QUICK REFERENCE GUIDE Ions and Ionic Compounds Acids and Bases - Acid (HA) and conjugate base (A ) concentrations, as a function of pH, Table 7.6 Interpreting equilibrium constants (Keq), Table 7.2 Ka and pKa values for selected acids, Table 7.4 Common acids and bases, Table 7.1 pH values of common solutions, Table 7.3 Relative strengths of some acids and their conjugate bases, Table 7.5 Common polyatomic ions, Table 3.2 Some transition metal ions, Table 3.1 The uses of some ionic compounds, Table 3.4 Amino Acids, Proteins, and Enzymes Math a-Amino acids present in proteins, Table 12.1 Amino acids that are essential for humans, Table 14.1 Selected enzyme cofactors, Table 12.2 Atoms Subatomic particles, Table 2.1 The ground state electron distribution for the first 20 elements, Table 2.6 Bonding Lipids Common fatty acids, Table 11.1 Key esters found in some waxes, Table 11.2 Conversion factors and the factor label method, Section 1.6 Logs and antilogs, Chapter Math Support Measurements and significant figures, Section 1.5 Scientific notation, SI and metric prefixes, Section 1.4 Significant figures, Table 1.5 Nucleic Acids Codons in the 5 to 3 sequence of mRNA, Table 13.1 Short tandem repeats (STRs) and the probability of their occurrence, Table 13.2 Bond types, Table 4.1 Organic Compounds Carbohydrates Common molecular shapes, Table 4.2 Formulas and names of alkyl groups, Table 8.3 Physical properties of selected alcohols, ethers, thiols, sulfides, and alkanes, Table 9.1 Physical properties of selected aldehydes and ketones, Table 9.2 Physical properties of selected amines, Table 8.7 Physical properties of selected phenols, Table 8.5 Physical properties of some small carboxylic acids, Table 8.4 Structure, name, and properties of selected hydrocarbons, Table 8.1 The first ten numbering prefixes for IUPAC naming, Table 8.2 Monosaccharides, Table 10.1 Relative sweetness, Table 10.2 Energy Specific heat, Table 1.8 Gases, Liquids, and Solids Density of common substances, Table 1.7 Solutions, colloids, and suspensions, Table 6.5 The solubility of ionic compounds in water, Table 6.3 The vapor pressure of water at various temperatures, Table 6.1 Health Adult body mass index, Table 1.6 Blood pressure guidelines, Table 6.2 Concentration ranges for some blood serum solutes, Table 6.4 Dietary reference intakes (DRIs) for some essential elements, Table 2.4 The biochemical significance of selected elements, Table 2.3 Radioactivity Common forms of nuclear radiation, Table 2.7 Half life and decay type for selected radioisotopes, Table 2.10 Health effects of short term exposure to radiation, Table 2.9 Some uses of radioisotopes in medicine, Table 2.11 Conversion Factors Mass kilogram pound milligram grain ounce = 1000 grams = 2.205 pounds = 453.59 grams = 16 ounces = 1000 micrograms = 65 milligrams = 28.3 grams Length meter kilometer mile inch = 1000 millimeters = 3.281 feet = 39.37 inches = 0.621 mile = 1.609 kilometers = 5280 feet = 2.54 centimeters Volume liter quart gallon = 1000 milliliters = 1.057 quarts = 0.946 liter = 3.785 liters milliliter teaspoon tablespoon = 1 centimeter3 = 1 cubic centimeter = 15 drops = 5 milliliters = 15 milliliters = 0.5 fluid ounces Energy calorie joule = 4.184 joule = 0.2390 calorie Temperature K K °C 0°C °F = °C + 273.15 = -273.15 °C = -459.67°F °F@32 = 1.8 = 273.15K = 32°F = (1.8 * °C) + 32 Pressure atmosphere = 14.7 pounds per square inch = 760 torr = 760 millimeters Hg SI and Metric Prefixes Prefix Symbol Multiplier giga G 1,000,000,000 mega M 1,000,000 kilo k 1,000 hecto h 100 deka da 10 deci d 0.1 centi c 0.01 milli m 0.001 m 0.000001 micro nano n 0.000000001 pico p 0.000000000001 = 109 = 106 = 103 = 102 = 101 = 100 = 10-1 = 10-2 = 10-3 = 10-6 = 10-9 = 10-12 GENERAL, ORGANIC, AND BIOLOGICAL CHEMISTRY An Integrated Approach F o u rth Kenneth W Raymond Eastern Washington University E dition VICE PRESIDENT AND PUBLISHER ASSOCIATE PUBLISHER ACQUISITIONS EDITOR SENIOR PROJECT EDITOR EDITORIAL ASSISTANT EXECUTIVE MARKETING MANAGER MARKETING MANAGER DESIGN DIRECTOR SENIOR PHOTO EDITOR DESIGNERS PRODUCT DESIGNER MEDIA SPECIALIST OUTSIDE PRODUCTION SENIOR PRODUCTION EDITOR Kaye Pace Petra Recter Nicholas Ferrari Jennifer Yee Ashley Gayle Christine Kushner Kristine Ruff Harry Nolan Lisa Gee Thomas Nery and Jim O'Shea Geraldine Osnato Evelyn Brigandi CMPreparé, Rebecca Dunn Elizabeth Swain COVER IMAGE © NREY/iStockphoto This book was set in 10.5/12 Adobe Garamond by CMPreparé and printed and bound by Courier/Kendallville The cover was printed by Courier/Kendallville This book is printed on acid-free paper.∞ Copyright © 2014, 2010, 2008, 2006 John Wiley & Sons, Inc All rights reserved No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 748-6008, website www.wiley.com/go/permissions Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year These copies are licensed and may not be sold or transferred to a third party Upon completion of the review period, please return the evaluation copy to Wiley Return instructions and a free of charge return shipping label are available at www.wiley.com/go/returnlabel Outside of the United States, please contact your local representative Library of Congress Cataloging-in-Publication Data Raymond, Kenneth William General, organic, and biological chemistry : an integrated approach / Kenneth W Raymond.–4th ed p cm Includes bibliographical references and index ISBN 978-1-118-835258-8 (cloth) Chemistry—Textbooks Chemistry, Organic–Textbooks Biochemistry–Textbooks I Title QD31.3.R39 2010 540—dc22 2009034009 ISBN: 978-1-118-35258-8 (Main Book) ISBN: 978-1-118-17219-3 (Binder-Ready Version) Printed in the United States of America 14.2 P iii PR E FA C E T his fourth edition of General, Organic, and Biological Chemistry: An Integrated Approach has, like the earlier editions, been written for students preparing for careers in health-related fields such as nursing, dental hygiene, nutrition, occupational therapy, athletic training, and medical technology The text is also suitable for students majoring in other fields where it is important to have an understanding of chemistry and its relationship to living things Students who use this text not need to have a previous background in chemistry but should possess basic math skills For those whose math is a bit rusty, the text provides reviews of the important material While designed for use in one-semester or two quarter General, Organic, and Biochemistry (GOB) courses, instructors have found that it also works well for one-year courses, especially when combined with the supplement Chemistry Case Studies for Allied Health Students by Colleen Kelley and Wendy Weeks In a GOB course it is essential to show how the subject matter relates to the students’ future careers For that reason, this text makes extensive use of real-life examples from the health sciences O R G A N I Z A T I O N Most GOB texts are divided into three distinct parts: general chemistry, organic chemistry, and biochemistry The integrated approach used in this text integrates these subject areas by juxtaposing chapters of related information For example, a study of bonding and compounds (Chapter 3) is followed by a first look at organic compounds (Chapter 4) and then an introduction to inorganic and organic reactions (Chapter 5) Other examples of this integration at the chapter level