Third Edition
`
`CHEMISTRY
`
`The Central Science
`
`THEODORE L. snown
`
`University of Illinois
`
`H. EUGENE LEMAY, JR.
`
`University of Nevada
`
`Page 1 of 12
`Prentice-Hall, Inc.
`
`Englewood Cliffs, NJ.
`
`07632
`
`FMC 1026
`
`

`

`
`
`Ii1Ii.,
`
`
`
`
`
`Library of Congress Cataloging in Publication Data
`
`Brown, Theodore L.
`Chemistry: the central science.
`
`Includes index.
`1. Chemistry.
`(date).
`II. Title.
`QD31.2.B78
`1985
`ISBN 0-13-128950-0
`
`I. LeMay, H. Eugene (Harold Eugene),
`
`540
`
`84—8413
`
`Development editor: Raymond Mullaney
`Editorial/production supervision: Karen ‘J. Clemments
`interior design: Levavi & Levavi
`Art direction and cover design: Janet Schmid
`Manufacturing buyer: Raymond Keating
`Page layout: Gail Collis
`Cover photograph: “Rainbow" (© Geoff Gove, The Image Bank)
`
`© 1985, 1981, 1977 by Prentice-Hall, Inc., Englewood Clifl's, New Jersey 07632
`
`All rights reserved. No part of this book may be
`reproduced,
`in any form or 192 my) means;
`without permission in writing from the publisher.
`
`Printed in the United States of America
`
`1098765432
`
`ISBN D-lB-lEBEIEEI-EI
`
`[IL
`
`Prentice—Hall International, Inc., London
`
`Prentice-Hall of Australia Pty. Limited, Sydney
`Editora Prentice-Hall do Brasil, Ltda., Rio de Jamiro
`Prentice-Hall Canada Inc., Toronto
`Page 2 of 12
`Prentice—Hall of India Private Limited, New Delhi
`Prentice—Hall of Japan, Inc., 771.190
`Prentice-Hall of Southeast Asia Pte. Ltd., Singapore
`Whitehall Books Limited, Wellington, New Zealand
`
`FMC 1026
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`
`
`3.1 LAW
`OF CONSERVATION
`OF MASS
`
`Stoichiometry
`
`Antoine Lavoisier (Section 1.1) was among the first to draw conclusions
`about chemical processes from careful, quantitative observations. His
`work laid the basis for the law of conservation of mass, one of the most
`fundamental laws of chemistry. In this chapter, we will consider many
`practical problems based on the law of conservation of mass. These prob—
`lems involve the quantitative relationships between substances undergo-
`ing chemical changes. The study of these quantitative relationships is
`known as stoichiometry (pronounced stoy-key-AHM-uh-tree), a word
`derived from the Greek words
`stoic/zeion
`(“element”) and metron
`(“measure” .
`
`Studies of countless chemical reactions have shown that the total mass of
`
`all substances present after a chemical reaction is the same as the total
`mass before the reaction. This observation is embodied in the law of
`
`conservation of mass: There are no detectable changes in mass in any
`chemical reaction. * More precisely, atoms are neither created nor destroyed
`during a chemical reaction; instead, they merely exchange partners or be—
`come otherwise rearranged. The simplicity with which this law can be
`stated should not mask its significance. As with'many other scientific
`laws, this law has implications far beyond the walls of the scientific
`laboratory.
`‘
`The law of conservation of mass reminds us that we really can’t throw
`anything away. If we discharge wastes into a lake to get rid of them, they
`are diluted and seem to disappear. However, they are part of the envi—
`
`*In Chapter 19, we will discuss the relationship between mass and energy summarized by the
`equation E : me2 (E is energy, In is mass, and c is the speed of light). We will find that whenever an
`Page 3 of 12
`object loses energy it loses mass, and whenever it gains energy it gains mass. These changes in mass
`are too small to detect in chemical reactions. However, for nuclear reactions, such as those involved
`in a nuclear reactor or in a hydrogen bomb, the energy changes are enormously larger; in these
`reactions there are detectable changes in mass.
`
`
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`3.2 CHEMICAL
`EQUATIONS
`
`ronment.Theymayundergochemicalchangesorremaininactive;they
`
`may reappear as toxic contaminants in fish or in water supplies or lie on
`the bottom unnoticed. Whatever their fates, the atoms are not destroyed.
`The law of conservation of mass suggests that we are converters, not
`consumers. In drawing upon nature’s storehouse of iron ore to build the
`myriad iron—containing objects used in modern society, we are not reduc-
`ing the number of iron atoms on the planet. We may, however, be con—
`verting the iron to less useful, less available forms from which it will not
`be practical to recover it later. For example, consider the millions of old
`washing machines that lie buried in dumps. Of course, if we expend
`enough energy, we can bring off almost any chemical conversions we
`choose. We have learned in recent years, however, that energy itself is a
`limited resource. Whether we like it or not, we must learn to conserve all
`
`our energy and material resources.
`
`We have seen (in Sections 2.2 and 2.6) that chemical substances can be
`represented by symbols and formulas. These chemical symbols and for—
`mulas can be combined to form a kind of statement, called a chemical
`equation, that represents or describes a chemical reaction. For example,
`the combustion of carbon involves a reaction with oxygen (02) in the air
`to form gaseous carbon dioxide ((302). This reaction is represented as
`
`c+o2 % co2
`
`[3.1]
`
`We read the + sign to mean “reacts with” and the arrow as “produces.”
`Carbon and oxygen are referred to as reactants and carbon dioxide as the
`product of the reaction.
`It is important to keep in mind that a chemical equation is a descrip-
`'tion of a chemical process. Before you can write a complete equation you
`must know what happens in the reaction or be prepared to predict the
`products. In this sense, a chemical equation has qualitative significance;
`it identifies the reactants and products in a chemical process. In addi—
`tion, a chemical equation is a quantitative statement; it must be consis—
`tent with the law of conservation of mass. This means that the equation
`must contain equal numbers of each type of atom on each side of the
`equation. When this condition is met the equation is said to be balanced.
`For example, Equation 3.1 is balanced because there are equal numbers
`of carbon and oxygen atoms on each side.
`A slightly more complicated situation is encountered when methane
`(CH4), the principal component of natural gas, burns and produces car—
`bon dioxide (C102) and water (H20). The combustion is “supported by”
`oxygen (02): meaning that oxygen is involved as a reactant. The unbal—
`anced equation is
`
`CH4 + 02 ——+ co2 + H20
`
`[3.2]
`
`The reactants are shown to the left of the arrow, the products to the
`Page 4 of 12
`right. Notice that the reactants and products both contain one carbon
`atom. However, the reactants contain more hydrogen atoms (four) than
`the products (two). If we place a coefficient 2 in front of H20, indicating
`
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`3.2 CHEMICAL EQUATIONS
`
`59
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`

