University Chemistry

Lai69040_fm_i-xxiv.indd Page i 1/30/08 10:23:06 PM elhi University Chemistry /Volumes/108/MHIA037/mhLai1/Lai1chfm%0 ...

Author: Brian B. Laird

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University Chemistry

/Volumes/108/MHIA037/mhLai1/Lai1chfm%0

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University Chemistry

Brian B. Laird University of Kansas

With significant contributions by

Raymond Chang Williams College

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UNIVERSITY CHEMISTRY Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © 2009 by The McGraw-Hill Companies, Inc. All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 DOW/DOW 0 9 8 ISBN 978–0–07–296904–7 MHID 0–07–296904–0 Publisher: Thomas Timp Senior Sponsoring Editor: Tamara L. Hodge Vice-President New Product Launches: Michael Lange Senior Developmental Editor: Shirley R. Oberbroeckling Marketing Manager: Todd L. Turner Senior Project Manager: Gloria G. Schiesl Senior Production Supervisor: Kara Kudronowicz Lead Media Project Manager: Judi David Senior Designer: David W. Hash Cover/Interior Designer: Elise Lansdon (USE) Cover Image: Water Droplets, ©Masato Tokiwa/Amana Images/Getty Images Senior Photo Research Coordinator: John C. Leland Photo Research: David Tietz/Editorial Image, LLC Supplement Producer: Mary Jane Lampe Compositor: Aptara, Inc. Typeface: 10/12 Times Roman Printer: R. R. Donnelley Willard, OH The credits section for this book begins on page C-1 and is considered an extension of the copyright page. Library of Congress Cataloging-in-Publication Data Laird, Brian B., 1960University chemistry / Brian B. Laird. p. cm. Includes index. ISBN 978–0–07–296904–7 — ISBN 0–07–296904–0 (hard copy : alk. paper) 1. Chemistry—Study and teaching (Higher) 2. Chemistry—Textbooks. I. Title. QD40.L275 2009 540—dc22 2007052540

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About the Author Brian B. Laird,

a native of Port Arthur, Texas, is currently a Professor of Chemistry at the University of Kansas in Lawrence, Kansas. He received Bachelor of Science degrees in Chemistry and Mathematics from the University of Texas, Austin, in 1982, and a Ph.D. in Theoretical Chemistry from the University of California, Berkeley, in 1987. Prior to his current position, he held postdoctoral and lecturer appointments at Columbia University, Forschungszentrum Jülich, Germany (NATO Fellowship), University of Utah, University of Sydney, and the University of Wisconsin. His research interests involve the application of statistical mechanics and computer simulation to the determination of properties of liquid and solids. In addition to honors general chemistry, he regularly teaches undergraduate physical chemistry and graduate courses in quantum and statistical mechanics. In his spare time, he enjoys golfing, bicycling, playing the piano, and traveling.

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I dedicate this work to my wife, Uschi, and to the memory of my parents, Don and Nanci Laird.

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Brief Contents 0

The Language of Chemistry ...............................................................

1

The Quantum Theory of the Submicroscopic World ........................ 71

2

Many-Electron Atoms and the Periodic Table ................................... 126

3

The Chemical Bond ............................................................................ 170

4

Molecular Structure and Interaction ................................................... 222

5

The States of Matter I: Phase Diagrams and Gases ......................... 281

6

The States of Matter II: Liquids and Solids ...................................... 333

7

Thermochemistry: Energy in Chemical Reactions ............................ 364

8

Entropy, Free Energy, and the Second Law of Thermodynamics .... 423

9

Physical Equilibrium ........................................................................... 466

1

10 Chemical Equilibrium ......................................................................... 511 11 Acids and Bases .................................................................................. 556 12 Acid-Base Equilibria and Solubility .................................................. 611 13 Electrochemistry .................................................................................. 663 14 Chemical Kinetics ............................................................................... 712 15 The Chemistry of Transition Metals .................................................. 772 16 Organic and Polymer Chemistry ........................................................ 800 17 Nuclear Chemistry .............................................................................. 855 Appendix 1 Measurement and Mathematical Background ................... A-1 Appendix 2 Thermodynamic Data at 1 Bar and 258C ......................... A-14 Appendix 3 Derivation of the Names of Elements ............................... A-20 Appendix 4 Isotopes of the First Ten Elements .................................... A-26

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Expanded Contents List of Applications xiv Preface xv

0 The Language of Chemistry

..........................................................

1

0.1

Chemistry Is the Study of Matter and Change ............................................... 2

0.2

Matter Consists of Atoms and Molecules ..................................................... 11

0.3

Compounds Are Represented by Chemical Formulas .................................. 20

0.4

Reactions Are Represented by Balanced Chemical Equations ..................... 31

0.5

Quantities of Atoms or Molecules Can Be Described by Mass or Number ....................................................................................... 34

0.6

Stoichiometry Is the Quantitative Study of Mass and Mole Relationships in Chemical Reactions .................................................. 52

1 The Quantum Theory of the Submicroscopic World ................................................................................................................... 71 1.1

Classical Physics Does Not Adequately Describe the Interaction of Light with Matter ..................................................................................................... 72

1.2

The Bohr Model Was an Early Attempt to Formulate a Quantum Theory of Matter ......................................................................................................... 83

1.3

Matter Has Wavelike Properties .................................................................... 94

1.4

The Hydrogen Atom Is an Exactly Solvable Quantum-Mechanical System ...................................................................... 109

2 Many-Electron Atoms and the Periodic Table

viii

.................... 126

2.1

The Wavefunctions of Many-Electron Atoms Can Be Described to a Good Approximation Using Atomic Orbitals ........................................... 127

2.2

Electron Configurations of Many-Electron Atoms Are Constructed Using the Aufbau (or “Building-up”) Principle ........................................... 134

2.3

The Periodic Table Predates Quantum Mechanics ...................................... 143

2.4

Elements Can Be Classified by Their Position in the Periodic Table ................................................................................................ 146

2.5

The Properties of the Elements Vary Periodically Across the Periodic Table .......................................................................................... 149

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3 The Chemical Bond

.................................................................................. 170

3.1

Atoms in a Molecule Are Held Together by Chemical Bonds .................. 171

3.2

A Covalent Bond Involves the Sharing of Electrons Between Atoms in a Molecule .................................................................................... 173

3.3

Electronegativity Differences Determine the Polarity of Chemical Bonds ............................................................................................ 182

3.4

Drawing Correct Lewis Structures Is an Invaluable Skill for a Chemist ......................................................................................................... 189

3.5

Molecular Orbital Theory Provides a Detailed Description of Chemical Bonding ........................................................................................ 202

4 Molecular Structure and Interaction

.........................................

222

4.1

The Basic Three-Dimensional Structure of a Molecule Can Be Predicted Using the VSEPR Model ............................................................................. 223

4.2

The Polarity of a Molecule Can Be Described Quantitatively by Its Dipole Moment ....................................................................................... 234

4.3

Valence Bond Theory for Polyatomic Molecules Requires the Use of Hybrid Orbitals ................................................................................. 240

4.4

Isomers Are Compounds That Have the Same Molecular Formula but Different Atomic Arrangements ................................................................... 252

4.5

Bonding in Polyatomic Molecules Can Be Explained Using Molecular Orbitals ........................................................................................ 257

4.6

The Interactions Between Molecules Greatly Affect the Bulk Properties of Materials ................................................................................. 262

5 The States of Matter I: Phase Diagrams and Gases ....................................................................................................... 281 5.1

Pressure and Temperature Are Two Important Macroscopic Properties of Chemical Systems ........................................................................................ 282

5.2

Substances and Mixtures Can Exist as Solid, Liquid, or Gas, Depending upon the External Conditions ...................................................................... 286

5.3

The Ideal-Gas Equation Describes the Behavior of All Gases in the Limit of Low Pressure ................................................................................. 292

5.4

The Kinetic Theory of Gases Provides a Molecular Explanation for the Behavior of Gases ........................................................................................ 308

5.5

Real Gases Exhibit Deviations from Ideal Behavior at High Pressures .... 317

6 The States of Matter II: Liquids and Solids

........................ 333

6.1

The Structure and Properties of Liquids Are Governed by Intermolecular Interactions .................................................................................................... 334

6.2

Crystalline Solids Can Be Classified in Terms of Their Structure and Intermolecular Interactions ........................................................................... 341

6.3

The Properties of Crystalline Solids Are Determined Largely by Intermolecular Interactions ........................................................................... 351

6.4

Band Theory Accurately Explains the Conductivity of Metals, Semiconductors, and Insulators ................................................................... 356 ix

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7 Thermochemistry: Energy in Chemical Reactions

........... 364

7.1

Thermodynamics Is the Study of Energy and Its Transformations in Macroscopic Systems ................................................................................... 365

7.2

The Energy Absorbed by a System as Heat in a Constant-Pressure Process Is Equal to the Change in Enthalpy ............................................... 375

7.3

The Temperature Change of a System upon Heating Is Governed by Its Heat Capacity .......................................................................................... 381

7.4

The Enthalpy Changes for any Reaction Can Be Calculated Using Standard Enthalpies of Formation ................................................................ 395

7.5

The Reaction Enthalpies Can Be Estimated from Bond Enthalpies .......... 401

7.6

Enthalpy Changes Also Accompany Physical Transformations ................. 405

7.7

The Temperature Dependence of Reaction Enthalpies Can Be Determined from Heat Capacity Data ................................................... 412

8 Entropy, Free Energy, and the Second Law of Thermodynamics ........................................................................................ 423 8.1

The Entropy of an Isolated System Always Increases in Any Spontaneous Process .................................................................................... 424

8.2

The Entropy Change for a Process Can Be Calculated Using the Thermodynamic Definition of Entropy ....................................................... 432

8.3

The Third Law of Thermodynamics Allows Us to Determine Absolute Entropies ....................................................................................... 440

8.4

The Spontaneity of a Process at Constant Temperature and Pressure Is Governed by the Gibbs Free Energy ........................................................... 446

8.5

The Mixing of Pure Substances Leads to an Increase in the Entropy and a Decrease in the Gibbs Free Energy .................................................. 456

8.6

In Living Systems, Spontaneous Reactions Are Used to Drive Other Nonspontaneous, but Essential, Biochemical Processes ............................. 459

9 Physical Equilibrium 9.1

The Phase Boundaries in Pure Substances Can Be Predicted Using Thermodynamics .......................................................................................... 467

9.2

The Solubility of a Substance Is Determined by Temperature, Pressure, and Intermolecular Forces ............................................................................ 473

9.3

The Liquid-Vapor Phase Equilibrium of a Solution Can Be Understood in Terms of the Entropy of Mixing and the Intermolecular Forces ................................................................................... 483

9.4

Colligative Properties Are Properties of Solution Phase Equilibria That Depend Only upon the Number of Solute Molecules, Not Their Type ........ 491

10 Chemical Equilibrium

x

............................................................................... 466

............................................................................. 511

10.1

The Equilibrium Constant Governs the Concentration of Reactants and Products at Equilibrium ........................................................................ 512

10.2

The Equilibrium Constant Can Be Used to Predict the Direction and Equilibrium Concentrations of a Chemical Reaction ................................. 524

10.3

The Equilibrium Constant for a Reaction Can Be Determined from the Standard Gibbs Energy Change ................................................................... 531

10.4

The Response of an Equilibrium System to a Change in Conditions Can Be Determined Using Le Châtelier’s Principle ........................................... 536

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11 Acids and Bases

........................................................................................ 556

11.1

Many Processes in Chemistry Are Acid-Base Reactions ........................... 557

11.2

The Acid-Base Properties of Aqueous Solutions Are Governed by the Autoionization Equilibrium of Water .......................................................... 564

11.3

The Strengths of Acids and Bases Are Measured by Their Ionization Constants ....................................................................................................... 570

11.4

The pH of an Acid or Base Can Be Calculated If Its Ionization Constant Is Known ....................................................................................................... 579

11.5

The Strength of an Acid Is Determined in Part by Molecular Structure ...................................................................................... 590

11.6

Many Salts Have Acid-Base Properties in Aqueous Solution .................... 594

11.7

Oxide and Hydroxide Compounds Can Be Acidic or Basic in Aqueous Solution Depending on Their Composition ................................................. 600

12 Acid-Base Equilibria and Solubility

............................................ 611

12.1

Ionization of Weak Acids and Bases Is Suppressed by the Addition of a Common Ion ................................................................................................. 612

12.2

The pH of a Buffer Solution Is Resistant to Large Changes in pH .............................................................................................. 615

12.3

The Concentration of an Unknown Acid or Base Can Be Determined by Titration ................................................................................................... 622

12.4

An Acid-Base Indicator Is a Substance That Changes Color at a Specific pH ................................................................................................... 631

12.5

A Precipitation Reaction Occurs when a Reaction in Solution Leads to an Insoluble Product .......................................................................................... 633

12.6

The Solubility Product Is the Equilibrium Constant for the Dissolution Process ...................................................................................... 635

12.7

The Solubility of a Substance Is Affected by a Number of Factors ......... 644

12.8

The Solubility Product Principle Can Be Applied to Qualitative Analysis ...................................................................................... 653

13 Electrochemistry

......................................................................................... 663

13.1

Oxidation-Reduction (Redox) Reactions Involve a Transfer of Electrons from One Species to Another ...................................................................... 664

13.2

Redox Reactions Can Be Used to Generate Electric Current in a Galvanic Cell ................................................................................................ 671

13.3

The Standard Emf of Any Electrochemical Cell Can Be Determined If the Standard Reduction Potentials for the Half-Reactions Are Known ................................................................................................... 674

13.4

The Emf of an Electrochemical Cell Is Directly Related to the Gibbs Free-Energy Change of the Redox Reaction ................................... 681

13.5

The Concentration Dependence of the Emf Can Be Determined Using the Nernst Equation ...................................................................................... 686

13.6

Batteries Use Electrochemical Reactions to Produce a Ready Supply of Electric Current ........................................................................................ 692

13.7

In Electrolysis, an Electric Current Is Used to Drive a Nonspontaneous Reaction ........................................................................................................ 697 xi

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14 Chemical Kinetics

..................................................................................... 712

14.1

Chemical Kinetics Is the Study of the Rates at Which Chemical Reactions Occur ............................................................................................ 713

14.2

The Rate Law Gives the Dependence of the Reaction Rate on the Reactant Concentration ................................................................................ 720

14.3

Integrated Rate Laws Specify the Relationship Between Reactant Concentration and Time ............................................................................... 723

14.4

The Arrhenius Equation Gives the Temperature Dependence of Rate Constants ....................................................................................................... 736

14.5

The Reaction Mechanism Is the Sequence of Elementary Steps That Lead to Product Formation ................................................................. 744

14.6

Reaction Rates Can Often Be Increased by the Addition of a Catalyst ................................................................................................. 754

15 The Chemistry of Transition Metals 15.1

Transition Metals Have Electron Configurations with Incomplete d or f Shells .................................................................................................. 773

15.2

Transition Metals Can Form a Variety of Coordination Compounds ........ 777

15.3

Bonding in Coordination Compounds Can Be Described by Crystal Field Theory .................................................................................... 786

15.4

The Reactions of Coordination Compounds Have a Wide Number of Useful Applications ...................................................................................... 793

16 Organic and Polymer Chemistry

.................................................... 800

16.1

Hydrocarbons Are Organic Compounds Containing Only Hydrogen and Carbon .................................................................................................... 801

16.2

Hydrocarbons Undergo a Number of Important Chemical Reactions ....... 811

16.3

The Structure and Properties of Organic Compounds Are Greatly Influenced by the Presence of Functional Groups ...................................... 815

16.4

Polymers Are Large Molecular Weight Compounds Formed from the Joining Together of Many Subunits Called Monomers .............................. 826

16.5

Proteins Are Polymer Chains Composed of Amino Acid Monomers ....... 833

16.6

DNA and RNA Are Polymers Composed of Nucleic Acids ...................... 841

17 Nuclear Chemistry

xii

............................................ 772

.................................................................................... 855

17.1

Nuclear Chemistry Is the Study of Changes Involving Atomic Nuclei ..... 856

17.2

The Stability of a Nucleus Is Determined Primarily by Its Neutron-to-Proton Ratio ......................................................................... 860

17.3

Radioactive Decay Is a First-Order Kinetic Process .................................. 867

17.4

New Isotopes Can Be Produced Through the Process of Nuclear Transmutation ............................................................................................... 873

17.5

In Nuclear Fission, a Large Nucleus Is Split into Smaller Nuclei ............ 876

17.6

In Nuclear Fusion, Energy Is Produced When Light Nuclei Combine to Form Heavier Ones ...................................................................................... 882

17.7

Radioactive and Stable Isotopes Alike Have Many Applications in Science and Medicine .............................................................................. 884

17.8

The Biological Effects of Radiation Can Be Quite Dramatic .................... 886

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Appendix 1 Measurement and Mathematical Background .......................... A-1 A1.1

Measurement ................................................................................................. A-1

A1.2

Mathematical Background ............................................................................ A-7

Appendix 2 Thermodynamic Data at 1 Bar and 258C ......................... A-14 Appendix 3 Derivation of the Names of the Elements ........................ A-20 Appendix 4 Isotopes of the First Ten Elements ...................................... A-26 Glossary ............................................................................................................................... G-1 Answers to Even-Numbered Problems .......................................................................... AP-1 Credits .................................................................................................................................. C-1 Index ..................................................................................................................................... I-1

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List of Applications Distribution of Elements on Earth and in Living Systems .............................. 18 Important Experimental Technique: The Mass Spectrometer ......................... 46 Laser—The Splendid Light ............................................................................. 92 Important Experimental Technique: Electron Microscopy ........................... 109 The Third Liquid Element? ........................................................................... 156 Discovery of the Noble Gases ....................................................................... 163 Major Experimental Technique: Microwave Spectroscopy .......................... 186 Just Say NO ................................................................................................... 198 Major Experimental Technique: Infrared Spectroscopy ............................... 238 cis-trans Isomerization in the Vision Process ................................................ 254 Buckyball, Anyone? ...................................................................................... 262 Super-Cold Atoms ......................................................................................... 316 Why Do Lakes Freeze from the Top Down? ................................................. 340 High-Temperature Superconductors ............................................................. 358 Fuel Values of Foods and Other Substances ................................................. 390 The Efficiency of Heat Engines: The Carnot Cycle ...................................... 438 The Thermodynamics of a Rubber Band ...................................................... 456 The Killer Lake ............................................................................................. 483 Life at High Altitudes and Hemoglobin Production ...................................... 545 Antacids and the pH Balance in Your Stomach ............................................. 602 Maintaining the pH of Blood ........................................................................ 620 Dental Filling Discomfort ............................................................................. 680 Femtochemistry ............................................................................................. 753 Coordination Compounds in Living Systems ................................................ 784 Cisplatin—an Anticancer Drug ..................................................................... 795 Important Experimental Technique: Nuclear Magnetic Resonance Spectroscopy ............................................................................... 824 Sickle Cell Anemia: A Molecule Disease ..................................................... 840 DNA Fingerprinting ...................................................................................... 843 Nature’s Own Fission Reactor ....................................................................... 881 Food Irradiation ............................................................................................. 888 xiv

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Preface The concept of University Chemistry grew from my experiences in teaching Honors General Chemistry at the University of Kansas for a number of semesters. It is my attempt to inform and challenge the well-prepared student to discover and learn the diverse, but related, topics within general chemistry. This text includes the core topics that are necessary for a solid foundation of chemistry.

