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3.1 Chemical Equations

3.2 Some Simple Patterns of Chemical Reactivity

3.3 Formula Weights

3.4 Avogadro’s Number and the Mole

3.5 Empirical Formulas from Analyses

3.6 Quantitative Information from Balanced Equations

3.7 Limiting Reactants

WHAT’S AHEAD• We begin by considering how we

can use chemical formulas to writeequations that represent chemicalreactions. (Section 3.1)

• We then examine some simple kindsof chemical reactions: combinationreactions, decomposition reactions, andcombustion reactions. (Section 3.2)

• We then use chemical formulas torelate the masses of substances to thenumbers of atoms, molecules, orions contained in the substances, arelationship that leads to thecrucially important concept of amole. A mole is objects(atoms, molecules, ions, orwhatever). (Sections 3.3and 3.4)

• We will apply the moleconcept to determinechemical formulas fromthe masses of each elementin a given quantity of acompound. (Section 3.5)

• We will use the quantitativeinformation inherent in chemicalformulas and equations togetherwith the mole concept topredict the amounts ofsubstances consumedand/or produced inchemical reactions.(Section 3.6)

• A special situation arises whenone of the reactants is used up beforethe others and the reaction thereforestops, leaving some of the excessstarting material unreacted.(Section 3.7)

6.022 * 1023

CH

AP

TE

R 3

Daniel Q. Duffy, Stephanie A. Shaw, WilliamD. Bare and Kenneth A. Goldsby, “MoreChemistry in a Soda Bottle: A Conservationof Mass Activity,” J. Chem. Educ., Vol. 72,1995, 734–736.

Frederic L. Holmes, “Antoine Lavoisier andThe Conservation of Matter,” Chem. Eng.News, September 12, 1994, 38–45.

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Stoichiometry:Calculations withChemical Formulas and Equations

79

THE CHAPTER-OPENING PHOTOGRAPH shows the reactionbetween hydrogen and oxygen gases to form

water vapor, a process illustrated earlier from amolecular perspective in Figure 1.10. The

flame is the visual evidence that somethingis happening; and to an experienced eye, it indicates a chemical

change, a kind of change that we also call a chemical reaction.The study of chemical changes is at the heart of chemistry. Some

chemical changes are simple, and some are complex. Some aredramatic; some are very subtle. Even as you sit reading this

chapter, there are chemical changes occurring in you. Thechanges that occur in your eyes and brain, for example,

are what allows you to see these words and think aboutthem. Although such chemical changes are not as ob-

vious as the reaction shown in the chapter-openingphotograph, they are nevertheless remarkable for

how they allow us to function.In this chapter we begin to explore some

important aspects of chemical changes. Ourfocus will be both on the use of chemical for-

mulas to represent reactions and on thequantitative information we can obtain

about the amounts of substances in-volved in reactions. The area of study

that examines the quantities of sub-stances consumed and produced in

chemical reactions is known asstoichiometry (pronounced

stoy-key-OM-uh-tree), a namederived from the Greek

stoicheion (“element”) andmetron (“measure”). This

study provides anCOMBUSTION OF HYDROGEN GAS. The gas is bubbled through a soap solution forming hydrogen-filledbubbles. As the bubbles float upwards, they are ignited by a candle on a long pole. The red-orange flame isdue to the combustion of the soap bubbles as the hydrogen reacts with oxygen in the air.

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80 Chapter 3 | Stoichiometry: Calculations with Chemical Formulas and Equations

essential set of tools that are widely used in chemistry. Such diverse problems asmeasuring the concentration of ozone in the atmosphere, determining the potentialyield of gold from an ore, and assessing different processes for converting coal intogaseous fuels all use aspects of stoichiometry.

Stoichiometry is built on an understanding of atomic masses (Section 2.4),chemical formulas, and the law of conservation of mass. • (Section 2.1) TheFrench nobleman and scientist Antoine Lavoisier (Figure 3.1 «) discovered thisimportant chemical law in the late 1700s. In a chemistry text published in 1789,Lavoisier stated the law in this eloquent way: “We may lay it down as an incon-testable axiom that, in all the operations of art and nature, nothing is created; anequal quantity of matter exists both before and after the experiment. Upon thisprinciple, the whole art of performing chemical experiments depends.” With theadvent of Dalton’s atomic theory, chemists came to understand the basis for thislaw: Atoms are neither created nor destroyed during any chemical reaction. Thechanges that occur during any reaction merely rearrange the atoms. The samecollection of atoms is present both before and after the reaction.

3.1 Chemical EquationsChemical reactions are represented in a concise way by chemical equations.When hydrogen burns, for example, it reacts with oxygen in the air toform water (chapter-opening photograph). We write the chemical equa-tion for this reaction as follows:

[3.1]

We read the sign as “reacts with” and the arrow as “produces.” The chemicalformulas to the left of the arrow represent the starting substances, calledreactants. The chemical formulas to the right of the arrow represent substancesproduced in the reaction, called products. The numbers in front of the formulasare coefficients. (As in algebraic equations, the numeral 1 is usually not written.)The coefficients indicate the relative numbers of molecules of each kind in-volved in the reaction.

Because atoms are neither created nor destroyed in any reaction, a chemicalequation must have an equal number of atoms of each element on each side ofthe arrow. When this condition is met, the equation is said to be balanced. On theright side of Equation 3.1, for example, there are two molecules of eachcomposed of two atoms of hydrogen and one atom of oxygen. Thus, (read “two molecules of water”) contains atoms and Notice that the number of atoms is obtained by multiplying the coefficient andthe subscripts in the chemical formula. Because there are four H atoms and twoO atoms on each side of the equation, the equation is balanced. We can repre-sent the balanced equation by the following molecular models, which illustratethat the number of atoms of each kind is the same on both sides of the arrow:

GIVE IT SOME THOUGHT

How many atoms of Mg, O, and H are represented by 3 Mg(OH)2 ?

�

2 * 1 = 2 O.2 * 2 = 4 H2 H2O

H2O,

+

2 H2 + O2 ¡ 2 H2O

(H2O)(O2)(H2)

Á Figure 3.1 Antoine Lavoisier(1734–1794). Lavoisier conducted manyimportant studies on combustion reactions.Unfortunately, his career was cut short by theFrench Revolution. He was a member of theFrench nobility and a tax collector. He wasguillotined in 1794 during the final months of theReign of Terror. He is now generally considered tobe the father of modern chemistry because heconducted carefully controlled experiments andused quantitative measurements.

MOVIEFormation of water

Stoichiometry literally means measurementof elements. The relationship between thenumbers of reactant and product speciesproduces a balanced chemical equation.

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3.1 | Chemical Equations 81

Balancing EquationsOnce we know the formulas of the reactants and products in a reaction, we canwrite the unbalanced equation. We then balance the equation by determiningthe coefficients that provide equal numbers of each type of atom on each side ofthe equation. For most purposes, a balanced equation should contain the small-est possible whole-number coefficients.

In balancing equations, it is important to understand the difference between acoefficient in front of a formula and a subscript in a formula. Refer to Figure 3.2 Á.Notice that changing a subscript in a formula—from to for example—changes the identity of the chemical. The substance hydrogen peroxide, isquite different from the substance water. Subscripts should never be changed inbalancing an equation. In contrast, placing a coefficient in front of a formula changesonly the amount of the substance and not its identity. Thus, means two mol-ecules of water, means three molecules of water, and so forth.

To illustrate the process of balancing equations, consider the reaction thatoccurs when methane the principal component of natural gas, burns inair to produce carbon dioxide gas and water vapor (Figure 3.3 ¥).Both of these products contain oxygen atoms that come from in the air.Thus, is a reactant, and the unbalanced equation is

[3.2]

It is usually best to balance first those elements that occur in the fewestchemical formulas on each side of the equation. In our example both C and H

CH4 + O2 ¡ CO2 + H2O (unbalanced)

O2

O2

(H2O)(CO2)(CH4),

3 H2O2 H2O

H2O,H2O2,

H2O2,H2O

Chemicalsymbol Meaning Composition

Four H atoms and two O atoms2 H2O Two moleculesof water:

Two H atoms and two O atomsH2O2 One moleculeof hydrogenperoxide:

Two H atoms and one O atomH2O One moleculeof water:

« Figure 3.2 The difference between asubscript in a chemical formula and acoefficient in front of the formula. Noticehow adding the coefficient 2 in front of theformula (line 2) has a different effect on theimplied composition than adding the subscript 2to the formula (in line 3). The number of atomsof each type (listed under composition) isobtained by multiplying the coefficient and thesubscript associated with each element in the formula.

Reactants

O2

CH4

CO2 and H2O

Products « Figure 3.3 Methane reacts withoxygen to produce the flame in a Bunsenburner. The methane in natural gas andoxygen from the air are the reactants in thereaction, while carbon dioxide and water

are the products.(H2O)(CO2)

(O2)(CH4)

The coefficients of a balanced chemicalequation tell how many of each species areinvolved in the reaction.

Students often change the identities ofsubstances when balancing chemicalequations. When balancing a chemicalequation, you must not change the identities(the subscripts) of any of the species; youmay only change how many (thecoefficients) of the species participate.

ACTIVITYReading a Chemical Equation

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appear in only one reactant and, separately, in one product each, so we begin byfocusing on Let’s consider first carbon and then hydrogen.

