Third Edition
`
`CHEMISTRY
`
`The Central Science
`
`
`
`A:j;§_-.?__,_
`
`THEODORE L. BROWN
`
`University of Illinois
`
`H. EUGENE LEMAY. JR.
`
`University of Nevada
`
`Prentice-Hill, Inc.
`
`Englewood Cliffs, NJ. O7632
`
`1 of 12
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`
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`Librmy qf Congress Cataloging in Publication Data
`
`Brown, Theodore L.
`Chcrniatry: the central science.
`
`fncludcs index.
`1. Cllcmistry.
`(date).
`II. Title.
`QD31.2.B78
`1935
`ISBN u—13-12a95n-0
`
`I. LcMay, I-I. Eugcnc (Harold Eugcnc),
`
`540
`
`34-3413
`
`Development adnor: Flayrnond Mulianay
`Editorial/production supervision: Karen ‘J. Clarnments
`Interior design: Lava»-I E Lavavl
`An direction and cover design: Janet Sch:-nld
`Manufacturing buyer: Raymund Keeling
`Page layout: Gall Gollls
`Cover photograph: "Ha|nbow" (@ Gaol‘! Game, The Image Bank)
`
`@ 1935. 1931, 1977 by Prentice-I-lall, Inc., Englewood Cliffs, New jersey 07532
`
`Ail rights rmrwd, No part qf this book may bx
`refimdumtifi
`in aryzfnnn or [gr any means,
`wI't1'm:-It permfa-'.r£an in writing fiam Me publisher.
`
`Printed in the United States of America
`
`1098765432
`
`ISBN D-1.El-lE'E’:"|EilJ-[I
`
`III.
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`Stoichiomelry
`
`3.1 LAW
`
`OI‘ GOIISIIVNITON
`OF MA$S
`
`I 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 stoichiolnetry (pronounced stoy~ltey-AHM-uh-tree), a word
`derived from the Greek words .staz'chez'on
`(“element”) and matron
`(“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 new 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 irnplications 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 = rm.-3 (E is energy, in is mass, and e is the speed of light). We will find that whenever an
`object laser; 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
`reaclions them are detectable changes in mass.
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`3.2 GHIHIGII.
`IOIJATIOIS
`
`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-
`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 clumps. Of course, if we expend
`enough energy, we can bring oi? 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 (CO2). This reaction is represented as
`
`o+o,
`
`-—-—> Go,
`
`[‘J.l]
`
`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
`(GHQ, the principal component of natural gas, burns and produces car-
`bon dioxide (CO2) and water (H20). The combustion is “supported by”
`oxygen (02), meaning that oxygen is involved as a reactant. The unbal-
`anced equation is
`
`on, + 0, —> co, + 11,0
`
`[as]
`
`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 H20, indicating
`
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`formation of two molecules of water, there will be four hydrogens on
`each side of the equation:
`
`CH4 -1- O2 ~—s CD2 + 21-I20
`
`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 I-I20 to H202, changes
`the identitycf the chemical involved. The substance I-I202, hydrogen
`peroxide, is quite different from water. The subscripts in the chemir:alfimflu-
`‘ins should never be changes! 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; 21130 means two molecules of water,
`31-I20 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:
`
`ctr, + 20,
`
`.—s cc, + sH,o
`
`[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 (C,_,_H,3), a component of gasoline, produces 002 and H20. The
`balancedchemicalequationforthisreaction canbe determinedby using
`
`the following four steps.
`First, the reactants and products are written in the unbalanced equa-
`tion
`
`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.
`
`Chemical
`symbol
`
`Meaning
`
`Composition
`
`H 0
`2
`
`On: molecule
`of wan“
`
`-
`
`i
`Two H atoms and one 0 atom
`
`FIGURE 3.1 Illustration of the dilfercnce in
`meaning between a subscript in a chemical for-
`mole. and a coefficient in front of the formula.
`
`Notice that the number of atoms of each type
`(listed under composition) is obtained by mu1ti-
`plying the coefficient and the subscript assoc1-
`ated with each element in the formula.
`
`EH20
`
`H202
`
`-pm, ,,,,,gm,1¢,,
`flfwmn
`
`‘
`
`F “H t m dl
`°
`E °
`a“ W"
`
`O t
`a ma
`
`om, malecuk
`of hydrogen
`1:H:r0xid==
`
`.
`
`Two H atoms and two 0 atoms
`
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`ta),
`,0?
`‘3'..n‘"’o'
`
`Onemcthane + Two oxygen
`molecule
`molecules
`
`_’ Dru-. carbon
`dioxide molecule
`
`4, Two water
`molecules
`
`FIGURE 3.2 Balanced chemical equation for the com-
`bustion of-CH4. The drawings of the molecules involved
`call attention to the conservation of atoms through the
`reaction.
`'
`
`CH4
`1 C
`*1 H
`
`+
`
`"
`
`292
`H 0)
`
`G02
`1 G
`2 O
`
`-4-
`
`EH20
`2 0
`4 H
`
`Second, the number of atoms of each type on each side of the equation
`is determined. In the reaction above there are BC}, 18H, and 20 among
`the reactants, and 1C, 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.