The Basic Features Organization. In this text, I adopt a “Molecular to Macroscopic” approach, in which the quantum theory of atomic and molecular structure and interaction is outlined in Chapters 1– 4. Building on this molecular foundation, the presentation moves to the macroscopic concepts, such as states of matter, thermodynamics, physical and chemical equilibrium, and chemical kinetics. This organization is based on “natural prerequisites”; that is, each topic is positioned relative to what other topics are required to understand it. For example, knowledge of thermodynamics or equilibrium chemistry is not needed to understand the structure and interaction of atoms and molecules; whereas, to understand deeply the application of thermodynamics to chemical systems or the material properties of liquids and solids, knowing how energy is stored in chemical bonds and how molecular structure and bonding affect intermolecular forces is desirable. Mathmematical Level. The presentation in this text assumes that the student has a good working knowledge of algebra, trigonometry and coordinate geometry at the high school level. Knowledge of calculus, while advantageous, is not strictly required for full understanding. Integral and differential calculus is used, where appropriate, in intermediate steps of concept development in quantum theory, thermodynamics, and kinetics. However,

the final primary concepts and most major equations (denoted by a blue box) do not depend on an understanding of calculus, nor do the overwhelming majority of end-of-chapter problems. For the interested and advanced student, I have included a small number of calculus-based end-of-chapter problems in the relevant chapters. The level of calculus used in University Chemistry is similar to that used in other general chemistry texts at this level; however, it is not relegated to secondary boxed text, as is often done, but integrated into the primary discussion, so as not to disrupt the linear flow of the presentation. For the interested student, I have included in Appendix 1 a brief review/tutorial of the basic concepts in integral and differential calculus. Problem-Solving Model. Worked Examples are included in every chapter for students to use as a base for applying their problem-solving skills to the concept discussed. The examples present the problem, a strategy, a solution, a check, and a practice problem. Every problem is designed to challenge the student to think logically through the problem. This problem-solving approach is used throughout the text.

Organization and Presentation Review. Students with a strong background in high school chemistry have already been exposed to the concepts of the structure and classification of matter, chemical nomenclature, and stoichiometry. Because of this assumed background, I have condensed the standard introductory chapters of a typical general chemistry text into a single chapter (Chapter 0). Chapter 0 is intended to serve as a refresher of the subject matter students covered in their high school chemistry courses. Early Coverage of Quantum Theory. To provide a molecular-level foundation for the later chapters on xv

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states Lai69040_ch01_071-125.indd of matter, thermodynamics, and 8:56:35 equilibrium, Page 80 1/5/08 PM elhi the quantum theory of atoms and molecules is presented early in the text. In Chapters 1 and 2, elementary quantum theory is used to discuss the electronic structure of atoms and the construction of the periodic table. Chapters 3 and 4 cover molecular bonding, structure, and interaction, including molecular-orbital theory. Contrary to the organization of most general chemistry texts, intermolecular forces are discussed at the end of Chapter 4 on molecular structure instead of in a later chapter on liquids and solids. This is a more natural position, which allows for a molecular-level discussion of the forces that influence real gas behavior in Chapter 5. States of Matter. Phase diagrams, equations of state, and states of matter (gases, liquids and solids) are treated in a unified manner in Chapters 5 and 6, with an emphasis on the role of molecular interaction in the determination of material properties.

allows for a more sophisticated /Volumes/108/MHIA037/mhLai1/Lai1ch01%0 discussion of physical and chemical equilibrium from a thermodynamic perspective. In particular, the central role of the entropy and free energy of mixing in colligative properties and chemical equilibrium is explored in detail. Physical and Chemical Equilibrium. The principles of physical equilibrium (phase boundary prediction and solubility) are discussed in Chapter 9, followed by a discussion of chemical equilibrium in Chapter 10. Chapters 11, 12, and 13 present applications of chemical equilibrium to acid-base chemistry, aqueous equilibria, and electrochemistry, respectively. Chemical Kinetics. Unlike many general chemistry texts, discussion of chemical kinetics (Chapter 14) follows the presentation of chemical equilibrium, allowing for full discussion of transition-state theory and detailed balance. Final chapters. Chapter 15, “The Chemistry of Transition Metals,” Chapter 16, “Organic and Polymer Chemistry,” and Chapter 17 “Nuclear Chemistry” are each an entity in itself. Every instructor and student can choose to assign and study the chapters according to time and preference.

Thermochemistry, Entropy, and Free Energy. The basic principles of thermodynamics are treated together in Chapters 7 “Thermochemistry: Energy in Chemical Reactions,” and 8, “Entropy, Free Energy, and the Second Law of Thermodynamics.” This

Pedagogy Problem Solving The development of problem-solving skills is a major objective of this text. Each problem is broken down into learning steps to help students increase their logical critical thinking skills.

Example 1.2 Chlorophyll-a is green because it absorbs blue light at about 435 nm and red light at about 680 nm, so that mostly green light is transmitted. Calculate the energy per mole of photons at these wavelengths.

Strategy Planck’s equation (Equation 1.3) gives the relationship between energy and frequency (n). Because we are given wavelength (l), we must use Equation 1.2, in which u is replaced with c (the speed of light), to convert wavelength to frequency. Finally, the problem asks for the energy per mole, so we must multiply the result we get from Equation 1.3 by Avogadro’s number. Solution The energy of one photon with a wavelength of 435 nm is c 3.00 3 108 m s21 E 5 hn 5 h a b 5 (6.626 3 10234 J s) l 435 nm (1 3 1029 m nm21 ) 5 4.57 3 10219 J

For one mole of photons, we have E 5 (4.57 3 10219 J) (6.022 3 1023 mol21 ) 5 2.75 3 105 J mol21 5 275 kJ mol21 Using an identical approach for the photons at 680 nm, we get E 5 176 kJ mol−1.

Practice Exercise X-rays are convenient to study the structure of crystals because their wavelengths are comparable to the distances between near neighbor atoms (on the order of a few Ångstroms, where 1Å 5 1 3 10210 m). Calculate the energy of a photon of X-ray radiation with a wavelength of 2.00 Å.

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There are numerous end-of-chapter problems to continue skill building and then practice solving problems. Many of these same problems appear in the electronic homework program ARIS, providing a seamless homework solution for the student and the instructor.

1.58 Certain sunglasses have small crystals of silver chloride (AgCl) incorporated in the lenses. When the lenses are exposed to light of the appropriate wavelength, the following reaction occurs: AgCl ¡ Ag 1 Cl

The Ag atoms formed produce a uniform gray color that reduces the glare. If the energy required for the preceding reaction is 248 kJ mol21, calculate the maximum wavelength of light that can induce this process.

Text

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ARIS

/Volumes/108/MHIA037/mhLai1/Lai1ch01%0

End-of-Chapter Material At the end of every chapter, you will find a summary of all the material that was presented in the chapter to use as a study tool. The summary highlights each section within the chapter. Key words are also listed and include the page number where the term was introduced.

Summary of Facts and Concepts Section 1.1 c At the end of the nineteenth century, scientists began to realize that the laws of classical physics were incompatible with a number of new experiments that probed the nature of atoms and molecules and their interaction with light. Through the work of a number of scientists over the first three decades of the twentieth century, a new theory—quantum mechanics—was developed that was able to explain the behavior of objects on the atomic and molecular scale. c The quantum theory developed by Planck successfully explains the emission of radiation by heated solids. The quantum theory states that radiant energy is emitted by atoms and molecules in small discrete amounts (quanta), rather than over a continuous range. This behavior is

a moving particle of mass m and velocity u is given by the de Broglie equation l 5 h/mu (Equation 1.20). c The realization that matter at the atomic and subatomic scale possesses wavelike properties lead to the development of the Heisenberg uncertainty principle, which states that it is impossible to know simultaneously both the position (x) and the momentum (p) of a particle with certainty (see Equation 1.22). c The Schrödinger equation (Equation 1.24) describes the motions and energies of submicroscopic particles. This equation, in which the state of a quantum particle is described by its wavefunction, launched modern quantum mechanics and a new era in physics. The wavefunction contains information about the probability of finding a particle in a given region of space.

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Applications Throughout the text, applications are included to reinforce students’ grasp of concepts and principles and to provide grounding to real-world experiences. The applications focus on key industrial chemicals, drugs, and technological advances in chemistry.

Important Experimental Technique: Electron Microscopy

T

he electron microscope is an extremely valuable application of the wavelike properties of electrons because it produces images of objects that cannot be seen with the naked eye or with light microscopes. According to the laws of optics, it is impossible to form an image of an object that is smaller than half the wavelength of the light used for the observation. Because the range of visible light wavelengths starts at around 400 nm, or 4 3 1027 m, we cannot see anything smaller than 2 3 1027 m. In principle, we can see objects on the atomic and molecular scale by using X-rays, whose wavelengths range from about 0.01 nm to 10 nm. Xrays cannot be focused easily, however, so they do not produce crisp images. Electrons, on the other hand, are charged particles, which can be focused in the same way the image on a TV screen is focused (that is, by applying an electric field or a magnetic field). According to Equation 1.20, the wavelength of an electron is inversely proportional to its velocity. By accelerating electrons to very high velocities, we can obtain wavelengths as short as 0.004 nm. A different type of electron microscope, called the scanning tunneling microscope (STM), uses quantum mechanical

tunneling to produce an image of the atoms on the surface of a sample. Because of its extremely small mass, an electron is able to move or “tunnel” through an energy barrier (instead of going over it). The STM consists of a metal needle with a very fine point (the source of the tunneling electrons). A voltage is maintained between the needle and the surface of the sample to induce electrons to tunnel through space to the sample. As the needle moves over the sample at a distance of a few atomic diameters from the surface, the tunneling current is measured. This current decreases with increasing distance from the sample. By using a feedback loop, the vertical position of the tip can be adjusted to a constant distance from the surface. The extent of these adjustments, which profile the sample, is recorded and displayed as a three-dimensional false-colored image. Both the electron microscope and the STM are among the most powerful tools in chemical and biological research.

An electron micrograph showing a normal red blood cell and a sickled red blood cell from the same person.

STM image of iron atoms arranged to display the Chinese characters for atom on a copper surface.

3608 Development Process A key factor in developing any chemistry text is the ability to adapt to teaching specifications in a universal way. The only way to do so is by contacting those universal voices—and learning from their suggestions. We are confident that our book has the most current content the industry has to offer, thus pushing our desire for accuracy and up-to-date information to the highest standard possible. To accomplish this, we have moved along an arduous road to production. Extensive and openminded advice is critical in the production of a superior text. Following is a brief overview of the initiatives included in the 360⬚ Development Process of this first edition of University Chemistry, by Brian B. Laird.

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Symposia Every year McGraw-Hill conducts a general chemistry symposium, which is attended by instructors from across the country. These events provide an opportunity for the McGraw-Hill editors to gather information about the needs and challenges of instructors teaching these courses. The information gleaned from these events helped to create the book plan for University Chemistry. In addition, these symposia offer a forum for the attendees to exchange ideas and experiences with colleagues whom they might not have otherwise met.

ARIS

Manuscript Review Panels Over 50 teachers and academics from across the country and internationally reviewed the various drafts of the manuscript to give feedback on content, pedagogy, and organization. This feedback was summarized by the book team and used to guide the direction of the text.

Build Assignments Choose from prebuilt assignments or create your own custom content by importing your own content or editing an existing assignment from the prebuilt assignment. Assignments can include quiz questions, animations, and videos—anything found on the website. Create announcements and utilize full course or individual student communication tools. Assign questions that were developed using the same problem-solving strategy as in the textual material, thus allowing students to continue the learning process from the text into their homework assignments. Assign algorithmic questions that give students multiple chances to practice and gain skill at problem-solving the same concept.

Developmental Editing In addition to being influenced by a distinguished chemistry author, the development of this manuscript was impacted by three freelance developmental editors. The first edit in early draft stage was completed by an editor who holds a PhD in chemistry, John Murdzek. Katie Aiken and Lucy Mullins went through the manuscript line-by-line offering suggestions on writing style and pedagogy. Accuracy Check and Class Test Cindy Berrie at the University of Kansas worked closely with the author, checking his work and providing detailed feedback as she and her students did a two-semester class test of the manuscript. The students also provided the author with comments on how to improve the manuscript so that the presentation of content was compatible with their variety of learning styles. Shawn Phillips at Vanderbilt University reviewed the entire manuscript after the final developmental edit was completed, checked all the content for accuracy, and provided suggestions for further improvement to the author. A select group reviewed text and art manuscript in draft and final form, reviewed page proofs in first and revised rounds, and oversaw the writing and accuracy check of the instructor’s solutions manuals, test bank, and other ancillary materials.

Enhanced Support for the Instructor McGraw-Hill offers instructors various tools and technology products in support of University Chemistry.

Assessment, Review, and Instruction System, also known as ARIS, is an electronic homework and course management system designed for greater flexibility, power, and ease of use than any other system. Whether you are looking for a preplanned course or one you can customize to fit your course needs, ARIS is your solution. In addition to having access to all student digital learning objects, ARIS allows instructors to do the following.

Track Student Progress Assignments are automatically graded. Gradebook functionality allows full-course management including: —Dropping the lowest grades —Weighting grades and manually adjusting grades —Exporting your grade book to Excel®, WebCT® or BlackBoard® —Manipulating data so that you can track student progress through multiple reports Offer More Flexibility Sharing Course Materials with Colleagues. Instructors can create and share course materials and assignments with colleagues with a few clicks of the mouse, allowing for multiple section courses with many instructors and teaching assistants to continually be in synch, if desired. xix

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Integration with BlackBoard or WebCT. Once a student is registered in the course, all student activity within McGraw-Hill’s ARIS is automatically recorded and available to the instructor through a fully integrated grade book that can be downloaded to Excel, WebCT, or Blackboard.

Presentation Center The Presentation Center is a complete set of electronic book images and assets for instructors. You can build instructional materials wherever, whenever, and however you want! Accessed from your textbook’s ARIS website, the Presentation Center is an online digital library containing selected photos, artwork, animations, and other media types that can be used to create customized lectures, visually enhanced tests and quizzes, compelling course websites, or attractive printed support materials. All assets are copyrighted

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by McGraw-Hill Higher Education, but can be used by instructors for classroom purposes. The visual resources in this collection include: Art Full-color digital files of all illustrations in the book can be readily incorporated into lecture presentations, exams, or custom-made classroom materials. In addition, all files are preinserted into PowerPoint® slides for ease of lecture preparation. Animations Numerous full-color animations illustrating important processes are also provided. Harness the visual impact of concepts in motion by importing these files into classroom presentations or online course materials. PowerPoint Slides For instructors who prefer to create their lectures from scratch, all illustrations, selected photos, and tables are preinserted by chapter into blank PowerPoint slides.

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Animations and Media Player Content Topics in chemistry are available in Media Player format and can be viewed on the text ARIS site. For the instructor, all McGraw-Hill chemistry animations are also available on the Presentation Center for use in lecture. Access to your book, access to all books! The Presentation Center library includes thousands of assets from many McGraw-Hill titles. This ever-growing resource gives instructors the power to utilize assets specific to an adopted textbook as well as content from all other books in the library. Nothing could be easier! Accessed from the instructor side of your textbook’s ARIS website, Presentation Center’s dynamic search engine allows you to explore by discipline, course, textbook chapter, asset type, or keyword. Simply browse, select, and download the files you need to build engaging course materials. All assets are copyright by McGraw-Hill Higher Education but can be used by instructors for classroom purposes. Instructors: To access ARIS, request registration information from your McGraw-Hill sales representative.

Computerized Test Bank Online A comprehensive bank of test questions, created by Thomas Seery from University of Connecticut, is provided within a computerized test bank powered by McGraw-Hill’s flexible electronic testing program EZ Test Online (www.eztestonline.com). EZ Test Online allows you to create paper and online tests or quizzes in this easy to use program! Imagine being able to create and access your test or quiz anywhere, at any time without installing the testing software. Now, with EZ Test Online, instructors can select questions from multiple McGraw-Hill test banks or author their own, and then either print the test for paper distribution or give it online.

Test Creation Author or edit questions online using the 14 different question-type templates. Create printed tests or deliver online to get instant scoring and feedback. Create question pools to offer multiple versions online—great for practice. Export your tests for use in WebCT, Blackboard, PageOut and Apple’s iQuiz.


Create tests that are compatible with EZ Test Desktop tests you’ve already created. Share tests easily with colleagues, adjuncts, and TAs.

Online Test Management Set availability dates and time limits for your quiz or test. Control how your test will be presented. Assign points by question or question type with the drop-down menu. Provide immediate feedback to students or delay until all finish the test. Create practice tests online to enable student mastery. Upload your roster to enable student self-registration. Online Scoring and Reporting Provides automated scoring for most of EZ Test’s numerous question types. Allows manual scoring for essay and other open response questions. Allows manual rescoring and feedback. Is designed so that EZ Test’s grade book will easily export to your grade book. Displays basic statistical reports. Support and Help The following provide support and help: User’s Guide and built-in page-specific help Flash tutorials for getting started on the support site Support Website at www.mhhe.com/eztest

Student Response System Wireless technology brings interactivity into the classroom or lecture hall. Instructors and students receive immediate feedback through wireless response pads that are easy to use and engage students. This system can be used by instructors to:

Take attendance. Administer quizzes and tests. Create a lecture with intermittent questions. Manage lectures and student comprehension through the use of the grade book. Integrate interactivity into their PowerPoint presentations. xxi

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Content Delivery Flexibility

Acknowledgments

University Chemistry by Brian B. Laird is available in many formats in addition to the traditional textbook so that instructors and students have more choices when deciding on the format of their chemistry text. Choices include the following.