`

`
`
`formation of two molecules of water, there will be four hydrogens on
`each side of the equation:
`
`OH, + 02 ——a co, + 2H20
`
`[3.3]
`
`Before we continue to balance this equation, let’s make sure that we
`clearly understand the distinction between a coefficient in front of a
`formula and a subscript in a formula. Refer to Figure 3.1. Notice that
`changing a subscript in a formula, such as from H20 to H202, changes
`the identity .of the chemical involved. The substance H202, hydrogen
`peroxide, is quite different from water. The subscripts in the ehemicalfonnu—
`'las should never be changed in balancing an equation. On the other hand, plac—
`ing a coefficient in front of a formula merely changes the amount and not
`the identity of the substance; ZHZO means two molecules of water,
`3HZO means three molecules of water, and .so forth. Now let’s continue .
`balancing Equation 3.3. There are equal numbers of carbon and hydro-
`gen atoms on both sides of this equation; however, there are more oxygen
`atoms among the products (four) than among the reactants (two). If we
`place a coefficient 2 in front of 02 there will be equal numbers of oxygen
`atoms on both sides of the equation:
`
`CH4 + 202 ———> GO2 + 2H20
`
`[3.4]
`
`The equation is now balanced. There are four oxygen atoms, four hydro-
`gen atoms, and one carbon atom on each side of the equation. The.
`balanced equation is shown schematically in Figure 3.2.
`Now,
`let’s look at a slightly more complicated example, analyzing
`stepwise what we are doing as we balance the equation. Combustion of
`octane (081,118), a component of gasoline, produces CO2 and H20. The
`balanced chemical equation for this reaction can be determined by using
`the following four steps.
`First, the reactants and products are written in the unbalanced equa-
`tion
`
`I
`
`;
`
`i
`i
`’
`1
`
`,
`;
`'
`
`ong,, + 02 ——9 002 + H20
`
`[3.5]
`
`Before a chemical equation can be written the identities of the reactants
`and products must be determined. In the present example this informa-
`tion was given to us in the verbal description of the reaction.
`
`.
`
`FIGURE 3.1 Illustration of the difference in
`meaning between a subscript in a chemical for-
`mula and a coeflicient in front of the formula.
`
`Notice that the number of atoms of each type
`(listed under composmon) is obtained by multi—
`plying the coeffic1ent and the subscript assom—
`ated with each element‘in the formula.
`
`Chemical
`symbol
`/_—_\/
`
`Meaning
`
`_\/'
`
`Composition
`
`~_\
`
`H2O
`
`2H20
`
`H202
`
`gpagg'ecule 4 TwoHatomsand one0atom
`
`Two molecules .
`2'
`;
`of Water:
`' a '
`Page 5 of 12
`One molecule
`of hydrogen 3 _
`peroxide:
`a
`
`F
`
`our
`
`d t
`H t
`a 0m an wo
`
`O t
`a 0m
`
`Two H atoms and two 0 atoms
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`