include the study of acid–base chemistry (Chapter 7) followed by a chapter that includes organic acids and bases (Chapter 8), and the chemistry of alcohols, aldehydes, and ketones (Chapter 9) followed by that of carbohydrates (Chapter 10) Studies have shown that effective learning can take place when material is presented in a context that shows its relationship to the “big picture.” The arrangement of chapters in this text helps students to see how inorganic chemistry and organic chemistry are linked to the biochemistry and health sciences that are so important to their future careers TAKING AN INTEGRATED APPROACH Whether taught in one semester or two, the GOB curriculum is very full Using an integrated approach can shorten the cycle time for returning to similar themes from the different branches of chemistry Having a shorter time interval between when a topic is first presented and when it is reintroduced can help students assimilate the material more readily An added benefit of integrating GOB course material is that students get a better sense of how the chemistry being presented relates to their future careers, and as a result, their interest and motivation are enhanced BENEFITS OF AN INTEGRATED APPROACH iii iv preface TRANSITIONING TO AN INTEGRATED APPROACH For instructors, making the transition from the traditional approach to an integrated one should not pose a problem The integration of material takes place at the chapter level, and required introductory material is always presented before a new organic chemistry or biochemistry topic is begun For example, instead of introducing carboxylic acids, phenols, and amines in their traditional position—late in the group of chapters devoted to organic chemistry—this text places these organic acids and bases (Chapter 8) directly after the introduction to acids and bases (Chapter 7) Supplements to the text can also assist with making the transition to an integrated text These include Chemistry Case Studies for Allied Health Students, an instructor’s manual, an instructor’s solutions manual, PowerPoint lecture slides, and a test bank KEY FEATURES OF THE FOURTH EDITION In terms of organization, some major changes have been made to this edition of the text A number of these modifications were suggested by reviewers and by instructors who have used previous editions Many reviewers recommended moving the chemistry of hydrocarbons from Chapter to a chapter later in the text In this fourth edition, hydrocarbons appear in Chapter The latter part of Chapter now introduces the key organic families One new feature of the text is the “Did you Know?” paragraphs that briefly highlight topics that relate to the chemistry being presented in each chapter Numerous end of chapter problems, sample problems, and practice problems have been added or revised in the Fourth Edition Other changes that will be noted by those familiar with the text include: Chapter • A new chapter section titled “Measurement in General, Organic, and Biochemistry” shows how the topics presented in Chapter relate to these three fields of chemistry • This chapter now includes a discussion of the kinetic molecular theory, phase changes, heat of fusion, and heat of vaporization In previous editions of the text this material appeared in Chapter Chapter • A new chapter section related to trace elements has been added • In earlier editions, Chapter included a section on fission and fusion This chapter section has been dropped in the new edition • Three new Health Links were added: Stable Isotopes and Drug Testing, Lead, and Radioisotopes for Sale Chapter • The Health Link Pass the Salt, Please was added Chapter • The chapter-opening vignette was changed • Hydrocarbon chemistry (Sections 4.4–4.8 in earlier editions) has been moved to Chapter In its place, a new chapter section related to organic families was added Chapter • The Biochemistry Link The Henderson-Hasselbalch Equation was added Chapter • A new section (Section 8.8 Reactions of Hydrocarbons) gathers topics that, in previous editions, were presented in earlier chapters Section 8.8 also introduces the alkane halogenation and aromatic substitution • The topics of decarboxylation (Section 8.9) and phenol oxidation (Section 8.8) have been removed • There is now a greater emphasis on skeletal structures than in previous editions Chapter • The treatment of nucleophilic substitution (Section 9.2) was trimmed and is now tied to the alkane halogenation reactions introduced in Chapter Chapter 10 • Examples of simple glycosides have been added as part of the introduction to glycosidic bonds Chapter 11 • The structure of esters and their hydrolysis are reviewed just before the discussion of triglycerides and saponification Chapter 12 • The discussion of DG was expanded by introducing the concept of DGo′ Chapter • Two new Health Links were added: Tamiflu and Relenza; and Immunotherapy Chapter 13 • A discussion of inhaled anesthetics and their solubility in blood was added Chapter • The Health Link Lupus was added • The Biochemistry Link Glowing Cats was added Chapter 14 • The effect of pressure and temperature on equilibrium is now described • Figures were modified and sample and practice problems were added preface v PR O B L E M S O L V I N G Learning to anything requires practice, and in chemistry this practice involves solving problems This text offers students ample opportunities to so SAMPLE PROBLEM 1.13 Unit conversions a An over-the-counter (nonprescription) cough syrup contains 7.5 mg of dextromethorphan in every mL The recommended dose of dextromethorphan for a 44 lb child is 10.0 mg How many milliliters of cough syrup should be given? b For a 55 lb child, the recommended dose of dextromethorphan is 12.5 mg How many milliliters of cough syrup should be given? Sample Problems and Practice Problems Each major topic is followed by a sample problem and a related practice problem The solution to each sample problem is accompanied by a strategy to use when solving the problem The answers to practice problems are given at the end of the chapter STRATEGY In part a, you are being asked to convert from a 10 mg dose of dextromethorphan to milliliters of cough syrup For the cough syrup, the relationship between these units (7.5 mg dextromethorphan mL ) can be used to make a conversion factor SOLUTION a 10.0 mg dextromethorphan b 12.5 mg dextromethorphan PRACTICE PROBLEM mL cough syrup 7.5 mg dextromethorphan mL cough syrup 7.5 mg dextromethorphan mL cough syrup mL cough syrup 1.13 The 44 lb child is given a cold tablet that contains mg of dextromethorphan and is then given mL of the cough syrup mentioned in Sample Problem 1.13a Has the child received greater than the recommended dose? END OF CHAPTER PROBLEMS Answers to problems whose numbers are printed in color are given in Appendix C More challenging questions are marked with an asterisk 4.1 One of the alkenes is nonpolar and the other is polar Which is which? F C End of Chapter Problems H H F C H 4.2 F F C H One of the aromatic compounds is nonpolar and the other is polar Which is which? H F Problems are paired and Appendix C provides answers for the odd-numbered problems Each chapter includes multistep Learning Group problems designed to be worked with other students and Thinking It Through problems that ask students to go a bit further with one or more of the concepts presented in the chapter-opening vignette H C C C C C C F F F H C C C C C C F F molecules? a CH3CH2NH2 O CH3CH2NH2 O c 4.86 Which of the molecules in Problem 4.85 can form a hydrogen bond with water? 