`
`Third Edition
`
`CHEMISTRY
`
`The Central Science
`
`THEODORE L. BROWN
`
`University of Illinois
`
`H. EUGENE LEMAY. JR.
`
`"
`
`University of Nevada
`
`Prentice-Hell, Inc.
`
`Englewood Cliffs, NJ.
`
`07652
`
`
`
`44——s-.-_______—____=—j———
`
`n
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`
`
`Libraty qf Congress Cataloging in Publication Data
`
`Brown, Theodore. L.
`Chemistry: thc ccntral scicncc.
`
`Includes index.
`1. Chemistry.
`(date).
`11. Title.
`Quateem 1935
`ISBN 0—13.123950-0
`
`I. LeMay, H. Eugc'nc (Hamid Eugene),
`
`540
`
`34—3413
`
`Development editor: Raymond Muiieney
`Editorial/production supervision: Karen ‘J. Clemn‘mnte
`Interior design: Levevi it LeVavi
`An direction and cover design: Janet Schrnid
`Manufacturing buyer: Raymond Keeling
`Page layout: Gall Ceills
`Cover photograph: “Fialnbow” ((9') Gear! Game, The image Bank)
`
`@ 1935, 1931, [9'77 by Prentice-Hall, Inc, Engieweod Clifi's. New jermy 07532
`
`Ail rights reserved. No part qf this book may be
`refimdumi,
`in myfarm or [gr any means,
`without per-minim in writing film the publisher.
`
`Printed in the United States of America
`
`1098765432
`
`ISBN D-lEl-lEB‘IEiiJrU [IL
`
`Pmnticerflall International, Inc, Landau
`
`Prentice‘l-Iall of Australia Pty. Limited, Sydney
`Editora Prentice-Hall do Brasil, Lida, Rio de jamim
`Prentice-Hall Canada Inc., Erwin.)
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`Stoichiomelry
`
`
`
`3.1 LAW
`
`OF GONEIIVA‘I'IO’N
`0F MASS
`
`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~ltey-AHM-uh-tree), a word
`derived from the Greek words staicksz'on
`(“element”) and matron
`(“mcasure” .
`
`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 meted 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-
`
`l‘In Chapter 19, we will discuss the relationship between mass and energy summarized by the
`equation E = rat-2 (E is energy, at is mass, and t is the speed of light}. We will find that whenever an
`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 GHMAL
`“IMHO"!
`
`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 ironvcontaining 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 olT 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 (C(32). This reaction is represented as
`
`o+o.,
`
`~—-—> co2
`
`[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
`((3H,), 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 unbalr
`anced equation is
`
`on, + 0, —> co, + H,o
`
`[3.2]
`
`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 coelhcient 2 in front of H20, indicating
`
`:1 2 CHEMICAL EQUATIONS
`
`I.
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`