One molecule of the reactant contains the same number of C atoms(one) as one molecule of the product The coefficients for these substancesmust be the same, therefore, and we choose them both to be 1 as we start thebalancing process. However, one molecule of contains more H atoms(four) than one molecule of the product (two). If we place a coefficient 2 infront of there will be four H atoms on each side of the equation:

[3.3]

At this stage the products have more O atoms (four—two from andtwo from ) than the reactants (two). If we place a coefficient 2 in front ofthe reactant we complete the balancing by making the number of O atomsequal on both sides of the equation:

[3.4]

The molecular view of the balanced equation is shown in Figure 3.4 ¥. We seethat there are one C, four H, and four O atoms on both sides of the arrow, indi-cating the equation is balanced.

CH4 + 2 O2 ¡ CO2 + 2 H2O (balanced)

O2,2 H2O

CO2

CH4 + O2 ¡ CO2 + 2 H2O (unbalanced)

H2O,H2O

CH4

CO2.CH4

CH4.

� �

CH4 2 O2 CO2 2 H2O

1 C4 H (4 O)( ) 1 C

2 O( ) 2 O4 H( )

� �

» Figure 3.4 Balanced chemical equationfor the combustion of The drawings ofthe molecules involved call attention to theconservation of atoms through the reaction.

CH4 .

Zoltan Toth, “Balancing Chemical Equationsby Inspection,” J. Chem. Educ., Vol. 74, 1997,1363–1364.

Chunshi Guo, “A New Inspection Method forBalancing Redox Equations,” J. Chem. Educ.,Vol. 74, 1997, 1365–1366.

William C. Herndon, “On BalancingChemical Equations: Past and Present(A Critical Review and AnnotatedBibliography),” J. Chem. Educ., Vol. 74, 1997, 1359–1362.

Addison Ault, “How to Say How Much:Amounts and Stoichiometry,” J. Chem.Educ., Vol. 78, 2001, 1347–1348.

ACTIVITYCounting Atoms

ACTIVITYReading a Balanced Chemical Equation

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3.1 | Chemical Equations 83

The approach we have taken in arriving at balanced Equation 3.4 is largelytrial and error. We balance each kind of atom in succession, adjusting coeffi-cients as necessary. This approach works for most chemical equations.

SAMPLE EXERCISE 3.1 | Interpreting and Balancing Chemical EquationsThe following diagram represents a chemical reaction in which the red spheres areoxygen atoms and the blue spheres are nitrogen atoms. (a) Write the chemical formu-las for the reactants and products. (b) Write a balanced equation for the reaction. (c) Isthe diagram consistent with the law of conservation of mass?

Solution (a) The left box, which represents the reactants, contains two kinds of mole-cules, those composed of two oxygen atoms and those composed of one nitrogenatom and one oxygen atom (NO). The right box, which represents the products, containsonly molecules composed of one nitrogen atom and two oxygen atoms

(b) The unbalanced chemical equation is

In this equation there are three O atoms on the left side of the arrow and two O atomson the right side. We can increase the number of O atoms by placing a coefficient 2 onthe product side:

Now there are two N atoms and four O atoms on the right. Placing a coefficient 2 infront of NO brings both the N atoms and O atoms into balance:

(c) The left box (reactants) contains four molecules and eight NO molecules.Thus, the molecular ratio is one for each two NO as required by the balancedequation. The right box (products) contains eight molecules. The number of

molecules on the right equals the number of NO molecules on the left as the bal-anced equation requires. Counting the atoms, we find eight N atoms in the eight NOmolecules in the box on the left. There are also atoms in the mole-cules and eight O atoms in the NO molecules, giving a total of 16 O atoms. In the boxon the right, we find eight N atoms and atoms in the eight mole-cules. Because there are equal numbers of both N and O atoms in the two boxes, thedrawing is consistent with the law of conservation of mass.

PRACTICE EXERCISEIn order to be consistent with the law of conservation of mass, how many mole-cules should be shown in the right box of the following diagram?

Answer: Six moleculesNH3

?

NH3

NO28 * 2 = 16 O

O24 * 2 = 8 O

NO2

NO2

O2

O2

O2 + 2 NO ¡ 2 NO2 (balanced)

O2 + NO ¡ 2 NO2 (unbalanced)

O2 + NO ¡ NO2 (unbalanced)

(NO2).

(O2)

ACTIVITYBalancing Equations

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SAMPLE EXERCISE 3.2 | Balancing Chemical Equations

Balance this equation: Na(s) + H2O(l) ¡ NaOH(aq) + H2(g)

Solution We begin by counting theatoms of each kind on both sides ofthe arrow. The Na and O atoms arebalanced (one Na and one O on eachside), but there are two H atoms onthe left and three H atoms on theright. Thus, we need to increase thenumber of H atoms on the left. As atrial beginning in our effort to bal-ance H, let’s place a coefficient 2 infront of H2O : Na(s) + 2 H2O(l) ¡ NaOH(aq) + H2(g)

Beginning this way doesn’t balanceH, but introducing the coefficient 2does increase the number of Hatoms among the reactants, whichwe need to do. The fact that it causesO to be unbalanced is something wewill take care of after we balance H.Now that we have on the left,we can balance H by putting a coef-ficient 2 in front of NaOH on theright:

2 H2O

Na(s) + 2 H2O(l) ¡ 2 NaOH(aq) + H2(g)

Balancing H in this way fortuitouslybrings O into balance, but noticethat Na is now unbalanced, withone on the left but two on the right.To rebalance Na, we put a coefficient2 in front of the reactant: 2 Na(s) + 2 H2O(l) ¡ 2 NaOH(aq) + H2(g)

Finally, we check the number of atoms of each element and find that we have two Na atoms, four H atoms, and two O atoms oneach side of the equation. The equation is balanced.Comment: Notice that in balancing this equation, we moved back and forth placing a coefficient in front of then NaOH,and finally Na. In balancing equations, we often find ourselves following this pattern of moving back and forth from one side ofthe arrow to the other, placing coefficients first in front of a formula on one side and then in front of a formula on the other sideuntil the equation is balanced.

PRACTICE EXERCISEBalance the following equations by providing the missing coefficients:

(a)(b)(c)

Answers: (a) 4, 3, 2; (b) 1, 3, 2, 2; (c) 2, 6, 2, 3 Al(s) + HCl(aq) ¡ AlCl3(aq) + H2(g) C2H4(g) + O2(g) ¡ CO2(g) + H2O(g) Fe(s) + O2(g) ¡ Fe2O3(s)

H2O,

Indicating the States of Reactants and ProductsAdditional information is often added to the formulas in balanced equations toindicate the physical state of each reactant and product. We use the symbols (g),(l), (s), and (aq) for gas, liquid, solid, and aqueous (water) solution, respectively.Thus, Equation 3.4 can be written

[3.5]

Sometimes the conditions (such as temperature or pressure) under which thereaction proceeds appear above or below the reaction arrow. The symbol (theGreek uppercase letter delta) is often placed above the arrow to indicate the ad-dition of heat.

¢

CH4(g) + 2 O2(g) ¡ CO2(g) + 2 H2O(g)

MOVIESodium and Potassium in Water

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3.2 | Some Simple Patterns of Chemical Reactivity 85

TABLE 3.1 Combination and Decomposition Reactions

Combination Reactions

A � B CC(s) � O2(g) CO2(g)N2(g) � 3 H2(g) 2 NH3(g)CaO(s) � H2O(l) Ca(OH)2 (s)

Two reactants combine to form a single product. Many elements react with one another in this fashion to form compounds.

Decomposition Reactions

A � BC2 KClO3(s) 2 KCl(s) � 3 O2(g)PbCO3(s) PbO(s) � CO2(g)Cu(OH)2(s) CuO(s) � H2O(l)

A single reactant breaks apart to form two or more substances. Many compounds react this way when heated.

3.2 Some Simple Patterns of Chemical ReactivityIn this section we examine three simple kinds of reactions that we will see fre-quently throughout this chapter. Our first reason for examining these reactionsis merely to become better acquainted with chemical reactions and their bal-anced equations. Our second reason is to consider how we might predict theproducts of some of these reactions knowing only their reactants. The key topredicting the products formed by a given combination of reactants is recog-nizing general patterns of chemical reactivity. Recognizing a pattern of reactiv-ity for a class of substances gives you a broader understanding than merelymemorizing a large number of unrelated reactions.

Combination and Decomposition ReactionsTable 3.1 ¥ summarizes two simple types of reactions, combination and de-composition reactions. In combination reactions two or more substances reactto form one product. There are many examples of such reactions, especiallythose in which elements combine to form compounds. For example, magne-sium metal burns in air with a dazzling brilliance to produce magnesium oxide,as shown in Figure 3.5 »:

[3.6]

This reaction is used to produce the bright flame generated by flares.When a combination reaction occurs between a metal and a nonmetal, as in

Equation 3.6, the product is an ionic solid. Recall that the formula of an ioniccompound can be determined from the charges of the ions involved• (Section 2.7). When magnesium reacts with oxygen, for example, the mag-nesium loses electrons and forms the magnesium ion, The oxygen gainselectrons and forms the oxide ion, Thus, the reaction product is MgO. Youshould be able to recognize when a reaction is a combination reaction and topredict the products of a combination reaction in which the reactants are ametal and a nonmetal.


When Na and S combine in a combination reaction, what is the chemical formula ofthe product?

O2-.Mg2+.

2 Mg(s) + O2(g) ¡ 2 MgO(s)

Although they may differ in the vigor ofreaction, members of a family on theperiodic table tend to react in similar ways.

An article on some of the uses of quicklime(CaO) and hydrated lime .Kenneth W. Watkins, “Lime,” J. Chem. Educ.,Vol. 60, 1983, 60–63.

(Ca(OH)2)

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Á Figure 3.6 An automobile air bag.The decomposition of sodium azide, isused to inflate automobile air bags. Whenproperly ignited, the decomposes rapidly,forming nitrogen gas, which expands the air bag.

