneous solution with water. (In fact, the solubility of both butanol and pentanoic acid in water is quite low, only about 3 g per 100 g water at 25°C.)
  2. An electrolyte is any compound that can form ions when it dissolves in water. When a strong electrolyte dissolves in water, it dissociates completely to give the constituent ions. In contrast, when a weak electrolyte dissolves in water, it produces relatively few ions in solution.

4.2 Solution Concentrations

Learning Objective

  1. To describe the concentrations of solutions quantitatively.

All of us have a qualitative idea of what is meant by concentration. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the concentrationThe quantity of solute that is dissolved in a particular quantity of solvent or solution. of a solution describes the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for reactions that occur in solution. Chemists use many different ways to define concentrations, some of which are described in this section.

Molarity

The most common unit of concentration is molarity, which is also the most useful for calculations involving the stoichiometry of reactions in solution. The molarity (M)A common unit of concentration that is the number of moles of solute present in exactly 1 L of solution (mol/L). of a solution is the number of moles of solute present in exactly 1 L of solution. Molarity is also the number of millimoles of solute present in exactly 1 mL of solution:

Equation 4.4

molarity =moles of soluteliters of solution=mmoles of solutemilliliters of solution

The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as M. An aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the concentration of a solute. So

[sucrose] = 1.00 M

is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either

Equation 4.5

VLMmol/L=L(molL)= moles

or

Equation 4.6

VmLMmmol/mL=mL(mmolmL)= mmoles

Example 2 illustrates the use of Equation 4.5 and Equation 4.6.

Example 2

Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH.

Given: identity of solute and volume and molarity of solution

Asked for: amount of solute in moles

Strategy:

Use either Equation 4.5 or Equation 4.6, depending on the units given in the problem.

Solution:

Because we are given the volume of the solution in liters and are asked for the number of moles of substance, Equation 4.5 is more useful:

moles NaOH =VLMmol/L= (2.50 L)(0.100 molL)= 0.250 mol NaOH

Exercise

Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine.

Answer: 41.6 mmol

Concentrations are often reported on a mass-to-mass (m/m) basis or on a mass-to-volume (m/v) basis, particularly in clinical laboratories and engineering applications. A concentration expressed on an m/m basis is equal to the number of grams of solute per gram of solution; a concentration on an m/v basis is the number of grams of solute per milliliter of solution. Each measurement can be expressed as a percentage by multiplying the ratio by 100; the result is reported as percent m/m or percent m/v. The concentrations of very dilute solutions are often expressed in parts per million (ppm), which is grams of solute per 106 g of solution, or in parts per billion (ppb), which is grams of solute per 109 g of solution. For aqueous solutions at 20°C, 1 ppm corresponds to 1 μg per milliliter, and 1 ppb corresponds to 1 ng per milliliter. These concentrations and their units are summarized in Table 4.1 "Common Units of Concentration".

Table 4.1 Common Units of Concentration

Concentration Units
m/m g of solute/g of solution
m/v g of solute/mL of solution
ppm g of solute/106 g of solution
μg/mL
ppb g of solute/109 g of solution
ng/mL

The Preparation of Solutions

To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution. Figure 4.6 "Preparation of a Solution of Known Concentration Using a Solid Solute" illustrates this procedure for a solution of cobalt(II) chloride dihydrate in ethanol. Note that the volume of the solvent is not specified. Because the solute occupies space in the solution, the volume of the solvent needed is almost always less than the desired volume of solution. For example, if the desired volume were 1.00 L, it would be incorrect to add 1.00 L of water to 342 g of sucrose because that would produce more than 1.00 L of solution. As shown in Figure 4.7 "Preparation of 250 mL of a Solution of (NH", for some substances this effect can be significant, especially for concentrated solutions.

Figure 4.6 Preparation of a Solution of Known Concentration Using a Solid Solute

Figure 4.7 Preparation of 250 mL of a Solution of (NH4)2Cr2O7 in Water

The solute occupies space in the solution, so less than 250 mL of water are needed to make 250 mL of solution.

Example 3

The solution in Figure 4.6 "Preparation of a Solution of Known Concentration Using a Solid Solute" contains 10.0 g of cobalt(II) chloride dihydrate, CoCl2·2H2O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of CoCl2·2H2O?

Given: mass of solute and volume of solution

Asked for: concentration (M)

Strategy:

To find the number of moles of CoCl2·2H2O, divide the mass of the compound by its molar mass. Calculate the molarity of the solution by dividing the number of moles of solute by the volume of the solution in liters.

Solution:

The molar mass of CoCl2·2H2O is 165.87 g/mol. Therefore,

moles CoCl2•2H2=(10.0 g165.87 g/mol)= 0.0603 mol

The volume of the solution in liters is

volume = 500 mL(1 L1000 mL)= 0.500 L

Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is

molarity =0.0603 mol0.500 L= 0.121 M =CoCl2•H2O

Exercise

The solution shown in Figure 4.7 "Preparation of 250 mL of a Solution of (NH" contains 90.0 g of (NH4)2Cr2O7 in enough water to give a final volume of exactly 250 mL. What is the molar concentration of ammonium dichromate?

Answer: (NH4)2Cr2O7 = 1.43 M

To prepare a particular volume of a solution that contains a specified concentration of a solute, we first need to calculate the number of moles of solute in the desired volume of solution using the relationship shown in Equation 4.5. We then convert the number of moles of solute to the corresponding mass of solute needed. This procedure is illustrated in Example 4.

Example 4

The so-called D5W solution used for the intravenous replacement of body fluids contains 0.310 M glucose. (D5W is an approximately 5% solution of dextrose [the medical name for glucose] in water.) Calculate the mass of glucose necessary to prepare a 500 mL pouch of D5W. Glucose has a molar mass of 180.16 g/mol.

Given: molarity, volume, and molar mass of solute

Asked for: mass of solute

Strategy:

A Calculate the number of moles of glucose contained in the specified volume of solution by multiplying the volume of the solution by its molarity.

B Obtain the mass of glucose needed by multiplying the number of moles of the compound by its molar mass.

Solution:

A We must first calculate the number of moles of glucose contained in 500 mL of a 0.310 M solution:

VLMmol/L= moles500 mL(L1000 mL)(0.310 mol glucose1L)= 0.155 mol glucose

B We then convert the number of moles of glucose to the required mass of glucose:

mass of glucose = 0.155 mol glucose(180.16 g glucosemol glucose)= 27.9 g glucose

Exercise

Another solution commonly used for intravenous injections is normal saline, a 0.16 M solution of sodium chloride in water. Calculate the mass of sodium chloride needed to prepare 250 mL of normal saline solution.

Answer: 2.3 g NaCl

A solution of a desired concentration can also be prepared by diluting a small volume of a more concentrated solution with additional solvent. A stock solutionA commercially prepared solution of known concentration., which is a commercially prepared solution of known concentration, is often used for this purpose. Diluting a stock solution is preferred because the alternative method, weighing out tiny amounts of solute, is difficult to carry out with a high degree of accuracy. Dilution is also used to prepare solutions from substances that are sold as concentrated aqueous solutions, such as strong acids.

The procedure for preparing a solution of known concentration from a stock solution is shown in Figure 4.8 "Preparation of a Solution of Known Concentration by Diluting a Stock Solution". It requires calculating the number of moles of solute desired in the final volume of the more dilute solution and then calculating the volume of the stock solution that contains this amount of solute. Remember that diluting a given quantity of stock solution with solvent does not change the number of moles of solute present. The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is therefore

Equation 4.7

(Vs)(Ms) = moles of solute = (Vd)(Md)

where the subscripts s and d indicate the stock and dilute solutions, respectively. Example 5 demonstrates the calculations involved in diluting a concentrated stock solution.

Figure 4.8 Preparation of a Solution of Known Concentration by Diluting a Stock Solution

(a) A volume (Vs) containing the desired moles of solute (Ms) is measured from a stock solution of known concentration. (b) The measured volume of stock solution is transferred to a second volumetric flask. (c) The measured volume in the second flask is then diluted with solvent up to the volumetric mark [(Vs)(Ms) = (Vd)(Md)].

Example 5

What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of the D5W solution in Example 4?

Given: volume and molarity of dilute solution

Asked for: volume of stock solution

Strategy:

A Calculate the number of moles of glucose contained in the indicated volume of dilute solution by multiplying the volume of the solution by its molarity.

B To determine the volume of stock solution needed, divide the number of moles of glucose by the molarity of the stock solution.

Solution:

A The D5W solution in Example 4 was 0.310 M glucose. We begin by using Equation 4.7 to calculate the number of moles of glucose contained in 2500 mL of the solution:

moles glucose = 2500mL(L1000 mL)(0.310 mol glucoseL)= 0.775 mol glucose

B We must now determine the volume of the 3.00 M stock solution that contains this amount of glucose:

volume of stock soln = 0.775 mol glucose(1 L3.00 mol glucose)= 0.258 L or 258 mL

In determining the volume of stock solution that was needed, we had to divide the desired number of moles of glucose by the concentration of the stock solution to obtain the appropriate units. Also, the number of moles of solute in 258 mL of the stock solution is the same as the number of moles in 2500 mL of the more dilute solution; only the amount of solvent has changed. The answer we obtained makes sense: diluting the stock solution about tenfold increases its volume by about a factor of 10 (258 mL → 2500 mL). Consequently, the concentration of the solute must decrease by about a factor of 10, as it does (3.00 M → 0.310 M).

We could also have solved this problem in a single step by solving Equation 4.7 for Vs and substituting the appropriate values:

Vs=(Vd)(Md)Ms=(2.500 L)(0.310 M)3.00 M= 0.258 L

As we have noted, there is often more than one correct way to solve a problem.

Exercise

What volume of a 5.0 M NaCl stock solution is necessary to prepare 500 mL of normal saline solution (0.16 M NaCl)?

Answer: 16 mL

Ion Concentrations in Solution

In Example 3, you calculated that the concentration of a solution containing 90.00 g of ammonium dichromate in a final volume of 250 mL is 1.43 M. Let’s consider in more detail exactly what that means. Ammonium dichromate is an ionic compound that contains two NH4+ ions and one Cr2O72− ion per formula unit. Like other ionic compounds, it is a strong electrolyte that dissociates in aqueous solution to give hydrated NH4+ and Cr2O72− ions:

Equation 4.8

(NH4)2Cr2O7(s)H2O(l)2NH4+(aq)+Cr2O72(aq)

Thus 1 mol of ammonium dichromate formula units dissolves in water to produce 1 mol of Cr2O72− anions and 2 mol of NH4+ cations (see Figure 4.9 "Dissolution of 1 mol of an Ionic Compound").

Figure 4.9 Dissolution of 1 mol of an Ionic Compound

In this case, dissolving 1 mol of (NH4)2Cr2O7 produces a solution that contains 1 mol of Cr2O72− ions and 2 mol of NH4+ ions. (Water molecules are omitted from a molecular view of the solution for clarity.)

When we carry out a chemical reaction using a solution of a salt such as ammonium dichromate, we need to know the concentration of each ion present in the solution. If a solution contains 1.43 M (NH4)2Cr2O7, then the concentration of Cr2O72− must also be 1.43 M because there is one Cr2O72− ion per formula unit. However, there are two NH4+ ions per formula unit, so the concentration of NH4+ ions is 2 × 1.43 M = 2.86 M. Because each formula unit of (NH4)2Cr2O7 produces three ions when dissolved in water (2NH4+ + 1Cr2O72−), the total concentration of ions in the solution is 3 × 1.43 M = 4.29 M.

Example 6

What are the concentrations of all species derived from the solutes in these aqueous solutions?

  1. 0.21 M NaOH
  2. 3.7 M (CH3)CHOH
  3. 0.032 M In(NO3)3

Given: molarity

Asked for: concentrations

Strategy:

A Classify each compound as either a strong electrolyte or a nonelectrolyte.

B If the compound is a nonelectrolyte, its concentration is the same as the molarity of the solution. If the compound is a strong electrolyte, determine the number of each ion contained in one formula unit. Find the concentration of each species by multiplying the number of each ion by the molarity of the solution.

Solution:

  1. Sodium hydroxide is an ionic compound that is a strong electrolyte (and a strong base) in aqueous solution:

    NaOH(s)H2O(l)Na+(aq)+OH(aq)

    B Because each formula unit of NaOH produces one Na+ ion and one OH ion, the concentration of each ion is the same as the concentration of NaOH: [Na+] = 0.21 M and [OH] = 0.21 M.

  2. A The formula (CH3)2CHOH represents 2-propanol (isopropyl alcohol) and contains the –OH group, so it is an alcohol. Recall from Section 4.1 "Aqueous Solutions" that alcohols are covalent compounds that dissolve in water to give solutions of neutral molecules. Thus alcohols are nonelectrolytes.

    B The only solute species in solution is therefore (CH3)2CHOH molecules, so [(CH3)2CHOH] = 3.7 M.

  3. A Indium nitrate is an ionic compound that contains In3+ ions and NO3 ions, so we expect it to behave like a strong electrolyte in aqueous solution:

    In(NO3)3(s)H2O(l)In3+(aq)+3NO3(aq)

    B One formula unit of In(NO3)3 produces one In3+ ion and three NO3 ions, so a 0.032 M In(NO3)3 solution contains 0.032 M In3+ and 3 × 0.032 M = 0.096 M NO3—that is, [In3+] = 0.032 M and [NO3] = 0.096 M.

Exercise

What are the concentrations of all species derived from the solutes in these aqueous solutions?

  1. 0.0012 M Ba(OH)2
  2. 0.17 M Na2SO4
  3. 0.50 M (CH3)2CO, commonly known as acetone

Answer:

  1. [Ba2+] = 0.0012 M; [OH] = 0.0024 M
  2. [Na+] = 0.34 M; [SO42−] = 0.17 M
  3. [(CH3)2CO] = 0.50 M

Key Equations

definition of molarity

Equation 4.4: molarity =moles of soluteliters of solution=mmoles of solutemilliliters of solution

relationship among volume, molarity, and moles

Equation 4.5: VLMmol/L=L(molL)= moles

relationship between volume and concentration of stock and dilute solutions

Equation 4.7: (Vs)(Ms) = moles of solute = (Vd)(Md)

Summary

The concentration of a substance is the quantity of solute present in a given quantity of solution. Concentrations are usually expressed as molarity, the number of moles of solute in 1 L of solution. Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.

Key Takeaway

  • Solution concentrations are typically expressed as molarity and can be prepared by dissolving a known mass of solute in a solvent or diluting a stock solution.

Conceptual Problems

  1. Which of the representations best corresponds to a 1 M aqueous solution of each compound? Justify your answers.

    1. NH3
    2. HF
    3. CH3CH2CH2OH
    4. Na2SO4

  2. Which of the representations shown in Problem 1 best corresponds to a 1 M aqueous solution of each compound? Justify your answers.

    1. CH3CO2H
    2. NaCl
    3. Na2S
    4. Na3PO4
    5. acetaldehyde
  3. Would you expect a 1.0 M solution of CaCl2 to be a better conductor of electricity than a 1.0 M solution of NaCl? Why or why not?

  4. An alternative way to define the concentration of a solution is molality, abbreviated m. Molality is defined as the number of moles of solute in 1 kg of solvent. How is this different from molarity? Would you expect a 1 M solution of sucrose to be more or less concentrated than a 1 m solution of sucrose? Explain your answer.

  5. What are the advantages of using solutions for quantitative calculations?

Answer

  1. If the amount of a substance required for a reaction is too small to be weighed accurately, the use of a solution of the substance, in which the solute is dispersed in a much larger mass of solvent, allows chemists to measure the quantity of the substance more accurately.

Numerical Problems

  1. Calculate the number of grams of solute in 1.000 L of each solution.

    1. 0.2593 M NaBrO3
    2. 1.592 M KNO3
    3. 1.559 M acetic acid
    4. 0.943 M potassium iodate
  2. Calculate the number of grams of solute in 1.000 L of each solution.

    1. 0.1065 M BaI2
    2. 1.135 M Na2SO4
    3. 1.428 M NH4Br
    4. 0.889 M sodium acetate
  3. If all solutions contain the same solute, which solution contains the greater mass of solute?

    1. 1.40 L of a 0.334 M solution or 1.10 L of a 0.420 M solution
    2. 25.0 mL of a 0.134 M solution or 10.0 mL of a 0.295 M solution
    3. 250 mL of a 0.489 M solution or 150 mL of a 0.769 M solution
  4. Complete the following table for 500 mL of solution.

    Compound Mass (g) Moles Concentration (M)
    calcium sulfate 4.86
    acetic acid 3.62
    hydrogen iodide dihydrate 1.273
    barium bromide 3.92
    glucose 0.983
    sodium acetate 2.42
  5. What is the concentration of each species present in the following aqueous solutions?

    1. 0.489 mol of NiSO4 in 600 mL of solution
    2. 1.045 mol of magnesium bromide in 500 mL of solution
    3. 0.146 mol of glucose in 800 mL of solution
    4. 0.479 mol of CeCl3 in 700 mL of solution
  6. What is the concentration of each species present in the following aqueous solutions?

    1. 0.324 mol of K2MoO4 in 250 mL of solution
    2. 0.528 mol of potassium formate in 300 mL of solution
    3. 0.477 mol of KClO3 in 900 mL of solution
    4. 0.378 mol of potassium iodide in 750 mL of solution
  7. What is the molar concentration of each solution?

    1. 8.7 g of calcium bromide in 250 mL of solution
    2. 9.8 g of lithium sulfate in 300 mL of solution
    3. 12.4 g of sucrose (C12H22O11) in 750 mL of solution
    4. 14.2 g of iron(III) nitrate hexahydrate in 300 mL of solution
  8. What is the molar concentration of each solution?

    1. 12.8 g of sodium hydrogen sulfate in 400 mL of solution
    2. 7.5 g of potassium hydrogen phosphate in 250 mL of solution
    3. 11.4 g of barium chloride in 350 mL of solution
    4. 4.3 g of tartaric acid (C4H6O6) in 250 mL of solution
  9. Give the concentration of each reactant in the following equations, assuming 20.0 g of each and a solution volume of 250 mL for each reactant.

    1. BaCl2(aq) + Na2SO4(aq) →
    2. Ca(OH)2(aq) + H3PO4(aq) →
    3. Al(NO3)3(aq) + H2SO4(aq) →
    4. Pb(NO3)2(aq) + CuSO4(aq) →
    5. Al(CH3CO2)3(aq) + NaOH(aq) →
  10. An experiment required 200.0 mL of a 0.330 M solution of Na2CrO4. A stock solution of Na2CrO4 containing 20.0% solute by mass with a density of 1.19 g/cm3 was used to prepare this solution. Describe how to prepare 200.0 mL of a 0.330 M solution of Na2CrO4 using the stock solution.

  11. Calcium hypochlorite [Ca(OCl)2] is an effective disinfectant for clothing and bedding. If a solution has a Ca(OCl)2 concentration of 3.4 g per 100 mL of solution, what is the molarity of hypochlorite?

  12. Phenol (C6H5OH) is often used as an antiseptic in mouthwashes and throat lozenges. If a mouthwash has a phenol concentration of 1.5 g per 100 mL of solution, what is the molarity of phenol?

  13. If a tablet containing 100 mg of caffeine (C8H10N4O2) is dissolved in water to give 10.0 oz of solution, what is the molar concentration of caffeine in the solution?