`Tizira’, 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 coeflicients to try First, it is often
`convenient to focus attention on the molecule with the most atoms, in
`this case CSHIB. This molecule contains BC, all of which must end up in
`CD2 molecules. Therefore, we place a coefficient 8 in front of 002.
`Similarly, the 13H end up as QHQO. At this stage the equation reads
`
`CBHH, + O, ——> 800, + 911,0
`
`[3,5]
`
`Although the G 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:
`
`0,1-1,, + 12.50,, ——.> sco, + 911,0
`
`'33}
`
`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:
`
`15G, 361-1, 500
`
`16G, 36H, 500
`
`1500, + 1sH,o
`
`(3.31
`
`
`
`Rcactants
`
`Products
`
`The atoms are inventoried below 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
`
`3.2 c:HeMlcw. 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 parcnthetically using the symbols
`(g), (I), (5), and (rig) to indicate gas, liquid, solid, and aqueous (water)
`solution, respectively. Thus the balanced equation above can be written
`
`2CsH1s(i) -1- 9502(3) —"
`
`15C02(s) + I8Hz0UJ
`
`I3-‘ll
`
`Sometimes an upward arrow (T) is employed to indicate the escape of a
`gaseous product, whereas a downward arrow
`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 E.i.'ERC.‘lSE 3.!
`
`Balan.ce the following equation:
`
`Nam + H2O(l) —> Naontaqi + Hag)
`
`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 coeflicient 2 in
`Front of H20:
`New + 2H=0<0 —> Na0H<-49> + H2421
`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
`f N CH:
`3‘
`Na(.s‘) + 21-I,,o(z) —> 2NaOH(mI) + H,,(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
`coeflicient 2 in front of Na gives two Na atoms on
`each side of the equation:
`ENROHGEQ) + H (g)
`2N3“) + 2H Ow
`2
`2
`i~ilha‘i§.i?§f’Fsi?“£1‘1ZfS‘.§1il.f‘§‘3..332.”6°§?s‘§s’i’idcZ2'§
`side of -the equation. The equation is therefore bal-
`
`:3‘
`
`-‘I-3 QHIHICAI.
`REICTIOIIS
`
`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 fiazing 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 captured, it can be identified as H2 from its- chemical and
`physical properties. After the reaction is complete, a clear solution re-
`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- STOIGHIOMETFW
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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(.«)+2H,o(z) ——3- 2NaOH(aq)+H2(g)
`
`|s.m|
`
`Because sodium and potassium are in the same family of the periodic
`table (the alkali metal family, family IA), 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(.v) + 21-12007)
`
`-——3-
`
`2KOH(aq) + I-I2(g)
`
`['.H I]
`
`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. I..et’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 ouyuun
`
`We have already encountered three examples of combustion reactions:
`the combustion of carbon, Equation 3.1; of methane, Equation 3.4-; and
`of octane (C131-I18), 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 C3H3, 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
`
`Gsflslgi 4' 502$) 1"‘
`
`3GO2l£l + "=l!HsDlZl
`
`[312]
`
`If we looked at further examples, we would find that combustion of
`compounds containing oxygen atoms as well as carbon and hydrogen
`(for example, CI-ISOH) also produces CO2 and H20.
`
`Anllln, Inn, and Ihutullntlon
`
`Acids are substances that increase the H‘* ion concentration in aqueous
`solution. For example, hydrochloric acid, which we often represent as
`HC‘.1(cq), exists in water as I-I+(e,q) and Gl‘(aq) ions. Thus the process of
`dissolving hydrogen chloride in water to form hydrochloric acid can be
`represented as follows:
`
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`
`
`'1
`
`HCll(g) 5% HCl(ag)
`
`or
`
`[£1.13]
`
`Hole)
`
`-*1’-9+ H+<ao + area)
`
`The H20 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 I-1*‘ ions in two successive steps:
`
`H2804!)
`
`”"=—":» H+<aq)+ HS04'(aq)
`
`HSCJ,,‘(aq) Eli H+(aq) + so,2—(sq)
`
`I3-H1
`
`[3.l5]
`
`Thus, although we frequently represent aqueous solutions of sulfuric
`acid as H,.,SO,,(aq), these solutions actually contain a mixture of I-I"'(aq),
`HSC),,3'"(aq), and SO42"(eq).
`.
`Bases are compounds that increase the hydroxide ion, 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
`NaC)I-I dissolves in water, the cations and anions simply separate in the
`solution:
`
`NflOH(S) E» Na"(aq) + DH‘(aq)
`
`[116]
`
`Thus, although aqueous solutions of sodium hydroxide might be written
`as NaOI-I(aq), sodium hydroxide exists as Na+(eq) and OH‘(aq) ions.
`Many other bases such as Ga(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 increases 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:
`
`N1-Isis) +H,o(z) —_> NH,+(aq) + OH‘(eq)
`
`[am
`
`Solutions of ammonia in water are often labeled ammonium hydroxide,
`NI-I4OH, 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 NH4+ 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-
`tities.
