`
`The Central Science
`
`THEODORE L. BROWN
`
`University of Illinois
`
`H. EUGENE LEMAY, JR.
`
`University of Nevada
`
`Prentice-Hall, Inc.
`
`Englewood Cliffs, N.J. 07632
`
`BMW1078
`Page 1 of 12
`
`
`
`Librat:y of Coni.rress Catalogi11g in Publicatio11 Data
`
`Brown, Theodore L.
`Chemistry: the central science.
`
`Includes index.
`1. Chemistry.
`II. Title.
`(date).
`QP31.2.B78 1985
`ISBN 0-13-128950-0
`
`I. LeMay, H. Euge'ne (Harold Eugene),
`
`540
`
`84-8413
`
`Development editor: Raymond Mullaney
`Edllorlal/ productlon supervision: Karen ·J . Clemments
`Interior design: Levavl & Levavl
`Art direction and cover design: Janet Schmid
`Manufacturing buyer: Raymond Keating
`Page layout: Gall Collls
`Cover photograph: " Rainbow'' (@ Geoff Gove, The Image Bank)
`
`© 1985, 1981, 1977 by Prentice-Hall, Inc., Englewood C liffs, New Jersey 07632
`
`A It rig his reserved. No part of this book may be
`reproduced, in at!)i fam1 or i!J any means,
`without pem1is.rion irl writing .from the publisher.
`
`Printed in the United States of America
`
`10 9 8 7 6 5 4 S 2
`
`ISBN D-13-128950- 0 01
`
`Prentice-Hall International, Inc., Lo11do11
`Prentice-H all of Aush·alia Pty. Limited, Sydney
`Editora Prentice-Hall do Brasil, Ltda., Rio de Janeiro
`Prentice-Hall Canada Inc., Toronto
`Prentice-Hall of India Private Limited, New Delhi
`Prentice-Hall of J apan, Inc., 1bkyo
`Prentice-Hall of Southeast Asia Pte. Ltd., Singapore
`Whitehall Books Limited, Welli11gto11, New Zeala11d
`
`BMW1078
`Page 2 of 12
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`
`Stoichiometry
`
`Antoine Lavoisier ( Sec ti on 1.1) was among the first to draw conclusions
`about chemical processes fro:m 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(cid:173)
`lems involve the quantitative relationships between substances undergo(cid:173)
`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 stoicheion ("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(cid:173)
`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 = mc2 (E is energy, mis mass, and c is the speed of light). We will find that whenever an
`object loses enel'gy it loses mass, and whenever it gains ene.rgy it gains mass. These changes in mass
`ate 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.
`
`3. 1 LAW
`OF CONS RVATION
`OF MASS
`
`58
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`
`
`ICAL
`3.2 CHE
`QUATIONS
`
`ronment. They may undergo chemical changes or remain ,inactive; 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(cid:173)
`ing the number of iron atoms on lhe planet. We may, however, be con(cid:173)
`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 Hke 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(cid:173)
`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 (CO2). This reaction is represented as
`
`[3.1]
`We read the + sign to mean "reacts with" and .the arrow as "produces."
`Carbon and oxygen are referred to as reactant and carbon dioxide as the
`product of the reaction.
`It is important t<;> keep in mind that a chemical equation is a descrip(cid:173)
`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(cid:173)
`tion, a chemical equation is a quantitative statement; it must be consis(cid:173)
`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
`(CH~), the principal component of natural gas, burns and produces car(cid:173)
`bon dioxide (CO2) and water (H20). The combustion is "supported by"
`oxygen (02), meaning that oxygen is involved as a reactant. The unbal(cid:173)
`anced equation is
`
`The reactants are shown to the left of the arrow, the products to the
`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 H 20, indicating
`
`3 2 CHEMICAL EOUAilONS
`
`59
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`formation of two molecules of water, there will be four hydrogens on
`each side of the equation:
`
`(3.31
`
`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 H 20 to H 20 2, changes
`the identity .of the chemical involved. The substance H 20 2, hydrogen
`peroxide, is quite different from water. The subscripts in the chemicalformu(cid:173)
`·tas should never be changed in balancing an equation. On the other hand, plac(cid:173)
`ing a coefficient in front of a formula merely changes the amount and not
`the identity of the substance; 2H20 means two molecules of water,
`3H 20 means three molecules of water, and so forth. Now let's continue
`balancing Equation 3. 3. There are equal numbers of carbon and hydro(cid:173)
`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 0 2 there will be equal numbers of oxygen
`atoms on both sides of the equation:
`
`[3.il
`
`The equation is now balanced. There are four oxygen atoms, four hydro(cid:173)
`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 (C8H 18), a component of gasoline, produces CO2 and H 20 . 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(cid:173)
`tion
`
`Before a chemical equation can be written the identities of the reactants
`and products must be determined. In the present example this jnforma(cid:173)
`tion was given to us in the verbal description of the reaction.