I would like to thank the following instructors, symposium participants, and students, whose comments were very helpful to me in preparing my first edition text.

Color Custom by Chapter For greater flexibility, we offer University Chemistry in a full-color custom version that allows instructors to pick the chapters they want included. Students pay for only what the instructor chooses. Cooperative Chemistry Laboratory Manual This innovative guide by Melanie Cooper (Clemson University) features open-ended problems designed to simulate experience in a research lab. Working in groups, students investigate one problem over a period of several weeks. Thus, they might complete three or four projects during the semester, rather than one preprogrammed experiment per class. The emphasis here is on experimental design, analysis problem solving, and communication.

Enhanced Support for Students Designed to help students maximize their learning experience in chemistry, we offer the following options to students.

ARIS Assessment, Review, and Instruction System, known as ARIS, is an electronic study system that offers students a digital portal of knowledge. Students can readily access a variety of digital learning objects which include:

Chapter level quizzing. Animations. Interactives. MP3 and MP4 downloads of selected content.

Student Solutions Manual In this manual by Jay Shore (South Dakota State University), the student will find detailed solutions and explanations for the even-numbered problems in University Chemistry. xxii

Colin D. Abernethy Western Kentucky University Joseph J. BelBruno Dartmouth College Philip C. Bevilacqua The Pennsylvania State University Toby F. Block Georgia Institute of Technology Robert Bohn University of Connecticut B. Edward Cain Rochester Institute of Technology Michelle Chatellier University of Delaware Charles R. Cornett University of Wisconsin– Platteville Charles T. Cox Georgia Institute of Technology Darwin B. Dahl Western Kentucky University Stephen Drucker University of Wisconsin–Eau Claire Darcy J. Gentleman University of Toronto David O. Harris University of California– Santa Barbara J. Joseph Jesudason Acadia University Kirk T. Kawagoe Fresno City College Paul Kiprof University of Minnesota–Duluth Craig Martens University of California–Irvine Stephen Mezyk California State University at Long Beach Matthew L. Miller South Dakota State University Michael Mombourquette Queens University Shawn T. Phillips Vanderbilt University Rozana Abdul Razak MARA University of Technology Thomas Schleich University of California–Santa Cruz Thomas A. P. Seery University of Connecticut Jay S. Shore South Dakota University Michael S. Sommer University of Wyoming Larry Spreer University of the Pacific Marcus L. Steele Delta State University Mark Sulkes Tulane University Paul S. Szalay Muskingum College Michael Topp University of Pennsylvania Robert B. Towery Houston Baptist University Thomas R. Webb Auburn University Stephen H. Wentland Houston Baptist University

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John S. Winn Dartmouth College Paulos Yohannes Georgia Perimeter College Timothy Zauche University of Wisconsin–Platteville Lois Anne Zook-Gerdau Muskingum College

Student Class Test, University of Kansas I would like to thank the following students for using my manuscript for their course throughout the year and providing me insight on student use. Mirza Nayyar Ahmad Elizabeth Beech Amelia Bray Thomas Chantz Jason Christian Connor Dennis Andrew Dick Ryan Edward Dowell Hollie Farrahi Lindsey Fisher Stephen Folmsbee Megan Fowler Megan Fracol Jessie Garrett Casey Gee Andy Haverkamp Erica Henderson Armand Heyns Allison Ho Kalin Holthaus Michael Holtz Jake S. Hopkins Josh Istas Ladini Jayaratne Libby Johnson Sophia Kaska Jim LaRocca Jenn Logue

Lizzy Mahoney Amber Markey Allison Martin Ian Mayhugh Nicole E. McClure Sara McElhaney Justin Moyes Chelsea Montgomery Ryan Murphy Thomas Northup Matthew Oliva Jace Parkhurst Sweta Patel Megan L. Razak Kate Remley Thomas Reynolds Richard Robinson Lauren N. Schimming Alan Schurle Amy Soules Jen Strande Sharayah Stitt Jessica Stogsdill Joanna Marie Wakeman Andre W. Wendorff Thomas K. Whitson Daniel Zehr Simon Zhang


I wish to acknowledge my colleagues at the University of Kansas for their help and support in preparing this text, with a special thanks to Professor Cindy Berrie, who has been using drafts of the text in her Honors General Chemistry course. The feedback from her and her students was invaluable. Thanks also to Craig Lunte and Robert Dunn who helped me keep my sanity during this project by enticing me to the golf course on many a sunny day. This text would have not become a reality without the extremely dedicated and competent team at McGraw-Hill Higher Education. For their generous support, I wish to acknowledge Thomas Timp (Publisher), Tami Hodge (Senior Sponsoring Editor), Gloria Schiesl (Senior Project Manager), John Leland (Senior Photo Research Coordinator) and especially my patiently persistent taskmaster Shirley Oberbroeckling (Senior Developmental Editor) for her continual advice, pep talks, and general support at all stages of the project. I would also like to thank former publisher Kent Peterson (VP—Director of Marketing) for talking me into pursuing this project on a balmy (if somewhat blurry) evening in Key West. Finally, I would like to acknowledge the significant contributions and sage advice of Professor Raymond Chang of Williams College, without which this book would not have been possible. Brian B. Laird Lawrence, Kansas February 2008

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0


C h a p t e r

The Language of Chemistry 0.1 Chemistry Is the Study of Matter and Change 2 0.2 Matter Consists of Atoms and Molecules 11 0.3 Compounds Are Represented by Chemical Formulas 20 0.4 Reactions Are Represented by Balanced Chemical Equations 31 0.5 Quantities of Atoms or Molecules Can Be Described by Mass or Number 34 0.6 Stoichiometry Is the Quantitative Study of Mass and Mole Relationships in Chemical Reactions 52

Since ancient times humans have pondered the nature of matter. Our modern ideas of the structure of matter began to take shape in the early nineteenth century with Dalton’s atomic theory. We now know that all ordinary matter is made up of atoms, molecules, and ions. All of chemistry is concerned with the nature of these species and their transformations. Over the past two centuries, chemists have developed a lexicon of terms and concepts that allows them to accurately and efficiently discuss chemical ideas among themselves and communicate these ideas to others. This language of chemistry provides us a way to visualize and quantify chemical transformations at the molecular level while simultaneously understanding the consequences of these transformations in the macroscopic world in which we live.

1


2

Chapter 0

0.1

The Chinese characters for chemistry mean “The study of change.”


The Language of Chemistry

Chemistry Is the Study of Matter and Change

Chemistry is the study of matter and the changes it undergoes. Because a basic knowledge of chemistry is essential for students of biology, physics, astronomy, geology, nutrition, ecology, and many other subjects, chemistry is often referred to as the central science. Indeed, the products of chemistry are central to our way of life; without them, we would be living shorter lives in what we would consider primitive conditions, without automobiles, electricity, computers, DVDs, and numerous other everyday conveniences. Although chemistry is an ancient science, its modern foundation was laid in the nineteenth and twentieth centuries when intellectual and technological advances enabled scientists to break down substances into ever smaller components and consequently to explain many of their physical and chemical characteristics. The rapid development of increasingly sophisticated technology throughout the twentieth century has given us even greater means to study things that cannot be seen with the naked eye. Using computers and special microscopes, for example, chemists can analyze the structure of atoms and molecules—the fundamental units on which chemistry is based—and rationally design new substances with specific properties, such as pharmaceuticals and environmentally friendly consumer products. In the twenty-first century, chemistry will remain an important element in science and technology. This is especially true in emerging fields such as nanotechnology and molecular biology, where the quantum-mechanical behavior of matter cannot be ignored at the molecular level, and in environmental science, where a fundamental understanding of complex chemical-reaction kinetics is crucial to understanding and solving pollution problems. Whatever your reasons for taking college chemistry, knowledge of the subject will better enable you to appreciate its impact on society and the individual. Chemistry is commonly perceived to be more difficult than other subjects, at least at the introductory level. There is some justification for this perception; for one thing, chemistry has a specialized vocabulary, which to a beginning student may seem quite abstract. Chemistry, however, is so deeply embedded in everyday experience that we all are familiar with the effects of chemical processes, even if we lack precise chemical language to describe them. For example, if you cook, then you are a practicing chemist! From experience gained in the kitchen, you know that oil and water do not mix and that boiling water left on the stove will evaporate. You apply chemical and physical principles when you use baking soda to leaven bread, choose a pressure cooker to shorten cooking time, add meat tenderizer to a pot roast, squeeze lemon juice over sliced pears to prevent them from turning brown or over fish to minimize its odor, and add vinegar to the water in which you are going to poach eggs. Every day we observe such changes without thinking about their chemical nature. The purpose of this course is help you learn to think like a chemist, to look at the macroscopic world—the things we can see, touch, and measure directly—and visualize the particles and events of the molecular world that we cannot perceive without modern technology and our imaginations. At first, some students find it confusing when their chemistry instructor and textbook seem to continually shift back and forth between the macroscopic and molecular worlds. Just keep in mind that the data for chemical investigations most often come from observations of largescale phenomena, but the explanations frequently lie in the unseen submicroscopic world of atoms and molecules. In other words, chemists often see one thing (in the macroscopic world) and think another (in the submicroscopic world). Looking at the rusted nails in Figure 0.1, for example, a chemist might think about the properties of individual atoms of iron and how these units interact with other atoms and molecules to produce the observed change.



3

0.1 Chemistry Is the Study of Matter and Change

O2

88n Fe2O3

Fe

Figure 0.1 A simplified molecular view of rust (Fe2O3) formation from iron atoms (Fe) and oxygen molecules (O2). In reality, the process requires the presence of water and rust also contains water molecules.

Classifications of Matter We defined chemistry as the study of matter and the changes it undergoes. Matter is anything that occupies space and has mass and includes things we can see and touch (such as water, earth, and trees), as well as things we cannot (such as air). Thus, everything in the universe has a “chemical” connection. Chemists distinguish among several subcategories of matter based on composition and properties. The classifications of matter include substances, mixtures, elements, and compounds, as well as atoms and molecules, which we will consider in Section 0.2.

Substances and Mixtures A substance is a form of matter that has a definite (constant) composition and distinct properties. Examples include water, ammonia, table salt, gold, and oxygen. Substances differ from one another in composition and can be identified by their appearance, smell, taste, and other properties. A mixture is a combination of two or more substances in which the substances retain their distinct identities. Some familiar examples are air, soft drinks, milk, and cement. Mixtures do not have constant composition. Samples of air collected in different cities differ in composition because of differences in altitude, pollution, weather conditions, and so on. Mixtures are either homogeneous or heterogeneous. When a spoonful of sugar dissolves in water we obtain a homogeneous mixture in which the composition of the mixture is the same throughout the sample. A homogeneous mixture is also called a solution. If one substance in a solution is present in significantly larger amounts than the other components of the mixture, we refer to the dominant substance as the solvent. The other components of the solution, present in smaller amounts, are referred to as solutes. A solution may be gaseous (such as air), solid (such as an alloy), or liquid (seawater, for example). If sand is mixed with iron filings, however, the sand grains and the iron filings remain separate [Figure 0.2(a)]. This type of mixture is called a heterogeneous mixture because the composition is not uniform. Any mixture, whether homogeneous or heterogeneous, can be created and then separated by physical means into pure components without changing the identities of the components. Thus, sugar can be recovered from a water solution by heating the


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Figure 0.2 (a) The mixture contains iron filings and sand. (b) A magnet separates the iron filings from the mixture. The same technique is used on a larger scale to separate iron and steel from nonmagnetic substances such as aluminum, glass, and plastics.

(a)

(b)

solution and evaporating it to dryness. Condensing the vapor will give us back the water component. To separate the iron-sand mixture, we can use a magnet to remove the iron filings from the sand, because sand is not attracted to the magnet [see Figure 0.2(b)]. After separation, the components of the mixture will have the same composition and properties with which they started.

Elements and Compounds Substances can be either elements or compounds. An element is a substance that cannot be separated into simpler substances by chemical means. To date, 117 elements have been positively identified. Most of them occur naturally on Earth, but scientists have created others artificially via nuclear processes (see Chapter 17). For convenience, chemists use symbols of one or two letters to represent the elements. The first letter of a symbol is always capitalized, but any following letters are not. For example, Co is the symbol for the element cobalt, whereas CO is the formula for the carbon monoxide molecule. Table 0.1 lists the names and symbols of some common elements; a complete list of the elements and their symbols appears inside the front cover of this book. Although Table 0.1 Name Aluminum

Some Common Elements and Their Symbols Symbol Al

Name Fluorine

Symbol F

Name Oxygen

Symbol O

Arsenic

As

Gold

Au

Phosphorus

P

Barium

Ba

Hydrogen

H

Platinum

Pt

Bismuth

Bi

Iodine

I

Potassium

K

Bromine

Br

Iron

Fe

Silicon

Si

Calcium

Ca

Lead

Pb

Silver

Ag

Carbon

C

Magnesium

Mg

Sodium

Na

Chlorine

Cl

Manganese

Mn

Sulfur

S

Chromium

Cr

Mercury

Hg

Tin

Sn

Cobalt

Co

Nickel

Ni

Tungsten

W

Copper

Cu

Nitrogen

N

Zinc

Zn



5


Matter

Separation by physical methods

Mixtures

Homogeneous mixtures

Heterogeneous mixtures

Pure substances

Compounds

Separation by chemical methods

Figure 0.3 Classification of matter by composition.

most symbols for elements are consistent with their English names, some elements have symbols derived from Latin, for example, Au from aurum (gold), Fe from ferrum (iron), and Na from natrium (sodium). In another exception, the symbol W for tungsten is derived from its German name Wolfram. Appendix 3 gives the origin of the names and lists the discoverers of most of the elements. Elements may combine with one another to form compounds. Hydrogen gas, for example, burns in oxygen gas to form water, which has properties that are distinctly different from the elements hydrogen and oxygen. Water is made up of two parts hydrogen and one part oxygen. This composition does not change, regardless of whether the water comes from a faucet in the United States, a lake in Outer Mongolia, or the ice caps on Mars. Thus, water is a compound, a substance composed of atoms of two or more elements chemically united in fixed proportions. Unlike mixtures, compounds can be separated only by chemical means into their elemental components. The relationships among elements, compounds, and other categories of matter are summarized in Figure 0.3.

The Three States of Matter In addition to composition, matter can also be classified according to its physical state. All matter can, in principle, exist in three physical states: solid, liquid, and gas (or vapor). A solid is a material that resists changes in both volume and shape—the force required to deform a block of solid steel is quite substantial. Solids can either be crystalline, possessing a highly ordered periodic array of closely spaced molecules (for example, table sugar), or amorphous, with a dense, but disordered, packing of molecules (for example, window glass). A liquid also resists changes in volume, but not of shape. When a liquid is poured from one container to another, the volume of the liquid does not change, but the shape of the liquid adapts to match the new container. In both solids and liquids, the space between molecules is similar to the sizes of the molecules themselves. In a gas, though, the distances between atoms or molecules are large compared to molecular size. As a result, a gas resists neither changes in volume nor changes in shape. Both gases and liquids are collectively known as fluids. The structural differences between crystalline solids, liquids, and gases are illustrated in Figure 0.4. The three states of matter can interconvert without changing the composition of the substance. Upon heating, a solid, such as ice, melts to form a liquid. (The

Elements


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Figure 0.4 Microscopic views of a solid, a liquid, and a gas.

Solid

Liquid

Gas

temperature at which this transition occurs is called the melting point.) Further heating converts the liquid into a gas. (The conversion of a liquid to a gas takes place at the boiling point of the liquid.) On the other hand, cooling a gas below the boiling point of the substance causes it to condense into a liquid. When the liquid is cooled further, below the melting point of the substance, it freezes to form the solid. Under proper conditions, some solids can convert directly into a gas; this process is called sublimation. For example, solid carbon dioxide, commonly known as dry ice, readily sublimes to carbon dioxide gas unless maintained below a temperature of 278.58C.

Physical and Chemical Properties of Matter H2O

H2

H2 O2

Hydrogen burning in air to form water

Substances are identified by their properties as well as by their composition. Color, melting point, and boiling point are physical properties. A physical property can be measured and observed without changing the composition or identity of a substance. For example, we can measure the melting point of ice by heating a block of ice and recording the temperature at which the ice is converted to water. Water differs from ice only in appearance, not in composition, so this is a physical change; we can freeze the water to recover the ice. Therefore, the melting point of a substance is a physical property. Similarly, when we say that helium gas is less dense than air, we are referring to a physical property. On the other hand, the statement “Hydrogen gas burns in oxygen gas to form water” describes a chemical property of hydrogen, because to observe this property we must carry out a chemical change (in this case, burning). After the change, the original chemical substance, the hydrogen gas, will have vanished, and all that will be left is a different chemical substance—water. We cannot recover the hydrogen from the water by means of a physical change, such as boiling or freezing. Every time we hard-boil an egg, we bring about a chemical change. When subjected to a temperature of about 1008C, the yolk and the egg white undergo changes that alter not only their physical appearance but their chemical makeup as well. When eaten, substances in our bodies called enzymes facilitate additional chemical transformations of the egg. This digestive action is another example of a chemical change. What happens during digestion depends on the chemical properties of both the enzymes and the food.




All measurable properties of matter may be additionally categorized as extensive or intensive. The measured value of an extensive property depends on how much matter is being considered. Mass, which is the quantity of matter in a given sample of a substance, is an extensive property. More matter means more mass. Values of the same extensive property can be added together. For example, two copper pennies have a combined mass that is the sum of the mass of each penny, and the length of two tennis courts is the sum of the length of each tennis court. Volume, defined as length cubed, is another extensive property. The value of an extensive property depends on the amount of matter. The measured value of an intensive property does not depend on how much matter is being considered. Density, defined as the mass of an object divided by its volume, is an intensive property. So is temperature. Suppose that we have two beakers of water at the same temperature. If we combine them to make a single quantity of water in a larger beaker, the temperature of the larger quantity of water will be the same as it was in the two separate beakers. Unlike mass, length, volume, and energy, temperature and other intensive properties are nonadditive.

Force and Energy We have discussed the properties of matter, but chemistry also studies the changes that matter undergoes. These changes are brought about by forces and resulting energy changes, which we examine now.

Force Anything that happens in the universe is the result of the action of a force. Force is the quantity that causes an object to change its course of motion (either in direction or speed). Isaac Newton1 first quantified the relationship between force and the motion of material objects in his second law of motion: force ⫽ mass ⫻ acceleration

(0.1)

where the acceleration of an object is the rate of change (derivative) of the velocity of the object with respect to time. Given the forces on a particle and its initial position and velocity, Equation 0.1 can be used to determine the future motion of the system. The SI2 units for mass and acceleration are kg and m s22, respectively, so the SI unit of force is kg m s22, or the newton (N): 1 N 5 1 kg m s22. Physicists have identified the following four fundamental forces in the universe: 䉴 The electromagnetic force is the force between electrically charged objects or magnetic materials. 䉴 The gravitational force is the attractive force between objects caused by their masses.