`

` +
`
`—>
`
`One methane
`molecule
`
`Two oxygen
`molecules
`
`_> One carbon
`dioxide molecule
`
`+ Two water
`molecules
`
`FIGURE 3.2 Balanced chemical equation for the com-
`bustlon of.CH4. The draw1ngs of the molecules Involved
`call attentlon to the conservatlon of atoms through the
`reaction.
`'
`
`CH4
`(1 C
`4H
`
`+
`
`—’
`
`202
`(4 O)
`
`C02
`1 C
`2 o
`
`+
`
`2H20
`2 0
`4 H
`
`Seconafl the number of atoms of each type on each side of the equation
`is determined. In the reaction above there are SO, 18H, and 20 among
`the reactants, and IC, 2H, and 30 among the products; clearly, the
`equation is not balanced, because the number of atoms of each type
`differs from one side of the equation to the other.
`Third: to balance the equation, coefficients are placed in front of the
`chemical formulas to indicate different quantities of reactants and prod-
`ucts, so that the same number of atoms of each type appears on both
`sides of the equation. To decide what coefficients to try first, it is often
`convenient to focus attention on the molecule with the most atoms, in
`this case GsHls- This molecule contains 80, all of which must end up in
`C02 molecules. Therefore, we place a coefficient 8 in front of 002.
`Similarly, the 18H end up as QHZO. At this stage the equation reads
`
`ch18 + 02 —> 8002 + 9H20
`
`[3.6]
`
`Although the C and H atoms are now balanced, the O atoms are not;
`there are 250 atoms among the products but only 2 among the reactants.
`It takes 12.502 to produce 250 atoms among the reactants:
`
`08ng + 12.50, —> 8002 + 9H20
`
`[3.7]
`
`However, this equation is not in its most conventional form, because it
`contains a fractional coefficient. Therefore, we must go on to the next
`step.
`Fourth, for most purposes a balanced equation should contain the
`smallest possible whole-number coefficients. Therefore, we multiply each
`side of the equation above by 2, removing the fraction and achieving the
`following balanced equation:
`
`208ng + 2502 —1—> 16002 + 18H20
`I
`
`[3.3]
`
`16G, 36H, 500
`
`Reactants
`
`I
`|
`I
`
`160, 36H, 500
`
`Products
`
`The atoms are inventoried below the equation to show graphically that
`the equation is indeed balanced. You might note that although atoms
`Page 6 of 12
`are conserved, molecules are not—the reactants contain 27 molecules
`while the products contain 34. All in all, this approach to balancing
`equations is largely trial and error. It is much easier to verify that an
`
`3.2 CHEMICAL EQUATIONS
`
`61 '
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`equation is balanced than actually to balance one, so practice in balanc-
`ing equations is essential.
`It should also be noted that the physical state of each chemical in a
`chemical equation is often indicated parenthetically using the symbols
`(g), (l), (s), and (aq) to indicate gas, liquid, solid, and aqueous (water)
`solution, respectively. Thus the balanced equation above can be written
`
`208H18(1) +2502(g) —> 16COz(g)+18H20(l)
`
`.
`
`[3.9]
`
`Sometimes an upward arrow (1‘) is employed to indicate the escape of a
`gaseous product, whereas a downward arrow (l) indicates a precipitating
`solid (that is, a solid that separates from solution during the reaction).
`Often the conditions under which the reaction proceeds are indicated
`above the arrow between the two sides of the equation. For example, the
`temperature or pressure at which the reaction occurs could be so indi-
`cated. The symbol A is often placed above the arrow to indicate the
`addition of heat.
`'
`
`!
`
`SAMPLE EXERCISE 3.]
`
`Balance the following equation:
`
`Na(s) + H200) ——> NaOH(aq) + H2(g)
`
`Solution: A quick inventory of atoms reveals that
`there are equal numbers of Na and O atoms on
`both sides of the equation, but that there are two H
`atoms among reactants and three H atoms among
`products. To increase the number of H atoms
`among reactants, we might place a coefficient 2 in
`front of H20:
`
`Na(s) + 2H20(1) —> NaOH(aq) + H2(g)
`
`Now we have four H atoms among reactants but
`only three H atoms among the products. The H
`
`anced.
`
`atoms can be balanced with a coefficient 2 in front
`of NaOH:
`
`Na(s) + 2H20(l) ——»
`
`2NaOH(aq) + H2(g)
`