  14. A certain drug label carries instructions to add 10.0 mL of sterile water, stating that each milliliter of the resulting solution will contain 0.500 g of medication. If a patient has a prescribed dose of 900.0 mg, how many milliliters of the solution should be administered?

Answers

  1. 0.48 M ClO

  2. 1.74 × 10−3 M caffeine

4.3 Stoichiometry of Reactions in Solution

Learning Objectives

  1. To balance equations that describe reactions in solution.
  1. To solve quantitative problems involving the stoichiometry of reactions in solution.

Quantitative calculations involving reactions in solution are carried out in the same manner as we discussed in Chapter 3 "Chemical Reactions". Instead of masses, however, we use volumes of solutions of known concentration to determine the number of moles of reactants. Whether we are dealing with volumes of solutions of reactants or masses of reactants, the coefficients in the balanced chemical equation tell us the number of moles of each reactant needed and the number of moles of each product that can be produced.

Calculating Moles from Volume

An expanded version of the flowchart for stoichiometric calculations illustrated in Figure 3.5 "Steps for Obtaining an Empirical Formula from Combustion Analysis" is shown in Figure 4.10 "An Expanded Flowchart for Stoichiometric Calculations". We can use the balanced chemical equation for the reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products to determine the amounts of other species, as illustrated in Example 7, Example 8, and Example 9.

Figure 4.10 An Expanded Flowchart for Stoichiometric Calculations

Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in a balanced chemical equation.

Note the Pattern

The balanced chemical equation for a reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used in stoichiometric calculations.

Example 7

Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN)2] ion. Gold is then recovered by reduction with metallic zinc according to the following equation:

Zn(s) + 2[Au(CN)2](aq) → [Zn(CN)4]2−(aq) + 2Au(s)

What mass of gold would you expect to recover from 400.0 L of a 3.30 × 10−4 M solution of [Au(CN)2]?

Given: chemical equation and molarity and volume of reactant

Asked for: mass of product

Strategy:

A Check the chemical equation to make sure it is balanced as written; balance if necessary. Then calculate the number of moles of [Au(CN)2] present by multiplying the volume of the solution by its concentration.

B From the balanced chemical equation, use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, multiply the number of moles of gold by its molar mass.

Solution:

A The equation is balanced as written, so we can proceed to the stoichiometric calculation. We can adapt Figure 4.10 "An Expanded Flowchart for Stoichiometric Calculations" for this particular problem as follows:

As indicated in the strategy, we start by calculating the number of moles of [Au(CN)2] present in the solution from the volume and concentration of the [Au(CN)2] solution:

moles [Au(CN)2]=VLMmol/L= 400.0 L(3.30×104– mol [Au(CN)2]L)= 0.132 mol [Au(CN)2]

B Because the coefficients of gold and the [Au(CN)2] ion are the same in the balanced chemical equation, if we assume that Zn(s) is present in excess, the number of moles of gold produced is the same as the number of moles of [Au(CN)2] we started with (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so we need to convert the number of moles of gold to the corresponding mass using the molar mass of gold:

mass of Au = (moles Au)(molar mass Au)= 0.132 mol Au(196.97 g Aumol Au)= 26.0 g Au

At a 2011 market price of over $1400 per troy ounce (31.10 g), this amount of gold is worth $1170.

26.0 g Au×troy oz31.10 g×$1400troy oz Au= $1170

Exercise

What mass of solid lanthanum(III) oxalate nonahydrate [La2(C2O4)3·9H2O] can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl3 by adding a stoichiometric amount of sodium oxalate?

Answer: 3.89 g

Limiting Reactants in Solutions

The concept of limiting reactants applies to reactions that are carried out in solution as well as to reactions that involve pure substances. If all the reactants but one are present in excess, then the amount of the limiting reactant may be calculated as illustrated in Example 8.

Example 8

Because the consumption of alcoholic beverages adversely affects the performance of tasks that require skill and judgment, in most countries it is illegal to drive while under the influence of alcohol. In almost all US states, a blood alcohol level of 0.08% by volume is considered legally drunk. Higher levels cause acute intoxication (0.20%), unconsciousness (about 0.30%), and even death (about 0.50%). The Breathalyzer is a portable device that measures the ethanol concentration in a person’s breath, which is directly proportional to the blood alcohol level. The reaction used in the Breathalyzer is the oxidation of ethanol by the dichromate ion:

3CH3CH2OH(aq)+2Cr2O72yellow-orange(aq)+16H + (aq)H2SO4(aq)Ag + 3CH3CO2H(aq)+4Cr3+(aq)+11H2O(l)green

When a measured volume (52.5 mL) of a suspect’s breath is bubbled through a solution of excess potassium dichromate in dilute sulfuric acid, the ethanol is rapidly absorbed and oxidized to acetic acid by the dichromate ions. In the process, the chromium atoms in some of the Cr2O72− ions are reduced from Cr6+ to Cr3+. In the presence of Ag+ ions that act as a catalyst, the reaction is complete in less than a minute. Because the Cr2O72− ion (the reactant) is yellow-orange and the Cr3+ ion (the product) forms a green solution, the amount of ethanol in the person’s breath (the limiting reactant) can be determined quite accurately by comparing the color of the final solution with the colors of standard solutions prepared with known amounts of ethanol.

A Breathalyzer ampul before (a) and after (b) ethanol is added. When a measured volume of a suspect’s breath is bubbled through the solution, the ethanol is oxidized to acetic acid, and the solution changes color from yellow-orange to green. The intensity of the green color indicates the amount of ethanol in the sample.

A typical Breathalyzer ampul contains 3.0 mL of a 0.25 mg/mL solution of K2Cr2O7 in 50% H2SO4 as well as a fixed concentration of AgNO3 (typically 0.25 mg/mL is used for this purpose). How many grams of ethanol must be present in 52.5 mL of a person’s breath to convert all the Cr6+ to Cr3+?

Given: volume and concentration of one reactant

Asked for: mass of other reactant needed for complete reaction

Strategy:

A Calculate the number of moles of Cr2O72− ion in 1 mL of the Breathalyzer solution by dividing the mass of K2Cr2O7 by its molar mass.

B Find the total number of moles of Cr2O72− ion in the Breathalyzer ampul by multiplying the number of moles contained in 1 mL by the total volume of the Breathalyzer solution (3.0 mL).

C Use the mole ratios from the balanced chemical equation to calculate the number of moles of C2H5OH needed to react completely with the number of moles of Cr2O72− ions present. Then find the mass of C2H5OH needed by multiplying the number of moles of C2H5OH by its molar mass.

Solution:

A In any stoichiometry problem, the first step is always to calculate the number of moles of each reactant present. In this case, we are given the mass of K2Cr2O7 in 1 mL of solution, which we can use to calculate the number of moles of K2Cr2O7 contained in 1 mL:

moles K2Cr2O71 mL=(0.25 mgK2Cr2O7)mL(g1000 mg)(1 mol294.18 gK2Cr2O7)=8.5×107 moles

B Because 1 mol of K2Cr2O7 produces 1 mol of Cr2O72− when it dissolves, each milliliter of solution contains 8.5 × 10−7 mol of Cr2O72−. The total number of moles of Cr2O72− in a 3.0 mL Breathalyzer ampul is thus

moles Cr2O72=(8.5×107 molmL)(3.0 mL)= 2.6×106 mol Cr2O72

C The balanced chemical equation tells us that 3 mol of C2H5OH is needed to consume 2 mol of Cr2O72− ion, so the total number of moles of C2H5OH required for complete reaction is

moles of C2H5OH =(2.6×106 mol Cr2O72)(3 mol C2H5OHmol Cr2O72)= 3.9×106 mol C2H5OH

As indicated in the strategy, this number can be converted to the mass of C2H5OH using its molar mass:

mass C2H5OH =(3.9×106 mol C2H5OH)(46.07 gmol C2H5OH)= 1.8×104 g C2H5OH

Thus 1.8 × 10−4 g or 0.18 mg of C2H5OH must be present. Experimentally, it is found that this value corresponds to a blood alcohol level of 0.7%, which is usually fatal.

Exercise

The compound para-nitrophenol (molar mass = 139 g/mol) reacts with sodium hydroxide in aqueous solution to generate a yellow anion via the reaction

Because the amount of para-nitrophenol is easily estimated from the intensity of the yellow color that results when excess NaOH is added, reactions that produce para-nitrophenol are commonly used to measure the activity of enzymes, the catalysts in biological systems. What volume of 0.105 M NaOH must be added to 50.0 mL of a solution containing 7.20 × 10−4 g of para-nitrophenol to ensure that formation of the yellow anion is complete?

Answer: 4.93 × 10−5 L or 49.3 μL

In Example 7 and Example 8, the identity of the limiting reactant has been apparent: [Au(CN)2], LaCl3, ethanol, and para-nitrophenol. When the limiting reactant is not apparent, we can determine which reactant is limiting by comparing the molar amounts of the reactants with their coefficients in the balanced chemical equation, just as we did in Chapter 3 "Chemical Reactions", Section 3.4 "Mass Relationships in Chemical Equations". The only difference is that now we use the volumes and concentrations of solutions of reactants rather than the masses of reactants to calculate the number of moles of reactants, as illustrated in Example 9.

Example 9

When aqueous solutions of silver nitrate and potassium dichromate are mixed, an exchange reaction occurs, and silver dichromate is obtained as a red solid. The overall chemical equation for the reaction is as follows:

2AgNO3(aq) + K2Cr2O7(aq) → Ag2Cr2O7(s) + 2KNO3(aq)

What mass of Ag2Cr2O7 is formed when 500 mL of 0.17 M K2Cr2O7 are mixed with 250 mL of 0.57 M AgNO3?

Given: balanced chemical equation and volume and concentration of each reactant

Asked for: mass of product

Strategy:

A Calculate the number of moles of each reactant by multiplying the volume of each solution by its molarity.

B Determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation.

C Use mole ratios to calculate the number of moles of product that can be formed from the limiting reactant. Multiply the number of moles of the product by its molar mass to obtain the corresponding mass of product.

Solution:

A The balanced chemical equation tells us that 2 mol of AgNO3(aq) reacts with 1 mol of K2Cr2O7(aq) to form 1 mol of Ag2Cr2O7(s) (Figure 4.11 "What Happens at the Molecular Level When Solutions of AgNO"). The first step is to calculate the number of moles of each reactant in the specified volumes:

moles K2Cr2O7= 500 mL(L1000 mL)(0.17 mol K2Cr2O7L)= 0.085 mol K2Cr2O7moles AgNO3= 250 mL(L1000 mL)(0.57 mol AgNO3L)= 0.14 mol AgNO3

B Now we can determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient:

K2Cr2O70.085 mol1 mol= 0.085AgNO30.14 mol2 mol= 0.070

Because 0.070 < 0.085, we know that AgNO3 is the limiting reactant.

C Each mole of Ag2Cr2O7 formed requires 2 mol of the limiting reactant (AgNO3), so we can obtain only 0.14/2 = 0.070 mol of Ag2Cr2O7. Finally, we convert the number of moles of Ag2Cr2O7 to the corresponding mass:

mass of Ag2Cr2O7= 0.070 mol (431.72 gmol)=30 g Ag2Cr2O7

Figure 4.11 What Happens at the Molecular Level When Solutions of AgNO3 and K2Cr2O7 Are Mixed

The Ag+ and Cr2O72− ions form a red precipitate of solid Ag2Cr2O7, while the K+ and NO3 ions remain in solution. (Water molecules are omitted from molecular views of the solutions for clarity.)

Exercise

Aqueous solutions of sodium bicarbonate and sulfuric acid react to produce carbon dioxide according to the following equation:

2NaHCO3(aq) + H2SO4(aq) → 2CO2(g) + Na2SO4(aq) + 2H2O(l)

If 13.0 mL of 3.0 M H2SO4 are added to 732 mL of 0.112 M NaHCO3, what mass of CO2 is produced?

Answer: 3.4 g

Summary

Quantitative calculations that involve the stoichiometry of reactions in solution use volumes of solutions of known concentration instead of masses of reactants or products. The coefficients in the balanced chemical equation tell how many moles of reactants are needed and how many moles of product can be produced.

Key Takeaway

  • Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in the balanced chemical equation.

Conceptual Problems

  1. What information is required to determine the mass of solute in a solution if you know the molar concentration of the solution?

  2. Is it possible for one reactant to be limiting in a reaction that does not go to completion?

Numerical Problems

  1. Refer to the Breathalyzer test described in Example 8. How much ethanol must be present in 89.5 mL of a person’s breath to consume all the potassium dichromate in a Breathalyzer ampul containing 3.0 mL of a 0.40 mg/mL solution of potassium dichromate?

  2. Phosphoric acid and magnesium hydroxide react to produce magnesium phosphate and water. If 45.00 mL of 1.50 M phosphoric acid are used in the reaction, how many grams of magnesium hydroxide are needed for the reaction to go to completion?

  3. Barium chloride and sodium sulfate react to produce sodium chloride and barium sulfate. If 50.00 mL of 2.55 M barium chloride are used in the reaction, how many grams of sodium sulfate are needed for the reaction to go to completion?

  4. How many grams of sodium phosphate are obtained in solution from the reaction of 75.00 mL of 2.80 M sodium carbonate with a stoichiometric amount of phosphoric acid? A second product is water; what is the third product? How many grams of the third product are obtained?

  5. How many grams of ammonium bromide are produced from the reaction of 50.00 mL of 2.08 M iron(II) bromide with a stoichiometric amount of ammonium sulfide? What is the second product? How many grams of the second product are produced?

  6. Lead(II) nitrate and hydroiodic acid react to produce lead(II) iodide and nitric acid. If 3.25 g of lead(II) iodide were obtained by adding excess HI to 150.0 mL of lead(II) nitrate, what was the molarity of the lead(II) nitrate solution?

  7. Silver nitrate and sodium chloride react to produce sodium nitrate and silver chloride. If 2.60 g of AgCl was obtained by adding excess NaCl to 100 mL of AgNO3, what was the molarity of the silver nitrate solution?

4.4 Ionic Equations

Learning Objective

  1. To understand what information is obtained by each type of ionic equation.

The chemical equations discussed in Chapter 3 "Chemical Reactions" showed the identities of the reactants and the products and gave the stoichiometries of the reactions, but they told us very little about what was occurring in solution. In contrast, equations that show only the hydrated species focus our attention on the chemistry that is taking place and allow us to see similarities between reactions that might not otherwise be apparent.

Let’s consider the reaction of silver nitrate with potassium dichromate. As you learned in Example 9, when aqueous solutions of silver nitrate and potassium dichromate are mixed, silver dichromate forms as a red solid. The overall chemical equationA chemical equation that shows all the reactants and products as undissociated, electrically neutral compounds. for the reaction shows each reactant and product as undissociated, electrically neutral compounds:

Equation 4.9

2AgNO3(aq) + K2Cr2O7(aq) → Ag2Cr2O7(s) + 2KNO3(aq)

Although Equation 4.9 gives the identity of the reactants and the products, it does not show the identities of the actual species in solution. Because ionic substances such as AgNO3 and K2Cr2O7 are strong electrolytes, they dissociate completely in aqueous solution to form ions. In contrast, because Ag2Cr2O7 is not very soluble, it separates from the solution as a solid. To find out what is actually occurring in solution, it is more informative to write the reaction as a complete ionic equationA chemical equation that shows which ions and molecules are hydrated and which are present in other forms and phases., showing which ions and molecules are hydrated and which are present in other forms and phases:

Equation 4.10

2Ag+(aq) + 2NO3(aq) + 2K+(aq) + Cr2O72−(aq) → Ag2Cr2O7(s) + 2K+(aq) + 2NO3(aq)

Note that K+(aq) and NO3(aq) ions are present on both sides of the equation, and their coefficients are the same on both sides. These ions are called spectator ionsIons that do not participate in the actual reaction. because they do not participate in the actual reaction. Canceling the spectator ions gives the net ionic equationA chemical equation that shows only those species that participate in the chemical reaction., which shows only those species that participate in the chemical reaction:

Equation 4.11

2Ag+(aq) + Cr2O72−(aq) → Ag2Cr2O7(s)

Both mass and charge must be conserved in chemical reactions because the numbers of electrons and protons do not change. For charge to be conserved, the sum of the charges of the ions multiplied by their coefficients must be the same on both sides of the equation. In Equation 4.11, the charge on the left side is 2(+1) + 1(−2) = 0, which is the same as the charge of a neutral Ag2Cr2O7 formula unit.

By eliminating the spectator ions, we can focus on the chemistry that takes place in a solution. For example, the overall chemical equation for the reaction between silver fluoride and ammonium dichromate is as follows:

Equation 4.12

2AgF(aq) + (NH4)2Cr2O7(aq) → Ag2Cr2O7(s) + 2NH4F(aq)

The complete ionic equation for this reaction is as follows:

Equation 4.13

2Ag+(aq) + 2F(aq) + 2NH4+(aq) + Cr2O72−(aq) → Ag2Cr2O7(s) + 2NH4+(aq) + 2F(aq)

Because two NH4+(aq) and two F(aq) ions appear on both sides of Equation 4.13, they are spectator ions. They can therefore be canceled to give the net ionic equation (Equation 4.14), which is identical to Equation 4.11:

Equation 4.14

2Ag+(aq) + Cr2O72−(aq) → Ag2Cr2O7(s)

If we look at net ionic equations, it becomes apparent that many different combinations of reactants can result in the same net chemical reaction. For example, we can predict that silver fluoride could be replaced by silver nitrate in the preceding reaction without affecting the outcome of the reaction.

Example 10

Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous barium nitrate with aqueous sodium phosphate to give solid barium phosphate and a solution of sodium nitrate.

Given: reactants and products

Asked for: overall, complete ionic, and net ionic equations

Strategy:

Write and balance the overall chemical equation. Write all the soluble reactants and products in their dissociated form to give the complete ionic equation; then cancel species that appear on both sides of the complete ionic equation to give the net ionic equation.

Solution:

From the information given, we can write the unbalanced chemical equation for the reaction:

Ba(NO3)2(aq) + Na3PO4(aq) → Ba3(PO4)2(s) + NaNO3(aq)

Because the product is Ba3(PO4)2, which contains three Ba2+ ions and two PO43− ions per formula unit, we can balance the equation by inspection:

3Ba(NO3)2(aq) + 2Na3PO4(aq) → Ba3(PO4)2(s) + 6NaNO3(aq)

This is the overall balanced chemical equation for the reaction, showing the reactants and products in their undissociated form. To obtain the complete ionic equation, we write each soluble reactant and product in dissociated form:

3Ba2+(aq) + 6NO3(aq) + 6Na+(aq) + 2PO43−(aq) → Ba3(PO4)2(s) + 6Na+(aq) + 6NO3(aq)

The six NO3(aq) ions and the six Na+(aq) ions that appear on both sides of the equation are spectator ions that can be canceled to give the net ionic equation:

3Ba2+(aq) + 2PO43−(aq) → Ba3(PO4)2(s)

Exercise

Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous silver fluoride with aqueous sodium phosphate to give solid silver phosphate and a solution of sodium fluoride.