1. Sir Isaac Newton (1643–1727). English physicist and mathematician. One of the most brilliant scientists in history, he founded the fields of classical mechanics and the differential and integral calculus, as well as made major contributions to the field of optics. 2. The International System of Units (abbreviated SI, from the French Système Internationale d’Unites), based on the metric system, was adopted in 1960 by the General Conference of Weights and Measures, the international authority on units. With a few exceptions, we will use SI units throughout this book. A discussion of SI units can be found in Appendix 1.

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c The weak force is the force responsible for some forms of radioactive decay. c The strong force is the force binding the protons and neutrons in a nucleus together, overcoming the powerful electromagnetic repulsion between positively charged protons. In chemistry, by far the most important of these forces is the electromagnetic force. With the exception of nuclear chemistry (discussed in Chapter 17), which involves the strong and weak forces, all of chemistry is a direct result of the action of electromagnetic forces between electrons and protons, and between matter and light (electromagnetic radiation). The gravitational force is far too weak to have an impact at the atomic and molecular scale.

Energy All chemical or physical transformations in nature are driven by the release, absorption, or redistribution of energy. A detailed description of any such transformation is not possible without an understanding of the role of energy. From a mechanical perspective, energy is the capacity to do work. Work is done whenever an object is moved from one place to another in the presence of some external force. For example, the process of lifting a ball off of the floor and placing it on a table requires work against the gravitational force between Earth and the ball. For a mechanical system, work is the product of the distance d that the object is moved times the applied force F (measured along the object’s direction of motion3): work ⫽ force ⫻ distance ⫽ F ⫻ d

(0.2)

The SI unit of energy is the joule (J), which is the work done in moving an object a distance of 1 m against a force of 1 N. Thus, 1 J ; 1 N m 5 1 kg m2 s22. The energy possessed by a material object is composed of two basic forms: kinetic and potential. The energy that an object has as a result of its motion is its kinetic energy. For an object with mass m and velocity u, the kinetic energy is given by kinetic energy ⫽

1 2 mu 2

(0.3)

Example 0.1 Calculate the kinetic energy of a 150-g baseball traveling at a velocity of 50 m s21.

Solution Substitute the mass and velocity of the baseball into Equation 0.3 but be careful to convert grams to kilograms, so that the final value is in joules. 1 kg 1 1 d (50 m s21)2 kinetic energy 5 mu2 5 c(150 g) 1000 g 2 2 5 1.9 3 102 kg m2 s22 5 1.9 3 102 J

Practice Exercise Calculate the kinetic energy (in joules) of a car with mass 1000 kg traveling at a velocity of 130 km per hour.

3. Both force and distance are vectors, so a more general definition would be work 5 F ⴢ d, where “?” denotes the usual vector dot product.

Lai69040_ch00_001-070.indd Page 9 2/6/08 2:16:14 PM user1



9

Objects can also possess energy as a result of their position in space. This type of energy is called potential energy (V ). The equation for potential energy depends upon the type of material and its specific environment. Examples include the following: 1. The gravitational potential energy of an object with mass m at a height h above the surface of Earth: potential energy ⫽ V ⫽ mgh

2.

where g 5 9.80665 m s21 is the standard terrestrial gravitational acceleration constant. The electrostatic potential energy of interaction between two particles with electric charges q1 and q2 separated by a distance r, which is given by Coulomb’s law: V⫽

3.

(0.4)

q1q2 4pe0r

(0.5)

where e0 5 8.8541878 3 10212 C2 J21 m21 is a fundamental physical constant called the permittivity of the vacuum, and q1 and q2 are measured in the SI unit of charge, the coulomb (C). (The presence of the factor 4pe0 is due to the use of the SI system of units.) The potential energy of a spring obeying Hooke’s law 1 V ⫽ k(x ⫺ x0 ) 2 2

(0.6)

where x is the length of the spring, x0 is the equilibrium value of the length, and k is called the spring constant and is a measure of the stiffness of the spring. This potential energy function also represents a good approximation to the potential energy of molecular bond vibrations. The potential energy is important in determining the future motion of an object because it is directly related to the force on that object. For a one-dimensional potential energy function V(x), the force is given by the negative of the derivative (see Appendix 1) of the potential energy: F (x) ⫽ ⫺

dV ( x ) dx

(0.7)

From this equation, the force on the object is in the direction of decreasing potential energy. Thus, an object placed at the top of a hill will roll down the hill to decrease its gravitational potential energy (Equation 0.4). Equation 0.7 can be used to show that the decrease in potential energy in a process is equal to the work done in that process: work done in process ⫽ ⫺3Vfinal ⫺ Vinitial 4 ⫽ ⫺¢V

(0.8)

(The symbol Δ is used to indicate the change in a quantity.) The total energy of an object is the sum of its kinetic and potential energies. As an object is accelerated (or deaccelerated) under the influence of external forces, its kinetic energy changes with time. Likewise, as the position of the object changes along its trajectory, the potential energy also varies. The total energy, however, remains constant. As the kinetic energy of an object increases, the potential energy must decrease by exactly the same amount to maintain constant total energy. The law of conservation of energy summarizes this principle: the total quantity of energy in the universe is assumed constant.

As the water falls from the dam, its potential energy is converted into kinetic energy. Use of this energy to generate electricity is called hydroelectric power.


10

Chapter 0



Example 0.2 A 0.295-kg ball, initially at rest, is released from a height of 29.3 m from the ground. How fast will the ball be traveling when it hits the ground? (Ignore any air resistance.)

Strategy The speed of the ball is related to the kinetic energy through Equation 0.3. From the principle of the conservation of energy, the gain in kinetic energy must exactly equal the loss of potential energy, which can be calculated from the change in height using Equation 0.4. Solution Use Equation 0.4 to calculate the change in potential energy: D(potential energy) 5 5 5 5

mghfinal 2 mghinitial mg(hfinal 2 hinitial) (0.295 kg)(9.807 m s22)(0 m 2 29.3 m) 284.8 kg m2 s22 5 284.8 J

From the conservation of energy, D(kinetic energy) 5 2D(potential energy), so 1 1 2 mufinal 2 mu2initial 5 2D(potential energy) 2 2 1 1 (0.295 kg)u2final 2 (0.295 kg)(0)2 5 2(284.8 kg m2 s22) 2 2 2 22 s ) 2(84.8 kg m 5 575 m2 s22 u2final 5 0.295 kg ufinal 5 24.0 m s21

Practice Exercise How high above the surface of Earth would you have to drop a ball of mass 10.0 kg for it to reach a speed of 20 m s21 before it hit the ground?

Although all energy can be ultimately identified as kinetic or potential energy or a combination of the two, it is convenient in chemistry to define additional classifications of energy that depend upon the system. Important examples include the following: c Radiant energy is the energy contained in electromagnetic radiation, such as X-rays, radio waves, and visible light. On Earth, the primary energy source is radiant energy coming from the sun, which is called solar energy. Solar energy heats the atmosphere and surface of Earth, stimulates the growth of vegetation through the process known as photosynthesis, and influences global climate patterns. Radiant energy is a combination of the potential and kinetic energy of the electromagnetic fields that make up light. c Thermal energy is the energy associated with the random motion of atoms and molecules. Thermal energy can be generally calculated from temperature measurements. The more vigorous the motion of the atoms and molecules in a sample of matter, the hotter the sample is and the greater its thermal energy. Keep in mind, however, thermal energy and temperature are different. A cup of coffee at 708C has a higher temperature than a bathtub filled with water at 408C, but the bathtub stores much more thermal energy because it has a much larger volume and greater mass than the coffee. The bathtub water has more water molecules and, therefore, more molecular motion. Put another way, temperature is an intensive property (does not depend upon the amount of



0.2 Matter Consists of Atoms and Molecules

matter), whereas thermal energy is an extensive property (does depend proportionally on the amount of matter). c Chemical energy is a form of potential energy stored within the structural units of chemical substances; its quantity is determined by the type and arrangement of the constituent atoms. When substances participate in chemical reactions, chemical energy is released, stored, or converted to other forms of energy. All forms of energy can be converted (at least in principle) from one form to another. We feel warm when we stand in sunlight because radiant energy is converted to thermal energy on our skin. When we exercise, chemical energy stored in the molecules within our bodies is used to produce kinetic energy. When a ball starts to roll downhill, its potential energy is converted to kinetic energy.

0.2

Matter Consists of Atoms and Molecules

Atomic Theory In the fifth century B.C. the Greek philosopher Democritus proposed that all matter consisted of very small, indivisible particles, which he named atomos (meaning indivisible). Although Democritus’ idea was not accepted by many of his contemporaries (notably Plato and Aristotle), it nevertheless endured. Experimental evidence from early scientific investigations provided support for the notion of “atomism” and gradually gave rise to the modern definitions of elements and compounds. In 1808 an English scientist and schoolteacher, John Dalton,4 formulated a more precise definition of the indivisible building blocks of matter that we call atoms. Dalton’s work marked the beginning of the modern era of chemistry. The hypotheses about the nature of matter on which Dalton’s atomic theory is based can be summarized as follows: 1.

2.

3.

Elements are composed of extremely small particles called atoms. All atoms of a given element are identical, having the same size, mass, and chemical properties. The atoms of one element are different from the atoms of all other elements. Compounds are composed of atoms of more than one element. In any compound, the ratio of the numbers of atoms of any two of the elements present is either an integer or a simple fraction. A chemical reaction involves only the separation, combination, or rearrangement of atoms; it does not result in the creation or destruction of atoms.

Figure 0.5 shows a schematic representation of the first two hypotheses. Dalton recognized that atoms of one element were different from atoms of all other elements (the first hypothesis), but he made no attempt to describe the structure or composition of atoms because he had no idea what an atom was really like. He did realize, however, that the different properties shown by elements such as hydrogen and oxygen could be explained by assuming that hydrogen atoms were not the same as oxygen atoms. The second hypothesis suggests that, to form a certain compound, we need not only atoms of the right kinds of elements, but specific numbers of these atoms as

4. John Dalton (1766–1844). English chemist, mathematician, and philosopher. In addition to the atomic theory, he also formulated several gas laws and gave the first detailed description of color blindness, from which he suffered. Dalton was described as an indifferent experimenter, and singularly wanting in the language and power of illustration. His only recreation was lawn bowling on Thursday afternoons. Perhaps it was the sight of those wooden balls that provided him with the idea of atomic theory.

11


12

Figure 0.5 (a) According to Dalton’s atomic theory, atoms of the same element are identical, but atoms of one element are different from atoms of other elements. (b) Compounds formed from atoms of elements X and Y. In this case, the ratio of the atoms of element X to the atoms of element Y is 2:1. Note that a chemical reaction results only in the redistribution of atoms, not in their destruction or creation.

Chapter 0



Atoms of element Y

Atoms of element X (a)

Compounds formed from elements X and Y (b)

well. This idea is an extension of the law of definite proportions, first published in 1799 by Joseph Proust.5 According to the law of definite proportions, different samples of the same compound always contain its constituent elements in the same proportion by mass. Thus, if we were to analyze samples of carbon dioxide gas obtained from different sources, we would find in each sample the same ratio by mass of carbon to oxygen. It stands to reason, then, that if the ratio of the masses of different elements in a given compound is fixed, then the ratio of the atoms of these elements in the compound must also be constant. Dalton’s second hypothesis also supports the law of multiple proportions. According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers. Dalton’s theory explains the law of multiple proportions quite simply: Different compounds made up of the same elements differ in the number of atoms of each kind that combine. Carbon, for example, forms two stable compounds with oxygen, namely, carbon monoxide and carbon dioxide. Modern measurement techniques have shown that one atom of carbon combines with one atom of oxygen in carbon monoxide and with two atoms of oxygen in carbon dioxide. Thus, the ratio of oxygen in carbon monoxide to oxygen in carbon dioxide is 1:2. This result is consistent with the law of multiple proportions. Dalton’s third hypothesis is another way of stating the law of conservation of mass6—that is, matter can be neither created nor destroyed. For chemical reactions this principle had been demonstrated experimentally by Antoine Lavoisier7 who heated mercury in air to form mercury(II) oxide and showed that the increase in mass of the oxide over the pure mercury was exactly equal to the decrease in the mass of the gas. Because matter is made of atoms that are unchanged in a chemical reaction, it follows that mass must be conserved as well. Dalton’s brilliant insight into the nature of matter was the main stimulus for the rapid progress of chemistry during the nineteenth century. 5. Joseph Louis Proust (1754–1826). French chemist. Proust was the first person to isolate sugar from grapes. 6. According to Albert Einstein, mass and energy are alternate aspects of a single entity called mass–energy. Chemical reactions usually involve a gain or loss of heat and other forms of energy. Thus, when energy is lost in a reaction, mass is lost, too. This is only meaningful, however, for nuclear reactions (see Chapter 17). For chemical reactions the changes of mass are too small to detect. For all practical purposes, therefore, mass is conserved. 7. Antoine Laurent Lavoisier (1743–1794). French chemist, sometimes referred to as the “father of modern chemistry.” In addition to his role in verifying the principle of the conservation of mass, he made important contributions to chemical nomenclature and to the understanding of the role of oxygen in combustion, respiration, and acidity. For his role as a tax collector for the king, Lavoisier was guillotined during the French Revolution.




(a)

(b)

13

(c)

Figure 0.6 (a) A cathode ray produced in a discharge tube travels from left to right. The ray itself is invisible, but the fluorescence of a zinc sulfide coating on the glass causes it to appear green. (b) The cathode ray is bent downward when the south pole of the bar magnet is brought toward it. (c) When the polarity of the magnet is reversed, the ray bends in the opposite direction.

The Building Blocks of the Atom Based on Dalton’s atomic theory, an atom is the basic unit of an element that can enter into chemical combination. Dalton imagined an atom that was both extremely small and indivisible. However, a series of investigations that began in the 1850s and extended into the twentieth century clearly demonstrated that atoms actually possess internal structure; that is, they are made up of even smaller subatomic particles. Particle physicists have discovered a complex hierarchy (or “zoo”) of such particles, but only electrons, protons, and neutrons are of primary importance in chemical reactions. The electron, discovered by J. J. Thomson8 in 1897, is a tiny, negatively charged particle with a charge of 21.602177 3 10219 C, where C stands for coulomb, the SI unit of electric charge, and a mass me 5 9.109383 3 10231 kg. (The magnitude of the electron charge, 1.602177 3 10219 C, is a fundamental physical constant and is given the symbol e while its charge is denoted as 2e.) The discovery was made using a device known as a cathode ray tube (Figure 0.6), which consists of two metal plates inside an evacuated glass tube. When the two metal plates are connected to a high-voltage source, the negatively charged plate, called the cathode, emits an invisible ray. This cathode ray is drawn to the positively charged plate, called the anode, where it passes through a hole and continues traveling to the other end of the tube. When the ray strikes the specially coated surface, it produces a strong fluorescence, or bright green light. Thomson showed that the cathode ray was actually a beam of negatively charged particles (electrons). By observing the degree to which the beam was deflected when placed in a magnetic field (see Figure 0.6), Thompson was able to determine the magnitude of the electron chargeto-mass ratio, eyme. A decade later, R. A. Millikan9 succeeded in directly measuring e with great precision by examining the motion of individual tiny drops of oil that picked up static charge from the air. Using the charge-to-mass ratio determined earlier by Thomson, Millikan was also able to calculate the mass of the electron, me. Because atoms are electrically neutral, they must also contain positively charged components in addition to electrons. Thomson proposed that an atom consisted of a diffuse sphere of uniform positively charged matter in which electrons were imbedded like raisins in a plum pudding (a traditional English dessert). 8. Joseph John Thomson (1856–1940). British physicist who received the Nobel Prize in Physics in 1906 for the discovery of the electron. 9. Robert Andrews Millikan (1868–1953). American physicist who was awarded the Nobel Prize in Physics in 1923 for determining the charge of the electron.


14

Figure 0.7 (a) Rutherford’s experimental design for measuring the scattering of particles by a piece of gold foil. The vast majority of the particles passed through the gold foil with little or no deflection; however, a few were deflected at wide angles and, occasionally, a particle was turned back. (b) Magnified view of particles passing through and being deflected by nuclei.

A picometer (pm) 5 10212 m and is a convenient unit for measuring atomic and molecular distances. Another commonly used unit for measuring molecular scale distances is the angstrom (Å), which is equal to 100 pm or 10 210 m.

If the size of an atom were expanded to that of this sports stadium, the size of the nucleus would be that of a marble.

Chapter 0



Gold foil α –Particle emitter

Slit

Detecting screen (a)

(b)

Thompsons’s “plum pudding” model of atomic structure was disproven by a series of experiments performed in 1910 by the physicist Ernest Rutherford.10 Together with his associate Hans Geiger11 and undergraduate student Ernest Marsden,12 Rutherford bombarded very thin gold foil with alpha particles (a), a recently discovered form of radiation that consists of positively charged helium nuclei (Figure 0.7). They observed that the overwhelming majority of a particles passed through the foil with little or no deflection. However, a small number of a particles were strongly scattered through large angles—some as large as 180°. This was inconsistent with Thomson’s model of the atom because the positive charge of the atom should have been so diffuse that the a particles should have passed through the foil with very little deflection. Rutherford’s initial reaction, when told of this discovery, was to say, “It was as incredible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you.” Rutherford was later able to explain these experimental results in terms of a new model for the atom. According to Rutherford, most of the atom must be empty space, with the positive charges concentrated within a very small, but dense, central core called the nucleus. Whenever an a particle neared the nucleus in the scattering experiment, it experienced a large repulsive force and therefore a large deflection. Moreover, an a particle traveling directly toward a nucleus would be completely repelled and its direction would be reversed. The positively charged particles in the nucleus are called protons. In separate experiments, it has been found that each proton carries a charge of 1e and has a mass mp 5 1.672622 3 10227 kg, or about 1840 times the mass of the oppositely charged electron. At this stage of the investigation, scientists perceived the atom as follows: The mass of a nucleus constitutes most of the mass of the entire atom, but the nucleus occupies only about 1y1013 of the volume of the atom. A typical atomic radius is about 100 pm, whereas the radius of an atomic nucleus is only about 5 3 1023 pm. You can appreciate the relative size of an atomic nucleus by imagining that if an atom were the size of a sports stadium, the volume of its nucleus would be comparable to a small marble. Although protons are confined to the nucleus of the atom, electrons are thought of as being spread out about the nucleus at some distance from it. 10. Ernest Rutherford (1871–1937). New Zealand physicist and former graduate student of J. J. Thomson. Rutherford did most of his work in England (Manchester and Cambridge Universities). He received the Nobel Prize in Chemistry in 1908 for his investigations into the structure of the atomic nucleus. His oftenquoted comment to his students was that “All science is either physics or stamp-collecting.” 11. Johannes Hans Wilhelm Geiger (1882–1945). German physicist. Geiger’s work focused on the structure of the atomic nucleus and on radioactivity. He invented a device for measuring radiation that is now commonly called the Geiger counter. 12. Ernest Marsden (1889–1970). English physicist. As an undergraduate he performed many of the experiments that led to Rutherford’s Nobel Prize. Marsden went on to contribute significantly to the development of science in New Zealand.