`
`Chcmic:11.I
`symbol
`r---\
`
`Meaning
`
`Composition
`
`~IOURE 3 1 lllustration of the difference in
`meaning between a subscript in a chemical for(cid:173)
`mula and a coefficient in front of the formula.
`Notice that the number of atoms of each type
`(listed under composition) is obtained by multi(cid:173)
`plying the coefficient and the subscript associ(cid:173)
`ated with each clement in the formula.
`
`6Q
`
`3 STOICHIOMETRY
`
`One molecule
`of WIiler:
`
`Two molecules
`of water:
`
`One molecule
`of hydrogen
`p eroxide:
`
`Two H ntoma and one O alom
`
`Four H nloms and lwo O atoms
`
`Two H ntoms and two O alorm
`
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`
`
`+
`
`.....
`
`+
`
`FIGURE 3.2 Balanced chemical equation for the com(cid:173)
`bustion of CH4• The drawings of the molecules involved
`call attention to the conservation of atoms through the
`I
`reaction.
`
`+
`
`One: methane + Two oxygen
`molecules
`molecule
`ca.
`U~)
`
`202
`
`(40)
`
`..... One carbon
`dioxjdc molecule
`....
`
`CO2
`
`Gg)
`
`+ Two water
`molecules
`
`+
`
`2~0
`
`(!~)
`
`Second, the number of atoms of each type on each side of the equation
`is determined. In the reaction above there arc BC, 18H, and 20 among
`the reactants, and l C, 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(cid:173)
`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 C8H 18. This molecule contains 8C, all of which must end up in
`CO2 molecules. Therefore, we place a coefficient 8 in front of CO2.
`Similarly, the 18H end up as 9H20 . At this stage the equation reads
`
`Although the C and H atoms arc 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:
`
`l3,7J
`
`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:
`
`2C8H 18 + 2502
`16C, 36H, 500
`Reactants
`
`16C02 + 18H20
`16C, 36H, 500
`Products
`
`13.8]
`
`The atoms are inventoried b elow the equation to show graphically that
`the equation is indeed balanced. You might note that although atoms
`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
`
`J.2 CttEMICAL EQUATIONS
`
`6 1
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`equation is balanced than actually to balance one, so practice in balanc(cid:173)
`ing equations is essential.
`It should also b e noted that the physical state of each chemical in a
`chemical equation is often indicated parentheticaUy 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
`
`Sometimes an upward arrow (t ) is employed to indicate the escape of a
`gaseous product, whereas a downward arrow (-1.) indicates a precipitating
`solid (that is, a solid that separates from solution during ~ e reaction).
`Often the conditions under which the reaction proceeds arc 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(cid:173)
`cated. The symbol ~ is often placed above the arrow to indicate the
`addition of heat.
`
`'U ,\f PLE EXERCISE 3. I
`
`Balance the following equation:
`Na(s) + H 20(l) ~ NaOH(aq) + H 2(g)
`
`Solution! A quick inventory of atoms reveals that
`there are equal numbers of N a 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 H 20:
`Na(s) + 2H20(l) ~ NaOH(aq) + H 2(g)
`Now we have four H a toms among reactants but
`only three H atoms among the products. The H
`
`atoms can be balanced with a coefficient 2 in front
`ofNaOH:
`Na(s) + 2H20(/) ~ 2NaOH(aq) + H 2(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) + 2H20 (l) ~ 2NaOH(aq) + Hi(g)
`If the atoms are inventoried once more we find two
`Na atoms, four H atoms, and two O atoms on each
`side of the equation. The equation is therefore bal(cid:173)
`anced.
`
`3.3 CH MICAL
`REACTIONS
`
`Our discussion in Section 3.2 focused on how to balance chemical equa(cid:173)
`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 H 2 ~cl 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 reaction. is
`quite violent, so small quantities would be used in our experiment). If
`the gas is captured1 it can be identified as H 2 from its chemical and
`physical properties. After the reaction is complete, a clear solution re(cid:173)
`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(cid:173)
`sary to perform an experiment every time we wish to write a reaction. We
`
`82
`
`3 STOICHIOMETRY
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`ca~ predict what will happen if we have seen the reaction or a similar
`one before. So far we have seen too Ii ttle chemistry to predict the prod(cid:173)
`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) + 2H20(l) ~ 2NaOH(aq) + Hig)
`
`[!5.H)J
`
`Because sodium and potassium are in the same family of the periodic
`table (the alkali metal family, family lA), 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) ~ 2KOH(aq) + H 2(g)
`
`[3.1 1 J
`
`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(cid:173)
`tory work and in the chapters ahead.