4.87 Of the pairs of molecules in Problem 4.83, which interact primarily through London forces? 4.88 Of the pairs of molecules in Problem 4.85, which interact primarily through London forces? 4.89 Of the pairs of molecules in Problem 4.81, which can interact through dipole–dipole forces, but not hydrogen bonds? 4.90 Of the pairs of molecules in Problem 4.82, which can interact through dipole–dipole forces, but not hydrogen bonds? HealthLink | PRION DISEASES 4.91 Are covalent bonds broken when PrPc is converted into PrPsc? Explain 4.92 Suggest a way to reduce the spread of mad cow disease between cattle BiochemistryLink | ETHYLENE, A PLANT HORMONE 4.93 During ripening, bananas produce small amounts of ethylene When bananas are shipped, why should they not be shipped in closed containers? 4.94 Ethylene gas can be produced from petroleum and then stored in metal cylinders When food processors want to ripen bananas, they expose the fruit to this manufactured ethylene Would you expect plants to react differently to ethylene made from petroleum than to ethylene that they have produced themselves? HealthLink | SUNSCREENS 4.95 What properties are important for molecules used as sunscreens? 4.96 When applied to the skin of mice, forskolin, a compound present in an Asian plant, was shown to hydrogen, should have an octet of valence electrons a OH b NH4 c CN 4.14 Draw each polyatomic ion Each atom, except for hydrogen, should have an octet of valence electrons b HPO42 c H2PO4 4.15 Draw each of the following Each atom should have an octet of valence electrons b SO32 4.16 Draw each of the following Each atom should have an octet of valence electrons b SH a PO33 4.17 Draw two different molecules that have the formula 4.1 STRUCTURAL FORMULAS 4.85 Will hydrogen bonds form between each pair of c C3H4 4.13 Draw each polyatomic ion Each atom, except for a SO3 H C2H6O 4.3 hydrogen bond with water? c C2H2 4.12 Draw each molecule a C3H8 b C3H6 a PO43 H H 4.84 Which of the molecules in Problem 4.83 can form a b C 4.11 Draw each molecule a C2H6 b C2H4 Indicate the number of covalent bonds that each nonmetal atom is expected to form a C b O c P d Br 4.4 Indicate the number of covalent bonds that each nonmetal atom is expected to form a Se b H c I d N 4.5 Draw the structural formula of the molecule that contains atoms increasethe thefollowing production of melanin Which, you a.suppose, one oxygen atom twoofhydrogen atomsstudy? were the and results this scientific b.a.one atommore and one iodine atom Thehydrogen mice tanned quickly c.b.one nitrogen atom three hydrogen The mice did notand sunburn as easily atoms 4.6 Draw themice structural formula of the molecule that c The were less susceptible to skin cancer contains the following atoms seleniumGatom two fluorine atoms 4.5a one LEARNING ROUPand PROBLEMS b one phosphorus atom and three hydrogen atoms 4.97 c.a.one Tohydrogen which organic family the molecule atom and onedoes bromine atom belong? CH3CH CH 2CH2of 2CH 2OH 4.7 Draw the Lewis structure each molecule formula of the molecule in a.b.F2Give the molecular b O2 part a 4.8 Draw the Lewis structure of each molecule c Can two of the molecules in part a interact through a I2 b N2 London forces? 4.9 Draw thetwo Lewis structure of each molecule d Can of the molecules in part a interact through a CH b NFforces? 2S dipole–dipole 4.10 Draw thetwo Lewis structure of each molecule e Can of the molecules in part a interact through a OCl b CS2 hydrogen bonds? f Draw a molecule that has the same molecular formula as the molecule in part a but belongs to a different family of organic compounds g Can two of the molecules in part f interact through London forces? h Can two of the molecules in part f interact through dipole–dipole forces? i Can two of the molecules in part f interact through hydrogen bonds? 4.98 a To which organic family does the molecule belong? O CH3CH2CH2CH2C OH b Give the molecular formula of the molecule in part a c Can two of the molecules in part a interact through London forces? d Can two of the molecules in part a interact through dipole–dipole forces? e Can two of the molecules in part a interact through hydrogen bonds? f Draw a molecule that has the same molecular formula as the molecule in part a but is an ester g Can two of the molecules in part f interact through London forces? h Can two of the molecules in part f interact through dipole–dipole forces? i Can two of the molecules in part f interact through hydrogen bonds? j Draw a molecule that has the same molecular formula as the molecule in part a but is both an aldehyde and an ether 4.18 Draw three different molecules that have the formula C3H9N 4.19 Write a condensed structural formula for each molecule H H H H ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ a H¬ C ¬ C ¬ C ¬ C ¬ H H H H H H H ƒ ƒ ƒ ƒ b H¬ C ¬ C ¬ N¬ H ƒ H H H 4.20 Write a condensed structural formula for each molecule H H H ƒ ƒ ƒ ƒ ƒ ƒ H ƒ a H¬ C ¬ C ¬ C ¬ O¬ C ¬ H H H H H H H ƒ ƒ ƒ ƒ ƒ ƒ b H¬ C ¬ C ¬ C ¬ F H H H ƒ H 62 Chapter 2 Atoms and Elements 2.7 n The Arrangement of Electrons The ground state is an atom’s most stable electron arrangement ■ ■ FIGURE 2.17 Hydrogen emission spectrum Light from a hydro- gen lamp, when passed through a prism, gives an emission spectrum that consists of just a few different colors of visible light Earlier we saw that if we know an atom’s atomic number and mass number, we can determine the number of protons and neutrons that it has (73Li has protons and neutrons in its nucleus) Further, we learned that a neutral (uncharged) atom has the same number of protons and electrons (a neutral atom of 73Li has electrons in the space that surrounds the nucleus) Making sense of much of the chemistry to follow in this text requires an understanding of how electrons are placed around the nucleus of an atom We will begin by looking at the Bohr model of the atom, an early attempt to describe electron arrangements When an electric current is passed through a tube containing hydrogen gas, the light emitted by the hydrogen atoms can be separated by a prism and viewed as an emission spectrum—a series of colored lines (Figure 2.17) At the beginning of the last century, the Danish physicist Niels Bohr proposed that the colors of light in the emission spectrum of hydrogen are directly related to the movement of a hydrogen atom’s electron between different energy levels In Bohr’s model of the atom, electrons circle the nucleus in specific orbits, with each orbit corresponding to a different energy level (Figure 2.18) An atom is in its ground state (most stable electron arrangement) when its electrons are in energy levels as near as possible to the nucleus For a hydrogen atom, the ground state has the lone electron sitting in energy level If a ground state hydrogen atom absorbs energy, its electron is pushed to an orbit farther from the nucleus, placing the atom into an excited state According to Bohr, the emission spectrum of hydrogen is produced when hydrogen atoms move from various excited states back to more stable states, releasing their energy in the process This energy is