N2(g),NaN3

NaN3(s),

In a decomposition reaction one substance undergoes a reaction to pro-duce two or more other substances. Many compounds undergo decompositionreactions when heated. For example, many metal carbonates decompose toform metal oxides and carbon dioxide when heated:

[3.7]

The decomposition of is an important commercial process. Limestoneor seashells, which are both primarily are heated to prepare CaO,

which is known as lime or quicklime. About (20 million tons) ofCaO is used in the United States each year, principally in making glass, in ob-taining iron from its ores, and in making mortar to bind bricks.

The decomposition of sodium azide rapidly releases so thisreaction is used to inflate safety air bags in automobiles (Figure 3.6 «):

[3.8]

The system is designed so that an impact ignites a detonator cap, which in turncauses to decompose explosively. A small quantity of (about 100 g)forms a large quantity of gas (about 50 L). We will consider the volumes ofgases produced in chemical reactions in Section 10.5.

NaN3NaN3

2 NaN3(s) ¡ 2 Na(s) + 3 N2(g)

N2(g),(NaN3)

2 * 1010 kg

CaCO3,CaCO3

CaCO3(s) ¡ CaO(s) + CO2(g)

COMBINATION REACTIONIn combination reactions, two or more substances react to form one product.

The ribbon of magnesium metal is surrounded by oxygen

in the air, and as it burns, an intense flame is produced.

When magnesium metal burns, the Mg atoms react with O2 molecules from the air to form magnesium

oxide, MgO, an ionic solid.

At the end of the reaction, a rather fragile ribbon of white

solid, MgO, remains.

�2 Mg(s) O2(g) 2 MgO(s)

Mg Mg OO

O2�

O2�

Mg2�

Mg2�

Á Figure 3.5 Combustion of magnesium metal in air.

MOVIEReactions with Oxygen

ANIMATIONAir Bags

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3.2 | Some Simple Patterns of Chemical Reactivity 87

* When there is an insufficient quantity of present, carbon monoxide (CO) will be produced alongwith the this is called incomplete combustion. If the amount of is severely restricted, fine par-ticles of carbon that we call soot will be produced. Complete combustion produces only and Unless specifically stated to the contrary, we will always take combustion to mean complete combustion.

H2O.CO2

O2CO2 ;O2

SAMPLE EXERCISE 3.3 | Writing Balanced Equations for Combination andDecomposition Reactions

Write balanced equations for the following reactions: (a) The combination reactionthat occurs when lithium metal and fluorine gas react. (b) The decomposition reac-tion that occurs when solid barium carbonate is heated. (Two products form: a solidand a gas.)

Solution (a) The symbol for lithium is Li. With the exception of mercury, all metalsare solids at room temperature. Fluorine occurs as a diatomic molecule (seeFigure 2.19). Thus, the reactants are Li(s) and The product will consist of a metaland a nonmetal, so we expect it to be an ionic solid. Lithium ions have a charge,

whereas fluoride ions have a charge, Thus, the chemical formula for theproduct is LiF. The balanced chemical equation is

(b) The chemical formula for barium carbonate is As noted in the text,many metal carbonates decompose to form metal oxides and carbon dioxide whenheated. In Equation 3.7, for example, decomposes to form CaO and Thus, we would expect that decomposes to form BaO and Barium andcalcium are both in group 2A in the periodic table, moreover, which further suggeststhey would react in the same way:

PRACTICE EXERCISEWrite balanced chemical equations for the following reactions: (a) Solid mercury(II)sulfide decomposes into its component elements when heated. (b) The surface of alu-minum metal undergoes a combination reaction with oxygen in the air.Answers: (a) (b)

Combustion in AirCombustion reactions are rapid reactions that produce a flame. Most of thecombustion reactions we observe involve from air as a reactant. Equation 3.5and Practice Exercise 3.1(b) illustrate a general class of reactions involving theburning, or combustion, of hydrocarbon compounds (compounds that containonly carbon and hydrogen, such as and ). • (Section 2.9)

When hydrocarbons are combusted in air, they react with to form and * The number of molecules of required in the reaction and thenumber of molecules of and formed depend on the composition ofthe hydrocarbon, which acts as the fuel in the reaction. For example, the com-bustion of propane a gas used for cooking and home heating, is de-scribed by the following equation:

[3.9]

The state of the water, or depends on the conditions of the reac-tion. Water vapor, is formed at high temperature in an open container.The blue flame produced when propane burns is shown in Figure 3.7 ».

Combustion of oxygen-containing derivatives of hydrocarbons, such asalso produces and The simple rule that hydrocarbons and

related oxygen-containing derivatives of hydrocarbons form and when they burn in air summarizes the behavior of about 3 million compounds.Many substances that our bodies use as energy sources, such as the sugar glu-cose similarly react in our bodies with to form and H2O.CO2O2(C6H12O6),

H2OCO2

H2O.CO2CH3OH,

H2O(g),H2O(l),H2O(g)

C3H8(g) + 5 O2(g) ¡ 3 CO2(g) + 4 H2O(g)

(C3H8),

H2OCO2

O2H2O.CO2O2

C2H4CH4

O2

4 Al(s) + 3 O2(g) ¡ 2 Al2O3(s)HgS(s) ¡ Hg(l) + S(s);

BaCO3(s) ¡ BaO(s) + CO2(g)

CO2.BaCO3

CO2.CaCO3

BaCO3.

2 Li(s) + F2(g) ¡ 2 LiF(s)

F-.1-Li+,1+

F2(g).

Á Figure 3.7 Propane burning in air.The liquid propane, vaporizes and mixeswith air as it escapes through the nozzle. Thecombustion reaction of and producesa blue flame.

O2C3H8

C3H8 ,

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In our bodies, however, the reactions take place in a series of steps that occur atbody temperature. The reactions are then described as oxidation reactions ratherthan combustion reactions.

SAMPLE EXERCISE 3.4 | Writing Balanced Equations for CombustionReactions

Write the balanced equation for the reaction that occurs when methanol, is burned in air.

Solution When any compound containing C, H, and O is combusted, it reacts withthe in air to produce and Thus, the unbalanced equation is

The C atoms are balanced with one on each side of the arrow. Because hasfour H atoms, we place a coefficient 2 in front of to balance the H atoms:

This balances H but gives four O atoms in the products. Because there are only threeO atoms in the reactants (one in and two in ), we are not finished yet. Wecan place the fractional coefficient in front of to give a total of four O atoms in thereactants (there are atoms in ):

Although the equation is now balanced, it is not in its most conventional form be-cause it contains a fractional coefficient. If we multiply each side of the equation by 2,we will remove the fraction and achieve the following balanced equation:

PRACTICE EXERCISEWrite the balanced equation for the reaction that occurs when ethanol, isburned in air.Answer:

3.3 Formula WeightsChemical formulas and chemical equations both have a quantitative signifi-cance; the subscripts in formulas and the coefficients in equations representprecise quantities. The formula indicates that a molecule of this substancecontains exactly two atoms of hydrogen and one atom of oxygen. Similarly, thecoefficients in a balanced chemical equation indicate the relative quantities ofreactants and products. But how do we relate the numbers of atoms or mole-cules to the amounts we measure out in the laboratory? Although we cannot di-rectly count atoms or molecules, we can indirectly determine their numbers ifwe know their masses. Therefore, before we can pursue the quantitative aspectsof chemical formulas or equations, we must examine the masses of atoms andmolecules, which we do in this section and the next.

Formula and Molecular WeightsThe formula weight of a substance is the sum of the atomic weights of eachatom in its chemical formula. Using atomic masses from a periodic table, wefind, for example, that the formula weight of sulfuric acid is 98.1 amu:*(H2SO4)

H2O

C2H5OH(l) + 3 O2(g) ¡ 2 CO2(g) + 3 H2O(g)

C2H5OH(l),

2 CH3OH(l) + 3 O2(g) ¡ 2 CO2(g) + 4 H2O(g)

CH3OH(l) + 32 O2(g) ¡ CO2(g) + 2 H2O(g)

32 O2

32 * 2 = 3 O

O232

O2CH3OH

CH3OH(l) + O2(g) ¡ CO2(g) + 2 H2O(g)

H2OCH3OH

CH3OH(l) + O2(g) ¡ CO2(g) + H2O(g)

H2O(g).CO2(g)O2(g)

CH3OH(l),

* The abbreviation AW is used for atomic weight, FW for formula weight, and MW for molecularweight.

Atomic weights can be obtained with manysignificant figures. Advise students to use asufficient number of significant figures suchthat the number of significant figures in theanswer to any calculations is not limited bythe number of significant figures in theformula weights they use.

Wayne L. Felty, “Gram Formula Weights andFruit Salad,” J. Chem. Educ., Vol. 62, 1985, 61.

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3.3 | Formula Weights 89

For convenience, we have rounded off all the atomic weights to one place be-yond the decimal point. We will round off the atomic weights in this way formost problems.