Answer:

overall chemical equation: 3AgF(aq) + Na3PO4(aq) → Ag3PO4(s) + 3NaF(aq)

complete ionic equation: 3Ag+(aq) + 3F(aq) + 3Na+(aq) + PO43−(aq) → Ag3PO4(s) + 3Na+(aq) + 3F(aq)

net ionic equation: 3Ag+(aq) + PO43−(aq) → Ag3PO4(s)

So far, we have always indicated whether a reaction will occur when solutions are mixed and, if so, what products will form. As you advance in chemistry, however, you will need to predict the results of mixing solutions of compounds, anticipate what kind of reaction (if any) will occur, and predict the identities of the products. Students tend to think that this means they are supposed to “just know” what will happen when two substances are mixed. Nothing could be further from the truth: an infinite number of chemical reactions is possible, and neither you nor anyone else could possibly memorize them all. Instead, you must begin by identifying the various reactions that could occur and then assessing which is the most probable (or least improbable) outcome.

The most important step in analyzing an unknown reaction is to write down all the species—whether molecules or dissociated ions—that are actually present in the solution (not forgetting the solvent itself) so that you can assess which species are most likely to react with one another. The easiest way to make that kind of prediction is to attempt to place the reaction into one of several familiar classifications, refinements of the five general kinds of reactions introduced in Chapter 3 "Chemical Reactions" (acid–base, exchange, condensation, cleavage, and oxidation–reduction reactions). In the sections that follow, we discuss three of the most important kinds of reactions that occur in aqueous solutions: precipitation reactions (also known as exchange reactions), acid–base reactions, and oxidation–reduction reactions.

Summary

The chemical equation for a reaction in solution can be written in three ways. The overall chemical equation shows all the substances present in their undissociated forms; the complete ionic equation shows all the substances present in the form in which they actually exist in solution; and the net ionic equation is derived from the complete ionic equation by omitting all spectator ions, ions that occur on both sides of the equation with the same coefficients. Net ionic equations demonstrate that many different combinations of reactants can give the same net chemical reaction.

Key Takeaway

  • A complete ionic equation consists of the net ionic equation and spectator ions.

Conceptual Problem

  1. What information can be obtained from a complete ionic equation that cannot be obtained from the overall chemical equation?

4.5 Precipitation Reactions

Learning Objective

  1. To identify a precipitation reaction and predict solubilities.

A precipitation reactionA subclass of an exchange reaction that yields an insoluble product (a precipitate) when two solutions are mixed. is a reaction that yields an insoluble product—a precipitateThe insoluble product that forms in a precipitation reaction.—when two solutions are mixed. In , we described a precipitation reaction in which a colorless solution of silver nitrate was mixed with a yellow-orange solution of potassium dichromate to give a reddish precipitate of silver dichromate:

Equation 4.15

AgNO3(aq) + K2Cr2O7(aq) → Ag2Cr2O7(s) + KNO3(aq)

This equation has the general form of an exchange reaction:

Equation 4.16

AC + BDADinsoluble+BC

Thus precipitation reactions are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble. Because both components of each compound change partners, such reactions are sometimes called double-displacement reactions. Two important uses of precipitation reactions are to isolate metals that have been extracted from their ores and to recover precious metals for recycling.

Note the Pattern

Precipitation reactions are a subclass of exchange reactions.

Predicting Solubilities

gives guidelines for predicting the solubility of a wide variety of ionic compounds. To determine whether a precipitation reaction will occur, we identify each species in the solution and then refer to to see which, if any, combination(s) of cation and anion are likely to produce an insoluble salt. In doing so, it is important to recognize that soluble and insoluble are relative terms that span a wide range of actual solubilities. We will discuss solubilities in more detail in , where you will learn that very small amounts of the constituent ions remain in solution even after precipitation of an “insoluble” salt. For our purposes, however, we will assume that precipitation of an insoluble salt is complete.

Table 4.2 Guidelines for Predicting the Solubility of Ionic Compounds in Water

Soluble Exceptions
Rule 1 most salts that contain an alkali metal (Li+, Na+, K+, Rb+, and Cs+) and ammonium (NH4+)
Rule 2 most salts that contain the nitrate (NO3) anion
Rule 3 most salts of anions derived from monocarboxylic acids (e.g., CH3CO2) but not silver acetate and salts of long-chain carboxylates
Rule 4 most chloride, bromide, and iodide salts but not salts of metal ions located on the lower right side of the periodic table (e.g., Cu+, Ag+, Pb2+, and Hg22+).
Insoluble Exceptions
Rule 5 most salts that contain the hydroxide (OH) and sulfide (S2−) anions but not salts of the alkali metals (group 1), the heavier alkaline earths (Ca2+, Sr2+, and Ba2+ in group 2), and the NH4+ ion.
Rule 6 most carbonate (CO32−) and phosphate (PO43−) salts but not salts of the alkali metals or the NH4+ ion.
Rule 7 most sulfate (SO42−) salts that contain main group cations with a charge ≥ +2 but not salts of +1 cations, Mg2+, and dipositive transition metal cations (e.g., Ni2+)

Just as important as predicting the product of a reaction is knowing when a chemical reaction will not occur. Simply mixing solutions of two different chemical substances does not guarantee that a reaction will take place. For example, if 500 mL of a 1.0 M aqueous NaCl solution is mixed with 500 mL of a 1.0 M aqueous KBr solution, the final solution has a volume of 1.00 L and contains 0.50 M Na+(aq), 0.50 M Cl(aq), 0.50 M K+(aq), and 0.50 M Br(aq). As you will see in the following sections, none of these species reacts with any of the others. When these solutions are mixed, the only effect is to dilute each solution with the other ().

Figure 4.12 The Effect of Mixing Aqueous KBr and NaCl Solutions

Because no net reaction occurs, the only effect is to dilute each solution with the other. (Water molecules are omitted from molecular views of the solutions for clarity.)

Example 11

Using the information in , predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs.

  1. Aqueous solutions of barium chloride and lithium sulfate are mixed.
  2. Aqueous solutions of rubidium hydroxide and cobalt(II) chloride are mixed.
  3. Aqueous solutions of strontium bromide and aluminum nitrate are mixed.
  4. Solid lead(II) acetate is added to an aqueous solution of ammonium iodide.

Given: reactants

Asked for: reaction and net ionic equation

Strategy:

A Identify the ions present in solution and write the products of each possible exchange reaction.

B Refer to to determine which, if any, of the products is insoluble and will therefore form a precipitate. If a precipitate forms, write the net ionic equation for the reaction.

Solution:

  1. A Because barium chloride and lithium sulfate are strong electrolytes, each dissociates completely in water to give a solution that contains the constituent anions and cations. Mixing the two solutions initially gives an aqueous solution that contains Ba2+, Cl, Li+, and SO42− ions. The only possible exchange reaction is to form LiCl and BaSO4:

     

    B We now need to decide whether either of these products is insoluble. shows that LiCl is soluble in water (rules 1 and 4), but BaSO4 is not soluble in water (rule 5). Thus BaSO4 will precipitate according to the net ionic equation

    Ba2+(aq) + SO42−(aq) → BaSO4(s)

    Although soluble barium salts are toxic, BaSO4 is so insoluble that it can be used to diagnose stomach and intestinal problems without being absorbed into tissues. An outline of the digestive organs appears on x-rays of patients who have been given a “barium milkshake” or a “barium enema”—a suspension of very fine BaSO4 particles in water.

    An x-ray of the digestive organs of a patient who has swallowed a “barium milkshake.” A barium milkshake is a suspension of very fine BaSO4 particles in water; the high atomic mass of barium makes it opaque to x-rays.

  2. A Rubidium hydroxide and cobalt(II) chloride are strong electrolytes, so when aqueous solutions of these compounds are mixed, the resulting solution initially contains Rb+, OH, Co2+, and Cl ions. The possible products of an exchange reaction are rubidium chloride and cobalt(II) hydroxide):

     

    B According to , RbCl is soluble (rules 1 and 4), but Co(OH)2 is not soluble (rule 5). Hence Co(OH)2 will precipitate according to the following net ionic equation:

    Co2+(aq) + 2OH(aq) → Co(OH)2(s)
  3. A When aqueous solutions of strontium bromide and aluminum nitrate are mixed, we initially obtain a solution that contains Sr2+, Br, Al3+, and NO3 ions. The two possible products from an exchange reaction are aluminum bromide and strontium nitrate:

     

    B According to , both AlBr3 (rule 4) and Sr(NO3)2 (rule 2) are soluble. Thus no net reaction will occur.

  4. A According to , lead acetate is soluble (rule 3). Thus solid lead acetate dissolves in water to give Pb2+ and CH3CO2 ions. Because the solution also contains NH4+ and I ions, the possible products of an exchange reaction are ammonium acetate and lead(II) iodide:

     

    B According to , ammonium acetate is soluble (rules 1 and 3), but PbI2 is insoluble (rule 4). Thus Pb(C2H3O2)2 will dissolve, and PbI2 will precipitate. The net ionic equation is as follows:

    Pb2+ (aq) + 2I(aq) → PbI2(s)

Exercise

Using the information in , predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs.

  1. An aqueous solution of strontium hydroxide is added to an aqueous solution of iron(II) chloride.
  2. Solid potassium phosphate is added to an aqueous solution of mercury(II) perchlorate.
  3. Solid sodium fluoride is added to an aqueous solution of ammonium formate.
  4. Aqueous solutions of calcium bromide and cesium carbonate are mixed.

Answer:

  1. Fe2+(aq) + 2OH(aq) → Fe(OH)2(s)
  2. 2PO43−(aq) + 3Hg2+(aq) → Hg3(PO4)2(s)
  3. NaF(s) dissolves; no net reaction
  4. Ca2+(aq) + CO32−(aq) → CaCO3(s)

Precipitation Reactions in Photography

Precipitation reactions can be used to recover silver from solutions used to develop conventional photographic film. Although largely supplanted by digital photography, conventional methods are often used for artistic purposes. Silver bromide is an off-white solid that turns black when exposed to light, which is due to the formation of small particles of silver metal. Black-and-white photography uses this reaction to capture images in shades of gray, with the darkest areas of the film corresponding to the areas that received the most light. The first step in film processing is to enhance the black/white contrast by using a developer to increase the amount of black. The developer is a reductant: because silver atoms catalyze the reduction reaction, grains of silver bromide that have already been partially reduced by exposure to light react with the reductant much more rapidly than unexposed grains.

Darkening of silver bromide crystals by exposure to light. The top image shows AgBr before exposure to light, and the bottom image after exposure.

After the film is developed, any unexposed silver bromide must be removed by a process called “fixing”; otherwise, the entire film would turn black with additional exposure to light. Although silver bromide is insoluble in water, it is soluble in a dilute solution of sodium thiosulfate (Na2S2O3; photographer’s hypo) because of the formation of [Ag(S2O3)2]3− ions. Thus washing the film with thiosulfate solution dissolves unexposed silver bromide and leaves a pattern of metallic silver granules that constitutes the negative. This procedure is summarized in . The negative image is then projected onto paper coated with silver halides, and the developing and fixing processes are repeated to give a positive image. (Color photography works in much the same way, with a combination of silver halides and organic dyes superimposed in layers.) “Instant photo” operations can generate more than a hundred gallons of dilute silver waste solution per day. Recovery of silver from thiosulfate fixing solutions involves first removing the thiosulfate by oxidation and then precipitating Ag+ ions with excess chloride ions.

Figure 4.13 Outline of the Steps Involved in Producing a Black-and-White Photograph

Example 12

A silver recovery unit can process 1500 L of photographic silver waste solution per day. Adding excess solid sodium chloride to a 500 mL sample of the waste (after removing the thiosulfate as described previously) gives a white precipitate that, after filtration and drying, consists of 3.73 g of AgCl. What mass of NaCl must be added to the 1500 L of silver waste to ensure that all the Ag+ ions precipitate?

Given: volume of solution of one reactant and mass of product from a sample of reactant solution

Asked for: mass of second reactant needed for complete reaction

Strategy:

A Write the net ionic equation for the reaction. Calculate the number of moles of AgCl obtained from the 500 mL sample and then determine the concentration of Ag+ in the sample by dividing the number of moles of AgCl formed by the volume of solution.

B Determine the total number of moles of Ag+ in the 1500 L solution by multiplying the Ag+ concentration by the total volume.

C Use mole ratios to calculate the number of moles of chloride needed to react with Ag+. Obtain the mass of NaCl by multiplying the number of moles of NaCl needed by its molar mass.

Solution:

We can use the data provided to determine the concentration of Ag+ ions in the waste, from which the number of moles of Ag+ in the entire waste solution can be calculated. From the net ionic equation, we can determine how many moles of Cl are needed, which in turn will give us the mass of NaCl necessary.

A The first step is to write the net ionic equation for the reaction:

Cl(aq) + Ag+(aq) → AgCl(s)

We know that 500 mL of solution produced 3.73 g of AgCl. We can convert this value to the number of moles of AgCl as follows:

moles AgCl =grams AgClmolar mass AgCl= 3.73 g AgCl(1 mol AgCl143.32 g AgCl)= 0.0260 mol AgCl

Therefore, the 500 mL sample of the solution contained 0.0260 mol of Ag+. The Ag+ concentration is determined as follows:

[Ag + =moles Ag+liters soln=0.0260 mol AgCl0.500 L= 0.0520 M

B The total number of moles of Ag+ present in 1500 L of solution is as follows:

moles Ag+= 1500 L(0.520 molL)= 78.1 mol Ag+

C According to the net ionic equation, one Cl ion is required for each Ag+ ion. Thus 78.1 mol of NaCl are needed to precipitate the silver. The corresponding mass of NaCl is

mass NaCl = 78.1 mol NaCl(58.44 g NaClmol NaCl)= 4560 g NaCl = 4.56 kg NaCl

Note that 78.1 mol of AgCl correspond to 8.43 kg of metallic silver, which is worth about $7983 at 2011 prices ($32.84 per troy ounce). Silver recovery may be economically attractive as well as ecologically sound, although the procedure outlined is becoming nearly obsolete for all but artistic purposes with the growth of digital photography.

Exercise

Because of its toxicity, arsenic is the active ingredient in many pesticides. The arsenic content of a pesticide can be measured by oxidizing arsenic compounds to the arsenate ion (AsO43−), which forms an insoluble silver salt (Ag3AsO4). Suppose you are asked to assess the purity of technical grade sodium arsenite (NaAsO2), the active ingredient in a pesticide used against termites. You dissolve a 10.00 g sample in water, oxidize it to arsenate, and dilute it with water to a final volume of 500 mL. You then add excess AgNO3 solution to a 50.0 mL sample of the arsenate solution. The resulting precipitate of Ag3AsO4 has a mass of 3.24 g after drying. What is the percentage by mass of NaAsO2 in the original sample?

Answer: 91.0%

Summary

In a precipitation reaction, a subclass of exchange reactions, an insoluble material (a precipitate) forms when solutions of two substances are mixed. To predict the product of a precipitation reaction, all species initially present in the solutions are identified, as are any combinations likely to produce an insoluble salt.

Key Takeaway

  • Predicting the solubility of ionic compounds in water can give insight into whether or not a reaction will occur.

Conceptual Problems

  1. Predict whether mixing each pair of solutions will result in the formation of a precipitate. If so, identify the precipitate.

    1. FeCl2(aq) + Na2S(aq)
    2. NaOH(aq) + H3PO4(aq)
    3. ZnCl2(aq) + (NH4)2S(aq)
  2. Predict whether mixing each pair of solutions will result in the formation of a precipitate. If so, identify the precipitate.

    1. KOH(aq) + H3PO4(aq)
    2. K2CO3(aq) + BaCl2(aq)
    3. Ba(NO3)2(aq) + Na2SO4(aq)
  3. Which representation best corresponds to an aqueous solution originally containing each of the following?

    1. 1 M NH4Cl
    2. 1 M NaO2CCH3
    3. 1 M NaOH + 1 M HCl
    4. 1 M Ba(OH)2 + 1 M H2SO4

  4. Which representation in Problem 3 best corresponds to an aqueous solution originally containing each of the following?

    1. 1 M CH3CO2H + 1 M NaOH
    2. 1 M NH3 + 1 M HCl
    3. 1 M Na2CO3 + 1 M H2SO4
    4. 1 M CaCl2 + 1 M H3PO4

Answer

    1. 1
    2. 1
    3. 1
    4. 2

Numerical Problems

  1. What mass of precipitate would you expect to obtain by mixing 250 mL of a solution containing 4.88 g of Na2CrO4 with 200 mL of a solution containing 3.84 g of AgNO3? What is the final nitrate ion concentration?

  2. Adding 10.0 mL of a dilute solution of zinc nitrate to 246 mL of 2.00 M sodium sulfide produced 0.279 g of a precipitate. How many grams of zinc(II) nitrate and sodium sulfide were consumed to produce this quantity of product? What was the concentration of each ion in the original solutions? What is the concentration of the sulfide ion in solution after the precipitation reaction, assuming no further reaction?

Answer

  1. 3.75 g Ag2CrO4; 5.02 × 10−2 M nitrate

4.6 Acid–Base Reactions

Learning Objective

  1. To know the characteristic properties of acids and bases.

Acid–base reactions are essential in both biochemistry and industrial chemistry. Moreover, many of the substances we encounter in our homes, the supermarket, and the pharmacy are acids or bases. For example, aspirin is an acid (acetylsalicylic acid), and antacids are bases. In fact, every amateur chef who has prepared mayonnaise or squeezed a wedge of lemon to marinate a piece of fish has carried out an acid–base reaction. Before we discuss the characteristics of such reactions, let’s first describe some of the properties of acids and bases.

Definitions of Acids and Bases

In , we defined acids as substances that dissolve in water to produce H+ ions, whereas bases were defined as substances that dissolve in water to produce OH ions. In fact, this is only one possible set of definitions. Although the general properties of acids and bases have been known for more than a thousand years, the definitions of acid and base have changed dramatically as scientists have learned more about them. In ancient times, an acid was any substance that had a sour taste (e.g., vinegar or lemon juice), caused consistent color changes in dyes derived from plants (e.g., turning blue litmus paper red), reacted with certain metals to produce hydrogen gas and a solution of a salt containing a metal cation, and dissolved carbonate salts such as limestone (CaCO3) with the evolution of carbon dioxide. In contrast, a base was any substance that had a bitter taste, felt slippery to the touch, and caused color changes in plant dyes that differed diametrically from the changes caused by acids (e.g., turning red litmus paper blue). Although these definitions were useful, they were entirely descriptive.

The Arrhenius Definition of Acids and Bases

The first person to define acids and bases in detail was the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry, 1903). According to the Arrhenius definition, an acid is a substance like hydrochloric acid that dissolves in water to produce H+ ions (protons; ), and a base is a substance like sodium hydroxide that dissolves in water to produce hydroxide (OH) ions ():

Equation 4.17

HCl(g)an Arrhenius acidH2O(l)H + (aq)+Cl(aq)

Equation 4.18

NaOH(s)an Arrhenius baseH2O(l)Na + (aq)+OH(aq)

According to Arrhenius, the characteristic properties of acids and bases are due exclusively to the presence of H+ and OH ions, respectively, in solution.

Although Arrhenius’s ideas were widely accepted, his definition of acids and bases had two major limitations. First, because acids and bases were defined in terms of ions obtained from water, the Arrhenius concept applied only to substances in aqueous solution. Second, and more important, the Arrhenius definition predicted that only substances that dissolve in water to produce H+ and OH ions should exhibit the properties of acids and bases, respectively. For example, according to the Arrhenius definition, the reaction of ammonia (a base) with gaseous HCl (an acid) to give ammonium chloride () is not an acid–base reaction because it does not involve H+ and OH:

Equation 4.19

NH3(g) + HCl(g) → NH4Cl(s)

The Brønsted–Lowry Definition of Acids and Bases

Because of the limitations of the Arrhenius definition, a more general definition of acids and bases was needed. One was proposed independently in 1923 by the Danish chemist J. N. Brønsted (1879–1947) and the British chemist T. M. Lowry (1874–1936), who defined acid–base reactions in terms of the transfer of a proton (H+ ion) from one substance to another.