15

Figure 0.8 The protons and neutrons of an atom are packed in an extremely small nucleus. Electrons are shown as “clouds” around the nucleus.

Proton Neutron

Rutherford’s model of atomic structure left one major problem unsolved. It was known that hydrogen, the simplest atom, contained only one proton and that the helium atom contained two protons. Therefore, the ratio of the mass of a helium atom to that of a hydrogen atom should be 2:1. (Because electrons are much lighter than protons, their contribution to atomic mass can be ignored.) In reality, however, the ratio is 4:1. Rutherford and others postulated that there must exist yet another type of subatomic particle in the atomic nucleus, a particle that had a mass similar to that of a proton, but was electrically neutral. In 1932, James Chadwick13 provided experimental evidence for the existence of this particle. When Chadwick bombarded a thin sheet of beryllium with a particles, the metal emitted very high-energy radiation. Later experiments showed that the “radiation” consisted of a third type of subatomic particles, which Chadwick named neutrons, because they proved to be electrically neutral particles slightly more massive than protons. The mystery of the mass ratio could now be explained. In the helium nucleus there are two protons and two neutrons, but in the hydrogen nucleus there is only one proton (no neutrons); therefore, the ratio is 4:1. Figure 0.8 shows the location of the protons, neutrons, and electrons in an atom. Table 0.2 lists the masses and charges of these three elementary particles. The number of protons in the nucleus of an atom is called the atomic number (Z), which, in a neutral atom, is also equal to the number of electrons. The atomic number determines the chemical properties of an atom and the identity of the element; 13. James Chadwick (1891–1972). English physicist. In 1935 he received the Nobel Prize in Physics for proving the existence of the neutron.

Table 0.2 Particle

Masses and Charges of Subatomic Particles Mass

Charge

electron

me 5 9.109383 3 10

kg

21.602177 3 10219 C 5 2e

proton

mp 5 1.672622 3 10227 kg

11.602177 3 10219 C 5 1e

neutron

mn 5 1.674927 3 10

231

227

kg

0


16

Chapter 0



that is, all atoms of a given element have the same atomic number. For example, the atomic number of nitrogen is 7; therefore, each nitrogen atom has seven protons (and seven electrons because it is neutral). The converse is also true—every atom in the universe that contains seven protons is correctly named “nitrogen.” With the exception of the most common form of hydrogen, which has one proton and no neutrons, all atomic nuclei contain both protons and neutrons. Although each element is specified by the number of protons (atomic number), the mass of an atom can only be determined if we also know the number of neutrons. The mass number (A) is the total number of neutrons and protons present in the nucleus of an atom. Thus, the number of neutrons in an atom is equal to the difference between the mass number and the atomic number. For example, the mass number of fluorine is 19 and the atomic number is 9 (indicating 9 protons in the nucleus). The number of neutrons in an atom of fluorine is 19 2 9 5 10. Note that the atomic number and mass number must be positive integers. Except in the case of common hydrogen, where they are equal, the mass number is always larger than the atomic number. Atoms of a given element do not all have the same mass. Most elements have two or more isotopes, atoms that have the same atomic number but different mass numbers. There are, for example, three isotopes of hydrogen. One, known simply as hydrogen (or, less commonly, as protium), has one proton and no neutrons. The deuterium isotope contains one proton and one neutron, and tritium has one proton and two neutrons. The accepted way to denote the atomic number and mass number of an atom of an element (X) is as follows: A Z

X

Thus, for the isotopes of hydrogen, we write 1 1 1 1H

2 1H

H

2 1

H

3 1

H

Hydrogen Deuterium Tritium (or protium)

3 1H

To a very good approximation, different isotopes of the same element have identical chemical properties14; however, they can differ in some physical properties, such as density, which depends on mass, and radioactivity, which depends on the ratio of neutrons to protons in the nucleus, as will be discussed in Chapter 17. Two common isotopes of uranium (Z 5 92), for example, have mass numbers 235 and 238. The 238 lighter isotope (235 92U) is used in nuclear reactors and atomic bombs, whereas ( 92U) lacks the nuclear properties necessary for these applications. With the exception of hydrogen, which has different names for each of its isotopes, isotopes of elements are identified by their mass numbers. Thus, the two common isotopes of uranium are called uranium-235 and uranium-238. A list of the stable and radioactive isotopes for the first 10 elements is given in Appendix 4.

The Periodic Table More than half of the elements known today were discovered in the nineteenth century. During this period, chemists noted that many elements showed strong similarities to one another. Recognition of periodic regularities in physical and chemical behavior

14. The existence of isotopes (atoms of the same element that differ in atomic mass) implies that Proust’s law of definite proportions, which states that different sample of the same compound always contain elements in the same proportion by mass, is in reality only an approximation. Two samples of carbon dioxide (CO2) would not have exactly the same mass ratio of carbon to oxygen if one of the samples were richer in carbon-13.



17


and the need to organize the volume of available information about the structure and properties of elemental substances led to the development of the periodic table, a chart in which elements having similar chemical and physical properties are grouped together. Figure 0.9 shows the modern periodic table, in which the elements are arranged by atomic number (shown above the element symbol) in horizontal rows called periods and in vertical columns known as groups or families, according to similarities in their chemical properties. Elements 112–116 and 118 have recently been synthesized, although they have not yet been named. The elements can be categorized as metals, nonmetals, or metalloids. A metal is a good conductor of heat and electricity, whereas a nonmetal is usually a poor conductor of heat and electricity. A metalloid has properties that are intermediate between those of metals and nonmetals. Figure 0.9 shows that the majority of the known elements are metals; only 17 elements are nonmetals, and 8 are metalloids. From left to right across any period, the physical and chemical properties of the elements change gradually from metallic to nonmetallic. Elements are often classified by their periodic table group number—Group 1A, Group 2A, and so on. For convenience, some element groups have special names. The Group 1A elements (Li, Na, K, Rb, Cs, and Fr) are called alkali metals, and the Group 2A elements (Be, Mg, Ca, Sr, Ba, and Ra) are 1 1A 1

H

18 8A 2 2A

13 3A

14 4A

15 5A

16 6A

17 7A

2

He

3

4

5

6

7

8

9

10

Li

Be

B

C

N

O

F

Ne

13

14

15

16

17

18

Al

Si

P

S

Cl

Ar

11

12

Na

Mg

3 3B

4 4B

5 5B

6 6B

7 7B

8

9 8B

10

11 1B

12 2B

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

K

Ca

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

Ga

Ge

As

Se

Br

Kr

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

Rb

Sr

Y

Zr

Nb

Mo

Tc

Ru

Rh

Pd

Ag

Cd

In

Sn

Sb

Te

I

Xe

55

56

57

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

Cs

Ba

La

Hf

Ta

W

Re

Os

Ir

Pt

Au

Hg

Tl

Pb

Bi

Po

At

Rn

112

113

114

115

116

(117)

118

87

88

89

104

105

106

107

108

109

110

111

Fr

Ra

Ac

Rf

Db

Sg

Bh

Hs

Mt

Ds

Rg

Metals

58

59

60

61

62

63

64

65

66

67

68

69

70

71

Ce

Pr

Nd

Pm

Sm

Eu

Gd

Tb

Dy

Ho

Er

Tm

Yb

Lu

Metalloids

90

91

92

93

94

95

96

97

98

99

100

101

102

103

Th

Pa

U

Np

Pu

Am

Cm

Bk

Cf

Es

Fm

Md

No

Lr

Nonmetals

Figure 0.9 The modern periodic table. The elements are arranged according to the atomic numbers above their symbols. With the exception of hydrogen (H), nonmetals appear at the far right of the table. The two rows of metals beneath the main body of the table are conventionally set apart to keep the table from being too wide. Actually, cerium (Ce) should follow lanthanum (La), and thorium (Th) should come right after actinium (Ac). The 1–18 group designation has been recommended by the International Union of Pure and Applied Chemistry (IUPAC). In this text we use the standard U.S. notation for group numbers (1A–8A and 1B–8B) still in wide use. Element 117 has not yet been synthesized.


18

Chapter 0



Distribution of Elements on Earth and in Living Systems he majority of elements are naturally occurring. How are these elements distributed on Earth, and which are essential to living systems? Earth’s crust extends from the surface to a depth of about 40 km. Because of technical difficulties, scientists have not been able to study the inner portions of Earth as easily as the crust. Nevertheless, it is believed that there is a solid core consisting mostly of iron at the center of Earth. Surrounding the core is a layer called the mantle, which consists of hot fluid containing iron, carbon, silicon, and sulfur. Of the 83 elements that are found in nature, 12 make up 99.7 percent of Earth’s crust by mass. They are, in decreasing order of natural abundance, oxygen (O), silicon (Si), aluminum (Al), iron (Fe), calcium (Ca), magnesium (Mg), sodium (Na), potassium (K), titanium (Ti), hydrogen (H), phosphorus (P), and manganese (Mn)—see the following figure. In discussing

T

the natural abundance of the elements, we should keep in mind that (1) the elements are not evenly distributed throughout Earth’s crust, and (2) most elements occur in combined forms. These facts provide the basis for most methods of obtaining pure elements from their compounds, as we will see in later chapters. The accompanying table lists the essential elements in the human body. Of special interest are the trace elements, such as iron (Fe), copper (Cu), zinc (Zn), iodine (I), and cobalt (Co), which together make up about 0.1 percent of the mass of the body. These elements are necessary for biological functions such as growth, transport of oxygen for metabolism, and defense against disease. There is a delicate balance in the amounts of these elements in our bodies. Too much or too little over an extended period can lead to serious illness, retardation, or even death.

Essential Elements in the Human Body

Mantle Crust

Element

Percent by Mass*

Oxygen Carbon Hydrogen Nitrogen Calcium Phosphorus Potassium Sulfur Chlorine

Core

2900 km 3480 km

Structure of the interior of the Earth

65 18 10 3 1.6 1.2 0.2 0.2 0.2

Element

Percent by Mass*

Sodium Magnesium Iron Cobalt Copper Zinc Iodine Selenium Fluorine

0.1 0.05 ,0.05 ,0.05 ,0.05 ,0.05 ,0.05 ,0.01 ,0.01

*Percent by mass gives the mass of the element in grams present in a 100-g sample.

(a) Natural abundance of the elements in percent by mass. For example, the abundance of oxygen is 45.5 percent. This means that in a 100-g sample of Earth’s crust there are, on average, 45.5 g of the element oxygen. (b) Abundance of elements in the human body in percent by mass.

All others 5.3% Magnesium 2.8% Calcium 4.7%

Oxygen 45.5%

Iron 6.2%

Silicon 27.2%

(a)

Aluminum 8.3%

Oxygen 65%

Carbon 18%

(b)

All others 1.2% Phosphorus 1.2% Calcium 1.6% Nitrogen 3%

Hydrogen 10%



19


called alkaline earth metals. Elements in Group 7A (F, Cl, Br, I, and At) are known as halogens, and elements in Group 8A (He, Ne, Ar, Kr, Xe, and Rn) are called noble gases, or rare gases. The periodic table is a handy tool that correlates the properties of the elements in a systematic way and helps us to predict chemical behavior. We will take a closer look at this keystone of chemistry in Chapter 2.

Molecules and Ions Of all the elements, only the six noble gases in Group 8A of the periodic table (He, Ne, Ar, Kr, Xe, and Rn) exist in nature as single atoms. For this reason, they are called monatomic (meaning a single atom) gases. Most matter is composed of molecules or ions formed by atoms. A molecule is an aggregate of two or more atoms in a definite arrangement held together by chemical forces (also called chemical bonds). A molecule may contain atoms of the same element or atoms of two or more elements joined in a fixed ratio, in accordance with the law of definite proportions. Thus, a molecule is not necessarily a compound, which, by definition, is made up of two or more elements (see Section 0.1). Hydrogen gas, for example, is a pure element, but it consists of molecules made up of two H atoms each. Water, on the other hand, is a molecular compound that contains hydrogen and oxygen in a ratio of two H atoms to one O atom. Like atoms, molecules are electrically neutral. The hydrogen molecule, symbolized as H2, is a diatomic molecule because it contains only two atoms. Other elements that normally exist as diatomic molecules are nitrogen (N2), oxygen (O2), and the Group 7A elements, except astatine (At); that is, the halogens: fluorine (F2), chlorine (Cl2), bromine (Br2), and iodine (I2). A diatomic molecule can contain atoms of different elements, too, such as hydrogen chloride (HCl) and carbon monoxide (CO). The vast majority of molecules contain more than two atoms. They can be atoms of the same element, as in ozone (O3), which consists of three atoms of oxygen, or they can be combinations of two or more different elements. Molecules containing more than two atoms are polyatomic molecules. Like ozone, water (H2O) and ammonia (NH3) are polyatomic molecules. An ion is an atom or a group of atoms that has a net positive or negative charge. The number of protons in the nucleus of an atom remains the same during ordinary chemical changes (called chemical reactions), but electrons may be lost or gained. The loss of one or more electrons from a neutral atom or molecule results in a cation, an ion with a net positive charge. For example, a sodium atom (Na) can readily lose an electron to become a sodium cation with a net charge of 1e. This ion is represented by the symbol Na1. On the other hand, an anion is an ion whose net charge is negative because of an increase in the number of electrons. A chlorine atom (Cl) can gain an electron to become the chloride ion, Cl2, which is an anion with net charge 2e. For simplicity, it is customary to state the charge of ions in units of the fundamental electron charge e; so we say that the sodium cation has a charge of 11 and the chloride anion has a charge of 21. An atom can lose or gain more than one electron to form an ion. Examples include Mg21, Fe31, S22, and N32. These ions, as well as Na1 and Cl2, are monatomic ions because they contain only one atom. Figure 0.10 shows the charges of a number of monatomic ions. With very few exceptions, metals tend to form cations and nonmetals tend to form anions. In addition, two or more atoms can form an ion that has a net positive or net negative charge. Polyatomic ions such as OH2 (hydroxide ion), CN2 (cyanide ion), and NH14 (ammonium ion) are ions containing more than one atom.

We will discuss the nature of chemical bonds in Chapters 3 and 4.

1A H 2A

8A 3A 4A 5A 6A 7A N O F Cl Br I

Elements that exist as diatomic molecules

In Chapter 2 we will see why atoms of different elements gain (or lose) a specific number of electrons.


20

Chapter 0



1 1A

18 8A 2 2A

13 3A

Li+ Na+

Mg2+

K+

Ca2+

Rb+ Cs+

3 3B

4 4B

5 5B

14 4A

15 5A

16 6A

17 7A

C4–

N3–

O2–

F–

P3–

S2–

Cl–

Se2–

Br–

Te2–

I–

Al3+

6 6B

7 7B

8

9 8B

10

11 1B

12 2B

Cr 2+ Cr 3+

Mn2+ Mn3+

Fe2+ Fe3+

Co2+ Co3+

Ni2+ Ni3+

Cu+ Cu2+

Zn2+

Sr2+

Ag+

Cd2+

Sn2+ Sn4+

Ba2+

Au+ Au3+

Hg2+ 2 Hg2+

Pb2+ Pb4+

Figure 0.10 Common monatomic ions arranged according to their positions in the periodic table. Note that the Hg21 2 ion is diatomic.

0.3

Compounds Are Represented by Chemical Formulas

Chemical Formulas Chemists use chemical formulas to express the composition of molecules and ionic compounds in terms of chemical symbols. By composition we mean not only the elements present, but also the ratios in which the atoms are combined. At present, we are concerned with three types of formulas: molecular formulas, structural formulas, and empirical formulas. A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance. In our discussion of molecules, each example was given with its molecular formula in parentheses. Thus, H2 is the molecular formula for hydrogen, O2 for oxygen, O3 for ozone, and H2O for water. The subscript numeral indicates the number of atoms of an element present. There is no subscript for O in H2O because there is only one atom of oxygen in a molecule of water, and so the number “1” is omitted from the formula. Some elements can exist in multiple molecular forms called allotropes. An allotrope is one of two or more distinct forms of an element. For example, diatomic oxygen (O2) and ozone (O3) are the two allotropic forms of oxygen. Two allotropes of the same element can have, despite their similar composition, dramatically different physical and chemical properties. Diamond and graphite, for example, are allotropic forms of the element carbon. Diamond is one of the hardest materials known, is transparent, does not conduct electricity, and is relatively expensive. Graphite, in contrast, is a soft, grey-black material that is electrically conductive and cheap. The molecular formula shows the number of atoms of each element in a molecule but does not indicate how the atoms are connected to each other in space. A structural formula shows how atoms in a molecule are bonded to one another. For example, each of the two H atoms is bonded to an O atom in the water molecule, so the structural formula of water is HOOOH. The lines connecting H and O (or any atomic symbols) represent chemical bonds. Some examples of molecular and

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21

0.3 Compounds Are Represented by Chemical Formulas

Molecular formula Structural formula

Hydrogen

Water

Ammonia

Methane

H2

H2O

NH3

CH4

H±N±H W H

H W H±C±H W H

H±H

H±O±H

Ball-and-stick model

Space-filling model

Figure 0.11 Molecular and structural formulas and molecular models of four common molecules.

structural formulas are shown in Figure 0.11. Also shown in Figure 0.11 are two types of three-dimensional molecular models. In the ball-and-stick models, molecules are represented by balls, which represent atoms, connected by sticks, which represent chemical bonds. The various elements are represented by balls of different colors and sizes. In the space-filling models, atoms are represented by overlapping spheres, each of which is proportional in size to the corresponding atoms. Both types of models are used extensively in this text. Many experiments designed to probe molecular composition can only determine the ratios of the numbers of atoms of each element. For example, chemical analysis of hydrogen peroxide (molecular formula H2O2), a substance used as an antiseptic and as a bleaching agent for textiles and hair, only indicates that there is a 1:1 ratio of hydrogen atoms to oxygen atoms. Such an experiment is said to yield only the empirical formula, which tells us which elements are present and the simplest wholenumber ratio of their atoms, and not necessarily the actual number of each kind of atom in the molecule. The empirical formula for hydrogen peroxide is HO. Hydrazine (N2H4), which has been used as rocket fuel, has the empirical formula NH2. Although the ratio of nitrogen to hydrogen is 1:2 in both the molecular formula (N2H4) and the empirical formula (NH2), only the molecular formula accounts for the actual number of N atoms (two) and H atoms (four) present. Empirical formulas are the simplest chemical formulas; they are written by reducing the subscripts in molecular formulas to the smallest possible whole numbers. Molecular formulas are the true formulas of molecules. For some compounds, such as water (H2O) and methane (CH4), the empirical and molecular formulas are identical. Determination of the molecular formula from the empirical formula requires either input from additional experiments or logical reasoning based on knowledge of the rules of chemical bonding (which we discuss in Chapters 3 and 4).