If the chemical formula is merely the chemical symbol of an element, suchas Na, then the formula weight equals the atomic weight of the element. If thechemical formula is that of a molecule, then the formula weight is also calledthe molecular weight. The molecular weight of glucose for exam-ple, is

Because ionic substances, such as NaCl, exist as three-dimensional arraysof ions (Figure 2.23), it is inappropriate to speak of molecules of NaCl. Instead,we speak of formula units, represented by the chemical formula of the sub-stance. The formula unit of NaCl consists of one ion and one ion. Thus,the formula weight of NaCl is the mass of one formula unit:

FW of NaCl = 23.0 amu + 35.5 amu = 58.5 amu

Cl-Na+

MW of C6H12O6 = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu) = 180.0 amu

(C6H12O6),

= 98.1 amu

= 2(1.0 amu) + 32.1 amu + 4(16.0 amu)

FW of H2SO4 = 2(AW of H) + (AW of S) + 4(AW of O)

SAMPLE EXERCISE 3.5 | Calculating Formula WeightsCalculate the formula weight of (a) sucrose, (table sugar), and (b) calcium nitrate, Ca(NO3)2 .C12H22O11

Solution (a) By adding the atomicweights of the atoms in sucrose, wefind it to have a formula weight of342.0 amu:

12 C atoms = 12(12.0 amu) = 44.0 amu22 H atoms = 22(1.0 amu) = 22.0 amu

11 O atoms = 11(16.0 amu) = 176.0 amu342.0 amu

(b) If a chemical formula hasparentheses, the subscript outsidethe parentheses is a multiplier for allatoms inside. Thus, for we have

Ca(NO3)2 ,

1 Ca atom = 1(40.1 amu) = 40.1 amu2 N atoms = 2(14.0 amu) = 28.0 amu6 O atoms = 6(16.0 amu) = 96.0 amu

164.1 amu

PRACTICE EXERCISECalculate the formula weight of (a) and (b)Answers: (a) 78.0 amu, (b) 32.0 amu

CH3OH.Al(OH)3

Percentage Composition from FormulasOccasionally we must calculate the percentage composition of a compound (thatis, the percentage by mass contributed by each element in the substance). Forexample, in order to verify the purity of a compound, we may wish to comparethe calculated percentage composition of the substance with that found experi-mentally. Calculating percentage composition is a straightforward matter if thechemical formula is known. The calculation depends on the formula weight ofthe substance, the atomic weight of the element of interest, and the number ofatoms of that element in the chemical formula:

[3.10]% element =(number of atoms of that element)(atomic weight of element)

formula weight of compound* 100%

George L. Gilbert, “Percentage Compositionand Empirical Formula—A New View,”J. Chem. Educ., Vol. 75, 1998, 851.

ACTIVITYMolecular Weight and Weight Percent

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SAMPLE EXERCISE 3.6 | Calculating Percentage CompositionCalculate the percentage of carbon, hydrogen, and oxygen (by mass) in

Solution Let’s examine this question using the problem-solving steps in the “Strate-gies in Chemistry: Problem Solving” essay.Analyze: We are given a chemical formula, and asked to calculate thepercentage by mass of its component elements (C, H, and O).Plan: We can use Equation 3.10, relying on a periodic table to obtain the atomicweight of each component element. The atomic weights are first used to determinethe formula weight of the compound. (The formula weight of 342.0 amu,was calculated in Sample Exercise 3.5.) We must then do three calculations, one foreach element.Solve: Using Equation 3.10, we have

Check: The percentages of the individual elements must add up to 100%, which theydo in this case. We could have used more significant figures for our atomic weights, giv-ing more significant figures for our percentage composition, but we have adhered to oursuggested guideline of rounding atomic weights to one digit beyond the decimal point.

PRACTICE EXERCISECalculate the percentage of nitrogen, by mass, in Answer: 17.1%

3.4 Avogadro’s Number and the MoleEven the smallest samples that we deal with in the laboratory contain enor-mous numbers of atoms, ions, or molecules. For example, a teaspoon of water(about 5 mL) contains water molecules, a number so large that it al-most defies comprehension. Chemists, therefore, have devised a special count-ing unit for describing such large numbers of atoms or molecules.

2 * 1023

Ca(NO3)2 .

%O =(11)(16.0 amu)

342.0 amu* 100% = 51.5%

%H =(22)(1.0 amu)

342.0 amu* 100% = 6.4%

%C =(12)(12.0 amu)

342.0 amu* 100% = 42.1%

C12H22O11,

C12H22O11,

C12H22O11.

which we will soon discuss) or look them up in tables(such as atomic weights). Recognize also that your planmay involve either a single step or a series of steps withintermediate answers.

Step 3: Solve the problem. Use the known information andsuitable equations or relationships to solve for the un-known. Dimensional analysis (Section 1.6) is a very usefultool for solving a great number of problems. Be carefulwith significant figures, signs, and units.

Step 4: Check the solution. Read the problem again tomake sure you have found all the solutions asked for inthe problem. Does your answer make sense? That is, isthe answer outrageously large or small, or is it in the ball-park? Finally, are the units and significant figures correct?

STRATEGIES IN CHEMISTRY | Problem Solving

The key to success in problem solving is practice. As youpractice, you will find that you can improve your skills by

following these steps:

Step 1: Analyze the problem. Read the problem carefullyfor understanding. What does it say? Draw any picture ordiagram that will help you visualize the problem. Writedown both the data you are given and the quantity thatyou need to obtain (the unknown).Step 2: Develop a plan for solving the problem. Considerthe possible paths between the given information and theunknown. What principles or equations relate the knowndata to the unknown? Recognize that some data may notbe given explicitly in the problem; you may be expectedto know certain quantities (such as Avogadro’s number,

George Gorin, “Mole, Mole per Liter, andMolar. A Primer on SI and Related Units forChemistry Students,” J. Chem. Educ., Vol. 80,2003, 103–104.

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3.4 | Avogadro’s Number and the Mole 91

* The term mole comes from the Latin word moles, meaning “a mass.” The term molecule is thediminutive form of this word and means “a small mass.”

In everyday life we use counting units like dozen (12 objects) and gross (144objects) to deal with modestly large quantities. In chemistry the unit for dealingwith the number of atoms, ions, or molecules in a common-sized sample is themole, abbreviated mol.* A mole is the amount of matter that contains as manyobjects (atoms, molecules, or whatever objects we are considering) as the num-ber of atoms in exactly 12 g of isotopically pure From experiments, scien-tists have determined this number to be Scientists call thisnumber Avogadro’s number, in honor of Amedeo Avogadro (1776–1856),an Italian scientist. For most purposes we will use or forAvogadro’s number throughout the text.

A mole of atoms, a mole of molecules, or a mole of anything else all containAvogadro’s number of these objects:

Avogadro’s number is so large that it is difficult to imagine. Spreadingmarbles over the entire surface of Earth would produce a layer

about 3 mi thick. If Avogadro’s number of pennies were placed side by side in astraight line, they would encircle Earth 300 trillion times.

SAMPLE EXERCISE 3.7 | Estimating Numbers of AtomsWithout using a calculator, arrange the following samples in order of increasing num-bers of carbon atoms: molecules of

SolutionAnalyze: We are given amounts of different substances expressed in grams, moles,and number of molecules and asked to arrange the samples in order of increasingnumbers of C atoms.Plan: To determine the number of C atoms in each sample, we must convert g mol and molecules all to numbers of C atoms. To do this converting, weuse the definition of mole and Avogadro’s number.Solve: A mole is defined as the amount of matter that contains as many units of thematter as there are C atoms in exactly 12 g of Thus, 12 g of contains 1 mol of Catoms (that is, atoms). In 1 mol there are molecules.Because there are two C atoms in each molecule, this sample contains atoms. Because each molecule contains one C atom, the sample of contains

atoms. Hence, the order is 12 g molecules Check: We can check our results by comparing the number of moles of C atoms ineach sample because the number of moles is proportional to the number of atoms.Thus, 12 g of is 1 mol C; 1 mol of contains 2 mol C, and moleculesof contain 1.5 mol C, giving the same order as above:

molecules (1.5 mol C) (2 mol C).

PRACTICE EXERCISEWithout using a calculator, arrange the following samples in order of increasing numberof O atoms: 1 mol 1 mol molecules Answer: 1 mol molecules Oatoms) (12 * 1023 O atoms)6 1 mol CO2

O3 (9 * 1023H2O (6 * 1023 O atoms) 6 3 * 1023O3.3 * 1023CO2,H2O,

C2H26 1 molCO2(1 mol C) 6 9 * 102312 g 12CCO2

9 * 1023C2H212C

C atoms).C2H2 (12 * 1023C atoms) 6 1 mol(9 * 1023CO2C atoms) 6 9 * 102312C (6 * 10239 * 1023 C

CO2CO2

12 * 1023 CC2H2

6 * 1023 C2H2C2H2,6.02 * 1023 C

12C12C.

CO2C2H2,

12C,

CO2.9 * 10231 mol C2H2,12 g 12C,

13 * 101426.02 * 1023

1 mol NO3

- ions = 6.02 * 1023 NO3

- ions

1 mol H2O molecules = 6.02 * 1023 H2O molecules

1 mol 12C atoms = 6.02 * 1023 12C atoms

6.022 * 10236.02 * 1023

6.0221421 * 1023.

12C.

The mole can be thought of as a collectioncontaining a very large number of objects,

, just as a dozen is a collectionof 12 objects.6.02 * 1023

Avogadro’s number is also the number of amuper gram, i.e., the conversion factor betweenamu and gram is .6.02 * 1023 amu>g

Dawn M. Wakeley and Hans de Grys,“Developing an Intuitive Approach to Moles,”J. Chem. Educ., Vol. 77, 2000, 1007–1009.

Mali Yin and Raymond S. Ochs, “The Mole,the Periodic Table, and Quantum Numbers:An Introductory Trio,” J. Chem. Educ.,Vol. 78, 2001, 1345–1347.

Miriam Toloudis, “The Size of a Mole,”J. Chem. Educ., Vol. 73, 1996, 348.

Sheryl Dominic, “What’s a Mole For?”J. Chem. Educ., Vol. 73, 1996, 309.