According to Brønsted and Lowry, an acidA substance with at least one hydrogen atom that can dissociate to form an anion and an H+ ion (a proton) in aqueous solution, thereby foming an acidic solution. is any substance that can donate a proton, and a baseA substance that produces one or more hydroxide ions (OH) and a cation when dissolved in aqueous solution, thereby forming a basic solution. is any substance that can accept a proton. The Brønsted–Lowry definition of an acid is essentially the same as the Arrhenius definition, except that it is not restricted to aqueous solutions. The Brønsted–Lowry definition of a base, however, is far more general because the hydroxide ion is just one of many substances that can accept a proton. Ammonia, for example, reacts with a proton to form NH4+, so in , NH3 is a Brønsted–Lowry base and HCl is a Brønsted–Lowry acid. Because of its more general nature, the Brønsted–Lowry definition is used throughout this text unless otherwise specified. We will present a third definition—Lewis acids and bases—in when we discuss molecular structure.

Polyprotic Acids

Acids differ in the number of protons they can donate. For example, monoprotic acidsA compound that is capable of donating one proton per molecule. are compounds that are capable of donating a single proton per molecule. Monoprotic acids include HF, HCl, HBr, HI, HNO3, and HNO2. All carboxylic acids that contain a single −CO2H group, such as acetic acid (CH3CO2H), are monoprotic acids, dissociating to form RCO2 and H+ (). Polyprotic acidsA compound that can donate more than one proton per molecule. can donate more than one proton per molecule. For example, H2SO4 can donate two H+ ions in separate steps, so it is a diprotic acidA compound that can donate two protons per molecule in separate steps., and H3PO4, which is capable of donating three protons in successive steps, is a triprotic acidA compound that can donate three protons per molecule in separate steps. (, , and ):

Equation 4.20

H3PO4(l)H2O(l)H + (aq)+H2PO4(aq)

Equation 4.21

H2PO4(aq)H + (aq)+HPO42(aq)

Equation 4.22

HPO42(aq)H + (aq)+PO43(aq)

In chemical equations such as these, a double arrow is used to indicate that both the forward and reverse reactions occur simultaneously, so the forward reaction does not go to completion. Instead, the solution contains significant amounts of both reactants and products. Over time, the reaction reaches a state in which the concentration of each species in solution remains constant. The reaction is then said to be in equilibriumThe point at which the rates of the forward and reverse reactions become the same, so that the net composition of the system no longer changes with time.. We will return to the concept of equilibrium in more detail in .

Strengths of Acids and Bases

We will not discuss the strengths of acids and bases quantitatively until . Qualitatively, however, we can state that strong acidsAn acid that reacts essentially completely with water to give H+ and the corresponding anion. react essentially completely with water to give H+ and the corresponding anion. Similarly, strong basesA base that dissociates essentially completely in water to give OH and the corresponding cation. dissociate essentially completely in water to give OH and the corresponding cation. Strong acids and strong bases are both strong electrolytes. In contrast, only a fraction of the molecules of weak acidsAn acid in which only a fraction of the molecules react with water to produce H+ and the corresponding anion. and weak basesA base in which only a fraction of the molecules react with water to produce OH and the corresponding cation. react with water to produce ions, so weak acids and weak bases are also weak electrolytes. Typically less than 5% of a weak electrolyte dissociates into ions in solution, whereas more than 95% is present in undissociated form.

In practice, only a few strong acids are commonly encountered: HCl, HBr, HI, HNO3, HClO4, and H2SO4 (H3PO4 is only moderately strong). The most common strong bases are ionic compounds that contain the hydroxide ion as the anion; three examples are NaOH, KOH, and Ca(OH)2. Common weak acids include HCN, H2S, HF, oxoacids such as HNO2 and HClO, and carboxylic acids such as acetic acid. The ionization reaction of acetic acid is as follows:

Equation 4.23

CH3CO2H(l)H2O(l)H + (aq)+CH3CO2(aq)

Although acetic acid is very soluble in water, almost all of the acetic acid in solution exists in the form of neutral molecules (less than 1% dissociates), as we stated in . Sulfuric acid is unusual in that it is a strong acid when it donates its first proton () but a weak acid when it donates its second proton () as indicated by the single and double arrows, respectively:

Equation 4.24

H2SO4(l)strong acidH2O(l)H + (aq)+HSO4(aq)

Equation 4.25

HSO4(aq)weak acidH + (aq)+SO42(aq)

Consequently, an aqueous solution of sulfuric acid contains H+(aq) ions and a mixture of HSO4(aq) and SO42−(aq) ions but no H2SO4 molecules.

The most common weak base is ammonia, which reacts with water to form small amounts of hydroxide ion:

Equation 4.26

NH3(g)+H2O(l)NH4 + (aq)+OH(aq)

Most of the ammonia (>99%) is present in the form of NH3(g). Amines, which are organic analogues of ammonia, are also weak bases, as are ionic compounds that contain anions derived from weak acids (such as S2−).

lists some common strong acids and bases. Acids other than the six common strong acids are almost invariably weak acids. The only common strong bases are the hydroxides of the alkali metals and the heavier alkaline earths (Ca, Sr, and Ba); any other bases you encounter are most likely weak. Remember that there is no correlation between solubility and whether a substance is a strong or a weak electrolyte! Many weak acids and bases are extremely soluble in water.

Note the Pattern

There is no correlation between the solubility of a substance and whether it is a strong electrolyte, a weak electrolyte, or a nonelectrolyte.

Table 4.3 Common Strong Acids and Bases

Strong Acids Strong Bases
Hydrogen Halides Oxoacids Group 1 Hydroxides Hydroxides of the Heavier Group 2 Elements
HCl HNO3 LiOH Ca(OH)2
HBr H2SO4 NaOH Sr(OH)2
HI HClO4 KOH Ba(OH)2
RbOH
CsOH

Example 13

Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these.

  1. CH3CH2CO2H
  2. CH3OH
  3. Sr(OH)2
  4. CH3CH2NH2
  5. HBrO4

Given: compound

Asked for: acid or base strength

Strategy:

A Determine whether the compound is organic or inorganic.

B If inorganic, determine whether the compound is acidic or basic by the presence of dissociable H+ or OH ions, respectively. If organic, identify the compound as a weak base or a weak acid by the presence of an amine or a carboxylic acid group, respectively. Recall that all polyprotic acids except H2SO4 are weak acids.

Solution:

  1. A This compound is propionic acid, which is organic. B It contains a carboxylic acid group analogous to that in acetic acid, so it must be a weak acid.
  2. A CH3OH is methanol, an organic compound that contains the −OH group. B As a covalent compound, it does not dissociate to form the OH ion. Because it does not contain a carboxylic acid (−CO2H) group, methanol also cannot dissociate to form H+(aq) ions. Thus we predict that in aqueous solution methanol is neither an acid nor a base.
  3. A Sr(OH)2 is an inorganic compound that contains one Sr2+ and two OH ions per formula unit. B We therefore expect it to be a strong base, similar to Ca(OH)2.
  4. A CH3CH2NH2 is an amine (ethylamine), an organic compound in which one hydrogen of ammonia has been replaced by an R group. B Consequently, we expect it to behave similarly to ammonia (), reacting with water to produce small amounts of the OH ion. Ethylamine is therefore a weak base.
  5. A HBrO4 is perbromic acid, an inorganic compound. B It is not listed in as one of the common strong acids, but that does not necessarily mean that it is a weak acid. If you examine the periodic table, you can see that Br lies directly below Cl in group 17. We might therefore expect that HBrO4 is chemically similar to HClO4, a strong acid—and, in fact, it is.

Exercise

Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these.

  1. Ba(OH)2
  2. HIO4
  3. CH3CH2CH2CO2H
  4. (CH3)2NH
  5. CH2O

Answer:

  1. strong base
  2. strong acid
  3. weak acid
  4. weak base
  5. none of these; formaldehyde is a neutral molecule

The Hydronium Ion

Because isolated protons are very unstable and hence very reactive, an acid never simply “loses” an H+ ion. Instead, the proton is always transferred to another substance, which acts as a base in the Brønsted–Lowry definition. Thus in every acid–base reaction, one species acts as an acid and one species acts as a base. Occasionally, the same substance performs both roles, as you will see later. When a strong acid dissolves in water, the proton that is released is transferred to a water molecule that acts as a proton acceptor or base, as shown for the dissociation of sulfuric acid:

Equation 4.27

H2SO4(l)acid(proton donor)+H2O(l)base(proton acceptor)H3O + (aq)acid+HSO4(aq)base

Technically, therefore, it is imprecise to describe the dissociation of a strong acid as producing H+(aq) ions, as we have been doing. The resulting H3O+ ion, called the hydronium ionThe H3O+ ion, represented as H+(aq)., is a more accurate representation of H+(aq). For the sake of brevity, however, in discussing acid dissociation reactions, we will often show the product as H+(aq) (as in ) with the understanding that the product is actually the H3O+(aq) ion.

Conversely, bases that do not contain the hydroxide ion accept a proton from water, so small amounts of OH are produced, as in the following:

Equation 4.28

NH3(g)base+H2O(l)acidNH4 + (aq)acid+OH(aq)base

Again, the double arrow indicates that the reaction does not go to completion but rather reaches a state of equilibrium. In this reaction, water acts as an acid by donating a proton to ammonia, and ammonia acts as a base by accepting a proton from water. Thus water can act as either an acid or a base by donating a proton to a base or by accepting a proton from an acid. Substances that can behave as both an acid and a base are said to be amphotericWhen substances can behave as both an acid and a base..

The products of an acid–base reaction are also an acid and a base. In , for example, the products of the reaction are the hydronium ion, here an acid, and the hydrogen sulfate ion, here a weak base. In , the products are NH4+, an acid, and OH, a base. The product NH4+ is called the conjugate acidThe substance formed when a Brønsted–Lowry base accepts a proton. of the base NH3, and the product OH is called the conjugate baseThe substance formed when a Brønsted–Lowry acid donates a proton. of the acid H2O. Thus all acid–base reactions actually involve two conjugate acid–base pairsAn acid and a base that differ by only one hydrogen ion. All acid–base reactions involve two conjugate acid–base pairs, the Brønsted–Lowry acid and the base it forms after donating its proton, and the Brønsted–Lowry base and the acid it forms after accepting a proton.; in , they are NH4+/NH3 and H2O/OH. We will describe the relationship between conjugate acid–base pairs in more detail in .

Neutralization Reactions

A neutralization reactionA chemical reaction in which an acid and a base react in stoichiometric amounts to produce water and a salt. is one in which an acid and a base react in stoichiometric amounts to produce water and a saltThe general term for any ionic substance that does not have OH as the anion or H+ as the cation., the general term for any ionic substance that does not have OH as the anion or H+ as the cation. If the base is a metal hydroxide, then the general formula for the reaction of an acid with a base is described as follows: Acid plus base yields water plus salt. For example, the reaction of equimolar amounts of HBr and NaOH to give water and a salt (NaBr) is a neutralization reaction:

Equation 4.29

HBr(aq)acid+NaOH(aq)baseH2O(l)water+NaBr(aq)salt

Note the Pattern

Acid plus base yields water plus salt.

If we write the complete ionic equation for the reaction in , we see that Na+(aq) and Br(aq) are spectator ions and are not involved in the reaction:

Equation 4.30

H + (aq)+Br(aq)+Na + (aq)+OH(aq)H2O(l)+Na + (aq)+Br(aq)

The overall reaction is therefore simply the combination of H+(aq) and OH(aq) to produce H2O, as shown in the net ionic equation:

Equation 4.31

H+(aq) + OH(aq) → H2O(l)

The net ionic equation for the reaction of any strong acid with any strong base is identical to .

The strengths of the acid and the base generally determine whether the reaction goes to completion. The reaction of any strong acid with any strong base goes essentially to completion, as does the reaction of a strong acid with a weak base, and a weak acid with a strong base. Examples of the last two are as follows:

Equation 4.32

HCl(aq)strong acid+NH3(aq)weak baseNH4Cl(aq)salt

Equation 4.33

CH3CO2H(aq)weak acid+NaOH(aq)strong baseCH3CO2Na(aq)salt+H2O(l)

Sodium acetate is written with the organic component first followed by the cation, as is usual for organic salts. Most reactions of a weak acid with a weak base also go essentially to completion. One example is the reaction of acetic acid with ammonia:

Equation 4.34

CH3CO2H(aq)weak acid+NH3(aq)weak baseCH3CO2NH4(aq)salt

An example of an acid–base reaction that does not go to completion is the reaction of a weak acid or a weak base with water, which is both an extremely weak acid and an extremely weak base. We will discuss these reactions in more detail in .

Note the Pattern

Except for the reaction of a weak acid or a weak base with water, acid–base reactions essentially go to completion.

In some cases, the reaction of an acid with an anion derived from a weak acid (such as HS) produces a gas (in this case, H2S). Because the gaseous product escapes from solution in the form of bubbles, the reverse reaction cannot occur. Therefore, these reactions tend to be forced, or driven, to completion. Examples include reactions in which an acid is added to ionic compounds that contain the HCO3, CN, or S2− anions, all of which are driven to completion ():

Equation 4.35

HCO3(aq)+H + (aq)H2CO3(aq)H2CO3(aq)CO2(g)+H2O(l)

Equation 4.36

CN(aq)+H + (aq) → HCN(g)

Equation 4.37

S2−(aq)+H + (aq)HS(aq)HS(aq)+H + (aq)H2S(g)

Figure 4.14 The Reaction of Dilute Aqueous HNO3 with a Solution of Na2CO3

Note the vigorous formation of gaseous CO2.

The reactions in are responsible for the rotten egg smell that is produced when metal sulfides come in contact with acids.

Example 14

Calcium propionate is used to inhibit the growth of molds in foods, tobacco, and some medicines. Write a balanced chemical equation for the reaction of aqueous propionic acid (CH3CH2CO2H) with aqueous calcium hydroxide [Ca(OH)2] to give calcium propionate. Do you expect this reaction to go to completion, making it a feasible method for the preparation of calcium propionate?

Given: reactants and product

Asked for: balanced chemical equation and whether the reaction will go to completion

Strategy:

Write the balanced chemical equation for the reaction of propionic acid with calcium hydroxide. Based on their acid and base strengths, predict whether the reaction will go to completion.

Solution:

Propionic acid is an organic compound that is a weak acid, and calcium hydroxide is an inorganic compound that is a strong base. The balanced chemical equation is as follows:

2CH3CH2CO2H(aq) + Ca(OH)2(aq) → (CH3CH2CO2)2Ca(aq) + 2H2O(l)

The reaction of a weak acid and a strong base will go to completion, so it is reasonable to prepare calcium propionate by mixing solutions of propionic acid and calcium hydroxide in a 2:1 mole ratio.

Exercise

Write a balanced chemical equation for the reaction of solid sodium acetate with dilute sulfuric acid to give sodium sulfate.

Answer: 2CH3CO2Na(s) + H2SO4(aq) → Na2SO4(aq) + 2CH3CO2H(aq)

Stomach acid. An antacid tablet reacts with 0.1 M HCl (the approximate concentration found in the human stomach).

One of the most familiar and most heavily advertised applications of acid–base chemistry is antacids, which are bases that neutralize stomach acid. The human stomach contains an approximately 0.1 M solution of hydrochloric acid that helps digest foods. If the protective lining of the stomach breaks down, this acid can attack the stomach tissue, resulting in the formation of an ulcer. Because one factor that is believed to contribute to the formation of stomach ulcers is the production of excess acid in the stomach, many individuals routinely consume large quantities of antacids. The active ingredients in antacids include sodium bicarbonate and potassium bicarbonate (NaHCO3 and KHCO3; Alka-Seltzer); a mixture of magnesium hydroxide and aluminum hydroxide [Mg(OH)2 and Al(OH)3; Maalox, Mylanta]; calcium carbonate (CaCO3; Tums); and a complex salt, dihydroxyaluminum sodium carbonate [NaAl(OH)2CO3; original Rolaids]. Each has certain advantages and disadvantages. For example, Mg(OH)2 is a powerful laxative (it is the active ingredient in milk of magnesia), whereas Al(OH)3 causes constipation. When mixed, each tends to counteract the unwanted effects of the other. Although all antacids contain both an anionic base (OH, CO32−, or HCO3) and an appropriate cation, they differ substantially in the amount of active ingredient in a given mass of product.

Example 15

Assume that the stomach of someone suffering from acid indigestion contains 75 mL of 0.20 M HCl. How many Tums tablets are required to neutralize 90% of the stomach acid, if each tablet contains 500 mg of CaCO3? (Neutralizing all of the stomach acid is not desirable because that would completely shut down digestion.)

Given: volume and molarity of acid and mass of base in an antacid tablet

Asked for: number of tablets required for 90% neutralization

Strategy:

A Write the balanced chemical equation for the reaction and then decide whether the reaction will go to completion.

B Calculate the number of moles of acid present. Multiply the number of moles by the percentage to obtain the quantity of acid that must be neutralized. Using mole ratios, calculate the number of moles of base required to neutralize the acid.

C Calculate the number of moles of base contained in one tablet by dividing the mass of base by the corresponding molar mass. Calculate the number of tablets required by dividing the moles of base by the moles contained in one tablet.

Solution:

A We first write the balanced chemical equation for the reaction:

2HCl(aq) + CaCO3(s) → CaCl2(aq) + H2CO3(aq)

Each carbonate ion can react with 2 mol of H+ to produce H2CO3, which rapidly decomposes to H2O and CO2. Because HCl is a strong acid and CO32− is a weak base, the reaction will go to completion.

B Next we need to determine the number of moles of HCl present:

75 mL(L1000 mL)(0.20 mol HClL)=0.015 mol HCl

Because we want to neutralize only 90% of the acid present, we multiply the number of moles of HCl by 0.90:

(0.015 mol HCl)(0.90) = 0.014 mol HCl

We know from the stoichiometry of the reaction that each mole of CaCO3 reacts with 2 mol of HCl, so we need

moles CaCO3= 0.014 mol HCl(1 mol CaCO3mol HCl)= 0.0070 mol CaCO3

C Each Tums tablet contains

(500 mg CaCO31 Tums tablet)(g1000 mg CaCO3)(1 mol CaCO3100.1 g)= 0.00500 mol CaCO3

Thus we need 0.0070 mol CaCO30.00500 mol CaCO3= 1.4 Tums tablets.

Exercise

Assume that as a result of overeating, a person’s stomach contains 300 mL of 0.25 M HCl. How many Rolaids tablets must be consumed to neutralize 95% of the acid, if each tablet contains 400 mg of NaAl(OH)2CO3? The neutralization reaction can be written as follows:

NaAl(OH)2CO3(s) + 4HCl(aq) → AlCl3(aq) + NaCl(aq) + CO2(g) + 3H2O(l)

Answer: 6.4 tablets

The pH Scale

One of the key factors affecting reactions that occur in dilute solutions of acids and bases is the concentration of H+ and OH ions. The pH scaleA logarithmic scale used to express the hydrogen ion (H+) concentration of a solution, making it possible to describe acidity or basicity quantitatively. provides a convenient way of expressing the hydrogen ion (H+) concentration of a solution and enables us to describe acidity or basicity in quantitative terms.