given off in the form of electromagnetic radiation, a type of energy that travels as waves The more well-known types of electromagnetic radiation, listed in order of increasing energy, are radio waves, microwaves, infrared light, visible light, ultraviolet light, and x-rays In the case of the hydrogen atom, transitions back to the ground state from energy level (2 S 1) or other energy levels (e.g., S 1, S 1) are of high enough energy that ultraviolet light is released, while movement of electrons back to energy level (3 S 2, S 2, etc.) causes visible frequencies of light to be emitted (Figure 2.18) The colors of visible light have energies ranging from red (lowest energy) to violet (highest energy) Other electron transitions (4 S 3, S 4, etc.) release lower energy infrared light Bohr’s model was very good at explaining the emission spectrum of hydrogen, but not those of other elements As work in this field progressed, experiments provided new information regarding the behavior of electrons One remarkable piece of evidence was that, under certain experimental conditions, electrons can behave as energy waves, rather than particles In 1926, the Austrian physicist Erwin Schrödinger used these wave-like properties of electrons to devise a mathematical equation that described electron energy levels in a new way Like the Bohr model, this new approach, called quantum mechanics, viewed an atom as having a series of energy levels Instead of picturing electrons in fixed orbits about the nucleus, however, quantum mechanics assigns them to various atomic orbitals These orbitals, three-dimensional regions of space where there is a high probability of finding an electron, are the electron clouds referred to when 2.7 The Arrangement of Electrons 63 6S2 ■ ■ FIGURE 2.18 5S2 The Bohr atomic model In the Bohr model, electrons orbit the nucleus, with each orbit corresponding to a different energy level When an electron jumps to an orbit nearer the nucleus, energy is released 4S2 Nucleus 3S2 atomic structure was discussed in Section 2.1 Examples of these atomic orbitals include s and p orbitals (Figure 2.19) For our purposes, quantum mechanics is useful because it lets us calculate the maximum number of electrons that any particular energy level can hold The equation used to carry out this calculation is Maximum number of electrons per energy level = 2n2 where n is the number of the energy level For example, the first energy level (n = 1) holds just two electrons (2 * 12 = 2) and the second energy level (n = 2) holds up to eight electrons (2 * 22 = 8) Table 2.5 shows the maximum number of electrons that can be carried by energy levels through Knowing the maximum number of electrons that an energy level can hold allows ground state electron arrangements to be predicted The ground state of a hydrogen atom has one electron in the n = energy level, while the ground state of a helium atom has two electrons in this energy level Any other arrangement for a helium atom, say, one electron z (a) y x s orbital z z z ■ ■ FIGURE 2.19 (b) y y x px x py p orbitals y x pz Orbitals According to quantum mechanics, various orbitals make up the energy levels of atoms Among them are s orbitals (spherically shaped) and p orbitals (each with two teardrop-shaped lobes) 64 Chapter 2 Atoms and Elements Table | 2.5 The Maximum Number of Electrons Held in the First Four Electron Energy Levels Maximum Number of Electrons (2n ) Energy Level (n) 18 32 ■ ■ FIGURE 2.20 Higher energy Ground state electron arrangements for H, He, and Li The n = energy level holds a maximum of two electrons and all of the electrons of hydrogen and helium are held in this first energy level There is not room for all three of lithium’s electrons in the n = energy level, so one electron is placed in the n = energy level n=4 n=3 electron n=2 n=1 Lower energy electron electrons electrons Hydrogen Helium Lithium in the n = energy level and one electron in the n = energy level, is an excited state The ground state of a lithium atom has two electrons in the n = energy level and one electron in the n = energy level (Figure 2.20) Table 2.6 provides the ground state electron arrangements for the first 20 elements and shows that an energy level is not always filled when the next energy level begins filling with electrons In a potassium (K) atom the n = energy level has electron even though the n = energy level holds only of the allowed 18 electrons As we will see later, atoms are particularly stable when an energy level holds electrons Sample Problem 2.9 Energy levels a. How many electrons can the n = energy level hold? b. For elements in the second period, which electron energy level is being filled? c. How many elements are in the second period? Strategy Once you have solved part a, refer to Table 2.6 and notice how the number of electrons in an energy level changes, moving from group 1A to 8A Solution a. 2n2 = * 22 = b. n = c. Practice Problem 2.9 In terms of electron distribution, what alkaline earth metals have in common? 2.7 The Arrangement of Electrons 65 Table | 2.6 The Ground State Electron Distribution for the First 20 Elements a E lement H He a Group Number of Electrons in Energy Level n = n = n = n=4 1A 8A Li 1A Be 2A 2 B 3A C 4A N 5A O 6A F 7A Ne 8A Na 1A Mg 2A Al 3A Si 4A P 5A S 6A Cl 7A Ar 8A 8 K 1A 8 Ca 2A 8 Valence electrons are listed in bold Valence Electrons Understanding the arrangement of electrons about an atom gives us some insight as to how members of a given group or period in the periodic table are related Note in Table 2.6 that atoms of elements belonging to the same group have an identical number of electrons in their valence shell (the highest numbered, occupied energy level) For example, all group 1A atoms have one valence shell electron, or valence electron For a hydrogen (H) atom the valence electron is in the n = energy level, for a lithium (Li) atom it is in the n = energy level, for a sodium (Na) atom it is in the n = energy level, and for a potassium (K) atom, it is in the n = energy level The number of valence electrons for the atoms of the representative elements follows a periodic trend—group 1A with one valence electron, group 2A with two valence electrons, group 3A with three valence electrons, and so on through group 8A with eight valence electrons The one exception in group 8A, the inert gases, is the helium atom, which has just two valence electrons n Valence electrons are those in an atom’s outermost occupied energy level 66 Chapter 2 Atoms and Elements A correlation exists between the periods of the periodic table and energy levels For atoms of the two elements in period (H and He), valence electrons are held in energy level 1, which holds just two electrons Atoms of period elements (Li, Be, , Ne—there are eight in all) hold their valence electrons in energy level 2, which holds up to eight electrons This same principle holds true for periods 3, 4, and so on Sample Problem 2.10 Valence electrons For each atom, give the total number of electrons and the number of valence electrons a. calcium (Ca) c. argon (Ar) b. silicon (Si) d. radium (Ra) Strategy For these neutral atoms the total number of electrons is the same as the total number of protons (atomic number), which can be obtained from a periodic table The number of valence electrons in these atoms (all of which are