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SAMPLE EXERCISE 3.8 | Converting Moles to Number of AtomsCalculate the number of H atoms in 0.350 mol of

SolutionAnalyze: We are given both the amount of a substance (0.350 mol) and its chemicalformula The unknown is the number of H atoms in the sample.Plan: Avogadro’s number provides the conversion factor between the number ofmoles of and the number of molecules of Once we know thenumber of molecules of we can use the chemical formula, which tells usthat each molecule of contains 12 H atoms. Thus, we convert moles of

to molecules of and then determine the number of atoms of Hfrom the number of molecules of

Solve:

Check: The magnitude of our answer is reasonable: It is a large number about themagnitude of Avogadro’s number. We can also make the following ballpark calcula-tion: Multiplying gives about molecules. Multiplying thisresult by 12 gives atoms, which agrees with the previous,more detailed calculation. Because we were asked for the number of H atoms, theunits of our answer are correct. The given data had three significant figures, so ouranswer has three significant figures.

PRACTICE EXERCISEHow many oxygen atoms are in (a) and (b) 1.50 mol of sodiumcarbonate?Answers: (a) (b)

Molar MassA dozen is the same number (12) whether we have a dozen eggs or a dozen ele-phants. Clearly, however, a dozen eggs does not have the same mass as a dozenelephants. Similarly, a mole is always the same number but 1-molesamples of different substances will have different masses. Compare, for exam-ple, 1 mol of and 1 mol of A single atom has a mass of 12 amu,whereas a single atom is twice as massive, 24 amu (to two significant fig-ures). Because a mole always has the same number of particles, a mole of must be twice as massive as a mole of Because a mole of has a mass of12 g (by definition), then a mole of must have a mass of 24 g. This exam-ple illustrates a general rule relating the mass of an atom to the mass ofAvogadro’s number (1 mol) of these atoms: The mass of a single atom of an element(in amu) is numerically equal to the mass (in grams) of 1 mol of that element. Thisstatement is true regardless of the element:

24Mg

12C12C.

24Mg

24Mg

12C24Mg.12C

16.02 * 10232,

2.71 * 10249.0 * 1023,

0.25 mol Ca(NO3)2

2.4 * 1024 H=24 * 10232 * 10230.35 * 6 * 1023

= 2.53 * 1024 H atoms

(0.350 mol C6H12O6 )¢6.02 * 1023 molecules C6H12O6

1 mol C6H12O6 ≤ ¢ 12 H atoms

1 molecule C6H12O6 ≤

H atoms =

Moles C6H12O6 ¡ molecules C6H12O6 ¡ atoms H

C6H12O6 :C6H12O6C6H12O6

C6H12O6

C6H12O6,C6H12O6.C6H12O6

(C6H12O6).

C6H12O6.

1 atom of Au has an atomic weight of 197 amu Q 1 mol Au has a mass of 197 g

1 atom of Cl has an atomic weight of 35.5 amu Q 1 mol Cl has a mass of 35.5 g

1 atom of 12C has a mass of 12 amu Q 1 mol 12C has a mass of 12 g

Notice that when we are dealing with a particular isotope of an element, we usethe mass of that isotope; otherwise we use the atomic weight (the averageatomic mass) of the element.

Shanthi R. Krishnan and Ann C. Howe, “TheMole Concept: Developing An Instrument toAssess Conceptual Understanding,” J. Chem.Educ., Vol. 71, 1994, 653–655.

Carmela Merlo and Kathleen E. Turner,“A Mole of M&M’s,” J. Chem. Educ., Vol. 70,1993, 453.

Henk van Lubeck, “How to VisualizeAvogadro’s Number,” J. Chem. Educ., Vol.66, 1989, 762.

A monolayer of stearic acid on water is usedto estimate Avogadro’s number. SallySolomon and Chinhyu Hur, “MeasuringAvogadro’s Number on the OverheadProjector,” J. Chem. Educ., Vol. 70, 1993,252–253.

A demonstration of the measurement ofAvogadro’s number. William Laurita,“Demonstrations for Nonscience Majors:Using Common Objects to Illustrate AbstractConcepts,” J. Chem. Educ., Vol. 67, 1990,60–61.

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Avogadro’snumber ofmolecules

(6.02 � 1023)

Single molecule

1 molecule H2O(18.0 amu)

1 mol H2O(18.0 g)

Laboratory-sizesample

For other kinds of substances, the same numerical relationship exists betweenthe formula weight (in amu) and the mass (in grams) of one mole of that substance:

Figure 3.8 Á illustrates the relationship between the mass of a single moleculeof and that of a mole of

The mass in grams of one mole of a substance (that is, the mass in gramsper mol) is called the molar mass of the substance. The molar mass (in )of any substance is always numerically equal to its formula weight (in amu).The substance NaCl, for example, has formula weight of 58.5 amu and amolar mass of Further examples of mole relationships are shownin Table 3.2 ¥. Figure 3.9 » shows 1-mole quantities of several commonsubstances.

The entries in Table 3.2 for N and point out the importance of stating thechemical form of a substance exactly when we use the mole concept. Supposeyou read that 1 mol of nitrogen is produced in a particular reaction. You mightinterpret this statement to mean 1 mol of nitrogen atoms (14.0 g). Unless other-wise stated, however, what is probably meant is 1 mol of nitrogen molecules,

(28.0 g), because is the usual chemical form of the element. To avoid am-biguity, it is important to state explicitly the chemical form being discussed.Using the chemical formula avoids ambiguity.N2

N2N2

N2

58.5 g>mol.

g>mol

H2O.H2O

1 NaCl unit has a mass of 58.5 amu Q 1 mol NaCl has a mass of 58.5 g

1 NO3

- ion has a mass of 62.0 amu Q 1 mol NO3

- has a mass of 62.0 g

1 H2O molecule has a mass of 18.0 amu Q 1 mol H2O has a mass of 18.0 g

Á Figure 3.9 One mole each of a solid,a liquid, and a gas. One mole of NaCl, the solid,has a mass of 58.45 g. One mole of theliquid, has a mass of 18.0 g and occupies avolume of 18.0 mL. One mole of the gas, hasa mass of 32.0 g and occupies a balloon whosediameter is 35 cm.

O2 ,

H2O,

TABLE 3.2 Mole Relationships

Name of substance

Atomic nitrogen Molecular nitrogen

Silver Silver ions

Barium chloride

FormulaFormula Weight (amu)

Molar Mass(g/mol)

Number and Kind ofParticles in One Mole

NN2

AgAg�

BaCl2

14.028.0

107.9107.9a

6.022 � 1023 N atoms6.022 � 1023 N2 molecules

2(6.022 � 1023) N atoms

2(6.022 � 1023) Cl� ions

6.022 � 1023 Ag atoms6.022 � 1023 Ag� ions6.022 � 1023 BaCl2 units6.022 � 1023 Ba2� ions208.2

14.028.0

107.9107.9

208.2

aRecall that the electron has negligible mass; thus, ions and atoms have essentially the same mass.

Á Figure 3.8 Comparing the mass of 1 molecule and 1 mol Notice that the masses arenumerically equal but have different units (18.0 amu compared to 18.0 g) representing the huge difference in mass.

H2O.H2O

Damon Diemente, “Demonstrations of theEnormity of Avogadro’s Number,” J. Chem.Educ., Vol. 75, 1998, 1565–1566.

R. E. Uthe, “For Mole Problems, CallAvogadro: 602–1023,” J. Chem. Educ.,Vol. 79, 2002, 1213.

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SAMPLE EXERCISE 3.9 | Calculating Molar MassWhat is the mass in grams of 1.000 mol of glucose,

SolutionAnalyze: We are given a chemical formula and asked to determine its molar mass.Plan: The molar mass of a substance is found by adding the atomic weights of itscomponent atoms.Solve:

Because glucose has a formula weight of 180.0 amu, one mole of this substance has amass of 180.0 g. In other words, has a molar mass of Check: The magnitude of our answer seems reasonable, and is the appropri-ate unit for the molar mass.Comment: Glucose is sometimes called dextrose. Also known as blood sugar, glu-cose is found widely in nature, occurring, for example, in honey and fruits. Othertypes of sugars used as food are converted into glucose in the stomach or liver beforethey are used by the body as energy sources. Because glucose requires no conversion,it is often given intravenously to patients who need immediate nourishment.

PRACTICE EXERCISECalculate the molar mass of Answer:

Interconverting Masses and MolesConversions of mass to moles and of moles to mass are frequently encounteredin calculations using the mole concept. These calculations are made easythrough dimensional analysis, as shown in Sample Exercises 3.10 and 3.11.

SAMPLE EXERCISE 3.10 | Converting Grams to MolesCalculate the number of moles of glucose in 5.380 g of

SolutionAnalyze: We are given the number of grams of a substance and its chemical formulaand asked to calculate the number of moles.Plan: The molar mass of a substance provides the factor for converting grams tomoles. The molar mass of is (Sample Exercise 3.9).Solve: Using to write the appropriate conversionfactor, we have

Check: Because 5.380 g is less than the molar mass, it is reasonable that our answer isless than one mole. The units of our answer (mol) are appropriate. The original datahad four significant figures, so our answer has four significant figures.

PRACTICE EXERCISEHow many moles of sodium bicarbonate are there in 508 g of Answer:

SAMPLE EXERCISE 3.11 | Converting Moles to GramsCalculate the mass, in grams, of 0.433 mol of calcium nitrate.

SolutionAnalyze: We are given the number of moles and name of a substance and asked tocalculate the number of grams in the sample.Plan: In order to convert moles to grams, we need the molar mass, which we can cal-culate using the chemical formula and atomic weights.