Pure liquid water contains extremely low but measurable concentrations of H3O+(aq) and OH(aq) ions produced via an autoionization reaction, in which water acts simultaneously as an acid and as a base:

Equation 4.38

H2O(l) + H2O(l) ⇌ H3O+(aq) + OH(aq)

The concentration of hydrogen ions in pure water is only 1.0 × 10−7 M at 25°C. Because the autoionization reaction produces both a proton and a hydroxide ion, the OH concentration in pure water is also 1.0 × 10−7 M. Pure water is a neutral solutionA solution in which the total positive charge from all the cations is matched by an identical total negative charge from all the anions., in which [H+] = [OH] = 1.0 × 10−7 M.

The pH scale describes the hydrogen ion concentration of a solution in a way that avoids the use of exponential notation; pHThe negative base-10 logarithm of the hydrogen ion concentration: pH =log[H+]. is defined as the negative base-10 logarithm of the hydrogen ion concentration:pH is actually defined as the negative base-10 logarithm of hydrogen ion activity. As you will learn in a more advanced course, the activity of a substance in solution is related to its concentration. For dilute solutions such as those we are discussing, the activity and the concentration are approximately the same.

Equation 4.39

pH = −log[H+]

Conversely,

Equation 4.40

[H+] = 10−pH

(If you are not familiar with logarithms or using a calculator to obtain logarithms and antilogarithms, consult Essential Skills 3 in 0.)

Because the hydrogen ion concentration is 1.0 × 10−7 M in pure water at 25°C, the pH of pure liquid water (and, by extension, of any neutral solution) is

Equation 4.41

pH = −log[1.0 × 10−7] = 7.00

Adding an acid to pure water increases the hydrogen ion concentration and decreases the hydroxide ion concentration because a neutralization reaction occurs, such as that shown in . Because the negative exponent of [H+] becomes smaller as [H+] increases, the pH decreases with increasing [H+]. For example, a 1.0 M solution of a strong monoprotic acid such as HCl or HNO3 has a pH of 0.00:

Equation 4.42

pH = −log[1.0] = 0.00

Note the Pattern

pH decreases with increasing [H+].

Conversely, adding a base to pure water increases the hydroxide ion concentration and decreases the hydrogen ion concentration. Because the autoionization reaction of water does not go to completion, neither does the neutralization reaction. Even a strongly basic solution contains a detectable amount of H+ ions. For example, a 1.0 M OH solution has [H+] = 1.0 × 10−14 M. The pH of a 1.0 M NaOH solution is therefore

Equation 4.43

pH = −log[1.0 × 10−14] = 14.00

For practical purposes, the pH scale runs from pH = 0 (corresponding to 1 M H+) to pH 14 (corresponding to 1 M OH), although pH values less than 0 or greater than 14 are possible.

We can summarize the relationships between acidity, basicity, and pH as follows:

  • If pH = 7.0, the solution is neutral.
  • If pH < 7.0, the solution is acidic.
  • If pH > 7.0, the solution is basic.

Keep in mind that the pH scale is logarithmic, so a change of 1.0 in the pH of a solution corresponds to a tenfold change in the hydrogen ion concentration. The foods and consumer products we encounter daily represent a wide range of pH values, as shown in .

Figure 4.15 A Plot of pH versus [H+] for Some Common Aqueous Solutions

Although many substances exist in a range of pH values (indicated in parentheses), they are plotted using typical values.

Example 16

  1. What is the pH of a 2.1 × 10−2 M aqueous solution of HClO4?
  2. The pH of a vinegar sample is 3.80. What is its hydrogen ion concentration?

Given: molarity of acid or pH

Asked for: pH or [H+]

Strategy:

Using the balanced chemical equation for the acid dissociation reaction and or 4.40, determine [H+] and convert it to pH or vice versa.

Solution:

  1. HClO4 (perchloric acid) is a strong acid, so it dissociates completely into H+ ions and ClO4 ions:

    HClO4(l) → H+(aq) + ClO4(aq)

    The H+ ion concentration is therefore the same as the perchloric acid concentration. The pH of the perchloric acid solution is thus

    pH = −log[H+] = −log(2.1 × 10−2) = 1.68

    The result makes sense: the H+ ion concentration is between 10−1 M and 10−2 M, so the pH must be between 1 and 2.

    Note: The assumption that [H+] is the same as the concentration of the acid is valid for only strong acids. Because weak acids do not dissociate completely in aqueous solution, a more complex procedure is needed to calculate the pH of their solutions, which we will describe in .

  2. We are given the pH and asked to calculate the hydrogen ion concentration. From ,

    10−pH = [H+]

Thus [H+] = 10−3.80 = 1.6 × 10−4 M.

Exercise

  1. What is the pH of a 3.0 × 10−5 M aqueous solution of HNO3?
  2. What is the hydrogen ion concentration of turnip juice, which has a pH of 5.41?

Answer:

  1. pH = 4.52
  2. [H+] = 3.9 × 10−6 M

Tools have been developed that make the measurement of pH simple and convenient (). For example, pH paper consists of strips of paper impregnated with one or more acid–base indicatorsAn intensely colored organic molecule whose color changes dramatically depending on the pH of the solution., which are intensely colored organic molecules whose colors change dramatically depending on the pH of the solution. Placing a drop of a solution on a strip of pH paper and comparing its color with standards give the solution’s approximate pH. A more accurate tool, the pH meter, uses a glass electrode, a device whose voltage depends on the H+ ion concentration.

Figure 4.16 Two Ways of Measuring the pH of a Solution: pH Paper and a pH Meter

Note that both show that the pH is 1.7, but the pH meter gives a more precise value.

Key Equations

definition of pH

: pH = −log[H+]

: [H+] = 10−pH

Summary

Acid–base reactions require both an acid and a base. In Brønsted–Lowry terms, an acid is a substance that can donate a proton (H+), and a base is a substance that can accept a proton. All acid–base reactions contain two acid–base pairs: the reactants and the products. Acids can donate one proton (monoprotic acids), two protons (diprotic acids), or three protons (triprotic acids). Compounds that are capable of donating more than one proton are generally called polyprotic acids. Acids also differ in their tendency to donate a proton, a measure of their acid strength. Strong acids react completely with water to produce H3O+(aq) (the hydronium ion), whereas weak acids dissociate only partially in water. Conversely, strong bases react completely with water to produce the hydroxide ion, whereas weak bases react only partially with water to form hydroxide ions. The reaction of a strong acid with a strong base is a neutralization reaction, which produces water plus a salt.

The acidity or basicity of an aqueous solution is described quantitatively using the pH scale. The pH of a solution is the negative logarithm of the H+ ion concentration and typically ranges from 0 for strongly acidic solutions to 14 for strongly basic ones. Because of the autoionization reaction of water, which produces small amounts of hydronium ions and hydroxide ions, a neutral solution of water contains 1 × 10−7 M H+ ions and has a pH of 7.0. An indicator is an intensely colored organic substance whose color is pH dependent; it is used to determine the pH of a solution.

Key Takeaway

  • An acidic solution and a basic solution react together in a neutralization reaction that also forms a salt.

Conceptual Problems

  1. Why was it necessary to expand on the Arrhenius definition of an acid and a base? What specific point does the Brønsted–Lowry definition address?

  2. State whether each compound is an acid, a base, or a salt.

    1. CaCO3
    2. NaHCO3
    3. H2SO4
    4. CaCl2
    5. Ba(OH)2
  3. State whether each compound is an acid, a base, or a salt.

    1. NH3
    2. NH4Cl
    3. H2CO3
    4. CH3COOH
    5. NaOH
  4. Classify each compound as a strong acid, a weak acid, a strong base, or a weak base in aqueous solution.

    1. sodium hydroxide
    2. acetic acid
    3. magnesium hydroxide
    4. tartaric acid
    5. sulfuric acid
    6. ammonia
    7. hydroxylamine (NH2OH)
    8. hydrocyanic acid
  5. Decide whether each compound forms an aqueous solution that is strongly acidic, weakly acidic, strongly basic, or weakly basic.

    1. propanoic acid
    2. hydrobromic acid
    3. methylamine
    4. lithium hydroxide
    5. citric acid
    6. sodium acetate
    7. ammonium chloride
    8. barium hydroxide
  6. What is the relationship between the strength of an acid and the strength of the conjugate base derived from that acid? Would you expect the CH3CO2 ion to be a strong base or a weak base? Why? Is the hydronium ion a strong acid or a weak acid? Explain your answer.

  7. What are the products of an acid–base reaction? Under what circumstances is one of the products a gas?

  8. Explain how an aqueous solution that is strongly basic can have a pH, which is a measure of the acidity of a solution.

Answer

    1. weakly acidic
    2. strongly acidic
    3. weakly basic
    4. strongly basic
    5. weakly acidic
    6. weakly basic
    7. weakly acidic
    8. strongly basic

Numerical Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 3 (0) before proceeding to the Numerical Problems.

  1. Derive an equation to relate the hydrogen ion concentration to the molarity of a solution of a strong monoprotic acid.

  2. Derive an equation to relate the hydroxide ion concentration to the molarity of a solution of

    1. a group I hydroxide.
    2. a group II hydroxide.
  3. Given the following salts, identify the acid and the base in the neutralization reactions and then write the complete ionic equation:

    1. barium sulfate
    2. lithium nitrate
    3. sodium bromide
    4. calcium perchlorate
  4. What is the pH of each solution?

    1. 5.8 × 10−3 mol of HNO3 in 257 mL of water
    2. 0.0079 mol of HI in 750 mL of water
    3. 0.011 mol of HClO4 in 500 mL of water
    4. 0.257 mol of HBr in 5.00 L of water
  5. What is the hydrogen ion concentration of each substance in the indicated pH range?

    1. black coffee (pH 5.10)
    2. milk (pH 6.30–7.60)
    3. tomatoes (pH 4.00–4.40)
  6. What is the hydrogen ion concentration of each substance in the indicated pH range?

    1. orange juice (pH 3–4)
    2. fresh egg white (pH 7.60–7.80)
    3. lemon juice (pH 2.20–2.40)
  7. What is the pH of a solution prepared by diluting 25.00 mL of 0.879 M HCl to a volume of 555 mL?

  8. Vinegar is primarily an aqueous solution of acetic acid. Commercial vinegar typically contains 5.0 g of acetic acid in 95.0 g of water. What is the concentration of commercial vinegar? If only 3.1% of the acetic acid dissociates to CH3CO2 and H+, what is the pH of the solution? (Assume the density of the solution is 1.00 g/mL.)

  9. If a typical household cleanser is 0.50 M in strong base, what volume of 0.998 M strong monoprotic acid is needed to neutralize 50.0 mL of the cleanser?

  10. A 25.00 mL sample of a 0.9005 M solution of HCl is diluted to 500.0 mL. What is the molarity of the final solution? How many milliliters of 0.223 M NaOH are needed to neutralize 25.00 mL of this final solution?

  11. If 20.0 mL of 0.10 M NaOH are needed to neutralize 15.0 mL of gastric fluid, what is the molarity of HCl in the fluid? (Assume all the acidity is due to the presence of HCl.) What other base might be used instead of NaOH?

  12. Malonic acid (C3H4O4) is a diprotic acid used in the manufacture of barbiturates. How many grams of malonic acid are in a 25.00 mL sample that requires 32.68 mL of 1.124 M KOH for complete neutralization to occur? Malonic acid is a dicarboxylic acid; propose a structure for malonic acid.

  13. Describe how you would prepare 500 mL of a 1.00 M stock solution of HCl from an HCl solution that is 12.11 M. Using your stock solution, how would you prepare 500 mL of a solution that is 0.012 M in HCl?

  14. Given a stock solution that is 8.52 M in HBr, describe how you would prepare a 500 mL solution with each concentration.

    1. 2.50 M
    2. 4.00 × 10−3 M
    3. 0.989 M
  15. How many moles of solute are contained in each?

    1. 25.00 mL of 1.86 M NaOH
    2. 50.00 mL of 0.0898 M HCl
    3. 13.89 mL of 0.102 M HBr
  16. A chemist needed a solution that was approximately 0.5 M in HCl but could measure only 10.00 mL samples into a 50.00 mL volumetric flask. Propose a method for preparing the solution. (Assume that concentrated HCl is 12.0 M.)

  17. Write the balanced chemical equation for each reaction.

    1. perchloric acid with potassium hydroxide
    2. nitric acid with calcium hydroxide
  18. Write the balanced chemical equation for each reaction.

    1. solid strontium hydroxide with hydrobromic acid
    2. aqueous sulfuric acid with solid sodium hydroxide
  19. A neutralization reaction gives calcium nitrate as one of the two products. Identify the acid and the base in this reaction. What is the second product? If the product had been cesium iodide, what would have been the acid and the base? What is the complete ionic equation for each reaction?

Answers

  1. [H3O+] = [HA] M

    1. H2SO4 and Ba(OH)2; 2H+ + SO42− + Ba2+ + 2OH → 2H2O + Ba2+ + SO42−
    2. HNO3 and LiOH; H+ + NO3 + Li+ + OH → H2O + Li+ + NO3
    3. HBr and NaOH; H+ + Br + Na+ + OH → H2O + Na+ + Br
    4. HClO4 and Ca(OH)2; 2H+ + 2ClO4 + Ca2+ + 2OH → 2H2O + Ca2+ + 2ClO4
    1. 7.9 × 10−6 M H+
    2. 5.0 × 10−7 to 2.5 × 10−8 M H+
    3. 1.0 × 10−4 to 4.0 × 10−5 M H+
  2. pH = 1.402

  3. 25 mL

  4. 0.13 M HCl; magnesium carbonate, MgCO3, or aluminum hydroxide, Al(OH)3

  5. 1.00 M solution: dilute 41.20 mL of the concentrated solution to a final volume of 500 mL. 0.012 M solution: dilute 12.0 mL of the 1.00 M stock solution to a final volume of 500 mL.

    1. 4.65 × 10−2 mol NaOH
    2. 4.49 × 10−3 mol HCl
    3. 1.42 × 10−3 mol HBr
    1. HClO4 + KOH → KClO4 + H2O
    2. 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O
  6. The acid is nitric acid, and the base is calcium hydroxide. The other product is water.

    2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O

    The acid is hydroiodic acid, and the base is cesium hydroxide. The other product is water.

    HI + CsOH → CsI + H2O

    The complete ionic equations are

    2H+ + 2NO3 + Ca2+ + 2OH → Ca2+ + 2NO3 + H2O H+ + I + Cs+ + OH → Cs+ + I + H2O

4.7 The Chemistry of Acid Rain

Learning Objective

  1. To understand the chemistry of acid rain.

Acid–base reactions can have a strong environmental impact. For example, a dramatic increase in the acidity of rain and snow over the past 150 years is dissolving marble and limestone surfaces, accelerating the corrosion of metal objects, and decreasing the pH of natural waters. This environmental problem is called acid rainPrecipitation that is dramatically more acidic because of human activities. and has significant consequences for all living organisms. To understand acid rain requires an understanding of acid–base reactions in aqueous solution.

The term acid rain is actually somewhat misleading because even pure rainwater collected in areas remote from civilization is slightly acidic (pH ≈ 5.6) due to dissolved carbon dioxide, which reacts with water to give carbonic acid, a weak acid:

Equation 4.44

CO2(g) + H2O(l) ⇌ H2CO3(aq) ⇌ H+(aq) + HCO3(aq)

The English chemist Robert Angus Smith is generally credited with coining the phrase acid rain in 1872 to describe the increased acidity of the rain in British industrial centers (such as Manchester), which was apparently caused by the unbridled excesses of the early Industrial Revolution, although the connection was not yet understood. At that time, there was no good way to measure hydrogen ion concentrations, so it is difficult to know the actual pH of the rain observed by Smith. Typical pH values for rain in the continental United States now range from 4 to 4.5, with values as low as 2.0 reported for areas such as Los Angeles. Recall from that rain with a pH of 2 is comparable in acidity to lemon juice, and even “normal” rain is now as acidic as tomato juice or black coffee.

What is the source of the increased acidity in rain and snow? Chemical analysis shows the presence of large quantities of sulfate (SO42−) and nitrate (NO3) ions, and a wide variety of evidence indicates that a significant fraction of these species come from nitrogen and sulfur oxides produced during the combustion of fossil fuels. At the high temperatures found in both internal combustion engines and lightning discharges, molecular nitrogen and molecular oxygen react to give nitric oxide:

Equation 4.45

N2(g) + O2(g) → 2NO(g)

Nitric oxide then reacts rapidly with excess oxygen to give nitrogen dioxide, the compound responsible for the brown color of smog:

Equation 4.46

2NO(g) + O2(g) → 2NO2(g)

When nitrogen dioxide dissolves in water, it forms a 1:1 mixture of nitrous acid and nitric acid:

Equation 4.47

2NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq)

Because molecular oxygen eventually oxidizes nitrous acid to nitric acid, the overall reaction is

Equation 4.48

2N2(g) + 5O2(g) + 2H2O(l) → 4HNO3(aq)

Large amounts of sulfur dioxide have always been released into the atmosphere by natural sources, such as volcanoes, forest fires, and the microbial decay of organic materials, but for most of Earth’s recorded history the natural cycling of sulfur from the atmosphere into oceans and rocks kept the acidity of rain and snow in check. Unfortunately, the burning of fossil fuels seems to have tipped the balance. Many coals contain as much as 5%–6% pyrite (FeS2) by mass, and fuel oils typically contain at least 0.5% sulfur by mass. Since the mid-19th century, these fuels have been burned on a huge scale to supply the energy needs of our modern industrial society, releasing tens of millions of tons of additional SO2 into the atmosphere annually. In addition, roasting sulfide ores to obtain metals such as zinc and copper produces large amounts of SO2 via reactions such as

Equation 4.49

2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)

Regardless of the source, the SO2 dissolves in rainwater to give sulfurous acid (), which is eventually oxidized by oxygen to sulfuric acid ():

Equation 4.50

SO2(g) + H2O(l) → H2SO3(aq)

Equation 4.51

2H2SO3(aq) + O2(g) → 2H2SO4(aq)

Concerns about the harmful effects of acid rain have led to strong pressure on industry to minimize the release of SO2 and NO. For example, coal-burning power plants now use SO2 “scrubbers,” which trap SO2 by its reaction with lime (CaO) to produce calcium sulfite dihydrate (CaSO3·2H2O; ).

Figure 4.17 Schematic Diagram of a Wet Scrubber System

In coal-burning power plants, SO2 can be removed (“scrubbed”) from exhaust gases by its reaction with a lime (CaO) and water spray to produce calcium sulfite dihydrate (CaSO3·2H2O). Removing SO2 from the gases prevents its conversion to SO3 and subsequent reaction with rainwater (acid rain). Scrubbing systems are now commonly used to minimize the environmental effects of large-scale fossil fuel combustion.