`
`Solutions of acids and bases have very different properties. Acids have
`
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`TABLE 3.1 US. production of some acids
`and bases, 1982
`
`
`Compound
`Fonnull
`Annual production (kg)
`Acids:
`
`Sulfuric
`Phosphoric
`Nitric
`Hydrochloric
`Bases;
`
`H230,
`I-l3PC|_,
`HND3
`I-I01
`
`3_o X 1010
`7.7 X [D9
`5.9 )4 10“
`2.3 x if)”
`
`8.3 X 109
`NaDH
`Sodium hydroxide
`1.3 x 1010
`eaten),
`Calcium hydroxide
`
`
`NI-1,,Ammonia 1.4 X 101”
`
`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 son
`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:
`
`HC.l(aq) +NaOH(aq)
`
`-—> H,o(z) + NaC1l(aq)
`
`[3.!B]
`
`When we write the reaction as we have here, it is important to keep in
`mind that the substances shown as (sq) 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 neutraliaation 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. I-Iowever, we have all had acids
`such as ascorbic acid (vitamin C), acctylsalloylic acid (aspirin), and citric acid (in citrus fruits) in our
`mouths, and we are Farniliar with the characteristic sour taste. It difiers from the taste of soaps,
`which are mostly basic.
`
`SAMPLE EXERCISE 3.2
`
`the cation of the base, Ba(CIH)3, and the anion of
`
`Write a balanced equation for the reaction of hy-
`drobromic acid, H]3r, with barium hydroxide,
`Ba(DH)2.
`
`Solution: The products of any acid-base reaction
`are a salt and water. The salt is that formed from
`
`3.3 CHEMICAL REACTIONS
`
`CI
`
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`the acid, I-lBr. 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 BaB1-2. The unbalanced
`equation for the neutralization reaction is therefore
`
`I-IBr(aq) + Ba(OH).,;(aq)
`
`:3-
`
`HZOU) + BaBr2(aq)
`
`To balance the equation we must provide two mol-
`ecules of I-IBr to furnish the two Br‘ ions and to
`supply the two H4’ ions needed to combine with the
`two OH“ ions of the base. The balanced equation is
`thus
`
`2HBr[aq) + Ba(OH)2(aq)
`
`-—-—-3-
`
`2H2O(l) + BaBr2(aq)
`
`Fnolpltlllun Iucllonc
`
`One very important class of reactions occurring in solution is the precipi-
`tation rcaction, 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. Wlien the two solutions are mixed, a finely di-
`vided white solid forms. Upon analysis this solid proves to be silver
`chloride, Agfll, a salt that has a very low solubility* in water. The reac-
`tion as just described can be represented by the equation
`
`HCll(.:zg) + AgNO3(aq) ——--3» AgC-l(r) + HNO3(oq)
`
`[EH9]
`
`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 (sq) may be present in the solu-
`tion as separated ions.
`The following equations provide further examples of precipitation
`reactions:
`
`Pb(N0a)2(aq) + Nr'3-gClrO,,(aq) —> _Pl)C.irC)4(.5') + 2NaND3(czq)
`
`CluCl2(»:zq) + 'ZNaOH(cq) _> C'»u(OH)2(s) + 2NaC1(aq)
`
`[120]
`
`[321]
`
`Notice that in each equation the positive ions (cations) and negative ions
`(anions) exchange partners. Reactions that lit this pattern of reactivity,
`whether they be precipitation reactions, neutralization reactions, or re-
`actions of some other sort, are called metathesis reactions (rnuh-TATH-
`uh-sis; Greek, “to transpose”).
`
`*Soluhility will be consiclencd 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).
`
`NA MPLE 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 tasl-: 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
`Na3PO,,. The barium ion is Ba“ and the nitrate
`ion is N03‘; thus barium nitrate is Ba(l‘~lO3)2. The
`Ba“ and P043“ ions combine to form the barium
`phosphate precipitate, Ba3(PO_,)2. The other ions,
`
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`
`

`
`sentecl as NaNO3(eq). The imbalanced equation for
`the reaction is thus
`
`Nasposlflfl + Ba(NO3lB2§:'E%Ofi)2(3) + NaNO3(aq)
`Because the N03“ and P045‘ ions maintain their
`identity through the reaction, we can treat them as
`units in balancing the equation. There are two
`
`Ba(NCJ_.,)2:
`
`2Na3PO4(aq) + 3Ba(NO3)2(aq) —)
`Baslposlsl-ll + 5N3ND3(¢3'§’l
`
`(P04) units on the right, so we place a coefficient 2
`
`in front of Na3PO,,. This then gives six Na atoms on
`the left, necessitating a cocfficient of 5 in front of
`NaNO . Finally the presence of six (NO ) units on
`
`the right requities a coefficient of 3 id front of
`
`3.4 ATOMIC
`AID HOIJGIILAII
`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 C'.3HB 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
`43 kg of oranges will then be the same as the number of plums in 12 kg of
`plums. Similarly, if you know that oxygen atoms were on the average 15
`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 any 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
`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.
`
`3.1 ATOMIC AND MOLECULAR WEIGHTS
`
`IT
`
`12 of 12
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