6.05 mol NaHCO3

NaHCO3 ?(NaHCO3)

Moles C6H12O6 = (5.380 g C6H12O6 )¢ 1 mol C6H12O6

180.0 g C6H12O6 ≤ = 0.02989 mol C6H12O6

1 mol C6H12O6 = 180.0 g C6H12O6

180.0 g>molC6H12O6

C6H12O6.(C6H12O6)

164.1 g>molCa(NO3)2 .

g>mol180.0 g>mol.C6H12O6

6 C atoms = 6(12.0 amu) = 72.0 amu12 H atoms = 12( 1.0 amu) = 12.0 amu

6 O atoms = 6(16.0 amu) = 96.0 amu180.0 amu

C6H12O6 ?

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Usemolarmass

UseAvogadro’s

numberGrams Moles Formula units

Á Figure 3.10 Procedure forinterconverting the mass and the numberof formula units of a substance. The numberof moles of the substance is central to thecalculation; thus, the mole concept can bethought of as the bridge between the mass of a substance in grams and the number of formula units.

Solve: Because the calcium ion is and the nitrate ion is calcium nitrate isAdding the atomic weights of the elements in the compound gives a for-

mula weight of 164.1 amu. Using 1 mol to write the ap-propriate conversion factor, we have

Check: The number of moles is less than 1, so the number of grams must be less than themolar mass, 164.1 g. Using rounded numbers to estimate, we have Thus, the magnitude of our answer is reasonable. Both the units (g) and the number ofsignificant figures (3) are correct.

PRACTICE EXERCISEWhat is the mass, in grams, of (a) 6.33 mol of and (b) ofsulfuric acid?Answers: (a) 532 g, (b)

Interconverting Masses and Numbers of ParticlesThe mole concept provides the bridge between masses and numbers of parti-cles. To illustrate how we can interconvert masses and numbers of particles,let’s calculate the number of copper atoms in an old copper penny. Such apenny weighs about 3 g, and we’ll assume that it is 100% copper:

Notice how dimensional analysis (Section 1.6) provides a straightforwardroute from grams to numbers of atoms. The molar mass and Avogadro’s num-ber are used as conversion factors to convert Notice also that our answer is a very large number. Any time you calculatethe number of atoms, molecules, or ions in an ordinary sample of matter, youcan expect the answer to be very large. In contrast, the number of moles in asample will usually be much smaller, often less than 1. The general procedurefor interconverting mass and number of formula units (atoms, molecules,ions, or whatever is represented by the chemical formula) of a substance issummarized in Figure 3.10 ¥.

grams ¡ moles ¡ atoms.

= 3 * 1022 Cu atoms

Cu atoms = (3 g Cu )a 1 mol Cu

63.5 g Cu b ¢6.02 * 1023 Cu atoms

1 mol Cu ≤

2.9 * 10-3 g

3.0 * 10-5 molNaHCO3

0.5 * 150 = 75 g.

Grams Ca(NO3)2 = A0.433 mol Ca(NO3)2 B ¢164.1 g Ca(NO3)2

1 mol Ca(NO3)2 ≤ = 71.1 g Ca(NO3)2

164.1 g Ca(NO3)2=Ca(NO3)2

Ca(NO3)2 .NO3

-,Ca2+

SAMPLE EXERCISE 3.12 | Calculating the Number of Molecules and Numberof Atoms from Mass

(a) How many glucose molecules are in 5.23 g of (b) How many oxygenatoms are in this sample?

SolutionAnalyze: We are given the number of grams and chemical formula and asked to cal-culate (a) the number of molecules and (b) the number of O atoms in the sample.(a) Plan: The strategy for determining the number of molecules in a given quantityof a substance is summarized in Figure 3.10. We must convert 5.23 g tomoles which can then be converted to molecules The first con-version uses the molar mass of The sec-ond conversion uses Avogadro’s number.

180.0 g C6H12O6.=C6H12O6C6H12O6 : 1 molC6H12O6.C6H12O6,

C6H12O6

C6H12O6 ?

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Solve: Molecules

Check: The magnitude of the answer is reasonable. Because the mass we began withis less than a mole, there should be less than molecules. We can make aballpark estimate of the answer:

molecules. The units (molecules) and significant figures(three) are appropriate.(b) Plan: To determine the number of O atoms, we use the fact that there are six O atomsin each molecule of Thus, multiplying the number of molecules bythe factor ( molecule ) gives the number of O atoms.Solve:

Check: The answer is simply 6 times as large as the answer to part (a). The numberof significant figures (three) and the units (atoms O) are correct.

PRACTICE EXERCISE(a) How many nitric acid molecules are in 4.20 g of (b) How many O atomsare in this sample?Answers: (a) (b) O

3.5 Empirical Formulas from AnalysesAs we learned in Section 2.6, the empirical formula for a substance tells us therelative number of atoms of each element it contains. Thus, the empirical for-mula indicates that water contains two H atoms for each O atom. Thisratio also applies on the molar level: 1 mol of contains 2 mol of H atomsand 1 mol of O atoms. Conversely, the ratio of the number of moles of each ele-ment in a compound gives the subscripts in a compound’s empirical formula.Thus, the mole concept provides a way of calculating the empirical formulas ofchemical substances, as shown in the following examples.

Mercury and chlorine combine to form a compound that is 73.9% mercuryand 26.1% chlorine by mass. This means that if we had a 100.0-g sample of thesolid, it would contain 73.9 g of mercury (Hg) and 26.1 g of chlorine (Cl). (Anysize sample can be used in problems of this type, but we will generally use100.0 g to simplify the calculation of mass from percentage.) Using the atomicweights of the elements to give us molar masses, we can calculate the numberof moles of each element in the sample:

We then divide the larger number of moles (0.735) by the smaller (0.368) to ob-tain a mole ratio of

moles of Clmoles of Hg

=0.735 mol Cl0.368 mol Hg

=1.99 mol Cl1 mol Hg

1.99 : 1:Cl : Hg

(26.1 g Cl )a 1 mol Cl35.5 g Cl

b = 0.735 mol Cl

(73.9 g Hg )a 1 mol Hg

200.6 g Hg b = 0.368 mol Hg

H2OH2O

1.20 * 1023 atoms4.01 * 1022 molecules HNO3,

HNO3 ?

= 1.05 * 1023 atoms O

Atoms O = (1.75 * 1022 molecules C6H12O6)¢ 6 atoms O1 molecule C6H12O6

≤C6H12O66 atoms O>1 C6H12O6C6H12O6.

15 * 1021 = 1.5 * 1022=2.5 * 10-2 * 6 * 10235>200 = 2.5 * 10-2 mol;

6.02 * 1023

= 1.75 * 1022 molecules C6H12O6

= (5.23 g C6H12O6 )¢ 1 mol C6H12O6

180.0 g C6H12O6 ≤ ¢6.022 * 1023 molecules C6H12O6

1 mol C6H12O6 ≤

C6H12O6


3-D MODELSWater, Hydrogen Peroxide

P. K. Thamburaj, “A Known-to-UnknownApproach to Teach About Empirical andMolecular Formulas,” J. Chem. Educ., Vol. 78,2001, 915–916.

Stephen DeMeo, “Making AssumptionsExplicit: How the Law of Conservation ofMatter Can Explain Empirical FormulaProblems,” J. Chem. Educ., Vol. 78, 2001,1050–1052.

An easy way to remember the strategy forconverting percentage composition to anempirical formula. “Percent to mass, Mass tomol, Divide by small, Multiply ’til whole.”Joel S. Thompson, “A Simple Rhyme for aSimple Formula,” J. Chem. Educ., Vol. 65,1988, 704.

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3.5 | Empirical Formulas from Analyses 97

Usemolarmass

Calculatemole ratio

Assume100 g

sample

Given: Find:

Mass %elements

Grams ofeach element

Moles ofeach element

Empiricalformula

Á Figure 3.11 Procedure for calculatingan empirical formula from percentagecomposition. The central part of the calculationis determining the number of moles of eachelement in the compound. The procedure is alsosummarized as “percent to mass, mass to mole,divide by small, multiply ’til whole.”

SAMPLE EXERCISE 3.13 | Calculating an Empirical FormulaAscorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid?

SolutionAnalyze: We are to determine an empirical formula of a compound from the mass percentages of its elements.Plan: The strategy for determining the empirical formula involves the three steps given in Figure 3.11.Solve: We first assume, for simplici-ty, that we have exactly 100 g of ma-terial (although any mass can beused). In 100 g of ascorbic acid, wehave 40.92 g C, 4.58 g H, and 54.50 g O.

Second, we calculate the number ofmoles of each element:

Moles O = (54.50 g O )a 1 mol O16.00 g O

b = 3.406 mol O

Moles H = (4.58 g H )a 1 mol H1.008 g H

b = 4.54 mol H

Moles C = (40.92 g C )a 1 mol C12.01 g C

b = 3.407 mol C

Third, we determine the simplestwhole-number ratio of moles by di-viding each number of moles by thesmallest number of moles, 3.406:

C :3.4073.406

= 1.000 H :4.54

3.406= 1.33 O :

3.4063.406

= 1.000

The ratio for H is too far from 1 to at-tribute the difference to experimen-tal error; in fact, it is quite close to

This suggests that if we multiplythe ratio by 3, we will obtain wholenumbers:

1 13 .

C : H : O = 3(1 : 1.33 : 1) = 3 : 4 : 3

The whole-number mole ratio givesus the subscripts for the empiricalformula: C3H4O3

Check: It is reassuring that the subscripts are moderately sized whole numbers. Otherwise, we have little by which to judge thereasonableness of our answer.