`If we again inventory the atoms on each side of the
`equation, we find that the H atoms and O atoms
`are balanced but not the Na atoms. However, a
`coefficient 2 in front of Na gives two Na atoms on
`each side of the equation:
`
`2Na(s) + 2H20U) —9 2NaOH(aq) + H2(g)
`
`If the atoms are inventoried once more we find two
`Na atoms, four H atoms, and two 0 atoms on each
`side ofthe equation. The equation is therefore bal-
`
`3.3 CHEMICAL
`
`REACTIONS
`
`Our discussion in Section 3.2 focused on how to balance chemical equa-
`tions given the reactants and products for the reactions. You were not
`asked to predict the products for a reaction. Students sometimes ask how
`the products are determined. For example, how do we know that sodium
`metal (Na) reacts with water (H20) to form H2 and NaOH as shown in
`Sample Exercise 3.1? These products are identified by experiment. As
`the reaction proceeds, there is a fizzing or bubbling where the sodium is
`in contact with the water (if too much sodium is used the reactionds
`quite violent, so small quantities would be used in our experiment). If
`the gas is captured, it can be identified as H2 from its chemical and
`physical properties. After the reaction is complete, a clear solution re—
`Page 7 of 12
`mains. If this is evaporated to dryness, a white solid will remain. From its
`properties this solid can be identified as NaOH. However, it is not neces—
`sary to perform an experiment every time we wish to write a reaction. We
`
`3 STOICHIOMETRY
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`can predict what will happen if we have seen the reaction or a similar
`one before. So far we have seen too little chemistry to predict the prod—
`ucts for many reactions. Nevertheless, even now you should be able to
`make some predictions. For example, what would you expect to happen
`when potassium metal is added to water? We have ‘just discussed the
`reaction of sodium metal with water, for which the balanced chemical
`equation is
`
`2Na(s) + 2HZO(Z) —> 2NaOH(aq) + H2(g)
`
`[3.10]
`
`Because sodium and potassium are in the same family of the periodic
`table (the alkali metal family, family 1A), we would expect them to
`behave similarly, producing the same types of products. Indeed,
`this
`prediction is correct, and the reaction of potassium metal with water is
`
`2K(s) + 2H20(l) a 2KOH(aq) +H2(g)
`
`[3.11]
`
`You can readily see that it will be helpful in your study of chemistry if
`you are able to classify chemical reactions into certain types. We have
`just considered two examples of a type we might call reaction of an
`active metal with water. Let’s briefly consider here a few of the more
`important and common types you will be encountering in your labora—
`tory work and in the chapters ahead.
`
`Combustion in Oxygen
`
`We have already encountered three examples of combustion reactions:
`the combustion of carbon, Equation 3.1; of methane, Equation 3.4; and
`of octane (08H18), Equation 3.8. Combustion is a rapid reaction that
`usually produces a flame. Most of the combustions we observe involve 02
`as a reactant. From the examples we have already seen it should be easy
`to predict the products of the combustion of propane C3H8. We expect
`that combustion of this compound would lead to carbon dioxide and
`water as products, by analogy with our previous examples. That expecta-
`tion is correct; propane is the major ingredient in LP (liquid propane)
`gas, used for cooking and home heating. It burns in air as described by
`the balanced equation
`
`CsHs(g) + 502(g) —> 3002(g) + 4111200)
`
`[3-12]
`
`If we looked at further examples, we would find that combustion of
`compounds containing oxygen atoms as well as carbon and hydrogen
`(for example, CHaOH) also produces CO2 and H20.
`
`Acids, Bases, and Neutralization
`
`Acids are substances that increase the H+ ion concentration in aqueous
`solution. For example, hydrochloric acid, which we often represent as
`Page 8 of 12
`HCl(aq), exists in water as H+(a‘q)‘ and Cl_(aq) ions. Thus the process of
`dissolving hydrogen chloride in water to form hydrochloric acid can be
`represented as follows:
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`