The damage that acid rain does to limestone and marble buildings and sculptures is due to a classic acid–base reaction. Marble and limestone both consist of calcium carbonate (CaCO3), a salt derived from the weak acid H2CO3. As we saw in , the reaction of a strong acid with a salt of a weak acid goes to completion. Thus we can write the reaction of limestone or marble with dilute sulfuric acid as follows:

Equation 4.52

CaCO3(s) + H2SO4(aq) → CaSO4(s) + H2O(l) + CO2(g)

Because CaSO4 is sparingly soluble in water, the net result of this reaction is to dissolve the marble or limestone. The Lincoln Memorial in Washington, DC, which was built in 1922, already shows significant damage from acid rain, and many older objects are exhibiting even greater damage (). Metal objects can also suffer damage from acid rain through oxidation–reduction reactions, which are discussed in .

Figure 4.18 Acid Rain Damage to a Statue of George Washington

Both marble and limestone consist of CaCO3, which reacts with acid rain in an acid–base reaction to produce CaSO4. Because CaSO4 is somewhat soluble in water, significant damage to the structure can result.

The biological effects of acid rain are more complex. As indicated in , biological fluids, such as blood, have a pH of 7–8. Organisms such as fish can maintain their internal pH in water that has a pH in the range of 6.5–8.5. If the external pH is too low, however, many aquatic organisms can no longer maintain their internal pH, so they die. A pH of 4 or lower is fatal for virtually all fish, most invertebrate animals, and many microorganisms. As a result of acid rain, the pH of some lakes in Europe and the United States has dropped below 4. Recent surveys suggest that up to 6% of the lakes in the Adirondack Mountains of upstate New York and 4% of the lakes in Sweden and Norway are essentially dead and contain no fish. Neither location contains large concentrations of industry, but New York lies downwind of the industrial Midwest, and Scandinavia is downwind of the most industrialized regions of western Europe. Both regions appear to have borne the brunt of the pollution produced by their upwind neighbors. One possible way to counter the effects of acid rain in isolated lakes is by adding large quantities of finely ground limestone, which neutralizes the acid via the reaction shown in (see 1, Problem 15).

A second major way in which acid rain can cause biological damage is less direct. Trees and many other plants are sensitive to the presence of aluminum and other metals in groundwater. Under normal circumstances, aluminum hydroxide [Al(OH)3], which is present in some soils, is insoluble. At lower pH values, however, Al(OH)3 dissolves via the following reaction:

Equation 4.53

Al(OH)3(s) + 3H+(aq) → Al3+(aq) + 3H2O(l)

The result is increased levels of Al3+ ions in groundwater. Because the Al3+ ion is toxic to plants, high concentrations can affect plant growth. Acid rain can also weaken the leaves and roots of plants so much that the plants are unable to withstand other stresses. The combination of the two effects can cause significant damage to established forests, such as the Black Forest in Germany and the forests of the northeastern United States and Canada and other countries ().

Figure 4.19 Acid Rain Damage to a Forest in the Czech Republic

Trees and many other plants are sensitive to aluminum and other metals in groundwater. Acid rain increases the concentration of Al3+ in groundwater, thereby adversely affecting plant growth. Large sections of established forests have been severely damaged.

Summary

Acid rain is rainfall whose pH is less than 5.6, the value typically observed, due to the presence of dissolved carbon dioxide. Acid rain is caused by nitrogen oxides and sulfur dioxide produced by both natural processes and the combustion of fossil fuels. Eventually, these oxides react with oxygen and water to give nitric acid and sulfuric acid.

Key Takeaway

  • The damaging effects of acid rain have led to strong pressure on industry to minimize the release of harmful reactants.

Conceptual Problems

  1. Why is it recommended that marble countertops not be used in kitchens? Marble is composed mostly of CaCO3.

  2. Explain why desulfurization of fossil fuels is an area of intense research.

  3. What is the role of NOx in the formation of acid rain?

4.8 Oxidation–Reduction Reactions in Solution

Learning Objective

  1. To identify oxidation–reduction reactions in solution.

We described the defining characteristics of oxidation–reduction, or redox, reactions in . Most of the reactions we considered there were relatively simple, and balancing them was straightforward. When oxidation–reduction reactions occur in aqueous solution, however, the equations are more complex and can be more difficult to balance by inspection. Because a balanced chemical equation is the most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactions in aqueous solution that is generally applicable. One such method uses oxidation states, and a second is referred to as the half-reaction method. We show you how to balance redox equations using oxidation states in this section; the half-reaction method is described in .

Balancing Redox Equations Using Oxidation States

To balance a redox equation using the oxidation state methodA procedure for balancing oxidation–reduction (redox) reactions in which the overall reaction is conceptually separated into two parts: an oxidation and a reduction., we conceptually separate the overall reaction into two parts: an oxidation—in which the atoms of one element lose electrons—and a reduction—in which the atoms of one element gain electrons. Consider, for example, the reaction of Cr2+(aq) with manganese dioxide (MnO2) in the presence of dilute acid. is the net ionic equation for this reaction before balancing; the oxidation state of each element in each species has been assigned using the procedure described in :

Equation 4.54

Cr2++2(aq)+Mn+4O22(s)+H + +1(aq)Cr3++3(aq)+Mn2++2(aq)+H2+1O2(l)

Notice that chromium is oxidized from the +2 to the +3 oxidation state, while manganese is reduced from the +4 to the +2 oxidation state. We can write an equation for this reaction that shows only the atoms that are oxidized and reduced:

Equation 4.55

Cr2+ + Mn4+ → Cr3+ + Mn2+

The oxidation can be written as

Equation 4.56

Cr2+ → Cr3+ + e

and the reduction as

Equation 4.57

Mn4+ + 2e → Mn2+

For the overall chemical equation to be balanced, the number of electrons lost by the reductant must equal the number gained by the oxidant. We must therefore multiply the oxidation and the reduction equations by appropriate coefficients to give us the same number of electrons in both. In this example, we must multiply the oxidation equation by 2 to give

Equation 4.58

2Cr2+ → 2Cr3+ + 2e

Note the Pattern

In a balanced redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant.

The number of electrons lost in the oxidation now equals the number of electrons gained in the reduction:

Equation 4.59

2Cr2+2Cr3++2eMn4++2eMn2+

We then add the equations for the oxidation and the reduction and cancel the electrons on both sides of the equation, using the actual chemical forms of the reactants and products:

Equation 4.60

2Cr2+(aq)2Cr3+(aq) + 2eMnO2(s) + 2eMn2 + (aq)2Cr2 + (aq) + MnO2(s)2Cr3 + (aq) + Mn2 + (aq)

Although the electrons cancel and the metal atoms are balanced, the total charge on the left side of the equation (+4) does not equal the charge on the right side (+8). Because the reaction is carried out in the presence of aqueous acid, we can add H+ as necessary to either side of the equation to balance the charge. By the same token, if the reaction were carried out in the presence of aqueous base, we could balance the charge by adding OH as necessary to either side of the equation to balance the charges. In this case, adding four H+ ions to the left side of the equation gives

Equation 4.61

2Cr2+(aq) + MnO2(s) + 4H+(aq) → 2Cr3+(aq) + Mn2+(aq)

Although the charges are now balanced, we have two oxygen atoms on the left side of the equation and none on the right. We can balance the oxygen atoms without affecting the overall charge balance by adding H2O as necessary to either side of the equation. Here, we need to add two H2O molecules to the right side:

Equation 4.62

2Cr2+(aq) + MnO2(s) + 4H+(aq) → 2Cr3+(aq) + Mn2+(aq) + 2H2O(l)

Although we did not explicitly balance the hydrogen atoms, we can see by inspection that the overall chemical equation is now balanced. All that remains is to check to make sure that we have not made a mistake. This procedure for balancing reactions is summarized in and illustrated in Example 17.

Table 4.4 Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method

  1. Write the unbalanced chemical equation for the reaction, showing the reactants and the products.
  2. Assign oxidation states to all atoms in the reactants and the products (see ) and determine which atoms change oxidation state.
  3. Write separate equations for oxidation and reduction, showing (a) the atom(s) that is (are) oxidized and reduced and (b) the number of electrons accepted or donated by each.
  4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons.
  5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting the coefficients as necessary to give the numbers of atoms in step 4.
  6. Add the two equations and cancel the electrons.
  7. Balance the charge by adding H+ or OH ions as necessary for reactions in acidic or basic solution, respectively.
  8. Balance the oxygen atoms by adding H2O molecules to one side of the equation.
  9. Check to make sure that the equation is balanced in both atoms and total charges.

Example 17

Arsenic acid (H3AsO4) is a highly poisonous substance that was once used as a pesticide. The reaction of elemental zinc with arsenic acid in acidic solution yields arsine (AsH3, a highly toxic and unstable gas) and Zn2+(aq). Balance the equation for this reaction using oxidation states:

H3AsO4(aq) + Zn(s) → AsH3(g) + Zn2+(aq)

Given: reactants and products in acidic solution

Asked for: balanced chemical equation using oxidation states

Strategy:

Follow the procedure given in for balancing a redox equation using oxidation states. When you are done, be certain to check that the equation is balanced.

Solution:

  1. Write a chemical equation showing the reactants and the products. Because we are given this information, we can skip this step.
  2. Assign oxidation states using the procedure described in and determine which atoms change oxidation state. The oxidation state of arsenic in arsenic acid is +6, and the oxidation state of arsenic in arsine is −3. Conversely, the oxidation state of zinc in elemental zinc is 0, and the oxidation state of zinc in Zn2+(aq) is +2:

    H3As+5O4(aq)+Zn0(s)As–3H3(g)+Zn2++2(aq)
  3. Write separate equations for oxidation and reduction. The arsenic atom in H3AsO4 is reduced from the +5 to the −3 oxidation state, which requires the addition of eight electrons:

    Reduction: As5++8eAs3––3

    Each zinc atom in elemental zinc is oxidized from 0 to +2, which requires the loss of two electrons per zinc atom:

    Oxidation: ZnZn2++2e
  4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons. The reduction equation has eight electrons, and the oxidation equation has two electrons, so we need to multiply the oxidation equation by 4 to obtain

    Reduction (× 1): As5+ + 8e → As3– Oxidation (× 4): 4Zn0 → 4Zn2+ + 8e
  5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting coefficients as necessary to give the numbers of atoms shown in step 4. Inserting the actual chemical forms of arsenic and zinc and adjusting the coefficients gives

    Reduction: H3AsO4(aq) + 8e → AsH3(g) Oxidation: 4Zn(s) → 4Zn2+(aq) + 8e
  6. Add the two equations and cancel the electrons. The sum of the two equations in step 5 is

    H3AsO4(aq) + 4Zn(s)+8eAsH3(g)+4Zn2+(aq)+8e

    which then yields

    H3AsO4(aq) + 4Zn(s) → AsH3(g) + 4Zn2+(aq)
  7. Balance the charge by adding H+or OHions as necessary for reactions in acidic or basic solution, respectively. Because the reaction is carried out in acidic solution, we can add H+ ions to whichever side of the equation requires them to balance the charge. The overall charge on the left side is zero, and the total charge on the right side is 4 × (+2) = +8. Adding eight H+ ions to the left side gives a charge of +8 on both sides of the equation:

    H3AsO4(aq) + 4Zn(s) + 8H+(aq) → AsH3(g) + 4Zn2+(aq)
  8. Balance the oxygen atoms by adding H2O molecules to one side of the equation. There are 4 O atoms on the left side of the equation. Adding 4 H2O molecules to the right side balances the O atoms:

    H3AsO4(aq) + 4Zn(s) + 8H+(aq) → AsH3(g) + 4Zn2+(aq) + 4H2O(l)

    Although we have not explicitly balanced H atoms, each side of the equation has 11 H atoms.

  9. Check to make sure that the equation is balanced in both atoms and total charges. To guard against careless errors, it is important to check that both the total number of atoms of each element and the total charges are the same on both sides of the equation:
Atoms: 1As + 4Zn + 4O + 11H = 1As + 4Zn + 4O + 11H Total charge: 8(+1) = 4(+2) = +8

The balanced chemical equation for the reaction is therefore:

H3AsO4(aq) + 4Zn(s) + 8H+(aq) → AsH3(g) + 4Zn2+(aq) + 4H2O(l)

Exercise

Copper commonly occurs as the sulfide mineral CuS. The first step in extracting copper from CuS is to dissolve the mineral in nitric acid, which oxidizes the sulfide to sulfate and reduces nitric acid to NO. Balance the equation for this reaction using oxidation states:

CuS(s) + H+(aq) + NO3(aq) → Cu2+(aq) + NO(g) + SO42−(aq)

Answer: 3CuS(s) + 8H+(aq) + 8NO3(aq) → 3Cu2+(aq) + 8NO(g) + 3SO42−(aq) + 4H2O(l)

Reactions in basic solutions are balanced in exactly the same manner. To make sure you understand the procedure, consider Example 18.

Example 18

The commercial solid drain cleaner, Drano, contains a mixture of sodium hydroxide and powdered aluminum. The sodium hydroxide dissolves in standing water to form a strongly basic solution, capable of slowly dissolving organic substances, such as hair, that may be clogging the drain. The aluminum dissolves in the strongly basic solution to produce bubbles of hydrogen gas that agitate the solution to help break up the clogs. The reaction is as follows:

Al(s) + H2O(aq) → [Al(OH)4](aq) + H2(g)

Balance this equation using oxidation states.

Given: reactants and products in a basic solution

Asked for: balanced chemical equation

Strategy:

Follow the procedure given in for balancing a redox reaction using oxidation states. When you are done, be certain to check that the equation is balanced.

Solution:

We will apply the same procedure used in Example 17 but in a more abbreviated form.

  1. The equation for the reaction is given, so we can skip this step.
  2. The oxidation state of aluminum changes from 0 in metallic Al to +3 in [Al(OH)4]. The oxidation state of hydrogen changes from +1 in H2O to 0 in H2. Aluminum is oxidized, while hydrogen is reduced:

    Al0(s)+H+12O(aq)[Al+3(OH)4](aq)+H20(g)
  3.  

    Reduction: Al0Al3++3eOxidation: H + +eH0 (in H2)
  4. Multiply the reduction equation by 3 to obtain an equation with the same number of electrons as the oxidation equation:

    Reduction: 3H+ + 3e → 3H0 (in H2) Oxidation: Al0 → Al3+ + 3e
  5. Insert the actual chemical forms of the reactants and products, adjusting the coefficients as necessary to obtain the correct numbers of atoms as in step 4. Because a molecule of H2O contains two protons, in this case, 3H+ corresponds to 3/2H2O. Similarly, each molecule of hydrogen gas contains two H atoms, so 3H corresponds to 3/2H2.

    Reduction: 32H2O+3e32H2Oxidation: Al[Al(OH)4]+3e
  6. Adding the equations and canceling the electrons gives

    Al+32H2O+3e[Al(OH)4]+32H2+3eAl+32H2O[Al(OH)4]+32H2 

    To remove the fractional coefficients, multiply both sides of the equation by 2:

    2Al + 3H2O → 2[Al(OH)4] + 3H2
  7. The right side of the equation has a total charge of −2, whereas the left side has a total charge of 0. Because the reaction is carried out in basic solution, we can balance the charge by adding two OH ions to the left side:

    2Al + 2OH + 3H2O → 2[Al(OH)4] + 3H2
  8. The left side of the equation contains five O atoms, and the right side contains eight O atoms. We can balance the O atoms by adding three H2O molecules to the left side:

    2Al + 2OH + 6H2O → 2[Al(OH)4] + 3H2
  9. Be sure the equation is balanced:

    Atoms: 2Al + 8O + 14H = 2Al + 8O + 14H Total charge: (2)(0) + (2)(–1) + (6)(0) = (2)(–1) + (3)(0) −2 = −2

    The balanced chemical equation is therefore

    2Al(s) + 2OH(aq) + 6H2O(l) → 2[Al(OH)4](aq) + 3H2(g)

    Thus 3 mol of H2 gas are produced for every 2 mol of Al.

Exercise

The permanganate ion reacts with nitrite ion in basic solution to produce manganese(IV) oxide and nitrate ion. Write a balanced chemical equation for the reaction.

Answer: 2MnO4(aq) + 3NO2(aq) + H2O(l) → 2MnO2(s) + 3NO3(aq) + 2OH(aq)

As suggested in Example 17 and Example 18, a wide variety of redox reactions are possible in aqueous solutions. The identity of the products obtained from a given set of reactants often depends on both the ratio of oxidant to reductant and whether the reaction is carried out in acidic or basic solution, which is one reason it can be difficult to predict the outcome of a reaction. Because oxidation–reduction reactions in solution are so common and so important, however, chemists have developed two general guidelines for predicting whether a redox reaction will occur and the identity of the products:

  1. Compounds of elements in high oxidation states (such as ClO4, NO3, MnO4, Cr2O72−, and UF6) tend to act as oxidants and become reduced in chemical reactions.
  2. Compounds of elements in low oxidation states (such as CH4, NH3, H2S, and HI) tend to act as reductants and become oxidized in chemical reactions.

Note the Pattern

Species in high oxidation states act as oxidants, whereas species in low oxidation states act as reductants.

When an aqueous solution of a compound that contains an element in a high oxidation state is mixed with an aqueous solution of a compound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur.

Redox Reactions of Solid Metals in Aqueous Solution

A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas:

Equation 4.63

Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)

In subsequent steps, FeCl2 undergoes oxidation to form a reddish-brown precipitate of Fe(OH)3.

Figure 4.20 Rust Formation

The corrosion process involves an oxidation–reduction reaction in which metallic iron is converted to Fe(OH)3, a reddish-brown solid.

Many metals dissolve through reactions of this type, which have the general form

Equation 4.64

metal + acid → salt + hydrogen

Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves:

Equation 4.65

Pb(s) + 2H+(aq) → Pb2+(aq) + H2(g)

Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul!

Single-Displacement Reactions

Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called single-displacement reactionsA chemical reaction in which an ion in solution is displaced through oxidation of a metal., in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc () and the reduction of silver salts by copper ( and ):

Equation 4.66

Zn(s) + Fe2+(aq) → Zn2+(aq) + Fe(s)

Equation 4.67

Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

The reaction in is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object.

Figure 4.21 The Single-Displacement Reaction of Metallic Copper with a Solution of Silver Nitrate

When a copper coil is placed in a solution of silver nitrate, silver ions are reduced to metallic silver on the copper surface, and some of the copper metal dissolves. Note the formation of a metallic silver precipitate on the copper coil and a blue color in the surrounding solution due to the presence of aqueous Cu2+ ions. (Water molecules and nitrate ions have been omitted from molecular views of the solution for clarity.)

The Activity Series

By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, we saw in and that metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing Zn2+. Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal:

Equation 4.68

Zn(s)+Mg2+(aq)Zn2+(aq)+Mg(s)

Equation 4.69

Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s)

Magnesium has a greater tendency to be oxidized than zinc does.

Pairwise reactions of this sort are the basis of the activity seriesA list of metals and hydrogen in order of their relative tendency to be oxidized. (), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metalsThe metals at the top of the activity series, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metalsThe metals at the bottom of the activity series, which have the least tendency to be oxidized. (at the bottom of the series).

Figure 4.22 The Activity Series

When using the activity series to predict the outcome of a reaction, keep in mind that any element will reduce compounds of the elements below it in the series. Because magnesium is above zinc in , magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Only those metals that lie above hydrogen in the activity series dissolve in acids to produce H2. Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example 19 demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions. We will return to the activity series when we discuss oxidation–reduction reactions in more detail in .

Example 19

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.