PRACTICE EXERCISEA 5.325-g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3.758 g of carbon,0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?Answer: C4H4O

ACTIVITYEmpirical Formula Determination: C8H6O

Because of experimental errors, the results may not lead to exact integers forthe ratios of moles. The number 1.99 is very close to 2, so we can confidentlyconclude that the empirical formula for the compound is This is theempirical formula because its subscripts are the smallest integers thatexpress the ratios of atoms present in the compound. • (Section 2.6) Thegeneral procedure for determining empirical formulas is outlined inFigure 3.11 Á.

HgCl2.

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Molecular Formula from Empirical FormulaFor any compound, the formula obtained from percentage compositions is al-ways the empirical formula. We can obtain the molecular formula from the em-pirical formula if we are given the molecular weight or molar mass of thecompound. The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in its empirical formula.• (Section 2.6) This multiple can be found by comparing the empirical for-mula weight with the molecular weight:

[3.11]

In Sample Exercise 3.13, for example, the empirical formula of ascorbic acidwas determined to be giving an empirical formula weight of

The experimentally de-termined molecular weight is 176 amu. Thus, the molecular weight is 2 timesthe empirical formula weight and the molecular formulamust therefore have twice as many of each kind of atom as the empirical for-mula. Consequently, we multiply the subscripts in the empirical formula by 2to obtain the molecular formula:

SAMPLE EXERCISE 3.14 | Determining a Molecular FormulaMesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empiri-cal formula of The experimentally determined molecular weight of this sub-stance is 121 amu. What is the molecular formula of mesitylene?

SolutionAnalyze: We are given an empirical formula and molecular weight and asked to de-termine a molecular formula.Plan: The subscripts in the molecular formula of a compound are whole-number mul-tiples of the subscripts in its empirical formula. To find the appropriate multiple, wemust compare the molecular weight with the formula weight of the empirical formula.Solve: First, we calculate the formula weight of the empirical formula,

Next, we divide the molecular weight by the empirical formula weight to obtain themultiple used to multiply the subscripts in

Only whole-number ratios make physical sense because we must be dealing withwhole atoms. The 3.02 in this case results from a small experimental error in the mol-ecular weight. We therefore multiply each subscript in the empirical formula by 3 togive the molecular formula: Check: We can have confidence in the result because dividing the molecular weightby the formula weight yields nearly a whole number.

PRACTICE EXERCISEEthylene glycol, the substance used in automobile antifreeze, is composed of 38.7%C, 9.7% H, and 51.6% O by mass. Its molar mass is (a) What is the empir-ical formula of ethylene glycol? (b) What is its molecular formula?Answers: (a) (b)

Combustion AnalysisThe empirical formula of a compound is based on experiments that give thenumber of moles of each element in a sample of the compound. That is why weuse the word “empirical,” which means “based on observation and experiment.”Chemists have devised a number of experimental techniques to determine

C2H6O2CH3O,

62.1 g>mol.

C9H12.

Molecular weight

Empirical formula weight=

12140.0

= 3.02

C3H4 :

3(12.0 amu) + 4(1.0 amu) = 40.0 amu

C3H4 :

C3H4.

C6H8O6.

1176>88.0 = 2.002,88.0 amu.=3(12.0 amu) + 4(1.0 amu) + 3(16.0 amu)

C3H4O3,

Whole-number multiple =molecular weight

empirical formula weight

An exploration of color change associatedwith the dehydration of copper sulfate. LeeR. Summerlin, Christie L. Borgford, and JulieB. Ealy, “Copper Sulfate: Blue to White,”Chemical Demonstrations, A Sourcebook forTeachers, Vol. 2 (Washington: AmericanChemical Society, 1988), pp. 69–70.

The combustion of methane, propane, andbutane are compared in this simpledemonstration ofstoichiometry. M. DaleAlexander and Wayne C. Wolsey,“Combustion of Hydrocarbons:A Stoichiometry Demonstration,” J. Chem.Educ., Vol. 70, 1993, 327–328.

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3.5 | Empirical Formulas from Analyses 99

O2

Furnacer

Á Figure 3.12 Apparatus to determine percentages of carbon and hydrogen in a compound.The compound is combusted to form and Copper oxide helps to oxidize traces of carbon and carbonmonoxide to carbon dioxide and to oxidize hydrogen to water.

H2O.CO2

empirical formulas. One of these is combustion analysis, which is commonlyused for compounds containing principally carbon and hydrogen as their com-ponent elements.

When a compound containing carbon and hydrogen is completely com-busted in an apparatus such as that shown in Figure 3.12 Á the carbon in thecompound is converted to and the hydrogen is converted to • (Section 3.2) The amounts of and produced are determined bymeasuring the mass increase in the and absorbers. From the massesof and we can calculate the number of moles of C and H in the origi-nal compound and thereby the empirical formula. If a third element is presentin the compound, its mass can be determined by subtracting the masses of Cand H from the compound’s original mass. Sample Exercise 3.15 shows how todetermine the empirical formula of a compound containing C, H, and O.

H2OCO2

H2OCO2

H2OCO2

H2O.CO2,

SAMPLE EXERCISE 3.15 | Determining Empirical Formula by Combustion AnalysisIsopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcoholproduces 0.561 g of and 0.306 g of Determine the empirical formula of isopropyl alcohol.

SolutionAnalyze: We are told that isopropyl alcohol contains C, H, and O atoms and given the quantities of and producedwhen a given quantity of the alcohol is combusted. We must use this information to determine the empirical formula for iso-propyl alcohol, a task that requires us to calculate the number of moles of C, H, and O in the sample.Plan: We can use the mole concept to calculate the number of grams of C present in the and the number of grams of H pre-sent in the These are the quantities of C and H present in the isopropyl alcohol before combustion. The number of grams ofO in the compound equals the mass of the isopropyl alcohol minus the sum of the C and H masses. Once we have the number ofgrams of C, H, and O in the sample, we can then proceed as in Sample Exercise 3.13: Calculate the number of moles of each ele-ment, and determine the mole ratio, which gives the subscripts in the empirical formula.

H2O.CO2

H2OCO2

H2O.CO2

Solve: To calculate the number ofgrams of C, we first use the molar massof toconvert grams of to moles of

Because there is only 1 C atom ineach molecule, there is 1 mol of Catoms per mole of molecules. Thisfact allows us to convert the moles of

to moles of C. Finally, we use themolar mass of C, 1 mol to convert moles of C to grams of C.Combining the three conversion fac-tors, we have:

C = 12.0 g C,CO2

CO2

CO2

CO2.CO2

1 mol CO2 = 44.0 g CO2,CO2,

Grams C = (0.561 g CO2 )¢ 1 mol CO2

44.0 g CO2 ≤ ¢ 1 mol C

1 mol CO2 ≤ a 12.0 g C

1 mol C b = 0.153 g C

The calculation of the number ofgrams of H from the grams of is similar, although we must remem-ber that there are 2 mol of H atomsper 1 mol of molecules:H2O

H2O

Grams H = (0.306 g H2O )¢ 1 mol H2O

18.0 g H2O ≤ ¢ 2 mol H

1 mol H2O ≤ a 1.01 g H

1 mol H b = 0.0343 g H

The total mass of the sample, 0.255g, is the sum of the masses of the C,H, and O. Thus, we can calculate themass of O as follows: = 0.255 g - (0.153 g + 0.0343 g) = 0.068 g O

Mass of O = mass of sample - (mass of C + mass of H)

Combustion analysis works only if completecombustion occurs:

.Fuel + O2 ¡ CO2 + H2O

If the compound contains only C, H, and O,the mass of the oxygen is the differencebetween the total mass and the masses of Cand H.

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3.6 Quantitative Information from Balanced EquationsThe coefficients in a chemical equation represent the relative numbers of mole-cules involved in a reaction. The mole concept allows us to convert this informa-tion to the masses of the substances. Consider the following balanced equation:

[3.12]

The coefficients indicate that two molecules of react with each molecule ofto form two molecules of It follows that the relative numbers of moles

are identical to the relative numbers of molecules:

2 molecules 1 molecule 2 molecules

2 mol 1 mol 2 mol

We can generalize this observation for all balanced chemical equations: The co-efficients in a balanced chemical equation indicate both the relative numbers of mole-cules (or formula units) involved in the reaction and the relative numbers of moles.Table 3.3 ¥ further summarizes this result and also shows how it corresponds

2(6.02 * 1023 molecules)1(6.02 * 1023 molecules)2(6.02 * 1023 molecules)

2 H2(g) + O2(g) ¡ 2 H2O(l)

H2O.O2

H2

2 H2(g) + O2(g) ¡ 2 H2O(l)

TABLE 3.3 Information from a Balanced Equation

Equation:

Molecules:

2 H2(g) O2(g)�

�

��

2 H2O(l)

2 molecules H2

4.0 amu H22 mol H24.0 g H2

1 molecule O2

32.0 amu O21 mol O232.0 g O2

36.0 amu H2O2 mol H2O36.0 g H2O

2 molecules H2O

Mass (amu):Amount (mol):Mass (g):


In Sample Exercise 3.15, how do you explain the fact that the ratios arerather than exact integers 3 : 8 : 1?2.98 : 7.91 : 1.00,

C : H : O

We then calculate the number ofmoles of C, H, and O in the sample:

Moles O = (0.068 g O )a 1 mol O16.0 g O

b = 0.0043 mol O

Moles H = (0.0343 g H )a 1 mol H1.01 g H

b = 0.0340 mol H

Moles C = (0.153 g C )a 1 mol C12.0 g C

b = 0.0128 mol C

To find the empirical formula, we must compare the relative number of moles of each element in the sample. The relative numberof moles of each element is found by dividing each number by the smallest number, 0.0043. The mole ratio of so obtainedis The first two numbers are very close to the whole numbers 3 and 8, giving the empirical formula Check: The subscripts work out to be moderately sized whole numbers, as expected.