`HCl(g) 2%) HCl(aq)
`
`or
`
`“
`
`(3-13]
`
`HCl(g) —O> H+(aq) + c1-(aq)
`
`The H20 given above the arrows in these equations is to remind us that
`the reaction medium is water. Pure sulfuric acid1s a liquid; when it
`dissolves1n water it releases HJrions in two successive steps:
`
`H20
`H2804!) —_+ H+(aq) +Hso4—(aq)
`
`‘
`
`[3.14]
`
`Hsor<aq> 3% WW) + SOs-(cw)
`
`[3.151
`
`Thus, although we frequently represent aqueous solutions of sulfuric
`acid as H2$040141), these solutions actually contain a mixture of H+(aq)
`HSO42(aq), and SO42—(aq).
`.
`Bases are compounds that1ncrease the hydroxide10H, OH,concen—
`tration in aqueous solution. A base such as sodium hydroxide does this
`because it is an ionic substance composed of Na+ and OH“ ions. When
`NaOH dissolves in water, the cations and anions simply separate in the
`solution:
`
`NaOH(s) Hi9 Na+(aq) + OH-(aq)
`
`[3.16]
`
`Thus, although aqueous solutions of sodium hydroxide might be written
`as NaOH(aq), sodium hydroxide exists as Na+(aq) and OH“(aq) ions.
`Many other bases such as Ca(OH)2 are also ionic hydroxide compounds.
`However, NH3 (ammonia) is a base although it is not a compound of this
`sort.
`
`It may seem odd' at first glance that ammonia is a base, because it
`contains no hydroxide ions. However, we must remember that the defi-
`nition of a base is that it‘z'nmgases the concentration of OH’ ions in water. .
`Ammonia does this by a reaction with water. We can represent the dis-
`solving of ammonia gas in water as follows:
`
`NH3(g) + H200) —,> NHJW) + OH’(aq)
`
`[3-17]
`
`Solutions of ammonia in water are often labeled ammonium hydroxide,
`NH4OH, to remind us that ammonia solutions are basic. (Ammonia is
`referred to as a weak base, which means that not all the NH3 that dis—
`solves in water goes on to form NH;r and OH‘ ions; but that is a matter
`for Chapter 15 and need not concern us here.)
`Acids and‘bases are among the most important compounds in indus—
`try and in the chemical laboratory. Table 3.1 lists several acids and- bases
`and the amount of each compound produced in the United States each
`year. You can see that these substances are produced in enormous quan-
`Page 9 of 12
`tities.
`Solutions of acids and bases have very different properties. Acids have
`
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`64
`
`3 STOICHIOMETRY
`
`