  1. A strip of aluminum foil is placed in an aqueous solution of silver nitrate.
  2. A few drops of liquid mercury are added to an aqueous solution of lead(II) acetate.
  3. Some sulfuric acid from a car battery is accidentally spilled on the lead cable terminals.

Given: reactants

Asked for: overall reaction and net ionic equation

Strategy:

A Locate the reactants in the activity series in and from their relative positions, predict whether a reaction will occur. If a reaction does occur, identify which metal is oxidized and which is reduced.

B Write the net ionic equation for the redox reaction.

Solution:

  1. A Aluminum is an active metal that lies above silver in the activity series, so we expect a reaction to occur. According to their relative positions, aluminum will be oxidized and dissolve, and silver ions will be reduced to silver metal. B The net ionic equation is as follows:

    Al(s) + 3Ag+(aq) → Al3+(aq) + 3Ag(s)

    Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction.

  2. A Mercury lies below lead in the activity series, so no reaction will occur.
  3. A Lead is above hydrogen in the activity series, so the lead terminals will be oxidized, and the acid will be reduced to form H2. B From our discussion of solubilities, recall that Pb2+ and SO42− form insoluble lead(II) sulfate. In this case, the sulfate ions are not spectator ions, and the reaction is as follows:

    Pb(s) + 2H+(aq) + SO42−(aq) → PbSO4(s) + H2(g)

    Lead(II) sulfate is the white solid that forms on corroded battery terminals.

Corroded battery terminals. The white solid is lead(II) sulfate, formed from the reaction of solid lead with a solution of sulfuric acid.

Exercise

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.

  1. A strip of chromium metal is placed in an aqueous solution of aluminum chloride.
  2. A strip of zinc is placed in an aqueous solution of chromium(III) nitrate.
  3. A piece of aluminum foil is dropped into a glass that contains vinegar (the active ingredient is acetic acid).

Answer:

  1. no reaction
  2. 3Zn(s) + 2Cr3+(aq) → 3Zn2+(aq) + 2Cr(s)
  3. 2Al(s) + 6CH3CO2H(aq) → 2Al3+(aq) + 6CH3CO2(aq) + 3H2(g)

Summary

In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method (), in which the overall reaction is separated into an oxidation equation and a reduction equation. Single-displacement reactions are reactions of metals with either acids or another metal salt that result in dissolution of the first metal and precipitation of a second (or evolution of hydrogen gas). The outcome of these reactions can be predicted using the activity series (), which arranges metals and H2 in decreasing order of their tendency to be oxidized. Any metal will reduce metal ions below it in the activity series. Active metals lie at the top of the activity series, whereas inert metals are at the bottom of the activity series.

Key Takeaway

  • Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation.

Conceptual Problems

  1. Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants?

  2. If two compounds are mixed, one containing an element that is a poor oxidant and one with an element that is a poor reductant, do you expect a redox reaction to occur? Explain your answer. What do you predict if one is a strong oxidant and the other is a weak reductant? Why?

  3. In each redox reaction, determine which species is oxidized and which is reduced:

    1. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
    2. Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
    3. BrO3(aq) + 2MnO2(s) + H2O(l) → Br(aq) + 2MnO4(aq) + 2H+(aq)
  4. Single-displacement reactions are a subset of redox reactions. In this subset, what is oxidized and what is reduced? Give an example of a redox reaction that is not a single-displacement reaction.

  5. Of the following elements, which would you expect to have the greatest tendency to be oxidized: Zn, Li, or S? Explain your reasoning.

  6. Of these elements, which would you expect to be easiest to reduce: Se, Sr, or Ni? Explain your reasoning.

  7. Which of these metals produces H2 in acidic solution?

    1. Ag
    2. Cd
    3. Ca
    4. Cu
  8. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.

    1. Mg(s) + Cu2+(aq) →
    2. Au(s) + Ag+(aq) →
    3. Cr(s) + Pb2+(aq) →
    4. K(s) + H2O(l) →
    5. Hg(l) + Pb2+(aq) →

Numerical Problems

  1. Balance each redox reaction under the conditions indicated.

    1. CuS(s) + NO3(aq) → Cu2+(aq) + SO42−(aq) + NO(g); acidic solution
    2. Ag(s) + HS(aq) + CrO42−(aq) → Ag2S(s) + Cr(OH)3(s); basic solution
    3. Zn(s) + H2O(l) → Zn2+(aq) + H2(g); acidic solution
    4. O2(g) + Sb(s) → H2O2(aq) + SbO2(aq); basic solution
    5. UO22+(aq) + Te(s) → U4+(aq) + TeO42−(aq); acidic solution
  2. Balance each redox reaction under the conditions indicated.

    1. MnO4(aq) + S2O32−(aq) → Mn2+(aq) + SO42−(aq); acidic solution
    2. Fe2+(aq) + Cr2O72−(aq) → Fe3+(aq) + Cr3+(aq); acidic solution
    3. Fe(s) + CrO42−(aq) → Fe2O3(s) + Cr2O3(s); basic solution
    4. Cl2(aq) → ClO3(aq) + Cl(aq); acidic solution
    5. CO32−(aq) + N2H4(aq) → CO(g) + N2(g); basic solution
  3. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.

    1. Platinum wire is dipped in hydrochloric acid.
    2. Manganese metal is added to a solution of iron(II) chloride.
    3. Tin is heated with steam.
    4. Hydrogen gas is bubbled through a solution of lead(II) nitrate.
  4. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.

    1. A few drops of NiBr2 are dropped onto a piece of iron.
    2. A strip of zinc is placed into a solution of HCl.
    3. Copper is dipped into a solution of ZnCl2.
    4. A solution of silver nitrate is dropped onto an aluminum plate.
  5. Dentists occasionally use metallic mixtures called amalgams for fillings. If an amalgam contains zinc, however, water can contaminate the amalgam as it is being manipulated, producing hydrogen gas under basic conditions. As the filling hardens, the gas can be released, causing pain and cracking the tooth. Write a balanced chemical equation for this reaction.

  6. Copper metal readily dissolves in dilute aqueous nitric acid to form blue Cu2+(aq) and nitric oxide gas.

    1. What has been oxidized? What has been reduced?
    2. Balance the chemical equation.
  7. Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:

    1. Pt2+(aq) + Ag(s) →
    2. HCN(aq) + NaOH(aq) →
    3. Fe(NO3)3(aq) + NaOH(aq) →
    4. CH4(g) + O2(g) →
  8. Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:

    1. Zn(s) + HCl(aq) →
    2. HNO3(aq) + AlCl3(aq) →
    3. K2CrO4(aq) + Ba(NO3)2(aq) →
    4. Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s)

4.9 Quantitative Analysis Using Titrations

Learning Objective

  1. To use titration methods to analyze solutions quantitatively.

To determine the amounts or concentrations of substances present in a sample, chemists use a combination of chemical reactions and stoichiometric calculations in a methodology called quantitative analysisA methodology that combines chemical reactions and stoichiometric calculations to determine the amounts or concentrations of substances present in a sample.. Suppose, for example, we know the identity of a certain compound in a solution but not its concentration. If the compound reacts rapidly and completely with another reactant, we may be able to use the reaction to determine the concentration of the compound of interest. In a titrationAn experimental procedure in which a carefully measured volume of a solution of known concentration is added to a measured volume of a solution containing a compound whose concentration is to be determined., a carefully measured volume of a solution of known concentration, called the titrantThe solution of known concentration that is reacted with a compound in a solution of unknown concentration in a titration., is added to a measured volume of a solution containing a compound whose concentration is to be determined (the unknown). The reaction used in a titration can be an acid–base reaction, a precipitation reaction, or an oxidation–reduction reaction. In all cases, the reaction chosen for the analysis must be fast, complete, and specific; that is, only the compound of interest should react with the titrant. The equivalence pointThe point in a titration where a stoichiometric amount (i.e., the amount required to react completely with the unknown) of the titrant has been added. is reached when a stoichiometric amount of the titrant has been added—the amount required to react completely with the unknown.

Determining the Concentration of an Unknown Solution Using a Titration

The chemical nature of the species present in the unknown dictates which type of reaction is most appropriate and also how to determine the equivalence point. The volume of titrant added, its concentration, and the coefficients from the balanced chemical equation for the reaction allow us to calculate the total number of moles of the unknown in the original solution. Because we have measured the volume of the solution that contains the unknown, we can calculate the molarity of the unknown substance. This procedure is summarized graphically here:

Example 20

The calcium salt of oxalic acid [Ca(O2CCO2)] is found in the sap and leaves of some vegetables, including spinach and rhubarb, and in many ornamental plants. Because oxalic acid and its salts are toxic, when a food such as rhubarb is processed commercially, the leaves must be removed, and the oxalate content carefully monitored.

The reaction of MnO4 with oxalic acid (HO2CCO2H) in acidic aqueous solution produces Mn2+ and CO2:

MnO4(aq)purple+HO2CCO2H(aq)+H + (aq)Mn2+(aq)colorless+CO2(g)+H2O(l)

Because this reaction is rapid and goes to completion, potassium permanganate (KMnO4) is widely used as a reactant for determining the concentration of oxalic acid.

Suppose you stirred a 10.0 g sample of canned rhubarb with enough dilute H2SO4(aq) to obtain 127 mL of colorless solution. Because the added permanganate is rapidly consumed, adding small volumes of a 0.0247 M KMnO4 solution, which has a deep purple color, to the rhubarb extract does not initially change the color of the extract. When 15.4 mL of the permanganate solution have been added, however, the solution becomes a faint purple due to the presence of a slight excess of permanganate (). If we assume that oxalic acid is the only species in solution that reacts with permanganate, what percentage of the mass of the original sample was calcium oxalate?

Figure 4.23 The Titration of Oxalic Acid with Permanganate

As small volumes of permanganate solution are added to the oxalate solution, a transient purple color appears and then disappears as the permanganate is consumed (a). As more permanganate is added, eventually all the oxalate is oxidized, and a faint purple color from the presence of excess permanganate appears, marking the endpoint (b).

Given: equation, mass of sample, volume of solution, and molarity and volume of titrant

Asked for: mass percentage of unknown in sample

Strategy:

A Balance the chemical equation for the reaction using oxidation states.

B Calculate the number of moles of permanganate consumed by multiplying the volume of the titrant by its molarity. Then calculate the number of moles of oxalate in the solution by multiplying by the ratio of the coefficients in the balanced chemical equation. Because calcium oxalate contains a 1:1 ratio of Ca2+:O2CCO2, the number of moles of oxalate in the solution is the same as the number of moles of calcium oxalate in the original sample.

C Find the mass of calcium oxalate by multiplying the number of moles of calcium oxalate in the sample by its molar mass. Divide the mass of calcium oxalate by the mass of the sample and convert to a percentage to calculate the percentage by mass of calcium oxalate in the original sample.

Solution:

A As in all other problems of this type, the first requirement is a balanced chemical equation for the reaction. Using oxidation states gives

2MnO4(aq) + 5HO2CCO2H(aq) + 6H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l)

Thus each mole of MnO4 added consumes 2.5 mol of oxalic acid.

B Because we know the concentration of permanganate (0.0247 M) and the volume of permanganate solution that was needed to consume all the oxalic acid (15.4 mL), we can calculate the number of moles of MnO4 consumed:

15.4 mL(L1000 mL)(0.0247 mol MnO4L)= 3.80×104 mol MnO4

The number of moles of oxalic acid, and thus oxalate, present can be calculated from the mole ratio of the reactants in the balanced chemical equation:

moles HO2CCO2= 3.80×104 mol MnO4(5 mol HO2CCO2Hmol MnO4)= 9.50×104 mol HO2CCO2H

C The problem asks for the percentage of calcium oxalate by mass in the original 10.0 g sample of rhubarb, so we need to know the mass of calcium oxalate that produced 9.50 × 10−4 mol of oxalic acid. Because calcium oxalate is Ca(O2CCO2), 1 mol of calcium oxalate gave 1 mol of oxalic acid in the initial acid extraction:

Ca(O2CCO2)(s) + 2H+(aq) → Ca2+(aq) + HO2CCO2H(aq)

The mass of calcium oxalate originally present was thus

mass of CaC2O4= 9.50×104 mol HO2CCO2H(mol CaC2O4mol HO2CCO2H)(128.10 g CaC2O4mol CaC2O4)= 0.122 g CaC2O4

The original sample contained 0.122 g of calcium oxalate per 10.0 g of rhubarb. The percentage of calcium oxalate by mass was thus

% CaC2O4=0.122 g10.0 g×100 = 1.22%

Because the problem asked for the percentage by mass of calcium oxalate in the original sample rather than for the concentration of oxalic acid in the extract, we do not need to know the volume of the oxalic acid solution for this calculation.

Exercise

Glutathione is a low-molecular-weight compound found in living cells that is produced naturally by the liver. Health-care providers give glutathione intravenously to prevent side effects of chemotherapy and to prevent kidney problems after heart bypass surgery. Its structure is as follows:

Glutathione is found in two forms: one abbreviated as GSH (indicating the presence of an –SH group) and the other as GSSG (the disulfide form, in which an S–S bond links two glutathione units). The GSH form is easily oxidized to GSSG by elemental iodine:

2GSH(aq) + I2(aq) → GSSG(aq) + 2HI(aq)

A small amount of soluble starch is added as an indicator. Because starch reacts with excess I2 to give an intense blue color, the appearance of a blue color indicates that the equivalence point of the reaction has been reached.

Adding small volumes of a 0.0031 M aqueous solution of I2 to 194 mL of a solution that contains glutathione and a trace of soluble starch initially causes no change. After 16.3 mL of iodine solution have been added, however, a permanent pale blue color appears because of the formation of the starch-iodine complex. What is the concentration of glutathione in the original solution?

Answer: 5.2 × 10−4 M

Standard Solutions

The reaction of KHP with NaOH. As with all acid-base reactions, a salt is formed.

In Example 20, the concentration of the titrant (I2) was accurately known. The accuracy of any titration analysis depends on an accurate knowledge of the concentration of the titrant. Most titrants are first standardized; that is, their concentration is measured by titration with a standard solutionA solution whose concentration is precisely known., which is a solution whose concentration is known precisely. Only pure crystalline compounds that do not react with water or carbon dioxide are suitable for use in preparing a standard solution. One such compound is potassium hydrogen phthalate (KHP), a weak monoprotic acid suitable for standardizing solutions of bases such as sodium hydroxide. The reaction of KHP with NaOH is a simple acid–base reaction. If the concentration of the KHP solution is known accurately and the titration of a NaOH solution with the KHP solution is carried out carefully, then the concentration of the NaOH solution can be calculated precisely. The standardized NaOH solution can then be used to titrate a solution of an acid whose concentration is unknown.

Acid–Base Titrations

Because most common acids and bases are not intensely colored, a small amount of an acid–base indicator is usually added to detect the equivalence point in an acid–base titration. The point in the titration at which an indicator changes color is called the endpointThe point in a titration at which an indicator changes color.. The procedure is illustrated in Example 21.

Example 21

The structure of vitamin C (ascorbic acid, a monoprotic acid) is as follows:

Ascorbic acid. The upper figure shows the three-dimensional representation of ascorbic acid. Hatched lines indicate bonds that are behind the plane of the paper, and wedged lines indicate bonds that are out of the plane of the paper.

An absence of vitamin C in the diet leads to the disease known as scurvy, a breakdown of connective tissue throughout the body and of dentin in the teeth. Because fresh fruits and vegetables rich in vitamin C are readily available in developed countries today, scurvy is not a major problem. In the days of slow voyages in wooden ships, however, scurvy was common. Ferdinand Magellan, the first person to sail around the world, lost more than 90% of his crew, many to scurvy. Although a diet rich in fruits and vegetables contains more than enough vitamin C to prevent scurvy, many people take supplemental doses of vitamin C, hoping that the extra amounts will help prevent colds and other illness.

Suppose a tablet advertised as containing 500 mg of vitamin C is dissolved in 100.0 mL of distilled water that contains a small amount of the acid–base indicator bromothymol blue, an indicator that is yellow in acid solution and blue in basic solution, to give a yellow solution. The addition of 53.5 mL of a 0.0520 M solution of NaOH results in a change to green at the endpoint, due to a mixture of the blue and yellow forms of the indicator (). What is the actual mass of vitamin C in the tablet? (The molar mass of ascorbic acid is 176.13 g/mol.)

Figure 4.24 The Titration of Ascorbic Acid with a Solution of NaOH

The solution, containing bromothymol blue as an indicator, is initially yellow (a). The addition of a trace excess of NaOH causes the solution to turn green at the endpoint (b) and then blue.

Given: reactant, volume of sample solution, and volume and molarity of titrant

Asked for: mass of unknown

Strategy:

A Write the balanced chemical equation for the reaction and calculate the number of moles of base needed to neutralize the ascorbic acid.

B Using mole ratios, determine the amount of ascorbic acid consumed. Calculate the mass of vitamin C by multiplying the number of moles of ascorbic acid by its molar mass.

Solution:

A Because ascorbic acid acts as a monoprotic acid, we can write the balanced chemical equation for the reaction as

HAsc(aq) + OH(aq) → Asc(aq) + H2O(l)

where HAsc is ascorbic acid and Asc is ascorbate. The number of moles of OH ions needed to neutralize the ascorbic acid is

moles OH= 53.5 mL(1 L1000 mL)(0.0520 mol OH1 L)= 2.78×103 mol OH

B The mole ratio of the base added to the acid consumed is 1:1, so the number of moles of OH added equals the number of moles of ascorbic acid present in the tablet:

mass ascorbic acid= 2.78×103 mol HAsc (176.13 g HAscmol HAsc)= 0.490 g HAsc

Because 0.490 g equals 490 mg, the tablet contains about 2% less vitamin C than advertised.

Exercise

Vinegar is essentially a dilute solution of acetic acid in water. Vinegar is usually produced in a concentrated form and then diluted with water to give a final concentration of 4%–7% acetic acid; that is, a 4% m/v solution contains 4.00 g of acetic acid per 100 mL of solution. If a drop of bromothymol blue indicator is added to 50.0 mL of concentrated vinegar stock and 31.0 mL of 2.51 M NaOH are needed to turn the solution from yellow to green, what is the percentage of acetic acid in the vinegar stock? (Assume that the density of the vinegar solution is 1.00 g/mL.)

Answer: 9.35%

Summary

The concentration of a species in solution can be determined by quantitative analysis. One such method is a titration, in which a measured volume of a solution of one substance, the titrant, is added to a solution of another substance to determine its concentration. The equivalence point in a titration is the point at which exactly enough reactant has been added for the reaction to go to completion. A standard solution, a solution whose concentration is known precisely, is used to determine the concentration of the titrant. Many titrations, especially those that involve acid–base reactions, rely on an indicator. The point at which a color change is observed is the endpoint, which is close to the equivalence point if the indicator is chosen properly.

Key Takeaway

  • Quantitative analysis of an unknown solution can be achieved using titration methods.

Conceptual Problems

  1. The titration procedure is an application of the use of limiting reactants. Explain why this is so.

  2. Explain how to determine the concentration of a substance using a titration.

  3. Following are two graphs that illustrate how the pH of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M acetic acid with 0.10 M NaOH, and the other corresponds to the titration of 100 mL 0.10 M NaOH with 0.10 M acetic acid. Which graph corresponds to which titration? Justify your answer.

  4. Following are two graphs that illustrate how the pH of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M ammonia with 0.10 M HCl, and the other corresponds to the titration of 100 mL 0.10 M NH4Cl with 0.10 M NaOH. Which graph corresponds to which titration? Justify your answer.