`
`Combu•tlon In 0 .Jtygen
`We have already encountered three examples of combustion reactions:
`the combustion of carbon, Equation 3.1; of methane, Equation 3.4; and
`of octane (C8H 18), Equation 3.8. Comb1:1stion is a rapid reaction that
`usually produces a flame. Most of the combustions we observe involve 0 2
`as a reactant. From the examples we have already seen it should be easy
`to predict the products of the combustion of propane C3H 8. We expect
`that combustion of this compound would lead to carbon dioxide and
`water as products, by analogy with our previous examples. That expecta(cid:173)
`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
`
`If we looked at further examples, we would find that combustion of
`compounds containing oxygen atoms as well as carbon and hydrogen
`(for example, CH30H) also produces CO2 and H 20.
`
`•clds, Bases, and Neutralhtallon
`
`Acids are substances that increase the H+ ion concentration in aqueous
`solution. For example, hydrochloric acid, which we often represent as
`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:
`
`3.3 CHEMICAL nEACTIONS
`
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`HCl(g) HiO > HCl(aq)
`
`or
`
`HCI(g) HP ) H+(aq) + Cl- (aq)
`
`[3. l ~i)
`
`The H 20 given above the arrows in these equations is to remind us that
`the reaction medium is water. Pure sulfuric acid is a liquid; when it
`dissolves in water it releases H+ ions in two successive steps:
`
`H 2S04(l) Hp > H+(aq) + HS04- (aq)
`HS04- (aq) HzO) H+(aq) + so..i2-(aq)
`
`l3,l1J
`
`f3.15]
`
`Thus, although we frequently represent aqueous solutions of sulfuric
`acid as H 2S0iaq), these solutions actually contain a mixture of H +(aq),
`HSOl- (aq), a nd S04
`2- (aq).
`Rases are compounds that increase the hydroxide ion, OH- , concen(cid:173)
`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) Hao ) 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.
`M any 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(cid:173)
`nition of a base is that it inm:_ases the concentration of OH- ions in water.
`Ammonia does this by a reaction with water. We can represent the dis(cid:173)
`solving of ammonia gas in water as follows:
`
`Solutions of ammonia in water are often labeled ammonium hydroxide,
`NH40H, to remind us that ammonia solutions are basic. (Ammonia is
`referred to as a weak base, which m eans that not all the NH3 that dis(cid:173)
`solves in water goes on to form NH/ 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(cid:173)
`try and in t he 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(cid:173)
`tities.
`Solutions of acids and bases have very different properties. Acids have
`
`3 STOICHIOMETRY
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`TABLE 3.1 U.S. production of some acids
`and bases, 1982
`
`Compound
`
`Formula
`
`Annual production (kg}
`
`Acids:
`Sulfuric
`Phosphoric
`Nitric
`Hydrochloric
`Ba.~es:
`Sodium hydroxide
`Calcium hydroxide
`Ammonia
`
`H2S0d
`HaP01
`HN03
`HCI
`
`3.0 X 1010
`7.7 X 109
`6.9 X 109
`2.3 X 109
`
`NaOH
`Ca(OH)2
`,NH a
`
`8.3 X 109
`1.3 X 1010
`1.4 X 10to
`
`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(cid:173)
`istic properties of either th e acid or b ase. For example, when a solution of
`hydrochloric acid is mixed with precisely the correct quantity of a so(cid:173)
`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 sali.s.) The neutralization
`reaction can be written as follows:
`HCl(aq) + NaOH(aq) ~ H 20(l) + NaCl(aq)
`
`[3.18]
`
`When we write tho 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 form a salt and waler. 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), acctylsalicylic acid (aspirin), and citric acid (in ci trus fruits) in our
`mouths, and we are fa.111iliar with the characteristic sour ta~te. It differs from the taste or soaps,
`which are mostly basic.
`
`SAMPLE EXERCISE 3. 2
`
`Write a balanced equation for the reaction of hy(cid:173)
`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
`the cation of the base, Ba(OH)2, and the anion of
`
`3 3 CHEMICAL REACTIONS
`
`65
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`the acid, HBr. The charge on the barium ion is 2 +
`(sec Table 2.5), and that on the bromide ion is 1- .