H2O2

N2H4


22

Chapter 0

(a)



(b)

(c)

Figure 0.12 (a) Structure of solid NaCl. (b) In reality, the cations are in contact with the anions. In both (a) and (b), the smaller spheres represent Na1 ions and the larger spheres, Cl2 ions. (c) Crystals of NaCl.

Formulas for Ionic Compounds

Sodium metal reacting with chlorine gas to form sodium chloride.

Ionic compounds are compounds that are formed from cations and anions. Such compounds are held together by the electrostatic attraction between opposite charges. An example of an ionic compound is sodium chloride (NaCl), common table salt. The formulas of ionic compounds are usually the same as their empirical formulas because ionic compounds do not consist of discrete molecular units. For example, a solid sample of sodium chloride (NaCl) consists of equal numbers of Na1 and Cl2 ions arranged in a three-dimensional network (Figure 0.12). There is a 1:1 ratio of sodium cations to chlorine anions so that the compound is electrically neutral. According to Figure 0.12, no Na1 ion in NaCl is associated with just one particular Cl2 ion. In fact, each Na1 ion is equally held by six surrounding Cl2 ions and vice versa. Thus, NaCl is the empirical formula for sodium chloride. (Note that the charges on the cation and anion are not shown in the formula for an ionic compound.) In other ionic compounds, the actual structure may be different, but the arrangement of cations and anions is such that the compounds are all electrically neutral. For the formula of an ionic compound to be electrically neutral, the sum of the charges on the cation and anion in each formula unit must be zero. If the magnitudes of the charges on the cation and anion are different, use the following rule to make the formula electrically neutral: The subscript of the cation is numerically equal to the charge on the anion, and the subscript of the anion is numerically equal to the charge on the cation. If the charges are numerically equal, then no subscripts are necessary. This rule is possible because the formulas of ionic compounds are empirical formulas, so the subscripts must always be reduced to the smallest ratios. Consider the following examples: 䉴 Potassium bromide. The potassium cation K1 and the bromine anion Br2 combine to form the ionic compound potassium bromide. The sum of the charges, measured in units of the electron charge is 11 1 (21) 5 0. The formula is KBr. 䉴 Zinc iodide. The zinc cation Zn21 and the iodine anion I2 combine to form zinc iodide. To make the charges add up to zero, there have to be twice as many I2 ions as Zn21 ions. Therefore, the formula for zinc iodide is ZnI2. 䉴 Aluminum oxide. The cation is Al31 and the oxygen anion is O22. To make a neutral compound, the ratio of Al31 to O22 must be 2:3, giving the formula Al2O3 for aluminum oxide.



23


Naming Compounds The number of known compounds is currently well over 20 million. Fortunately, it is unnecessary to memorize their names. Over the years, chemists have devised a system for naming chemical substances. The rules are accepted worldwide, facilitating communication among chemists and providing a useful way of labeling an overwhelming variety of substances. Mastering these rules now will prove beneficial almost immediately as we proceed with our study of chemistry. To begin our discussion of chemical nomenclature, the naming of chemical compounds, we must first distinguish between organic and inorganic compounds. Organic compounds contain carbon bonded to hydrogen, sometimes in combination with other elements such as oxygen, nitrogen, or sulfur. All other compounds are classified as inorganic compounds. The nomenclature of organic compounds is not discussed in detail until Chapter 16. To organize and simplify our venture into naming compounds, we can categorize inorganic compounds as ionic compounds, molecular compounds, acids and bases, and hydrates.

Naming Ionic Compounds As discussed earlier, ionic compounds consist of cations (positive ions) and anions (negative ions). With a few exceptions (such as the ammonium ion), all cations of interest to us are derived from metal atoms. Metal cations take their names from their respective elements. For example: Element

Name of Cation

Na (sodium) K (potassium) Mg (magnesium) Al (aluminum)

Na1 (sodium ion or sodium cation) K1 (potassium ion or potassium cation) Mg21 (magnesium ion or magnesium cation) Al31 (aluminum ion or aluminum cation)

Many ionic compounds, such as NaCl, KBr, ZnI2, and Al2O3, are binary compounds, that is, compounds formed from just two elements. For binary compounds, the first element named is the metal cation, followed by the nonmetallic anion. Thus, NaCl is sodium chloride. We name the anion by taking the first part of the element name (the “chlor-” of chlorine) and adding “-ide.” The names of KBr, ZnI2, and Al2O3 are potassium bromide, zinc iodide, and aluminum oxide, respectively. The “-ide” ending is also used in the names of some simple polyatomic anions, such as hydroxide (OH2) and cyanide (CN2). Thus, the compounds LiOH and KCN are named lithium hydroxide and potassium cyanide, respectively. These and a number of other such ionic substances are ternary compounds, meaning compounds consisting of three elements. Table 0.3 lists the names of some common cations and anions. Certain metals, especially the transition metals, can form more than one type of cation. For example, iron can form two cations: Fe21 and Fe31. An older nomenclature system that is still in limited use assigns the ending “-ous” to the cation with fewer positive charges and the ending “-ic” to the cation with more positive charges: Fe21 Fe31

8A

1A 3A 4A 5A 6A 7A N O F Al S Cl Br I

2A Li Na Mg K Ca Rb Sr Cs Ba

The most reactive metals (green) and the most reactive nonmetals (blue) combine to form ionic compounds.

3B 4B 5B 6B 7B

8B

1B 2B

The transition metals are the elements in Groups 1B and 3B–8B.

ferrous ion ferric ion

The names of the compounds that these iron ions form with chlorine would thus be FeCl2 FeCl3

ferrous chloride ferric chloride

FeCl2 (left) and FeCl3 (right)


24

Chapter 0



Table 0.3

Names and Formulas of Some Common Inorganic Cations and Anions

Cation

Anion 31

Aluminum (Al )

Bromide (Br2)

Ammonium (NH14 ) 21

Carbonate (CO22 3 ) Chlorate (ClO23 )

Barium (Ba )

21

Chromate (CrO22 4 )

Cadmium (Cd ) Cesium (Cs1)

Cyanide (CN2) 31

Chromium(III) or chromic (Cr )

Dichromate (Cr2O272)

Cobalt(II) or cobaltous (Co21)

Dihydrogen phosphate (H2PO24 )

1

Copper(I) or cuprous (Cu )

Fluoride (F2)

Copper(II) or cupric (Cu21)

Hydride (H2)

1

Hydrogen carbonate or bicarbonate (HCO23 )

Hydrogen (H ) Iron(II) or ferrous (Fe21)

Hydrogen phosphate (HPO22 4 )

31

Iron(III) or ferric (Fe )

Hydrogen sulfate or bisulfate (HSO24 )

Lead(II) or plumbous (Pb21)

Hydroxide (OH2)

1

Iodide (I2)

Lithium (Li ) Magnesium (Mg21)

Nitrate (NO23 ) 21

Manganese(II) or manganous (Mn )

Nitride (N32)

Mercury(I) or mercurous (Hg221)

Nitrite (NO22 )

21

Mercury(II) or mercuric (Hg )

Oxide (O22)

Potassium (K1)

Permanganate (MnO24 )

Rubidium (Rb )

Peroxide (O222)

Silver (Ag1)

Phosphate (PO342)

1

Sulfate (SO242)

1

Sodium (Na ) Strontium (Sr21)

Sulfide (S22) 21

Tin(II) or stannous (Sn )

Sulfite (SO22 3 )

Zinc (Zn21)

Thiocyanate (SCN2)

This method of naming ions has some distinct limitations. First, the “-ous” and “-ic” suffixes do not provide information regarding the actual charges of the two cations involved. Thus, the ferric ion is Fe31, but the cation of copper named cupric has the formula Cu21. In addition, the “-ous” and “-ic” designations provide names for only two different elemental cations. Some metallic elements can form three or more different cations, so it has become increasingly common to designate different cations with Roman numerals. This is called the Stock system after Alfred Stock,15 a German inorganic chemist. In this system, a Roman numeral indicates the number of positive charges, for example, II means two positive charges. The three possible cations of manganese (Mn) are: Mn21 Mn31 Mn41

MnO Mn2O3 MnO2

manganese(II) oxide manganese(III) oxide manganese(IV) oxide

15. Alfred E. Stock (1876–1946). German chemist. Stock did most of his research in the synthesis and characterization of boron, beryllium, and silicon compounds. He was the first scientist to explore the dangers of mercury poisoning.




Using the Stock system, the ferrous ion and the ferric ion are iron(II) and iron(III), respectively, so ferrous chloride becomes iron(II) chloride; and ferric chloride is called iron(III) chloride. In keeping with modern practice, we will use the Stock system of naming compounds in this textbook. Examples 0.3 and 0.4 illustrate how to name and write formulas for ionic compounds based on the information given in Table 0.3.

Example 0.3 Name the following compounds: (a) Cu(NO3)2, (b) KH2PO4, and (c) NH4ClO3.

Strategy Our reference for the names of cations and anions is Table 0.3. If a metal can form cations of different charges (see Figure 0.10), we need to use the Stock system.

Solution (a) Each of the two nitrate ions (NO23 ) bears one negative charge, so the copper ion must have a charge of 12. Because copper forms both Cu1 and Cu21 ions, we need to use the Stock system. The name of the compound is copper(II) nitrate. (b) The cation is K1 and the anion is H2PO24 (dihydrogen phosphate). Because potassium only forms one type of ion (K1), there is no need to use the Stock system. The name of the compound is potassium dihydrogen phosphate. (c) The cation is NH14 (ammonium ion) and the anion is ClO23 (chlorate). The name of the compound is ammonium chlorate. Practice Exercise Name the following compounds: (a) PbO, (b) Li2SO3, and (c) Fe2O3.

Example 0.4 Write chemical formulas for: (a) mercury(I) nitrite, (b) cesium sulfide, and (c) calcium phosphate.

Strategy Refer to Table 0.3 for the formulas of cations and anions. The Roman numerals indicate the charges on the cations.

Solution (a) The Roman numeral indicates that the mercury ion bears a 11 charge. According to Table 0.3, however, the mercury(I) ion is diatomic with a 12 charge (that is, Hg221) and the nitrite ion is NO22. Therefore, the formula is Hg2(NO2)2. (b) Each sulfide ion bears a 22 charge, whereas each cesium ion bears a 11 charge. Therefore, the formula is Cs2S. (c) Each calcium ion (Ca21) bears a 12 charge, and each phosphate ion (PO342) bears a 23 charge. To make the compound electrically neutral (that is, to make the sum of the charges equal zero), we must adjust the numbers of cations and anions 3(12) 1 2(23) 5 0, so that the formula is Ca3(PO4)2.

Practice Exercise Write formulas for the following ionic compounds: (a) rubidium sulfate, (b) chromium(VI) oxide, and (c) barium hydride.

Molecular Compounds Unlike ionic compounds, molecular compounds contain discrete molecular units. They are usually composed of nonmetallic elements (see Figure 0.9). Many molecular compounds are binary compounds. Naming binary molecular compounds is

25


26

Chapter 0


similar to naming binary ionic compounds. We place the name of the first element in the formula first, and the second element is named by adding the suffix “-ide” to the root of the element name:

Table 0.4 Greek Prefixes Prefix


Meaning

Mono-

1

Di-

2

Tri-

3

Tetra-

4

Penta-

5

Hexa-

6

Hepta-

7

Octa-

8

Nona-

9

Deca-

10

HCl HBr SiC

hydrogen chloride hydrogen bromide silicon carbide

It is quite common for one pair of elements to form several different compounds (for example, CO and CO2). In these cases, confusion in naming the compounds is avoided by the use of Greek prefixes (see Table 0.4) to denote the number of atoms of each element present. Consider the following examples: CO CO2 SO2 SO3 CCl4 N2O4 P2O5

carbon monoxide carbon dioxide sulfur dioxide sulfur trioxide carbon tetrachloride dinitrogen tetroxide diphosphorus pentoxide

The following guidelines are helpful in naming compounds with prefixes: c The prefix “mono-” can be omitted for the first element because the absence of a prefix is assumed to indicate that only one atom of that element is present. For example, PCl3 is named phosphorus trichloride, not monophosphorus trichloride. It can also be omitted for the second element if only one compound of that type can be formed. For example, HCl is hydrogen chloride, not hydrogen monochloride, because HCl is the only chloride of hydrogen possible. However, carbon has two oxide forms, CO and CO2, so CO must be named “carbon monoxide” to distinguish it from CO2. c For oxides, the ending “a” in the prefix is sometimes omitted. For example, N2O4 is named dinitrogen tetroxide rather than dinitrogen tetraoxide. Molecular compounds containing hydrogen are not named using Greek prefixes. Traditionally, many of these compounds are called either by their common, nonsystematic names or by names that do not specifically indicate the number of H atoms present: B2H6 CH4 SiH4 NH3 PH3 H2O H2S

diborane methane silane ammonia phosphine water hydrogen sulfide

Even the order of writing the elements in the formulas for hydrogen compounds is irregular. In water and hydrogen sulfide, for example, H is written first, whereas it appears last in the other compounds. Writing formulas for most molecular compounds is usually straightforward. Thus, the name arsenic trifluoride means that there is one As atom and three F atoms in each molecule, and the molecular formula is AsF3. Moreover, the order of the elements in the formula is the same as in its name.



27


Example 0.5 Name the following molecular compounds: (a) SiCl4 and (b) P4O10.

Strategy Refer to Table 0.4 for prefixes. In (a) there is only one Si atom so we do not use the prefix “mono.”

Solution (a) Because there are four chlorine atoms present, the compound is silicon tetrachloride. (b) There are four phosphorus atoms and ten oxygen atoms present, so the compound is tetraphosphorus decoxide. Note that the “a” is omitted from “deca.”

Practice Exercise Name the following molecular compounds: (a) NF3 and (b) Cl2O7.

Example 0.6 Write chemical formulas for the following molecular compounds: (a) carbon disulfide and (b) disilicon hexabromide.

Strategy Here we need to convert prefixes to numbers of atoms (see Table 0.4). Because there is no prefix for carbon in (a), it means that there is only one carbon atom present.

Solution (a) Because there are two sulfur atoms (from the prefix di) and one carbon

atom (implied mono-) present, the formula is CS2. (b) There are two (di-) silicon atoms and six (hexa-) bromine atoms present, so the formula is Si2Br6.

Practice Exercise Write chemical formulas for the following molecular compounds: (a) sulfur tetrafluoride and (b) dinitrogen pentoxide.

Figure 0.13 summarizes the steps for naming ionic and binary molecular compounds.

Naming Acids and Bases An acid can be described as a substance that yields hydrogen ion (H1) when dissolved in water (we will discuss more general definitions in Chapter 10). Formulas for acids contain one or more hydrogen atoms and an anionic group. Anions whose names end in “-ide” form acids with a “hydro-” prefix and an “-ic” ending, as shown in Table 0.5. In some cases, two different names seem to be assigned to the same chemical formula, for example: HCl HCl

hydrogen chloride hydrochloric acid

The name assigned to the compound depends on its physical state. In the gaseous or pure liquid state, HCl is a molecular compound called hydrogen chloride. When HCl is dissolved in water, the molecules break up into H1 and Cl2 ions; in this state, the substance is called hydrochloric acid. Oxoacids are acids that contain hydrogen, oxygen, and another element (the central element). The formulas of oxoacids are usually written with the H first, followed

HCl

H3O+

Cl–

When dissolved in water, the HCl molecule is converted to the H1 and Cl2 ions. The H1 ion is associated with one or more water molecules, and is usually represented as H3O1.

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28

Chapter 0



Compound

Ionic

Molecular

Cation: metal or NH+4 Anion: monatomic or polyatomic

• Binary compounds of nonmetals

Naming Cation has only one charge

• Use prefixes for both elements present (Prefix “mono–” usually omitted for the first element) • Add “–ide” to the root of the second element

Cation has more than one charge • Other metal cations

• Alkali metal cations • Alkaline earth metal cations • Ag+, Al3+, Cd2+, Zn2+

Naming Naming

• Name metal first • Specify charge of metal cation with Roman numeral in parentheses • If monatomic anion, add “–ide” to the root of the element name • If polyatomic anion, use name of anion (see Table 0.3)

• Name metal first • If monatomic anion, add “–ide” to the root of the element name • If polyatomic anion, use name of anion (see Table 0.3)

Figure 0.13 Steps for naming ionic and binary molecular compounds. H O

by the central element and then O, as illustrated by the following examples, which all have the suffix “-ic”: HNO3 H2CO3 H2SO4 HClO3

N

Table 0.5

HNO3

Anion

H

2

O C

Some Simple Acids Corresponding Acid

F (fluoride)

HF (hydrofluoric acid)

Cl2 (chloride)

HCl (hydrochloric acid)

2

Br (bromide)

HBr (hydrobromic acid)

I2 (iodide)

HI (hydroiodic acid)

2

H2CO3

nitric acid carbonic acid sulfuric acid chloric acid

CN (cyanide)

HCN (hydrocyanic acid)

S22 (sulfide)

H2S (hydrosulfuric acid)



29


Often two or more oxoacids have the same central atom but a different number of O atoms. Starting with the oxoacids above, whose names end with “-ic,” we use the following rules to name these compounds: 1.

2. 3.

Note that these acids all exist as molecular compounds in the gas phase.