representative elements) can be determined from the group number Solution a. 20 total electrons, valence electrons c. 18 total electrons, valence electrons b. 14 total electrons, valence electrons d. 88 total electrons, valence electrons Practice Problem 2.10 For each of the atoms in Sample Problem 2.10, which energy level holds the valence electrons? Electron Dot Structures In the early 1900s, the American chemist Gilbert N Lewis developed electron dot structures to show the number of valence electrons that an atom carries In these structures, valence electrons are represented by dots Lithium has one valence electron, so its electron dot structure is the symbol Li with a dot next to it (Figure 2.21) The electron dot structure of beryllium (Be) has two electron dots and that of fluorine (F) has seven ■ ■ FIGURE 2.21 Valence electrons Repre- sentative elements in the same group have the same number of valence electrons In the electron dot structures used here, valence electrons are shown as dots 1A 2A 3A 4A 5A 6A 7A H 8A He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca 2.7 The Arrangement of Electrons 67 Sample Problem 2.11 Drawing electron dot structures of atoms Draw the electron dot structure of a. a bromine (Br) atom b. a rubidium (Rb) atom Strategy To draw the electron dot structure of these atoms, you must know the number of valence electrons that they carry Solution a Br ả ê b Rb Practice Problem 2.11 Draw the electron dot structure of a. a krypton (Kr) atom b. a barium (Ba) atom Bioluminescence BiochemistryLink Experience tells us that a lightbulb gets hot when it is switched on This happens because the production of light is usually accompanied by the release of heat Sometimes, however, light can be produced without heat, in a process called luminescence When luminescence takes place in a living thing, it is called bioluminescence A number of bacteria, sponges, jellyfish, clams, insects, and fish are bioluminescent This production of light serves a number of different functions For fireflies, the ability to emit light plays a key role in attracting mates One species of squid ejects a glowing cloud of bioluminescent material that hides it from its attackers The anglerfish, which lives deep in the ocean, attracts prey using a bioluminescent appendage (Figure 2.22) The substance that all of these organisms use to produce light is called luciferin, after the Latin name Lucifer (bearer of light) Bioluminescence occurs when luciferin is acted on by the enzyme called luciferase, when oxygen gas and ATP (a biochemical energy source) are present Electrons in luciferin are pushed into an excited state and, when they return to the ground state, light is emitted Medical researchers have isolated the jellyfish gene (section of DNA) responsible for the production of a fluorescent green protein and have learned how to insert it into other animals, including mice One goal of this research is to find new treatments for cancer and other diseases Eye of Science/Photo Researchers, Inc Danté Fenolio/Photo Researchers, Inc ■ ■Figure 2.22 The anglerfish In the dark ocean depths, a bioluminescent appendage attracts prey toward the anglerfish ■ ■Figure 2.23 Glowing mice A jellyfish gene that produces a biolumine scent protein was inserted into fertilized mouse egg cells The mice that developed from these egg cells have a green glow 68 Chapter 2 Atoms and Elements 2.8 n Ra d i o a c t i v e I s o t o p e s Radioisotopes emit nuclear radiation (high energy particles and electromagnetic radiation) Earlier we were introduced to physical change, any change in which the chemical composition of matter is not changed (nothing new is created) Physical changes include melting ice, crushing sugar, and bending iron When matter undergoes a chemical change (chemical reaction), the arrangement of atoms is changed and something new is produced The Chapter Health Link Stable Isotopes and Drug Testing, for example, describes how the carbon atoms in atmospheric carbon dioxide become incorporated into plants, and how these same carbon atoms can end up in testosterone The third type of change that matter can undergo involves change to atomic nuclei This nuclear change is different than physical change and chemical change because atoms are altered during the process As we will see below, something new is usually created during nuclear change—when an atom’s nucleus changes, so does the identity of the atom For the 91 elements that occur naturally, more than a total of 300 isotopes have been identified Over 1000 additional artificial isotopes have been produced Studies of naturally occurring and artificial isotopes have shown that some have unstable nuclei that spontaneously disintegrate to become more stable, releasing high energy particles (individual or groups of subatomic particles) and/or high energy electromagnetic radiation The particles and energy released in this nuclear change are called nuclear radiation, and atoms that emit nuclear radiation are called radioactive isotopes or radioisotopes Of the three hydrogen isotopes, 11H and 21H are stable and emit no nuclear radiation, while 31H is a radioisotope Common Forms of Nuclear Radiation Radioisotopes typically emit one or more of the four common types of radiation: alpha particles, beta particles, positrons, and gamma rays An alpha particle (42a) is identical to the nucleus of a helium-4 atom (42He), in that each has two protons, two neutrons, and a 2+ charge The alpha particle has a much greater energy, however, due to its high speed—a typical alpha particle is ejected from the nucleus of a radioisotope at 5–10% of the speed of light (Table 2.7) When a radioisotope emits an alpha particle, the new atom that is formed has two fewer protons and two fewer neutrons than the original For example, when thorium-230 emits an alpha particle (Figure 2.24a), radium-226 is formed, as is represented by the following nuclear equation 230 90Th ¡ 226 88Ra + 42a In a nuclear equation an arrow separates the starting radioisotope (shown on the left) from the products (shown on the right) To be most useful, nuclear equations must be balanced, which means that the sum of the mass numbers and the sum of the charges on atomic nuclei and subatomic particles must be the same on both sides of the equation In the balanced equation above, thorium’s atomic number indicates a net nuclear charge of 90+ (from the 90 protons in its nucleus) The products also have a charge of 90+, due to radium’s 88 protons and the alpha particle’s protons The sum of the mass numbers is 230 on each side of the equation Table | 2.7 C ommon Forms of Nuclear Radiation Velocity a Penetrating Ability 2+ 5–10% of light speed Low Electron 1- Up to 90% of light speed Moderate Positively charged electron 1+ Up to 90% of light speed Moderate Electromagnetic radiation 0 Light speed High Name Symbol Alpha a protons + neutrons Beta b + Positron b Gamma g a Makeup Charge Light speed 3 10 m/s (meters/second) 186,000 mps (miles per second) 2.8 Radioactive Isotopes 69 Sample Problem 2.12 Balancing nuclear equations In targeted alpha therapy, a carrier substance is used to transport particular radioisotopes directly to cancer cells, which are then destroyed by alpha radiation Balance the nuclear equation for the loss of an alpha particle from actinium-225, a radioisotope used in