`Therefore, to maintain electrical neutrality, the for(cid:173)
`mula for the salt must be BaBr2. The unbalanced
`equation for the neutralization reaction is therefore
`HBr(aq) + Ba(OHh(aq) --+
`H 20(l) + BaBr2(aq)
`
`To balance the equation we must provide two mol(cid:173)
`ecules of HBr to furnish the two Br- ions and to
`supply the two 1-1+ ions needed to combine with the
`two OH- ions of the base. The balanced equation is
`thus
`2HBr(aq) + Ba(OH)2(aq) --+
`2H20(l) + BaBr2(aq)
`
`Precipitation Reaction•
`One very important class of reactions occurring in solution is the precipi(cid:173)
`tation reaction, in which one of the reaction products is insoluble. We
`will concern ourselves in this brief introduction with reactions bet-ween
`acids, bases, or salts in aqueous solution. As a simple example, consider
`the reaction bet'vl,'.een hydrochloric acid solution and a solution of the salt
`silver nitrate, AgN03• -When. the two solutions are mixed, a finely di(cid:173)
`vided white solid forms. Upon analysis this solid proves to be silver
`chloride, AgCl, a salt that bas a very low solubility* in water. The reac(cid:173)
`tion as j ust described can be represented by the equation
`HCl(aq) + AgN09(aq) ~ AgCl(s) + HN03(aq)
`
`[3.19]
`
`The formation of a precipitate in a chemical equation may be repre(cid:173)
`sented by a following (s), 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 t he solu(cid:173)
`tion as separated ions.
`The following equations provide further examples of precipitation
`reactions:
`Pb(N03)2(aq) + Na2Cr01(aq) ~ . PbCrOis) + 2NaNOg(aq)
`CuC12(aq) + 2NaOH(aq) ~ Cu(OHh(s) + 2NaCl(aq)
`Notice th at in each equation the positive ions (cations) and negative ions
`(anions) exchange partners. Reactions th at fit this pattern of reactivity,
`whether they be precipitation reactions, neutralization reactions, or re(cid:173)
`actions of some other sort, are called metathesis reactions (muh-TATH(cid:173)
`uh-sis; Greek, "to transpose").
`
`f3.2UI
`[3.2lJ
`
`*Solubility will be considered in some detail in Chapter 11. It is a measure of the amount ol'
`subslance thal can be dissolved in a given quantity of solvent (see Section 3.9).
`
`,\i i MPLE l!-'XERCl~E 3 .. 1
`
`When solutions of sodium phosphate and barium
`nitrate are mixed, a precipitate of barium phos(cid:173)
`phate forms. Write a balanced equation to describe
`the reaction.
`
`Solution: Our first task is to determine the formu-
`
`66
`
`3 STOICHIOMETRY
`
`las of the reactants. The sodium ion is Na+ and the
`phosphate ion is POl- ; thus sodium phosphate is
`Na3P04• The barium ion is Ba2+ and the nitrate
`ion is N03- ; thus barium nitrate is Ba(N03)z. The
`Ba2+ and Po/- ions combine to form the barium
`phosphate precipitate, Ba8(P04) 2. The othc1· ions,
`Na+ and N03-, remain in solution and are repre-
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`sented as NaN03(aq). The unbalanced equation for
`the rea~tion is thus
`Na3POiaq) + Ba(N03)z(aq) ~
`Bas(P04h(s) + NaN08(aq)
`3-
`ions maintain their
`Because the N03- and P04
`identity through the reaction, we can treat them as
`units in balancing the equation. There are two
`(P04) units on the right, so we place a coefficient 2
`
`in front ofNa3P01. This then gives six N a atoms on
`the left, necessitating a coefficient of 6 in front of
`NaN08• Finally, the presence of six (N03) u·nits on
`the right requires a coefficient of 3 in front of
`Ba(N03) 2 :
`2Na3P04,(aq) + 3Ba(N03)i(aq) ~
`Ba3(P04)i(s) + 6NaNOa(aq)
`
`A balanced equation implies a quantitative relation between the reac(cid:173)
`tants and the products involved in a chemical reaction. Thus complete
`combustion of a molecule of C3H 8 requires exactly five molecules of 0 2,
`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(cid:173)
`titative information about the amounts of substances involved in any
`chemical transformation. Therefore, before we can pursue the quantita(cid:173)
`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 ra tios 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(cid:173)
`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
`48 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 atom s in 1 g of hydrogen. Thus the problem of determining
`the combining ratios becomes one of determining the relative 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 a791 compound. Dalton
`thought that the formula for water was HO. However, the French scien(cid:173)
`tist Gay-Lussac showed in a brilliant set of measurements that it re(cid:173)
`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
`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 A ATOMIC ANO MOLECULAR WEIGHTS
`
`67
`
`3.4 ATOMIC
`AND MOLECULAR
`WEIGHTS
`
`I
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