`m.
`
`formation of two molecules of water, there will be four hydrogens on
`each side of the equation:
`
`CH, + 0, _—s co, + 2H20
`
`|s.s]
`
`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 11202, changes
`the identityof the chemical involved. The substance H202, hydrogen
`peroxide, is quite difierent from water. The subscripts in the chemicnlfiirmu-
`’iss shank! never be changed in balancing an equation. On the other hand, plac—
`ing a coefficient in front of a formula merelyr changes the amount and not
`the identity of the substance; 21-120 means two molecules of water,
`31-120 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:
`
`on, + so,
`
`-—1 co, + 21-120
`
`[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 (GEHIB), a component of gasoline, produces C102 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
`
`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
`
`age
`2
`
`2H 0
`
`H302
`
`FIGURE 3.1 Illustration of the dilfercnce in
`mula and a cocfficrent in front ofthe formula.
`meaning between a subscript in a chemical for—
`.
`.
`Notice that the number of atoms of each type
`(listed under composition) is obtained by multi-
`plying the coefficient and the subscript MSOClr
`ated with each element in the formula.
`
`Meaning
`
`
`‘
`
`Composition
`
`I
`
`‘
`
`gpggglfwlc a TwoHatomsand one0atom
`Diwali!“ J 4 “MEMO” and mnoatms
`‘
`
`Two molecules
`
`.
`
`om: Wham:
`of hydrogen
`peroxide:
`
`.
`
`Two H atoms and two 0 atoms
`
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`‘3' a) ”or
`
`On: urethane + Two city-gen
`molecule
`molecules
`
`.—
`
`flnr. carbon
`dioxide molecule
`
`+
`
`Two water
`molecules
`
`FIGURE 3.2 Balanced chemical equation for the com-
`bustion of 0H,. The drawings of the molecules involved
`call attention to the conservation of atoms through the
`react ion.
`
`CH4
`1 C
`4 H)
`
`+
`
`"
`
`202
`40
`
`J
`
`l
`
`GO!
`1 G
`(2 o
`
`'4‘
`
`2H20
`20
`4 H)
`
`Second, the number of atoms of each type on each side of the equation
`is determined. In the reaction above there are 80, 18H, and 20 among
`the reactants, and 10, 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 CaHm- This molecule contains BC, all of which must end up in
`CC), molecules. Therefore, we place a coefficient 8 in front of 002.
`Similarly, the 18H end up as QHQO. At this stage the equation reads
`
`c,H,, + 0, ——> 800, + 911,0
`
`{3.61
`
`Although the C3 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 12.50, to produce 250 atoms among the reactants:
`
`c,H,, + 12.50, ———> 300, + 911,0
`
`p.71
`
`However, this equation is not in its most conventional form, because it
`contains a fractional coeflicient. Therefore, we must go on to the next
`step.
`Fourth, for most purposes a balanced equation should contain the
`smallest possible whole-number coeflicients. Therefore, we multiply each
`side of the equation above by 2, removing the fraction and achieving the
`following balanced equation:
`
`2c,H,, + 250,
`
`1500, + 18H,0
`
`{3.31
`
`
`
`
`150, 36H, 500
`
`160, 36H, 500
`
`Reactants
`
`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 GHEMIGAL EQUATIONS
`
`B ‘I
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`
`9
`
`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), (I), (r), and (rig) to indicate gas, liquid, solid, and aqueous (water)
`solution, respectively. Thus the balanced equation above can be written
`
`268H13(l)+ 250,(g)
`
`-—) 16002(g)+18H20(!)
`
`[an]
`
`Sometimes an upward arrow (1‘) 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 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 seasons 3.:
`Balance the following equation:
`
`Nam + 1120(2) —} NaOH(eq) + H,(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:
`
`NaOH(aq) + 143(3)
`Na(s) + 21-120“)
`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‘) + smog!) —; smonoq) + 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
`coefficient 2 in front of Na gives two Na atoms on
`each side of the equation:
`
`2Na(s)+2H20(1) —> 2NaOI-I(aq) + H2(g)
`If the atoms are inventoricd once more we find two
`Na atoms, four H atoms, and two 0 atoms on each
`side of the equation. The equation is therefore bal-
`
`3.3 QHIHICAI.
`flflfi'flflfls
`
`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 showr1 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 1-12 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- STOIOHIOMETH’Y
`
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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(r) + smog) ——i 2Na0H(aq) + H,(g)
`
`IMO]
`
`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(.i)+ smog) _—i
`
`2KOH(aq) + H,(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 laborah
`tory work and in the chapters ahead.
`
`Communal! In Dun-n
`
`We have already encountered three examples of combustion reactions:
`the combustion of carbon, Equation 3.1; of methane, Equation 3.4-; and
`of octane (OBI-113), 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 CsHsv 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
`
`Cal-13%) + 503(g) a 3002(g) + 4HZDU)
`
`[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, CHBOH) also produces C02 and H20.
`
`ADIIII, In“, Ind Mauls-Illusion
`
`Acids are substances that increase the 11+ ion concentration in aqueous
`solution. For example, hydrochloric acid, which we often represent as
`HGl(aq), exists in water as H+(alq) and Cl_(eq) 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)
`
`339—) HCl(ag)
`
`or
`
`[3.13]
`
`Halo) 1% Haas + GI-(aq)
`
`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 Hl‘ inns in two successive steps:
`
`H2804!)
`
`”'=—°> H+(aq)+ Hsoi-(aq)
`
`Hsoqu)
`
`i=9.) H+(eq) + SO42‘{oq)
`
`[3-H]
`
`[3.15]
`
`Thus, although we frequently represent aqueous solutions of sulfuric
`acid as HfiOqu), these solutions actually contain a mixture of H+(aq),
`HSOfiTaq), and SOfTaq).
`.
`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
`NaOI-I dissolves in water, the cations and anions simply separate in the
`solution:
`
`NaOI—I(r) fl) Na+(aq) + (DH—(sq)
`
`[3.16]
`
`Thus, although aqueous solutions of sodium hydroxide might be written
`as NaOI-I(aq), sodium hydroxide exists as Naflaq) and (DH—(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 add 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:
`
`NHats) +H20(l) —.’ NH,+(aq) + OH-(aq)
`
`[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
`
`3 BTDIGHIDMETFW
`
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`