this procedure 225 89Ac ¡ ? + 42a Strategy To balance this nuclear equation you must come up with an atomic number for the missing atom that will make the sum of the atomic numbers the same on both sides of the reaction arrow (hint: 89 = x + 2) The same must be true of the mass numbers (225 = y + 4) Once you have decided on the atomic number for the missing element, you can refer to the periodic table to discover its identity Solution 225 89Ac ¡ 221 87Fr + 42a Balancing the charges on the nuclei and subatomic particles requires a sum of 89 on each side of the equation, so the missing product must have an atomic number of 87 (89 = 87 + 2), which corresponds to francium (Fr) Balancing the mass numbers requires a sum of 225 on each side of the equation, so the missing product must have a mass number of 221 (225 = 221 + 4) Practice Problem 2.12 Bismuth-213 is another radioisotope used in targeted alpha therapy Balance the nuclear equation for the loss of an a particle from this radioisotope 213 83Bi ¡ ? + 42a A beta particle (-10b) is an electron that is ejected from the nucleus of a radioisotope at up to 90% of the speed of light (Table 2.7) Since a beta particle is just a fast moving electron, it has the same charge and mass as an electron On release of a beta particle, the nucleus formed has one more proton and one less neutron than the original radioisotope Boron-12, for example, releases a beta particle to form carbon-12 (Figure 2.24b), as shown in the nuclear equation 12 5B ¡ 12 6C + -1b A beta particle carries a 1- charge and since it has no protons or neutrons, its mass number is The nuclear equation shown above is balanced, because the sum of the charges on the nuclei and subatomic particles (5) and the sum of the mass numbers (12) is the same on both sides of the arrow It may be puzzling to think that a nucleus composed of protons and neutrons can eject an electron What happens is that a neutron transforms into a proton and electron, and the electron is ejected from the nucleus at high speed as a beta particle A positron (01b+) is a subatomic particle that has the same mass as a beta particle but carries a 1+ charge Like beta particles, positrons are ejected from the nucleus of a radioisotope at speeds of up to 90% of the speed of light Positron radiation has an important use in a medical procedure called positron emission tomography (Section 2.9) The nucleus formed by release of a positron has one less proton and one more neutron than the original radioisotope For example, when fluorine-18 emits a positron (Figure 2.24c), oxygen-18 is produced, as is shown by the nuclear equation, 18 9F ¡ 18 8O + 01b+ This nuclear equation is balanced because the sum of the charges on the nuclei and subatomic particles (9) and the sum of the mass numbers (18) are the same on both sides of the equation n When using atomic notation to describe certain types of nuclear radiation, charge is written in the place normally reserved for atomic number Examples include -10b, 01b+, and 00g 70 Chapter 2 Atoms and Elements ■ ■Figure 2.24 Radioactive decay (a) Loss of an alpha particle from thorium-230 (b) Loss of a beta particle from boron-12 (c) Loss of a positron from fluorine-18 (d) Loss of a gamma ray from an unstable nucleus 88p 138n (a) 90p 140n 2p 2n alpha particle 6p 6n ? 14.11 Did You Know Fission and fusion are nuclear changes that release large amounts of energy In fission, an atom’s nucleus splits to produce two smaller nuclei, neutrons, and energy One example of a fission reaction is that of uranium-235, which fragments to produce barium-142, krypton-91, and neutrons 235 92U (b) 5p 7n beta particle 8p 10n (c) 9p 9n positron + 10n ¡ 142 56Ba + 91 36Kr + 10n This is an example of a chain reaction, one in which a reaction product (the neutrons) can be used to repeat the reaction In nuclear reactors, the heat released by fission is used to produce steam that spins a turbine and generates electricity In fusion reactions, energy is released when nuclei combine to make larger ones Our Sun is a giant fusion reactor One of the fusion reactions taking place in the Sun is the combination of two hydrogen-1 nuclei to produce deuterium and a positron 211H ¡ 21H + 01+ (d ) gamma ray The fourth common form of nuclear radiation, the gamma ray (00), is a very high energy form of electromagnetic radiation The release of alpha, beta, or positron radiation is often accompanied by the release of gamma rays because loss of any of these particles can leave the newly produced nucleus in an unstable state When this unstable nucleus rearranges to a more stable form, energy is given off as gamma rays (Figure 2.24d) Iodine-131 is an example of a beta and gamma emitting radioisotope 131 53I ¡ 131 54Xe + -1 + 00 A gamma ray has no charge and, since it has no protons or neutrons, its mass number is Sample Problem 2.13 Balancing nuclear equations To treat some forms of cancer, wire containing iridium-192 is surgically implanted in a tumor and removed at a later date Beta particles emitted by this radioisotope kill the cancer cells Balance the nuclear equation for the loss of a beta particle from this radioisotope 192 77Ir ¡ ? + -1 2.9 Radioisotopes in Medicine 71 Strategy To balance this nuclear equation you must find an atomic number for the missing atom that will make the sum of the charges on nuclei and subatomic particles the same on both sides of the reaction arrow (hint: 77 = x + -1) The same must be true of the mass numbers (192 = y + 0) Once you have decided on the atomic number for the missing element, you can refer to the periodic table to discover its identity Solution 192 77Ir ¡ 192 78Pt + -1 Practice Problem 2.13 Instead of releasing a beta particle, an iridium-192 nucleus may release a positron Write a balanced nuclear equation for this process 2.9 Radioisotopes in Medicine When nuclear radiation comes into contact with matter, including living tissue, its kinetic energy is transferred to the surrounding atoms The changes that follow can alter water and important biochemical substances (proteins, DNA, lipids, and others) involved in regulating processes that take place within cells, disrupting normal cellular functions (Figure 2.25) This accounts for the health risks associated with exposure to radiation Dosage Units To understand radiation exposure, it will help to learn something about commonly used radiation units These units are based on one of three properties: (1) how fast a sample of an unstable radioisotope disintegrates, (2) the energy transferred when radiation strikes matter, or (3) the biological effect of radiation exposure The curie (Ci), named after Marie Curie, who did much of the significant early work on radioactivity, is based on the speed at which a radioisotope decays One curie is defined as 3.7 * 1010 dps (disintegrations per second) Scientists chose this particular number because it is the number of disintegrations that take place per second in gram of radium, one of the elements discovered by Marie Curie The rad (for radiation absorbed dose) is related to the energy absorbed by an object exposed to nuclear radiation One rad is a dose of SPL/Photo Researchers, Inc 0.01 Joules per kilogram (0.01 J/kg) of the mass of the object The rem is a radiation unit that takes biological effect into account A dose in rems is calculated by multiplying