  5. Following are two graphs that illustrate how the electrical conductivity of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M Ba(OH)2 with 0.10 M H2SO4 , and the other corresponds to the titration of 100 mL of 0.10 M NaOH with 0.10 M H2SO4. Which graph corresponds to which titration? Justify your answer.

Answers

    1. titration of NaOH with acetic acid
    2. titration of acetic acid with NaOH
    1. titration of Ba(OH)2 with sulfuric acid
    2. titration of NaOH with sulfuric acid

Numerical Problems

  1. A 10.00 mL sample of a 1.07 M solution of potassium hydrogen phthalate (KHP, formula mass = 204.22 g/mol) is diluted to 250.0 mL. What is the molarity of the final solution? How many grams of KHP are in the 10.00 mL sample?

  2. What volume of a 0.978 M solution of NaOH must be added to 25.0 mL of 0.583 M HCl to completely neutralize the acid? How many moles of NaOH are needed for the neutralization?

  3. A student was titrating 25.00 mL of a basic solution with an HCl solution that was 0.281 M. The student ran out of the HCl solution after having added 32.46 mL, so she borrowed an HCl solution that was labeled as 0.317 M. An additional 11.5 mL of the second solution was needed to complete the titration. What was the concentration of the basic solution?

4.10 Essential Skills 3

Topics

  • Base-10 Logarithms
  • Calculations Using Common Logarithms

Essential Skills 1 in , and Essential Skills 2 in , described some fundamental mathematical operations used for solving problems in chemistry. This section introduces you to base-10 logarithms, a topic with which you must be familiar to do the Questions and Problems at the end of . We will return to the subject of logarithms in Essential Skills 6 in , .

Base-10 (Common) Logarithms

Essential Skills 1 introduced exponential notation, in which a base number is multiplied by itself the number of times indicated in the exponent. The number 103, for example, is the base 10 multiplied by itself three times (10 × 10 × 10 = 1000). Now suppose that we do not know what the exponent is—that we are given only a base of 10 and the final number. If our answer is 1000, the problem can be expressed as

10a = 1000

We can determine the value of a by using an operation called the base-10 logarithm, or common logarithm, abbreviated as log, that represents the power to which 10 is raised to give the number to the right of the equals sign. This relationship is stated as log 10a = a. In this case, the logarithm is 3 because 103 = 1000:

log 103 = 3 log 1000 = 3

Now suppose you are asked to find a when the final number is 659. The problem can be solved as follows (remember that any operation applied to one side of an equality must also be applied to the other side):

10a = 659 log 10a = log 659 a = log 659

If you enter 659 into your calculator and press the “log” key, you get 2.819, which means that a = 2.819 and 102.819 = 659. Conversely, if you enter the value 2.819 into your calculator and press the “10x” key, you get 659.

You can decide whether your answer is reasonable by comparing it with the results you get when a = 2 and a = 3:

a = 2: 102 = 100 a = 2.819: 102.819 = 659 a = 3: 103 = 1000

Because the number 659 is between 100 and 1000, a must be between 2 and 3, which is indeed the case.

lists some base-10 logarithms, their numerical values, and their exponential forms.

Table 4.5 Relationships in Base-10 Logarithms

Numerical Value Exponential Form Logarithm (a)
1000 103 3
100 102 2
10 101 1
1 100 0
0.1 10−1 −1
0.01 10−2 −2
0.001 10−3 −3

Base-10 logarithms may also be expressed as log10, in which the base is indicated as a subscript. We can write log 10a = a in either of two ways:

log 10a = a log10 = (10a) = a

The second equation explicitly indicates that we are solving for the base-10 logarithm of 10a.

The number of significant figures in a logarithmic value is the same as the number of digits after the decimal point in its logarithm, so log 62.2, a number with three significant figures, is 1.794, with three significant figures after the decimal point; that is, 101.794 = 62.2, not 62.23. Skill Builder ES1 provides practice converting a value to its exponential form and then calculating its logarithm.

Skill Builder ES1

Express each number as a power of 10 and then find the common logarithm.

  1. 10,000
  2. 0.00001
  3. 10.01
  4. 2.87
  5. 0.134

Solution

  1. 10,000 = 1 × 104; log 1 × 104 = 4.0
  2. 0.00001 = 1 × 10−5; log 1 × 10−5 = −5.0
  3. 10.01 = 1.001 × 10; log 10.01 = 1.0004 (enter 10.01 into your calculator and press the “log” key); 101.0004 = 10.01
  4. 2.87 = 2.87 × 100; log 2.87 = 0.458 (enter 2.87 into your calculator and press the “log” key); 100.458 = 2.87
  5. 0.134 = 1.34 × 10−1; log 0.134 = −0.873 (enter 0.134 into your calculator and press the “log” key); 10−0.873 = 0.134

Skill Builder ES2

Convert each base-10 logarithm to its numerical value.

  1. 3
  2. −2.0
  3. 1.62
  4. −0.23
  5. −4.872

Solution

  1. 103
  2. 10−2
  3. 101.62 = 42
  4. 10−0.23 = 0.59
  5. 10−4.872 = 1.34 × 10−5

Calculations Using Common Logarithms

Because logarithms are exponents, the properties of exponents that you learned in Essential Skills 1 apply to logarithms as well, which are summarized in . The logarithm of (4.08 × 20.67), for example, can be computed as follows:

log(4.08 × 20.67) = log 4.08 + log 20.67 = 0.611 + 1.3153 = 1.926

We can be sure that this answer is correct by checking that 101.926 is equal to 4.08 × 20.67, and it is.

In an alternative approach, we multiply the two values before computing the logarithm:

4.08 × 20.67 = 84.3 log 84.3 = 1.926

We could also have expressed 84.3 as a power of 10 and then calculated the logarithm:

log 84.3 = log(8.43 × 10) = log 8.43 + log 10 = 0.926 + 1 = 1.926

As you can see, there may be more than one way to correctly solve a problem.

We can use the properties of exponentials and logarithms to show that the logarithm of the inverse of a number (1/B) is the negative logarithm of that number (−log B):

log(1B)=log B

If we use the formula for division given and recognize that log 1 = 0, then the logarithm of 1/B is

log(1B)= log 1log B= 0log B=log B

Table 4.6 Properties of Logarithms

Operation Exponential Form Logarithm
multiplication (10a)(10b) = 10a+b log(ab) = log a + log b
division (10a10b)=10a  b log(ab)=log alog b

Skill Builder ES3

Convert each number to exponential form and then calculate the logarithm (assume all trailing zeros on whole numbers are not significant).

  1. 100 × 1000
  2. 0.100 ÷ 100
  3. 1000 × 0.010
  4. 200 × 3000
  5. 20.5 ÷ 0.026

Solution

  1. 100 × 1000 = (1 × 102)(1 × 103)

    log[(1 × 102)(1 × 103)] = 2.0 + 3.0 = 5.0

    Alternatively, (1 × 102)(1 × 103) = 1 × 102 + 3 = 1 × 105

    log(1 × 105) = 5.0

  2. 0.100 ÷ 100 = (1.00 × 10−1) ÷ (1 × 102)

    log[(1.00 × 10−1) ÷ (1 × 102)] = 1 × 10−1−2 = 1 × 10−3

    Alternatively, (1.00 × 10−1) ÷ (1 × 102) = 1 × 10[(−1) − 2] = 1 × 10−3

    log(1 × 10−3) = −3.0

  3. 1000 × 0.010 = (1 × 103)(1.0 × 10−2)

    log[(1 × 103)(1 × 10−2)] = 3.0 + (−2.0) = 1.0

    Alternatively, (1 × 103)(1.0 × 10−2) = 1 × 10[3 + (−2)] = 1 × 101

    log(1 × 101) = 1.0

  4. 200 × 3000 = (2 × 102)(3 × 103)

    log[(2 × 102)(3 × 103)] = log(2 × 102) + log(3 × 103)

    = (log 2 + log 102) + (log 3 + log 103)

    = 0.30 + 2 + 0.48 + 3 = 5.8

    Alternatively, (2 × 102)(3 × 103) = 6 × 102 + 3 = 6 × 105

    log(6 × 105) = log 6 + log 105 = 0.78 + 5 = 5.8

  5. 20.5 ÷ 0.026 = (2.05 × 10) ÷ (2.6 × 10−2)

    log[(2.05 × 10) ÷ (2.6 × 10−2)] = (log 2.05 + log 10) − (log 2.6 + log 10−2)

    = (0.3118 + 1) − [0.415 + (−2)]

    = 1.3118 + 1.585 = 2.90

    Alternatively, (2.05 × 10) ÷ (2.6 × 10−2) = 0.788 × 10[1 − (−2)] = 0.788 × 103

    log(0.79 × 103) = log 0.79 + log 103 = −0.102 + 3 = 2.90

Skill Builder ES4

Convert each number to exponential form and then calculate its logarithm (assume all trailing zeros on whole numbers are not significant).

  1. 10 × 100,000
  2. 1000 ÷ 0.10
  3. 25,000 × 150
  4. 658 ÷ 17

Solution

  1. (1 × 10)(1 × 105); logarithm = 6.0
  2. (1 × 103) ÷ (1.0 × 10−1); logarithm = 4.00
  3. (2.5 × 104)(1.50 × 102); logarithm = 6.57
  4. (6.58 × 102) ÷ (1.7 × 10); logarithm = 1.59

4.11 End-of-Chapter Material

Application Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 3 (Section 4.1 "Aqueous Solutions"0) before proceeding to the Application Problems.Problems marked with a ♦ involve multiple concepts.

  1. ♦ Acetaminophen (molar mass = 151 g/mol) is an analgesic used as a substitute for aspirin. If a child’s dose contains 80.0 mg of acetaminophen/5.00 mL of an ethanol-water solution, what is the molar concentration? Acetaminophen is frequently packaged as an ethanol-water solution rather than as an aqueous one. Why?

  2. ♦ Lead may have been the first metal ever recovered from its ore by humans. Its cation (Pb2+) forms a precipitate with Cl according to the following equation:

    Pb2+(aq) + 2Cl(aq) → PbCl2(s)

    When PbCl2 is dissolved in hot water, its presence can be confirmed by its reaction with CrO42−, with which it forms a yellow precipitate:

    Pb2+(aq) + CrO42−(aq) → PbCrO4(s)

    The precipitate is used as a rust inhibitor and in pigments.

    1. What type of reaction does each equation represent?
    2. If 100 mL of a Pb2+ solution produce 1.65 g of lead chromate, what was the concentration of the lead solution?
    3. What volume of a potassium chromate solution containing 0.503 g of solute per 250.0 mL is needed for this reaction?
    4. If all the PbCrO4 originated from PbCl2, what volume of a 1.463 M NaCl solution was needed for the initial reaction?
    5. Why is there environmental concern over the use of PbCrO4?
  3. ♦ Reactions that affect buried marble artifacts present a problem for archaeological chemists. Groundwater dissolves atmospheric carbon dioxide to produce an aqueous solution of carbonic acid:

    CO2(g) + H2O(l) → H2CO3(aq)

    This weakly acidic carbonic acid solution dissolves marble, converting it to soluble calcium bicarbonate:

    CaCO3(s) + H2CO3(aq) → Ca(HCO3)2(aq)

    Evaporation of water causes carbon dioxide to be driven off, resulting in the precipitation of calcium carbonate:

    Ca(HCO3)2(aq) → CaCO3(s) + H2O(l) + CO2(g)

    The reprecipitated calcium carbonate forms a hard scale, or incrustation, on the surface of the object.

    1. If 8.5 g of calcium carbonate were obtained by evaporating 250 mL of a solution of calcium bicarbonate followed by drying, what was the molarity of the initial calcium bicarbonate solution, assuming complete reaction?
    2. If the overall reaction sequence was 75% efficient, how many grams of carbonic acid were initially dissolved in the 250 mL to produce the calcium bicarbonate?
  4. How many Maalox tablets are needed to neutralize 5.00 mL of 0.100 M HCl stomach acid if each tablet contains 200 mg Mg(OH)2 + 200 mg Al(OH)3? Each Rolaids tablet contains 412 mg CaCO3 + 80.0 mg Mg(OH)2. How many Rolaids tablets are needed? Suggest another formula (and approximate composition) for an effective antacid tablet.

  5. Citric acid (C6H8O7, molar mass = 192.12 g/mol) is a triprotic acid extracted from citrus fruits and pineapple waste that provides tartness in beverages and jellies. How many grams of citric acid are contained in a 25.00 mL sample that requires 38.43 mL of 1.075 M NaOH for neutralization to occur? What is the formula of the calcium salt of this compound?

  6. ♦ A method for determining the molarity of a strongly acidic solution has been developed based on the fact that a standard solution of potassium iodide and potassium iodate yields iodine when treated with acid:

    IO3 + 5I + 6H+ → 3I2 + 3H2O

    Starch is used as the indicator in this titration because starch reacts with iodine in the presence of iodide to form an intense blue complex. The amount of iodine produced from this reaction can be determined by subsequent titration with thiosulfate:

    2S2O32+ + I2 → S4O62− + 2I

    The endpoint is reached when the solution becomes colorless.

    1. The thiosulfate solution was determined to be 1.023 M. If 37.63 mL of thiosulfate solution were needed to titrate a 25.00 mL sample of an acid, what was the H+ ion concentration of the acid?
    2. If the 25.00 mL sample that was titrated had been produced by dilution of a 10.00 mL sample of acid, what was the molarity of the acid in the original solution?
    3. Why might this be an effective method for determining the molarity of a strong acid, such as H2SO4?
  7. ♦ Sewage processing occurs in three stages. Primary treatment removes suspended solids, secondary treatment involves biological processes that decompose organic matter, and tertiary treatment removes specific pollutants that arise from secondary treatment (generally phosphates). Phosphate can be removed by treating the HPO42− solution produced in the second stage with lime (CaO) to precipitate hydroxyapatite [Ca5(PO4)3OH].

    1. Write a balanced chemical equation for the reaction that occurs in the tertiary treatment process.
    2. What has been neutralized in this process?
    3. Four pounds of hydroxyapatite precipitated from the water. What mass of lime was used in the reaction?
    4. Assuming a volume of water of 30 m3, what was the hydrogen phosphate anion concentration in the water?
  8. Calcium hydroxide and calcium carbonate are effective in neutralizing the effects of acid rain on lakes. Suggest other compounds that might be effective in treating lakes. Give a plausible reason to explain why Ca(OH)2 and CaCO3 are used.

  9. ♦ Approximately 95% of the chlorine produced industrially comes from the electrolysis of sodium chloride solutions (brine):

    NaCl(aq) + H2O(l) → Cl2(g) + H2(g) + NaOH(aq)

    Chlorine is a respiratory irritant whose presence is easily detected by its penetrating odor and greenish-yellow color. One use for the chlorine produced is in the manufacture of hydrochloric acid.

    1. In the chemical equation shown, what has been oxidized and what has been reduced?
    2. Write the oxidation and reduction equations for this reaction.
    3. Balance the net ionic equation.
    4. Name another salt that might produce chlorine by electrolysis and give the expected products. Identify those products as gases, liquids, or solids.
  10. ♦ The lead/acid battery used in automobiles consists of six cells that produce a 12 V electrical system. During discharge, lead(IV) oxide, lead, and aqueous sulfuric acid react to form lead(II) sulfate and water.

    1. What has been oxidized and what has been reduced?
    2. Write and balance the equation for the reaction.
    3. What is the net ionic equation?
    4. What is the complete ionic equation?
    5. What hazard is associated with handling automobile batteries?
  11. ♦ The use of iron, which is abundant in Earth’s crust, goes back to prehistoric times. In fact, it is believed that the Egyptians used iron implements approximately 5000 years ago. One method for quantifying the iron concentration in a sample involves three steps. The first step is to dissolve a portion of the sample in concentrated hydrochloric acid to produce ferric chloride; the second is to reduce Fe3+ to Fe2+ using zinc metal; and the third is to titrate Fe2+ with permanganate, producing Mn2+(aq) and ferric iron in the form of Fe2O3.

    1. Write chemical equations for all three steps.
    2. Write the net ionic equations for these three reactions.
    3. If 27.64 mL of a 1.032 M solution of permanganate are required to titrate 25.00 mL of Fe2+ in the third step, how many grams of Fe were in the original sample?
    4. Based on your answer to part c, if the original sample weighed 50.32 g, what was the percentage of iron?
  12. Baking powder, which is a mixture of tartaric acid and sodium bicarbonate, is used in baking cakes and bread. Why does bread rise when you use baking powder? What type of reaction is involved?

  13. An activity series exists for the halogens, which is based on the ease of reducing the diatomic halogen molecule (X2) to X. Experimentally, it is found that fluorine is the easiest halogen to reduce (i.e., F2 is the best oxidant), and iodine is the hardest halogen to reduce (i.e., I2 is the worst oxidant). Consequently, the addition of any diatomic halogen, Y2, to solutions containing a halide ion (X) that lies below Y in the periodic table will result in the reduction of Y2 to Y and the oxidation of X. Describe what you would expect to occur when

    1. chlorine is added to an aqueous solution of bromide.
    2. iodine crystals are added to a solution of potassium bromide.

    Bromide is present in naturally occurring salt solutions called brines. Based on your answers, propose an effective method to remove bromide from brine.

  14. ♦ Marble is composed of mostly calcium carbonate. Assuming that acid rain contains 4.0 × 10−5 M H2SO4, approximately what volume of rain is necessary to dissolve a 250 lb marble statue?

  15. ♦ One of the “first-aid” measures used to neutralize lakes whose pH has dropped to critical levels is to spray them with slaked lime (Ca(OH)2) or limestone (CaCO3). (A slower but effective alternative is to add limestone boulders.) How much slaked lime would be needed to neutralize the acid in a lake that contains 4.0 × 10−5 M H2SO4 and has a volume of 1.2 cubic miles (5.0 × 1012 L)?

  16. Recall from Section 4.3 "Stoichiometry of Reactions in Solution" that the reaction of ethanol with dichromate ion in acidic solution yields acetic acid and Cr3+(aq):

    Cr2O72−(aq) + CH3CH2OH(aq) → Cr3+(aq) + CH3CO2H(aq)

    Balance the equation for this reaction using oxidation states. (Hint: the oxidation state of carbon in the CH3 group remains unchanged, as do the oxidation states of hydrogen and oxygen.)

Answers

  1. 0.106 M acetaminophen; acetaminophen is an organic compound that is much more soluble in ethanol than water, so using an ethanol/water mixture as the solvent allows a higher concentration of the drug to be used.

    1. 0.34 M Ca(HCO3)2
    2. 7.0 g H2CO3
  2. 2.646 g citric acid, Ca3(C6H5O7)2

    1. 5CaO + 3HPO42− + 2H2O → Ca5(PO4)3OH + 6OH
    2. This is an acid–base reaction, in which the acid is the HPO42− ion and the base is CaO. Transferring a proton from the acid to the base produces the PO43− ion and the hydroxide ion.
    3. 2.2 lbs (1 kg) of lime
    4. 3.6 × 10−4 M HPO42−
    1. Chloride is oxidized, and protons are reduced.
    2.  

      Oxidation: 2Cl → Cl2 + 2e− Reduction: 2H2O + 2e− → H2 + 2OH−
    3.  

      2Cl + 2H2O → Cl2 + H2 + 2OH