`

`TABLE 3.1 U.S. production of some acids
`and bases, 1982
`
` Compound Formula Annual production (kg)
`
`
`Acids:
`
`
`
`Sulfuric
`Phosphoric
`Nitric
`Hydrochloric
`Bases: 7
`
`H2504
`H3P04
`I-INO3
`HCl
`
`3.0 X 1010
`7.7 X 109
`6.9 X 109
`2.3 X 109
`
`8.3 X 109
`NaOH
`Sodium hydroxide
`1.3 X 1010
`Ca(OH)2
`Calcium hydroxide
`
`
`
`NH3 ,Ammonia 1.4 X 1010
`
`a sour taste, whereas bases have a bitter taste. * Acids can change the
`colors of certain dyes in a specific way that differs from the effect of a base.
`For example, the dye known as litmus is changed from blue to red by an
`acid, and from red to blue by a base. In addition, acidic and basic
`solutions differ in chemical properties in several important ways. When a
`solution of an acid is mixed with a solution of a base, a neutralization
`reaction occurs. The products of the reaction have none of the character-
`istic properties of either the acid or base. For example, when a solution of
`hydrochloric acid is mixed with precisely the correct quantity of a so-
`dium hydroxide solution, the result is a solution of sodium chloride, a
`simple ionic compound possessing neither acidic nor basic properties. (In
`general, such ionic products are referred to as salts.) The neutralization
`reaction can be written as follows:
`
`HCl(aq) + NaOH(aq)
`
`_—> H200) + NaCl(aq)
`
`[3.18]
`
`When we write the reaction as we have here, it is important to keep in
`mind that the substances shown as (aq) are present in the form of the
`separated ions, as discussed above. Notice that the acid and base in
`Equation 3.18 have combined to form water as a product. The general
`description of an acid-base neutralization reaction in aqueous solution,
`then,
`is that an acid and base react
`to firm a salt and water. Using this
`general description we can predict the products formed in any acid-base
`neutralization reaction.
`
`*Tasting chemical solutions is, of course, not a good practice. However, we have all had acids
`such as ascorbic acid (vitamin C), acetylsalicylic acid (aspirin), and citric acid (in citrus fruits) in our
`mouths, and we are familiar with the characteristic sour taste. It differs from the taste of soaps,
`which are mostly basic.
`
`SAMPLE EXERCISE 3. 2
`
`the cation of the base, Ba(OH)2, and the anion of
`
`Write a balanced equation for the reaction of hy~
`drobromic acid, HBr, with barium hydroxide,
`Ba(OH)2.
`
`Solution: The products of any acid-base reaction
`are a salt and water. The salt is that formed from
`Page 10 of 12
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`3.3 CHEMICAL REACTIONS
`
`65
`
`

`

`2HZO(Z) + BaBr2(aq)
`
`To balance the equation we must provide two mol-
`ecules of HBr to furnish the two Br‘ ions and to
`
`supply the two H+ ions needed to combine with the
`two OH‘ ions of the base. The balanced equation is
`thus
`
`2HBr(aq) + Ba(OH)2(04) fl
`
`the acid, HBr. The charge on the barium ion is 2 +
`(see Table 2.5), and that on the bromide ion is 1—.
`Therefore, to maintain electrical neutrality, the for-
`mula for the salt must be BaBrz. The unbalanced
`equation for the neutralization reaction is therefore
`
`HBr(aq) + Ba(OH)2(aq) —>
`H200) + BaBrzOlq)
`
`Precipitation Reactions
`
`One very important class of reactions occurring in solution is the precipi—
`tation reaction, in which one of the reaction products is insoluble. We
`will concern ourselves in this brief introduction with reactions between
`acids, bases, or salts in aqueous solution. As a simple example, consider
`the reaction between hydrochloric acid solution and a solution of the salt
`silver nitrate, AgNO3. When. the two solutions are mixed, a finely di-
`vided white solid forms. Upon analysis this solid proves to be silver
`chloride, AgCl, a salt that has a very low solubility* in water. The reac—
`tion as just described can be represented by the equation
`
`HCl(aq) + AgNO3(aq) a AgCl(s) + HNO3(aq)
`
`[3.19]
`
`The formation of a precipitate in a chemical equation may be repre-
`sented by a following (5), by a downward arrow following the formula for
`the solid, or by underlining the formula for the solid. You are reminded
`once again that substances indicated by (aq) may be present in the solu-
`tion as separated ions.
`The following equations provide further examples of precipitation
`reactions:
`
`Pb(NO3)2(aq) + NaZCrO4(aq) ——> K PbCrO4(s) + 2NaNO3(aq)
`
`CuC12(aq) + 2NaOH(aq) ——9 Cu(OH)2(s) + 2NaCl(aq)
`
`[3.20]
`
`[321]
`
`Notice that in each equation the positive ions (cations) and negative ions
`(anions) exchange partners. Reactions that fit this pattern of reactivity,
`whether they be precipitation reactions, neutralization reactions, or re-
`actions of some other sort, are called metathesis reactions (muh—TATH—
`uh—sis; Greek, “to transpose”).
`
`*Solubility will be considered in some detail in Chapter 11. It is a measure of the amount of
`substance that can be dissolved in a given quantity of solvent (see Section 3.9).
`
`SAMPLE EXERCISE 3.3
`
`When solutions of sodium phosphate and barium
`nitrate are mixed; a precipitate of barium phos-
`phate forms. Write a balanced equation to describe
`the reaction.
`
`Solution: Our first task is to determine the formu—
`
`Na+ and N03“, remain in solution and are repre—
`
`las of the reactants. The sodium ion is Na+ and the
`
`phosphate ion is PO43_; thus sodium phosphate is
`Na3PO4. The barium ion is Ba2+ and the nitrate
`ion is NO3_; thus barium nitrate is Ba(NO3)2. The
`Ba2+ and P043“ ions combine to form the barium
`Page 11 of 12
`phosphate precipitate, Ba3(PO4)2. The other ions,
`
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`