Addition of one O atom to the “-ic” acid: The acid is called “per . . . -ic” acid. Thus, adding an O atom to HClO3 changes chloric acid to perchloric acid, HClO4. Removal of one O atom from the “-ic” acid: The acid is called “-ous” acid. Thus, nitric acid (HNO3) becomes nitrous acid (HNO2). Removal of two O atoms from the “-ic” acid: The acid is called “hypo . . . -ous” acid. Thus, bromic acid (HBrO3) becomes hypobromous acid (HBrO).

The rules for naming oxoanions, anions of oxoacids, are as follows: 1.

2. 3.

When all the H1 ions are removed from the “-ic” acid, the anion’s name ends with “-ate.” For example, CO2⫺ (the anion derived from H2CO3) is called 3 carbonate. When all the H1 ions are removed from the “-ous” acid, the anion’s name ends with “-ite.” Thus, ClO⫺ 2 , the anion derived from HClO2, is called chlorite. The names of anions in which one or more but not all the hydrogen ions have been removed must indicate the number of H1 ions present. Consider, for example, the anions derived from phosphoric acid: H3PO4 H2PO2 4 HPO422 PO432

phosphoric acid dihydrogen phosphate hydrogen phosphate phosphate

Analogous to the names of molecular compounds, we usually omit the prefix “mono-” when there is only one H in the anion. A base can be described as a substance that yields hydroxide ions (OH2) when dissolved in water (once again a more general definition is discussed in Chapter 11). Commonly, bases are ionic compounds of metal cations with the hydroxide anion and are named according to the usual rules of ionic compounds: NaOH KOH Ba(OH)2

sodium hydroxide potassium hydroxide barium hydroxide

Ammonia (NH3), a molecular compound in the gaseous or pure liquid state, is also classified as a common base. Although, at first glance this may seem to be an exception to the definition of a base just given, as long as the substance yields hydroxide ions when dissolved in water, it need not contain hydroxide ions in its structure to be considered a base. In fact, when ammonia dissolves in water, some of the NH3 2 molecules react with H2O to yield NH1 4 and OH ions. Thus, it is properly classified as a base.

Naming Hydrates Many solid ionic compounds have a specific number of water molecules incorporated into their crystal structure. Such compounds are called hydrates. For example, in its normal state, each unit of copper(II) sulfate has five water molecules associated with it. In the naming of hydrates, we use Greek prefixes (see Table 0.4) to denote the number of water molecules associated with each ionic compound formula unit,

O H P

H3PO4


30

Chapter 0



Figure 0.14 CuSO4 ? 5H2O (left) is blue; CuSO4 (right) is white.

followed by the suffix “-hydrate.” Thus, the systematic name for this compound is copper(II) sulfate pentahydrate, and its formula is written as CuSO4 ⴢ 5H2O. The water molecules can be driven off by heating. When this occurs, the resulting compound is CuSO4, which is sometimes called anhydrous copper(II) sulfate; “anhydrous” means that the compound no longer has water molecules associated with it (Figure 0.14). Some other hydrates are BaCl2 ⴢ 2H2O barium chloride dihydrate LiCl ⴢ H2O lithium chloride monohydrate MgSO4 ⴢ 7H2O magnesium sulfate heptahydrate Sr(NO3)2 ⴢ 4H2O strontium nitrate tetrahydrate

Familiar Inorganic Compounds Some compounds are better known by their common names than by their systematic chemical names. Familiar examples are listed in Table 0.6.

Table 0.6

Common and Systematic Names of Some Familiar Compounds

Formula

Common Name

Systematic Name

H2O

Water

Dihydrogen oxide

NH3

Ammonia

Trihydrogen nitride

CO2

Dry ice (in solid form)

Carbon dioxide

NaCl

Table salt

Sodium chloride

N2O

Laughing gas

Dinitrogen oxide

CaCO3

Marble, chalk, limestone

Calcium carbonate

CaO

Quicklime

Calcium oxide

Ca(OH)2

Slaked lime

Calcium hydroxide

NaHCO3

Baking soda

Sodium hydrogen carbonate

Na2CO3 ⴢ 10H2O

Washing soda

Sodium carbonate decahydrate

MgSO4 ⴢ 7H2O

Epsom salt

Magnesium sulfate heptahydrate

Mg(OH)2

Milk of magnesia

Magnesium hydroxide

CaSO4 ⴢ 2H2O

Gypsum

Calcium sulfate dihydrate

NaOH

Lye

Sodium hydroxide



31

0.4 Reactions Are Represented by Balanced Chemical Equations

0.4

Reactions Are Represented by Balanced Chemical Equations

Writing Chemical Equations According to the atomic theory of matter, atoms and molecules react with one another chemically as discrete units and the number of each type of atom does not change in a reaction. Therefore, the number of atoms of a given element must be the same in the products as in the reactants. Consider, for example, the reaction of hydrogen (H2) and oxygen (O2) to form water (H2O). Because a water molecule contains twice as many H atoms as O atoms, the ratio of hydrogen molecules to oxygen molecules participating in the reaction must be 2:1 to preserve atomic balance. Therefore, we could describe this reaction by saying “two hydrogen molecules react with one oxygen molecule to yield two molecules of water.” However, these kinds of descriptions become quite cumbersome for more complicated reactions and a more compact notation, called a chemical equation, has been developed to accurately describe chemical reactions. The chemical equation for the described reaction is 2H2 ⫹ O2 ¡ 2H2O where the “1” sign means “reacts with,” the arrow means “to yield,” and the reaction is implied to proceed from left to right; that is, the molecules on the left are the reactants and those on the right are the products: reactants ¡ products The numbers in front of each molecular species in a chemical equation are called stoichiometric coefficients, and the quantitative study of amounts of individual reactants and products in chemical reactions is called stoichiometry. Chemical equations often include the physical state of the reactants and products. For example, we can write 2H2 (g) ⫹ O2 (g) ¡ 2H2O(l) to indicate that in this reaction hydrogen and oxygen gas (denoted g) react to form liquid (l) water. Other abbreviations commonly used are (s) for solids and (aq) to denote aqueous species, which are substances dissolved in a water solution. (An aqueous solution is a solution in which water is the solvent.) Consider the combustion of propane (C3H8). Commonly used as fuel in gas grills, propane (a gas) combines with oxygen (O2, a gas) to form carbon dioxide (CO2, a gas) and water (H2O, a liquid). Because each propane molecule contains three carbon atoms and because carbon dioxide (with one carbon atom) is the only product that contains carbon, each propane molecule that reacts must produce three CO2 molecules. Similarly, the eight hydrogen atoms from each propane molecule should produce four water molecules. These data give ten oxygen atoms in the products (from three CO2 and four H2O molecules), so there must be five O2 molecules in the reactants. Thus, the chemical equation for this reaction is C3H8 (g) ⫹ 5O2 (g) ¡ 3CO2 (g) ⫹ 4H2O(l) In this equation there are three C atoms, eight H atoms, and ten O atoms on both the reactant and product sides of the equation. Such a chemical equation is balanced; that is, there are equal numbers of each type of atom on both the reactant and product sides of the equation. Before a chemical equation can be used in any stoichiometric

C3H8


32

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analysis, it must be balanced; therefore, learning how to balance a chemical equation is an essential skill for a chemist.

Balancing Chemical Equations Suppose we want to write an equation to describe a chemical reaction that we have just carried out in the laboratory. How should we go about doing it? Because we know the identities of the reactants, we can write their chemical formulas. The identities of products are more difficult to establish. For simple reactions, it is often possible to guess the product(s). For more complicated reactions involving three or more products, chemists may need to perform tests to establish the presence of specific compounds. Once we have identified all of the reactants and products and have written the correct formulas for each of them, we can assemble them in the conventional sequence—reactants on the left separated by an arrow from products on the right. The equation written at this point is likely to be unbalanced; that is, the number of each type of atom on one side of the arrow differs from the number on the other side. In general, we can balance a chemical equation by using the following steps: 1. 2.

3.

Identify all reactants and products and write their correct formulas on the left side and right side of the equation, respectively. Begin balancing the equation by trying different coefficients to make the number of atoms of each element the same on both sides of the equation. We can change the stoichiometric coefficients (the numbers preceding the formulas), but not the subscripts (the numbers within formulas). Changing the stoichiometric coefficients changes only the amounts of the substances, whereas changing subscripts changes the identity of the substance. For example, 2NO2 means “two molecules of nitrogen dioxide,” but if we double the subscripts, we have N2O4, which is the formula of dinitrogen tetroxide, a completely different compound. To do this, first look for elements that appear only once on each side of the equation with the same number of atoms on each side. The formulas containing these elements must have the same coefficient. Therefore, there is no need to adjust the coefficients of these elements at this point. Next, look for elements that appear only once on each side of the equation but in unequal numbers of atoms. Balance these elements. Finally, balance elements that appear in two or more formulas on the same side of the equation. Check your balanced equation to be sure that you have the same total number of each type of atom on both sides of the equation arrow.

Let’s consider a specific example. In the laboratory, small amounts of oxygen gas can be prepared by heating potassium chlorate (KClO3). The products are oxygen gas (O2) and potassium chloride (KCl). From this information, we write KClO3 ¡ KCl ⫹ O2

Heating potassium chlorate produces oxygen, which supports the combustion of wood splint.

(For simplicity, we omit the physical states of the reactants and products at this point.) All three elements (K, Cl, and O) appear only once on each side of the equation, but only for K and Cl do we have equal numbers of atoms on both sides. Thus, KClO3 and KCl must have the same coefficient. The next step is to make the number of O atoms the same on both sides of the equation. Because there are three O atoms on



33

0.4 Reactions Are Represented by Balanced Chemical Equations

the left and two O atoms on the right of the equation, we can balance the O atoms by placing a 2 in front of KClO3 and a 3 in front of O2. 2KClO3 ¡ KCl ⫹ 3O2 Finally, we balance the K and Cl atoms by placing a 2 in front of KCl: 2KClO3 ¡ 2KCl ⫹ 3O2 As a final check, confirm that the number of atoms of each element present is the same on both sides of the chemical equation. For the heating of KClO3, we have two K atoms, two Cl atoms, and six O atoms on the reactant side and two K atoms, two Cl atoms, and six O atoms on the product side, so the equation is balanced. This equation could also be balanced by multiplying all of the coefficients by a constant. For example, we could multiply all stoichiometric coefficients by 2 to obtain a balanced chemical equation: 4KClO3 ¡ 4KCl ⫹ 6O2 It is common practice, however, to use the smallest possible set of whole-number coefficients to balance the equation. Next consider the combustion (that is, the burning) of the natural gas component ethane (C2H6) in oxygen or air, which yields carbon dioxide (CO2) and water. The unbalanced equation is C2H6 ⫹ O2 ¡ CO2 ⫹ H2O The number of atoms is not the same on both sides of the equation for any of the elements (C, H, and O). In addition, C and H appear only once on each side of the equation, whereas O appears in two compounds on the right side (CO2 and H2O). To balance the C atoms, place a 2 in front of CO2: C2H6 ⫹ O2 ¡ 2CO2 ⫹ H2O To balance the H atoms, place a 3 in front of H2O: C2H6 ⫹ O2 ¡ 2CO2 ⫹ 3H2O At this stage, the C and H atoms are balanced, but the O atoms are not because there are seven O atoms on the right-hand (product) side and only two O atoms on the left-hand (reactant) side of the equation. The inequality of the oxygen atoms is eliminated by placing a stoichiometric coefficient of 7y2 in front of the reactant O2 molecule: C2H6 ⫹ 72 O2 ¡ 2CO2 ⫹ 3H2O Molecules exist as discrete units, however, so it is generally preferable to express the stoichiometric coefficients as whole numbers rather than as fractions; therefore, we multiply the entire equation by 2 to clear the fraction: 2C2H6 ⫹ 7O2 ¡ 4CO2 ⫹ 6H2O Quick inspection shows that there are 4 C atoms, 14 O atoms, and 12 H atoms on both sides of the equation, indicating that it is properly balanced. Example 0.7 gives you additional practice balancing chemical equations.

C2H6


34

Chapter 0



Example 0.7 When aluminum metal is exposed to air, a protective layer of aluminum oxide (Al2O3) forms on its surface. This layer prevents further reaction between aluminum and oxygen, and it is the reason that aluminum beverage cans and aluminum airplanes do not corrode. (In the case of iron exposed to air, the rust, or iron(III) oxide, that forms is too porous to protect the iron metal underneath, so rusting continues.) Write a balanced equation for the formation of Al2O3.

Strategy Remember that the formula of an element or compound cannot be changed when balancing a chemical equation. The equation is balanced by placing the appropriate coefficients in front of the formulas. Solution The unbalanced equation is Al 1 O2 ¡ Al2O3 In a balanced equation, the number and types of atoms on each side of the equation must be the same. There is one Al atom on the reactant side, but two on the product side, so the Al atoms can be balanced by placing a coefficient of 2 in front of reactant Al, yielding 2Al 1 O2 ¡ Al2O3 There are two O atoms on the reactant side, and three O atoms on the product side of the equation. We can balance the O atoms by placing a coefficient of 3y2 in front of O2 on the reactants side. This is a balanced equation. To eliminate the fraction (3y2), multiply both sides of the equation by 2 to obtain whole-number coefficients: 4Al 1 3O2 ¡ 2Al2O3

Check When you are finished balancing an equation, check your answer by making sure that the number of each type of atom is the same on both sides of the equation. For the formation of aluminum oxide, there are four Al and six O atoms on both sides, so the equation is balanced. Practice Exercise Balance the equation representing the reaction between iron(III) oxide (Fe2O3) and carbon monoxide (CO) to yield iron (Fe) and carbon dioxide (CO2).

0.5

Quantities of Atoms or Molecules Can Be Described by Mass or Number

Once we have a balanced chemical equation, we can begin to predict the numbers of each type of molecule being reacted or produced in a chemical reaction. For example, in the equation 2C2H6 ⫹ 7O2 ¡ 4CO2 ⫹ 6H2O we can infer that if eight ethane (C2H6) molecules are reacted, then the number of water molecules produced is 3 3 8 5 24 because three water molecules are produced for every ethane molecule that reacts; the multiplier 3 5 6y2 is simply the ratio of the stoichiometric coefficients. However, such predictive power is of somewhat limited practicality for two reasons: c First, chemists generally measure the mass of chemical samples involved in a reaction and not the number of molecules they contain, so some method for converting mass to number of atoms is necessary in practical calculations.



35

0.5 Quantities of Atoms or Molecules Can Be Described by Mass or Number

c Second, the number of molecules contained in a laboratory sample is exceedingly large. For example, 1 g of water contains more than 5 3 1022 molecules of water—that is, 5 followed by 22 zeros! Therefore, a simple method for dealing with such large numbers in an efficient way is needed. These limitations are overcome in modern chemistry through the introduction of two extremely important concepts: the atomic mass and the mole.

Atomic Mass The mass of an atom depends on the number of electrons, protons, and neutrons it contains and all atoms of a given isotope are identical in mass. The SI unit of mass (the kilogram) is too large to function as a convenient unit for the mass of an atom, thus a smaller unit is desirable. In 1961, the International Union of Pure and Applied Chemistry (IUPAC) defined the atomic mass unit (u)16 to be exactly equal to onetwelfth the mass of one carbon-12 atom.17 Carbon-12 (12C) is the carbon isotope that has six protons, six neutrons, and six electrons. Using this definition, we have that 1 u 5 1.660539 3 10227 kg. The atomic mass (sometimes called atomic weight) of an atom is then defined, relative to this standard, as the mass of the atom in atomic mass units (u). For example, the two naturally occurring isotopes of helium, 3He and 4 He, have atomic masses of 3.01602931 u and 4.00260324 u, respectively. This means that a helium-4 (4He) atom is 4.00260324y12 5 0.33355027 times as massive as a carbon-12 atom.18 When you look up the atomic mass of carbon in a table, such as the one on the inside front cover of this book, you will find that its value is not 12.00 u but 12.01 u. The reason for the difference is that most elements found in nature (including carbon) have more than one naturally occurring isotope. When measuring the mass of a collection of carbon atoms, we obtain not the individual masses of each atom, but an average of the masses of the different isotopes, weighted by their natural abundances. This weighted average is known as the average atomic mass and is the mass most frequently used by chemists. For example, the natural abundances of carbon-12 and carbon-13 are 98.90 percent and 1.10 percent, respectively. The atomic mass of 13C is 13.00335 u and that of 12C is defined to be exactly 12. We can calculate the average atomic mass of carbon by multiplying the atomic mass of each isotope by its relative abundance (expressed as a decimal fraction) and adding them together, as follows: Average atomic mass of C ⫽ (0.9890)(12.00000 u) ⫹ (0.0110)(13.00335 u) ⫽ 12.01 u

16. An earlier abbreviation for atomic mass unit (amu) is still in common usage. In biochemistry and molecular biology, the atomic mass unit is often called a dalton (Da) especially in reference to the atomic mass of proteins. 17. Before 1961, two definitions of the atomic mass unit were used. In physics, the atomic mass unit was defined as one-sixteenth the mass of one 16O atom. In chemistry, the atomic mass unit was defined as onesixteenth the average atomic mass of oxygen. These units were slightly smaller than the current carbon-12 based unit. 18. Helium-4 (4He) contains two protons, two neutrons, and two electrons, and 12C contains six protons, six neutrons, and six electrons. As a result, you would expect that the mass of 4He would be exactly 1/3 of the mass of 12C, however, three 4He atoms have a mass of 3 3 4.00260324 u 5 12.00780972 u—slightly more than a single 12C atom. As we will discuss in Chapter 17, this extra mass is converted into the energy necessary to bind the larger 12C nucleus together. (According to Einstein’s theory of relativity, mass can be converted to energy and vice versa.) This nuclear binding energy is the source of energy released in nuclear power and nuclear bombs.

Atomic number

6 C 12.01

12

C 98.90%

Atomic mass

13 C 1.10%

Natural abundances of C-12 and C-13 isotopes


36

Chapter 0



Because there are many more carbon-12 atoms than carbon-13 atoms in naturally occurring carbon, the average atomic mass is much closer to 12 u than to 13 u. For simplicity, we will generally use atomic mass to mean average atomic mass. It is important to understand that when we say that the atomic mass of carbon is 12.01 u, we are referring to the average value. If carbon atoms could be examined individually, we would find either an atom of atomic mass 12.00000 u or one of 13.00335 u, but never one of 12.01 u. Examples 0.8 and 0.9 illustrate some calculations involving average atomic mass.

Example 0.8 Copper, a metal known since ancient times, is used in electrical cables and pennies, among other things. The atomic masses of its two stable isotopes, 63 29Cu (69.09%) and 65 Cu (30.91%), are 62.93 u and 64.9278 u, respectively. Calculate the average atomic 29 masses of copper. The relative abundances are given in parentheses.