the dose in rads by a quality factor, QF, which depends on the type of radiation (Table 2.8) SPL/Photo Researchers, Inc dose (rad) * QF = dose (rem) The higher the QF value, the more harmful the (b) radiation If a source of radiation exposes people to (a) ■ ■ Figure 2.25 10 rad of gamma radiation (QF = 1), the corresponding 10 rem of gamma radiation might Radiation effects Nuclear cause a drop in the number of their infection-fighting white blood cells In contrast, 10 rad of radiation can disrupt normal alpha radiation (QF = 20) corresponds to 200 rem, possibly a lethal dose 10 rad gamma radiation * = 10 rem 10 rad alpha radiation * 20 = 200 rem cellular functions by damaging proteins, DNA, lipids, and other biochemical substances important to life (a) Normal cells (b) Radiation-damaged cells 72 Chapter 2 Atoms and Elements Table | 2.8 Quality Factors for Various Radiation Types Type of Radiation Quality Factor (QF) a X-rays, gamma rays, beta particles, positrons Neutrons, high energy protons 10 b Alpha particles, fragments of fission reactions 20 a The higher the quality factor, the more harmful the radiation exposure A radiation dose in rems is calculated by multiplying the dose in rads by the quality factor (rem = rad * QF) b ©Vladimir Repik/Reuters/Corbis Fission reactions are described on page 70 ■ ■Figure 2.26 Chernobyl In attempting to repair the damaged reactor at the Chernobyl power plant, some workers received a lethal dose of radiation The QF values given in Table 2.8 are only estimates, because the actual values depend upon a number of other factors, including the energy of the radiation, how rapidly a dose is absorbed, and which part of the body absorbs the dose In medical applications, SI units for radiation are often used These units are the gray (Gy) and the seivert (Sv): Gy = 100 rads, Sv = 100 rem The short-term effects of a single exposure to a high dose of nuclear radiation range from nausea to death within a few weeks, depending on the dose (Table 2.9) Some of the workers who tried to control the 1986 Chernobyl power plant disaster in the Soviet Union were exposed to extremely high levels of radiation in a very short time (Figure 2.26), and more than 30 of them died soon after The delayed effects of radiation exposure include an increased risk of cancer, cataracts, and mutations—changes in egg and sperm cells that can transmit genetic disorders to offspring The effects of long-term exposure to radiation at low levels are not well understood One factor that makes estimating the effects of low doses of radiation difficult is that we are continually exposed to naturally occurring radioisotopes that are present in and around us These levels of background radiation vary depending on where you live For example, in regions of Brazil, India, and China, where monazite (a mineral containing radioactive thorium-232 and radium-226) is found, the background radiation level is hundreds of times higher than the average level found in the United States The altitude that you live at determines how much exposure you will have to background radiation Table | 2.9 Health Effects of Short-Term Exposure to Radiation Exposure (rem) Health Effect Time to Onset (without treatment) 5–10 Changes in blood chemistry 50 Nausea 55 Fatigue 70 Vomiting 75 Hair loss 90 Diarrhea 100 Hemorrhage 400 Possible death within months 1000 Destruction of intestinal lining, internal bleeding, and death 1–2 weeks 2000 Damage to central nervous system and loss of consciousness Death minutes hours to days Source: http://www.epa.gov/rpdweb00/understand/health_effects.html hours 2–3 weeks 2.9 Radioisotopes in Medicine 73 from the cosmic rays (atomic nuclei) that constantly bombard the earth from outer space The higher the altitude, the less atmosphere there is above you to block the cosmic rays A significant portion of your exposure to background radiation may come from radon gas Radon-222, an alpha emitter, is produced as part of the radioactive breakdown of uranium-238 Granite and other rocks that contain uranium can be a source of radon Because it is radioactive, inhaling radon gas can lead to health problems—it is estimated that 12% of lung cancer deaths in the United States are due to radon exposure For this reason, the Environmental Protection Agency recommends that houses be tested for radon If radon levels are high, there are a number of actions that can be taken These include sealing cracks in the floor to prevent movement of radon gas into the house and installing pipes and exhaust fans to pull radon from the soil under the house X-rays are another common source of radiation exposure A standard dental x-ray gives a patient about 10 mrem, while the newer digital x-rays cut this exposure to about mrem A chest x-ray gives 2–3 mrem of radiation exposure and a CT scan (see Health Link CT and MRI Imaging) of the abdomen about 1000 mrem (1 rem) Note that all these values are below the radiation levels listed in Table 2.9 SAMPLE PROBLEM 2.14 Background radiation Of the potassium found in nature, about 0.02% is the positron emitter potassium-40 Of the typical person’s background radiation, about 11% or 0.4 mSv comes from potassium naturally present in the body Convert 0.4 mSv into (a) millirems and (b) millirads Strategy Look for the definition of these units in the paragraphs above Table 2.8 will help you to answer part b Solution a. There are two ways to approach this calculation One uses the definition 100 rem = Sv The other uses a shortcut: If 100 rem = Sv, then 100 mrem = mSv 0.4 mSv * * 10-3 Sv 100 rem mrem * * = 40 mrem mSv Sv * 10-3 rem or 0.4 mSv * 100 mrem = 40 mrem mSv b. From part a of this problem, 0.4 mSv = 40 mrem According to Table 2.8, rem of positron radiation corresponds to rad This means that mrem = mrad 40 mrem * * 10-3 rem rad mrad * * = 40 mrad mrem rem * 10-3 rad or 40 mrem * mrad = 40 mrad mrem PRACTICE PROBLEM 2.14 Each year the average U.S resident is exposed to mSv of background radiation from the alpha emitter radon Convert mSv into millirems, millirads, and milligrays Controlling Exposure to Radiation While we can’t much about background radiation, we can control our exposure to other sources of nuclear radiation One way to so is to use proper shielding (blocks radiation) when working with radioisotopes Although fast moving, alpha particles travel only to cm Did You Know ? 