`

`TABLE 3.! US. production of some acids
`and bases, 1982
`
`
`Compound
`Formula
`Annual production 0th
`Acids:
`
`Sulfuric
`Phosphoric
`Nitric
`Hydrochloric
`Bases;
`
`H2304
`1-131-"0,1
`HNDS
`H01
`
`3.0 x 1010
`7.7 X 1C)“
`6.9 x 10“
`2.3 x 109
`
`8.3 X 109
`NaOl-I
`Sodium hydroxide
`1.3 X 101“
`(infill-1).,
`Calcium hydroxide
`
`
`NH“Ammonia 1.4 x 10m
`
`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:
`
`Helm) +NaOH(aq) —-—} H200) + NaCl(aq)
`
`[3.13]
`
`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 neutralization reaction in aqueous solution,
`then,
`is that an arid and bare react
`to farm 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), acctylsaiieylic acid (aspirin), and citric acid ( in citrus fruits) in our
`mouths, and we are familiar with the characteristic sour taste. It difl‘crs from the taste of soaps,
`which are mostly basic.
`
`SAMPLE EXERCISE 3.?
`
`the cation of the base, Ba(C}H)2, and the anion of
`
`Write a balanced equation for the reaction of hy-
`drobromic acid, HBr, with barium hydroxide,
`Ba(0H),.
`
`Solution: The products of any acid-base reaction
`are a salt and water. The salt is that formed from
`
`as CHEMICAL REACTIONS
`
`85
`
`10 of 12
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`
`

`

`2H20(l) + BaBr2(aq)
`
`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
`Hana) + Ba(0H).,(sq) —>
`H,0(r) + Basr,(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 H+ ions needed to combine with the
`two OH“ ions of the base. The balanced equation is
`thus
`2HBr(aq) + Ba(0H)2(as) he
`
`Fnolplhuun Reaction.
`
`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, AgNOa. When the two solutions are mixed, a finely di-
`vided white solid forms. Upon analysis this solid proves to be silver
`chloride, AgGl, a salt that has a very low solubility* in water. The reac-
`tion as just described can be represented by the equation
`
`HCKGQ) + AgNOafizq) ——~+ AgCKs} + HN03(aq)
`
`[3.19]
`
`The formation of a precipitate in a chemical equation may be repre—
`sented by a following (3), 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 (eq) may be present in the solu-
`tion as separated ions.
`The following equations provide further examples of precipitation
`reactions:
`
`Pb(No,),(sq) + Na,oro,(ag) —> _Pboro,o) + 2NaN03(cq)
`
`CuClz(aq) + 2NaOl-I(aq) c_> ou(0H),(s) + 2NaGl(ag)
`
`[3.201
`
`[3.21 I
`
`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”).
`
`*Solubilily 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).
`
`SA MPLE EXERCISE 3.3
`
`Na+ and N03_, remain in solution and are repre-
`
`las of the reactants. The sodium ion is Na*' and the
`
`phosphate ion is P0431 thus sodium phosphate is
`NaaPOd. The barium ion is Ba2+ and the nitrate
`ion is 5103—; thus barium nitrate is Ba(N03)2. The
`Ba2+ and P043' ions combine to form the barium
`phosphate precipitate, Ba3(PO4)2. The other ions,
`
`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 taslc is to determine the formu-
`
`11 of 12
`
`FORD 1879
`
`

`

`sented as NaN03(eq). The imbalanced equation for
`the reaction is thus
`
`-——>
`Na3P04(eq) + Ba(N05)2(aq)
`BastPOslzv) + NaN03(aq)
`
`Because the NO; and P045” inns maintain their
`.
`_
`.
`.
`y
`.
`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 of NaaPOi. This then gives six Na atoms on
`the left, necessitating a coefficient of 5 in front of
`NaNO . Finally the presence of six (NO ) units on
`. 3
`.’
`.
`.
`. 3
`the right requires a coeflicmnt of 3 in front of
`Ba(NOS)2:
`
`2Na3Pngaq) + 3Ba(N03)2(aq) —)
`BaaEPOtlatsl + GNaNDa(eq)
`
`3.4 ATOMIC
`AID HOLIWLAR
`WIIGHTS
`
`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 (3ng 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
`4:3 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 an}: 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.
`
`an atomic AND MOLECULAR WEIGHTS
`
`I"
`
`12 of 12
`
`FORD 1879
`
`