Strategy Each isotope contributes to the average atomic mass based on its relative abundance. Multiplying the mass of an isotope by its fractional abundance (not percentage) will give the contribution to the average atomic mass of that particular isotope. Solution First, convert the percentages to decimal fractions:

69.09% 5 0.6909 and 100%

30.91% 5 0.3091. Next find the contribution to the average atomic mass for each 100% isotope and then add the contributions together to give the average atomic mass. (0.6909)(62.93 u) 1 (0.3091)(64.9278 u) 5 63.55 u

Check The average atomic mass should be between two atomic masses and closer to the mass of the more abundant isotope, copper-63. Therefore, the answer is reasonable. Practice Exercise The atomic masses of the two stable isotopes of boron, 105B

(19.78%) and 115B (80.22%), are 10.0129 u and 11.0093 u, respectively. Calculate the average atomic mass of boron.

Example 0.9 Chlorine has two naturally occurring isotopes, 35Cl and 37Cl, with atomic masses of 34.968852721 u and 36.96590262 u, respectively. The average atomic mass of chlorine is measured to be 35.453 u. Calculate the natural abundances of the two isotopes of Cl.

Strategy There are two unknown quantities in this problem: the relative abundances of 35Cl and 37Cl. A general rule of algebra is that you need as many equations as there are unknown variables to solve such a problem. For two unknowns, therefore, we need two equations that relate these variables. It is also useful to assign the unknowns simple algebraic symbols (such as x or y) to simplify the algebraic manipulations. Solution Let the fractional abundances of 35Cl and 37Cl be x and y, respectively. The average atomic mass (35.453 u) and the atomic masses of the two isotopes are given in the problem statement; therefore, one equation is given by the definition of the average atomic mass discussed previously. 35.453 u 5 x (34.968852721 u) 1 y (36.96590262 u)

(1) —Continued




37

Continued—

We need one more equation to solve this problem. This can be obtained by noting that we are assuming that all chlorine atoms are either chlorine-35 or chlorine-37, so their fractional abundances must add to 1: x1y51

(2)

Solving Equation (2) for y gives y 5 1 2 x. Substituting this value of y into Equation (1) gives 35.453 u 5 x (34.968852721 u) 1 (1 2 x)(36.96590262 u) Collecting like terms and rearranging gives (36.96590262 u 2 34.968852721 u) x 5 36.96590262 u 2 35.453 u 1.997049899 x 5 1.513 u yielding x 5 0.7576. (In the last two equations, we have used the standard rules of significant figures, as described in Appendix 1, to determine the number of significant figures in the final answer.) Substituting this back into Equation (2) gives y 5 1 2 x 5 1 2 0.7576 5 0.2424 Converting the fractions into percentages, the relative abundances of 75.76% and 24.24%, respectively.

35

Cl and

37

Cl are

Practice Exercise Gallium has two naturally occurring isotopes (69Ga and 71Ga) with atomic masses of 68.925581 u and 70.924701 u, respectively. Given that the average atomic mass of gallium is 69.72 u, determine the relative abundances of 69Ga and 71Ga.

The Mole Concept: Avagadro’s Number and the Molar Mass of an Element Atomic mass units provide a relative scale for the masses of the elements. But because atoms have such small masses, no usable scale can be devised to weigh them in calibrated units of atomic mass units. In any real situation, we deal with macroscopic samples containing enormous numbers of atoms. It is convenient, therefore, to have a special unit to describe a very large number of atoms. The idea of a unit to denote a particular number of objects is not new. For example, the pair (2 items), the dozen (12 items), and the gross (144 items) are all commonly used. Chemists measure atoms and molecules in moles. In the SI system the mole (mol) is the amount of a substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 g (or 0.012 kg) of the carbon-12 isotope. The actual number of atoms in 12 g of carbon-12 is determined experimentally. This number is called Avogadro’s number (NA ), in honor of the Italian scientist Amedeo Avogadro.19 The currently accepted value is NA ⫽ 6.0221415 ⫻ 1023 19. Lorenzo Romano Amedeo Carlo Avagadro di Quaregua e di Cerreto (1776–1856). Italian mathematical physicist. He practiced law for many years before he became interested in science. His most famous work, now known as Avagadro’s law (see Chapter 5), was largely ignored during his lifetime, although it became the basis for determining atomic masses in the late nineteenth century.

The adjective formed from the noun “mole” is “molar.”


38

Chapter 0



Figure 0.15 One mole each of several common elements: Carbon (black charcoal powder), sulfur (yellow powder), iron (as nails), copper wires, and mercury (shiny liquid metal).

For most purposes, we can round Avogadro’s number to 6.022 3 1023. Thus, just as one dozen oranges contains 12 oranges, 1 mole of hydrogen atoms contains 6.022 3 1023 H atoms. Figure 0.15 shows samples containing 1 mole each of several common elements. By definition, 1 mole of carbon-12 atoms has a mass of exactly 12 g and contains 6.022 3 1023 atoms. This mass of carbon-12 is its molar mass (m), defined as the mass (usually in grams or kilograms) of 1 mole of units (such as atoms or molecules) of a substance. The molar mass of carbon-12 (in grams) is equal to its atomic mass in u. Likewise, the atomic mass of sodium (Na) is 22.99 u and its molar mass is 22.99 g, the atomic mass of phosphorus is 30.97 u and its molar mass is 30.97 g, and so on. If we know the atomic mass of an element, we also know its molar mass. Because NA carbon-12 atoms have a mass of exactly 12 g, the mass of one carbon12 atom can be calculated by dividing 12 g by the numerical value of NA: mass of one

C atom 5 12 g/(6.0221367 3 1023) 5 1.9926467 3 10223 g

12

By definition, the mass of one 12C atom has a mass of exactly 12 u, so that one u corresponds to 1.9926482 3 10227 kg/12 5 1.6605369 3 10227 kg. The notions of Avogadro’s number and molar mass enable us to carry out conversions between mass and moles of atoms and between the number of atoms and mass and to calculate the mass of a single atom, as illustrated in Examples 0.10–0.12.

Example 0.10 Helium (He) is a valuable gas used in industry, low-temperature research, deep-sea diving tanks, and balloons. How many moles of He atoms are in 6.46 g of He?

Strategy We are given grams of helium and asked to solve for moles of helium. To do this we need to know the number of grams in one mole of helium; that is we need to know the molar mass of He, which is given in the table at the front of the book. A scientific research helium balloon

—Continued



39


Continued—

(Remember that both atomic mass in u and molar mass in grams have the same numerical value.)

Solution The molar mass of He is 4.003 g. Because 1 mole of He and 4.003 g of He represent the same quantity, their ratio is 1; that is 1 mol He 51 4.003 g He

and

4.003 g He 51 1 mol He

To convert from grams to moles, we multiply the given number of grams by 1 mol He , so that the grams cancel and we are left only with moles. 4.003 g He mol He 5 6.46 g He 3

1 mol He 5 1.61 mol He 4.003 g He

Thus, there are 1.61 mol of He atoms in 6.46 g of He. In this problem the ratio 1 mol He is a conversion factor to covert grams of He to moles of He. 4.003 g He

Check Because the given mass (6.46 g) is larger than the molar mass of He, we expect to have (and do have) more than 1 mole of He. Practice Exercise How many moles of magnesium (Mg) are there in 87.3 g of Mg?

Example 0.11 Zinc (Zn) is a silvery metal that is used in making brass (with copper) and in plating iron to prevent corrosion. How many grams of Zn are in 0.356 mole of Zn?

Strategy We are given the number of moles of zinc and need to find the number of grams. Multiplying the number of moles by the number of grams in a mole (molar mass) will give the mass of the sample in grams. To do this, we need to look up the molar mass of zinc. Solution According to the periodic table on the inside front cover, we see the molar mass of Zn is 65.39 g. Multiplying the number of moles in the sample by the following conversion factor 65.39 g Zn 1 mol Zn yields the mass in grams: 0.356 mol Zn 3

65.39 g Zn 5 23.3 g Zn 1 mol Zn

(Notice that we write the conversion factor in such a way that the mol unit will cancel, leaving only units of grams in the answer.) Thus, there are 23.3 g of Zn in 0.356 mole of Zn.

Check The mass of one mole of Zn is 65.39 g. The given amount (0.356 mol) is approximately one-third of a mole. 23.3 g is close to one-third of 65.39, so the answer is reasonable. Practice Exercise Calculate the number of grams of lead (Pb) in 12.4 moles of lead.

Zinc


40

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Example 0.12 Sulfur (S) is a nonmetallic element that is present in coal. When coal is burned, sulfur is converted to sulfur dioxide and eventually to sulfuric acid that gives rise to acid rain. How many atoms are in 16.3 g of S?

Strategy We can calculate the number of atoms of sulfur if we know the number of moles—we just multiply by Avogadro’s number. We are given the number of grams, however, so we must first convert 16.3 g of S to moles of S and then convert the moles to atoms using Avogadro’s number: grams S ¡ moles S ¡ atoms of S

Solution The molar mass of S is 32.07 g. To convert from grams of S to moles of S, we multiply by the conversion factor 1 mol S 32.07 g S This will give us the number of moles. Multiplying the number of moles by Avogadro’s number (which gives the conversion factor between atoms and moles), NA 5

6.022 3 1023 atoms of S 1 mol S

will give the desired answer in number of atoms of S. We can combine these conversions in one step as follows:

Elemental sulfur (S8) consists of eight S atoms joined in a ring.

atoms of S 5 16.3 g S 3

6.022 3 1023 atoms of S 1 mol S 5 3.06 3 1023 atoms of S 3 1 mol S 32.07 g S

Both the units “grams” and “mol” cancel, leaving only the desired unit “atoms” in the final answer. Thus, there are 3.06 3 1023 atoms in 16.3 g of S.

Check The mass of sulfur given (16.3 g) is about one-half of the mass of one mole of sulfur (32.07 g), which would contain Avagadro’s number of atoms, so we should expect the number of atoms in our sample to be equal to about one-half of Avagadro’s number, which is in agreement with our calculation. Our answer is reasonable. Practice Exercise Calculate the number of atoms in 0.551 g of potassium (K).

Molecular Mass If we know the atomic masses of the component atoms, we can calculate the mass of a molecule. The molecular mass (sometimes called molecular weight) is the sum of the atomic masses (in u) in the molecule. For example, the molecular mass of H2O is or

2(atomic mass of H) ⫹ the atomic mass of O 2(1.008 u) ⫹ 16.00 u ⫽ 18.02 u

In general, we need to multiply the atomic mass of each element by the number of atoms of that element present in the molecule and sum over all the elements. Example 0.13 illustrates this approach.



41


Example 0.13 Calculate the molecular masses (in u) of the following compounds: (a) sulfur dioxide (SO2) and (b) caffeine (C8H10N4O2).

Strategy Add the atomic masses of the elements that make up each compound, multiplying each by the number of atoms of that element in the compound (as indicated by the subscripts).

SO2

Solution We find atomic masses in the periodic table (inside front cover). (a) There are two O atoms (atomic mass 16.00 u) and one S atom (atomic mass 32.07 u) in SO2, so that molecular mass of SO2 5 32.07 u 1 2(16.00 u) 5 64.07 u (b) There are eight C atoms, ten H atoms, four N atoms, and two O atoms in caffeine, so the molecular mass of C8H10N4O2 is given by 8(12.01 u) 1 10(1.008 u) 1 4(14.01 u) 1 2(16.00 u) 5 194.20 u

Practice Exercise Calculate the molecular mass of (a) methanol (CH3OH) and (b) acetylsalicylic acid (C9H8O4), commonly known as aspirin.

From the molecular mass, we can determine the molar mass of a molecule or compound. The molar mass of a compound (in grams) is numerically equal to its molecular mass (in u). For example, the molecular mass of water is 18.02 u, so its molar mass is 18.02 g. Note that 1 mole of water has a mass of 18.02 g and contains 6.022 3 1023 H2O molecules, just as 1 mole (12.01 g) of elemental carbon contains 6.022 3 1023 carbon atoms. As Examples 0.14 shows, knowing the molar mass enables us to calculate the numbers of moles and individual atoms in a given quantity of a compound.

Example 0.14 Methane (CH4) is the principal component of natural gas. (a) How many moles of CH4 are present in 6.07 g of CH4? (b) To how many molecules of methane does this correspond?

Strategy To calculate moles of CH4 from grams of CH4 we must determine the conversion factor between grams and moles. To do that, we first need to calculate the molecular mass (and thus the molar mass) of methane by adding the appropriate atomic masses.

Solution (a) The molar mass of CH4 is determined by the procedure used in Example 0.11. In this case, we add the molar mass of carbon to four times the molar mass of hydrogen: molar mass of CH4 5 12.01 g 1 4(1.008 g) 5 16.04 g —Continued

CH4


42

Chapter 0



Continued—

Thus, 1 mol CH4 5 16.04 g CH4. Because we are converting grams to moles, the conversion factor we need should have grams in the denominator so that grams will cancel, leaving the moles in the numerator: mol CH4 5 6.07 g CH4 3

Methane gas burning on a cooking range

1 mol CH4 5 0.378 mol 16.04 g CH4

Thus, there are 0.378 moles of CH4 in 6.07 g of CH4. (b) Multiply the number of moles calculated in part (a) by Avogadro’s number to get the number of molecules: 6.022 3 1023 molecules CH4 1 mol CH4 molecules

number of CH4 molecules 5 0.378 mol 3 5 2.28 3 1023

Practice Exercise (a) Calculate the number of moles of chloroform (CHCl3) in 198 g of chloroform. (b) To how many molecules of CHCl3 does this correspond?

Finally, note that for ionic compounds like NaCl and MgO that do not contain discrete molecular units, we use the term formula mass instead. The formula mass of NaCl is the mass of one formula unit in u: formula mass of NaCl ⫽ 22.99 u ⫹ 35.45 u ⫽ 58.44 u and its molar mass is 58.44 g.

Percent Composition Knowing the chemical formula and the molecular mass of a compound enables us to calculate the percent composition by mass—the percent by mass of each element in a compound. It is useful to know the percent composition by mass if, for example, we needed to verify the purity of a compound for use in a laboratory experiment. From the formula we could calculate what percent of the total mass of the compound is contributed by each element. Then, by comparing the result to the percent composition obtained experimentally for our sample, we could determine the purity of the sample. Mathematically, the percent composition is obtained by dividing the mass of each element in 1 mole of the compound by the molar mass of the compound and multiplying by 100 percent: percent composition of an element ⫽

n ⫻ molar mass of element molar mass of compound

⫻ 100%

(0.9)

where n is the number of moles of the element in 1 mole of the compound. For example, in 1 mole of hydrogen peroxide (H2O2) there are 2 moles of H atoms and 2 moles of O atoms. The molar masses of H2O2, H, and O are 34.02 g, 1.008 g, and 16.00 g, respectively. Therefore, the percent composition of H2O2 is calculated as follows: %H ⫽

H2O2

%O ⫽

2 ⫻ 1.008 g H 34.02 g H2O2

⫻ 100% ⫽ 5.926%

2 ⫻ 16.00g H ⫻ 100% ⫽ 94.06% 34.02 g H2O2



43


The sum of the percentages is 5.926% 1 94.06% 5 99.99%. The small discrepancy from 100 percent is due to round-off error. Note, if we had used the empirical formula for hydrogen peroxide (HO) instead, we would get the same answer, because both the molecular formula and the empirical formula reveal the composition of a compound.

Example 0.15 Phosphoric acid (H3PO4) is a colorless, syrupy liquid used in detergents, fertilizers, toothpastes, and in carbonated beverages for a “tangy” flavor. Calculate the percent composition by mass of H, P, and O in H3PO4.

Strategy Use Equation 0.9 to calculate a percentage and assume that we have 1 mole of H3PO4. The percent by mass of each element (H, P, and O) is given by the combined molar mass of the atoms of the element in 1 mole of H3PO4 divided by the molar mass of H3PO4, then multiplied by 100 percent.

H3PO4

Solution The molar mass of H3PO4 is 97.99 g. The percent by mass of each of the elements in H3PO4 is calculated as follows: 3 3 1.008 g H 3 100% 5 3.086% 97.99 g H3PO4 30.97 g P %P 5 3 100% 5 31.61% 97.99 g H3PO4 4 3 16.00 g O 3 100% 5 65.31% %O 5 97.99 g H3PO4 %H 5

Check The percentages are reasonable because they sum to 100 percent: (3.086% 1 31.61% 1 65.31%) 5 100.01%. The small discrepancy from 100 percent is due to the way we rounded off in the calculation. Practice Exercise Calculate the percent composition by mass of each of the elements in sulfuric acid (H2SO4).

The procedure used in Example 0.15 can be reversed, if necessary, to calculate the empirical formula of a compound from the percent composition by mass of the compound (Figure 0.16). Because we are dealing with percentages and the sum of all the percentages is 100 percent, it is convenient to assume that we started with 100 g of a compound, as shown in Example 0.16.

Mass percent

Convert to grams and divide by molar mass

Moles of each element

Divide by the smallest number of moles

Mole ratios of elements

Change to integer subscripts

Empirical formula

Figure 0.16 Schematic diagram for calculating the empirical formula of a compound from its percent compositions.


44

Chapter 0



Example 0.16 Ascorbic acid (vitamin C) cures scurvy. It is composed of 40.92 percent carbon (C), 4.58 percent hydrogen (H), and 54.50 percent oxygen (O) by mass. Determine its empirical formula.

Strategy In a chemical formula, the subscripts represent the ratio of the number of moles of each element that combine to form one mole of the compound. If we assume exactly 100-g for the compound sample, we can use the percentage composition to determine the number of moles of each element in that sample. Examining the ratios of these molar quantities gives us the empirical formula. The molecular formula of ascorbic acid is C6H8O6.

Solution If we have 100 g of ascorbic acid, then each percentage can be converted directly to grams. In this sample, there will be 40.92 g of C, 4.58 g of H, and 54.50 g of O. Because the subscripts in the formula represent mole ratios, we need to convert the grams of each element to moles. The conversion factor needed is the molar mass of each element. Let n represent the number of moles of each element so that 1 mol C 5 3.407 mol C 12.01 g C 1 mol H 5 4.54 mol H nH 5 4.58 g 3 1.0079 g H 1 mol O 5 3.406 mol O nO 5 54.50 g 3 16.00 g O nC 5 40.92 g 3

Thus, we arrive at the formula C3.407H4.54O3.406, which gives the identity and the mole ratios of the atoms present. However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing all the subscripts by the smallest subscript (3.406): C:

3.407 K

G°(reactants) Q

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