14.11 The new full-body x-ray scanners being used at airports expose each traveler to 10 microrems of radiation This is 50 times less radiation than experienced during each hour of airplane flight and 100,000 times less exposure than from a medical CT scan 74 Chapter 2 Atoms and Elements ■ ■Figure 2.27 Shielding Alpha particles are blocked by paper, beta particles and positrons by a thin piece of plastic, and gamma rays by concrete Paper Plastic Concrete a a a b b b g g g (a) (b) (c) in air before losing their energy, and they can only slightly penetrate surrounding matter (Figure 2.27 and Table 2.7) Gloves, clothing, or a sheet of paper are usually sufficient to block alpha particles Being faster moving and smaller than alpha particles, beta particles and positrons can travel more than meter in air and are able to penetrate matter to a greater depth Blocking these particles requires more substantial shielding than needed for alpha particles, typically a thin sheet of plastic or metal Gamma rays are so energetic that they have a very high penetrating power—a thick slab of concrete or lead is required to stop them Half-life n After one half-life, half of the atoms in a radioisotope will have decayed Knowing how long a radioisotope will remain hazardous is also an important factor in controlling exposure to radiation This is determined from the half-life of a radioisotope, the time required for one-half of the atoms in a sample to decay (Figure 2.28) Take, for example, the radioactive decay of radon-222 into polonium-218 by emission of an alpha particle 222 86Rn ¡ 218 84Po + 42 Radon-222 has a half-life of 3.8 days, which means that beginning with a mg sample of this radioisotope, only mg remains after 3.8 days After another 3.8 days, 0.5 mg (onehalf of mg) of radon-222 remains The half-life for a radioactive decay is unrelated to the starting amount of a sample, so beginning with 10 mg of 222 86Rn, mg will remain after 3.8 days, and beginning with 0.01 mg, 0.005 mg will remain after 3.8 days Like radon-222, plutonium-239 is an alpha emitter, but its half-life of 24,000 years is considerably longer 239 94Pu Half-life After one half-life, a sample will have decayed to 50% of its original amount After two half-lives 25% will remain and after three half-lives 12.5% will remain 235 92U + 42 Beginning with g of plutonium-239 and waiting for one half-life (24,000 years), 0.5 g would remain Over this same period of time, g of radon-222 would disappear—after 71 half-lives (about months) only one atom would remain As Table 2.10 shows, halflives range from small fractions of a second to over a billion years Half-life also plays a role 100% in determining whether a particular isotope is long-lived 80% enough to be found in nature Radioisotopes with extremely 60% short half-lives, such as polonium-214 (0.0016 sec), decay so rapidly that they can be 40% observed only immediately after their formation Those 20% with very long half-lives, such as uranium-238 (4.5 billion 0% years), decay very slowly and 10 are therefore present in nature Number of half-lives Amount of sample remaining ■ ■Figure 2.28 ¡ 2.9 Radioisotopes in Medicine 75 Table | 2.10 Half-life and Decay Type for Selected Radioisotopes a Isotope Half-life Decay Type 214 84Po 0.0016 s a 11 6C 20 m b 24 11Na 15 h b 222 86Rn 3.8 d a 131 53I 8.1 d b 32 15P 14.3 d b 1H 12.3 y b 90 38Sr 28.1 y b 14 6C 5,730 y b 239 94Pu 24,000 y a 40 19K 1.3 * 109 y b and b+ 238 92U 4.5 * 109 y a a b214 222 40 131 239 24 84Po, 11Na, 86Rn, 53I, 94Pu, 19K, b + Units: seconds (s), minutes (m), hours (h), days (d), and years (y) and 238 92U also emit gamma radiation Sample Problem 2.15 Calculations involving half-life Rubidium-84, used to monitor cardiac output, has a half-life of 33 days How many milligrams of a 10.0 mg sample of this radioisotope remain after 99 days? Strategy To solve this problem it helps to remember that a half-life of 33 days means that whatever the starting amount of rubidium-84, after 33 days only half remains Of that remaining amount, in another 33 days, only half remains Solution 1.25 mg After half-life (33 days), 5.00 mg remain After half-lives (66 days), 2.50 mg remain After half-lives (99 days), 1.25 mg remain Practice Problem 2.15 How many days are required for a 1.00 mg sample of 84 37Rb to decay to 0.0625 mg? Diagnosis and Therapy The beginning of this section described the health problems that can follow exposure to nuclear radiation Interestingly enough, radioisotopes can also be used to diagnose and treat disease Not just any radioisotope is appropriate for medical use, however Half-life is one factor that must be taken into consideration Radioisotopes with half-lives ranging from hours to weeks are generally preferred for medical use, because they not decay before serving their purpose and they not pose problems by being present long after their use Did You Know ? 14.11 Half-life is not just used to describe radioactive decay When discussing herbicides and pesticides, half-life refers to the time that it takes one-half of an amount of a sample to break down once it has been applied to the soil The common weed killer 2,4-D has a half-life of 10 days When used to describe a drug, half-life is the time required for one-half of a dose in the body to be reduced by half The halflife of the dental anesthetic Novocaine is 1.5 to hours 76 Chapter 2 Atoms and Elements Table | 2.11 Some Uses of Radioisotopes in Medicine Radioisotope Use 24 11Na Detecting blood vessel obstruction 32 15P Treating leukemia; detecting eye tumors 51 24Cr Imaging the spleen; detecting gastrointestinal disorders 59 26Fe Detecting bone marrow disorders 60 27Co Treating cancer 67 31Ga Treating lymphomas; whole-body tumor scans 75 34Se Pancreas scans 84 37Rb Measuring cardiac output 85 38Sr Bone scans 125 53I Treating prostate and brain cancer 131 53I Treating and detecting thyroid disorders 133 54Xe Detecting lung malfunctions 137 55Cs Treating cancer 167 69Tm Bone and tumor scans 186 75Re Pain relief for bone, prostate, and breast cancer 192 77Ir Internal radiation therapy 197 80Hg Brain scans 213 83Bi Targeted alpha therapy 225 89Ac Targeted alpha therapy and 52 26Fe Radioisotopes allow clinicians to diagnose a wide range of health problems (Table 2.11) Iodine-131 (131 53I) can be used to monitor the state of the thyroid gland, because cancer, hyperthyroidism (an overactive thyroid), and other thyroid disorders cause characteristic changes in the rate at which iodine is taken up by this gland After a patient is given a dose of an 131 53I containing substance, external radiation detectors are used to measure the speed at which this radioisotope accumulates in the thyroid (Figure 2.29) The diagnostic tool called positron emission tomography (PET) also makes use of radioisotopes PET can be used to monitor glucose (sugar) metabolism, oxygen usage, blood flow, and other biochemical processes In this technique a positron-emitting radioisotope is introduced into the body Fluorine-18, one of the radioisotopes used in PET, decays to form oxygen-18 and a positron 18 9F ¡ 18 8O + 01 b+ When the released positron collides with an electron (-10e-) from a nearby atom, they destroy one another, producing two gamma rays in the process + 1b + -1e ¡ 00g The gamma rays are detected by the PET scanner, and a computer processes the information to construct an array of two-dimensional images Fluorine-18 is an artificial radioisotope with a half-life of 110 minutes To use PET for studying glucose metabolism, this rapidly decaying radioisotope must, within about three hours of being produced, be incorporated into glucose, delivered to a clinic, and used Some ... =? ?10 6 =? ?10 3 =? ?10 2 =? ?10 1 =? ?10 0 =? ?10 -1 =? ?10 -2 =? ?10 -3 =? ?10 -6 =? ?10 -9 = 10 -12 GENERAL, ORGANIC, AND BIOLOGICAL CHEMISTRY An Integrated Approach F o u rth Kenneth W Raymond Eastern Washington University... Exponent 0.00 01 * 10 -4 0.0 01 * 10 -3 -3 0. 01 * 10 -2 -2 0 .1 * 10 -1 -1 1 * 10 0 10 * 10 1 35000 = 3.5 * 10 4 10 0 * 10 2 285.2 = 2.852 * 10 2 10 00 * 10 3 10 000 * 10 4 8300000 = 8.3 * 10 6 For a number... M 1, 000,000 = 10 6 kilo k 10 00 = 10 3 hecto h 10 0 = 10 2 deka da 10 = 10 1 = 10 0 deci d 0 .1 = 10 -1 centi c 0. 01 = 10 -2 milli m 0.0 01 = 10 -3 micro m 0.0000 01 = 10 -6 nano n 0.0000000 01 = 10 -9