`

`sented as NaN03(aq). The unbalanced equation for
`the reaction is thus
`
`Na3P04(aq) + Ba(NO3)2(aq)
`333(PO4)2(5) + NaN03(aq)
`._
`3— I
`o
`n
`1
`
`135552535 $5353.11:33.3: .3212?SLZiaifiJmh:
`
`’
`units in balancing the equation. There are two
`
`(P04) units on the right, so we place a coefficient 2
`
`in front of Na3PO4. This then gives six Na atoms on
`the left, necessitating a coefficient of 6 in front of
`NaN03. Finally, the presence of six (N03) units on
`the right requires a coefficient of 3 in front of
`Ba(NO3)2:
`
`2Na3P04<aq> + 3Ba<N03)2<aq> —>
`
`Ba3<Posz<s> + 6NaNo3<aq>
`
`3.4 ATOMIC
`AND MOLECULAR
`WEIGHTS
`
`A balanced equation implies a quantitative relation between the reac-
`tants and the products involved in a chemical reaction. Thus complete
`combustion of a molecule of 0ng8 requires exactly five molecules of 02,
`no more and no less, as shown in Equation 3.12. Although it is not
`possible to count directly the number of molecules of each type in any
`reaction, this count can be made indirectly if the mass of each molecule
`is known. Indeed, this indirect approach is the one taken to obtain quan—
`titative information about the amounts of substances involved in any
`chemical transformation. Therefore, before we can pursue the quantita-
`tive aspects of chemical reactions further, we must explore the concept of
`atomic and molecular weights.
`(
`
`Dalton’s atomic theory led him and other scientists of the time to a new
`problem. If it is true that atoms combine with one another in the ratios of
`small whole numbers to form compounds, what are the ratios with which
`they combine? Atoms are too small to be measured individually by any
`means available in the early nineteenth century. However, if one knew
`the relative masses of the atoms, then by measuring out convenient quan—
`tities in the laboratory, one could determine ‘the relative numbers of
`atoms in a sample. Consider a simple analogy: Suppose that oranges are
`on the average four times heavier than plums; the number of oranges in
`4-8 kg of oranges will then be the same as the number of plums in 12 kg of
`plums. Similarly, if you knew that oxygen atoms were on the average 16
`times more massive than hydrogen atoms, then you would know that the
`number of oxygen atoms in 16 g of oxygen is the same as the number of
`hydrogen atoms in 1 g of hydrogen. Thus the problem of determining
`the combining ratios becomes one of determining therelative masses of
`the atoms of the elements.
`
`This is all very well, but there was great difficulty in getting started.
`Since atoms and molecules can’t be seen, there was no simple way to be
`sure about the relative numbers of atoms in (my compound. Dalton
`thought that the formula for water was HO. However, the French scien-
`tist Gay—Lussac showed in a brilliant set of measurements that it re-
`quired two volumes of hydrogen gas to react with one volume of oxygen
`to form two volumes of water vapor. This observation was inconsistent
`with Dalton’s formula for water. Furthermore, if oxygen were assumed to
`Page 12 of 12
`be a monatomic gas, as Dalton did, one could obtain two volumes of
`water vapor only by splitting the oxygen atoms in half, which of course
`violates the concept of the atom as indivisible in chemical reactions.
`
`3.4 ATOMIC AND MOLECULAR WEIGHTS
`
`.
`
`'
`
`67
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`