PRACTICE EXERCISE(a) Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225-gsample of this compound produces and What is the empirical formula of caproic acid? (b) Caproic acidhas a molar mass of What is its molecular formula?Answers: (a) (b) C6H12O2C3H6O,

116 g>mol.0.209 g H2O.0.512 g CO2

C3H8O.2.98 : 7.91 : 1.00.C : H : O

Christer Svensson, “How Many Digits ShouldWe Use in Formula or Molar MassCalculations,” J. Chem. Educ., Vol. 81, 2004,827–829.

Due to experimental or round-off errors,thecoefficients may come out close to wholenumbers (see text). You should then roundthem off to that whole number. However, ifthe coefficients come out close to commonfractions (e.g., ), the formulashould be multiplied by the least commondenominator (4, 3, 2, respectively) and notrounded.

1>2,1>3,1>4,

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3.6 | Quantitative Information from Balanced Equations 101

to the law of conservation of mass. Notice that the total mass of the reactantsequals the total mass of the products (36.0 g).

The quantities 2 mol and which are given by thecoefficients in Equation 3.12, are called stoichiometrically equivalent quantities.The relationship between these quantities can be represented as

where the symbol means “is stoichiometrically equivalent to.” In otherwords, Equation 3.12 shows 2 mol of and 1 mol of forming 2 mol of These stoichiometric relations can be used to convert between quantities of re-actants and products in a chemical reaction. For example, the number of molesof produced from 1.57 mol of can be calculated as follows:


When reacts with to form how many moles of are con-sumed in the process?

As an additional example, consider the combustion of butane thefuel in disposable cigarette lighters:

[3.13]

Let’s calculate the mass of produced when 1.00 g of is burned. The co-efficients in Equation 3.13 tell how the amount of consumed is related tothe amount of produced: In order to use this re-lationship, however, we must use the molar mass of to convert grams of

to moles of Because 1 mol we have

We can then use the stoichiometric factor from the balanced equation, 2 molto calculate moles of

Finally, we can calculate the mass of the in grams, using the molar mass of

Thus, the conversion sequence is

Gramsreactant

Gramsproduct

Molesproduct

Molesreactant

= 3.03 g CO2

Grams CO2 = (6.88 * 10-2 mol CO2 )¢ 44.0 g CO2

1 mol CO2 ≤

44.0 g CO2):=CO2 (1 mol CO2

CO2,

= 6.88 * 10-2 mol CO2

Moles CO2 = (1.72 * 10-2 mol C4H10 )¢ 8 mol CO2

2 mol C4H10 ≤

CO2 :C4H10 � 8 mol CO2,

= 1.72 * 10-2 mol C4H10

Moles C4H10 = (1.00 g C4H10 )¢ 1 mol C4H10

58.0 g C4H10 ≤

C4H10 = 58.0 g C4H10,C4H10.C4H10

C4H10

2 mol C4H10 � 8 mol CO2.CO2

C4H10

C4H10CO2

2 C4H10(l) + 13 O2(g) ¡ 8 CO2(g) + 10 H2O(g)

(C4H10),

H2H2O,H21.57 mol O2

Moles H2O = (1.57 mol O2 )a2 mol H2O1 mol O2

b = 3.14 mol H2O

O2H2O

H2O.O2H2

�

2 mol H2 � 1 mol O2 � 2 mol H2O

2 mol H2O,1 mol O2,H2,(4.0 g + 32.0 g)

John Olmsted III, “Amounts Tables as aDiagnostic Tool for Flawed StoichiometricReasoning,” J. Chem. Educ., Vol. 76, 1999,52–54.

Carla R. Krieger, “Stoogiometry: A CognitiveApproach to Teaching Stoichiometry,”J. Chem. Educ., Vol. 74, 1997, 306–309.

Richard L. Poole, “Teaching Stoichiometry:A Two-Cycle Approach,” J. Chem. Educ.,Vol. 66, 1989, 57.

John J. Fortman, “Pictorial Analogies XII:Stoichiometric Calculations,” J. Chem. Educ.,Vol. 71, 1994, 571–572.

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Usecoefficientsof A and B

frombalanced equation

Usemolar mass

of A

Usemolar mass

of B

Given: Find:

Grams ofsubstance A

Moles ofsubstance A

Moles ofsubstance B

Grams ofsubstance B

» Figure 3.13 The procedure forcalculating amounts of reactants orproducts in a reaction. The number of gramsof a reactant consumed or of a product formed ina reaction can be calculated, starting with thenumber of grams of one of the other reactants orproducts. Notice how molar masses and thecoefficients in the balanced equation are used.

SAMPLE EXERCISE 3.16 | Calculating Amounts of Reactants and Products How many grams of water are produced in the oxidation of 1.00 g of glucose,

SolutionAnalyze: We are given the mass of a reactant and are asked to determine the mass of a product in the given equation.Plan: The general strategy, as outlined in Figure 3.13, requires three steps. First, the amount of must be converted fromgrams to moles. We can then use the balanced equation, which relates the moles of to the moles of 1 mol

Finally, the moles of must be converted to grams.H2OC6H12O6 � 6 mol H2O.H2O:C6H12O6

C6H12O6

C6H12O6(s) + 6 O2(g) ¡ 6 CO2(g) + 6 H2O(l)

C6H12O6 ?

Solve: First, use the molar mass ofto convert from grams

to moles C6H12O6 :C6H12O6

C6H12O6 Moles C6H12O6 = (1.00 g C6H12O6 )¢ 1 mol C6H12O6

180.0 g C6H12O6 ≤

Second, use the balanced equationto convert moles of tomoles of H2O:

C6H12O6 Moles H2O = (1.00 g C6H12O6 )¢ 1 mol C6H12O6

180.0 g C6H12O6 ≤ ¢ 6 mol H2O

1 mol C6H12O6 ≤

Third, use the molar mass of toconvert from moles of to gramsof H2O:

H2OH2O

= 0.600 g H2O

Grams H2O = (1.00 g C6H12O6 )¢ 1 mol C6H12O6

180.0 g C6H12O6 ≤ ¢ 6 mol H2O

1 mol C6H12O6 ≤ ¢ 18.0 g H2O

1 mol H2O ≤

These steps can be combined in a single sequence of factors:

Similarly, we can calculate the amount of consumed or producedin this reaction. For example, to calculate the amount of consumed, we againrely on the coefficients in the balanced equation to give us the appropriate stoi-chiometric factor: 2 mol

Figure 3.13 Á summarizes the general approach used to calculate the quan-tities of substances consumed or produced in chemical reactions. The balancedchemical equation provides the relative numbers of moles of reactants andproducts involved in the reaction.

= 3.59 g O2

Grams O2 = (1.00 g C4H10 )¢ 1 mol C4H10

58.0 g C4H10 ≤ ¢ 13 mol O2

2 mol C4H10 ≤ ¢ 32.0 g O2

1 mol O2 ≤

C4H10 � 13 mol O2 :

O2

H2OO2

= 3.03 g CO2

Grams CO2 = (1.00 g C4H10 )¢ 1 mol C4H10

58.0 g C4H10 ≤ ¢ 8 mol CO2

2 mol C4H10 ≤ ¢ 44.0 g CO2

1 mol CO2 ≤

ACTIVITYStoichiometry Calculation

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3.6 | Quantitative Information from Balanced Equations 103

The steps can be summarized in adiagram like that in Figure 3.13:

5.56 � 10�3 mol C6H12O66 mol H2O

1 mol C6H12O6

no directcalculation

3.33 � 10�2 mol H2O

1.00 g C6H12O6 0.600 g H2O

180.0 g C6H12O6

18.0 g H2O1 mol H2O

1 mol C6H12O6�

�

�

Check: An estimate of the magnitude of our answer, and agrees with the exact calculation. The units,grams are correct. The initial data had three significant figures, so three significant figures for the answer is correct.Comment: An average person ingests 2 L of water daily and eliminates 2.4 L. The difference between 2 L and 2.4 L is producedin the metabolism of foodstuffs, such as in the oxidation of glucose. (Metabolism is a general term used to describe all the chemi-cal processes of a living animal or plant.) The desert rat (kangaroo rat), on the other hand, apparently never drinks water. It sur-vives on its metabolic water.

PRACTICE EXERCISEThe decomposition of is commonly used to prepare small amounts of in the laboratory:

How many grams of can be prepared from 4.50 g of Answer: 1.77 g

KClO3 ?O23 O2(g).+2 KClO3(s) ¡ 2 KCl(s)O2KClO3

H2O,0.1 * 6 = 0.6,18>180 = 0.1

of remained fairly constant from the last Ice Age, some10,000 years ago, until roughly the beginning of the IndustrialRevolution, about 300 years ago. Since that time the concentra-tion of has increased by about 25% (Figure 3.14 ¥).

Although is a minor component of the atmosphere, itplays a significant role by absorbing radiant heat, acting muchlike the glass of a greenhouse. For this reason, we often refer to

and other heat-trapping gases as greenhouse gases, andwe call the warming caused by these gases the greenhouse effect.Most atmospheric scientists believe that the accumulation of

and other heat-trapping gases has begun to change the cli-mate of our planet. Still, it is recognized that the factors affect-ing climate are complex and incompletely understood.

We will examine the greenhouse effect more closely inChapter 18.Related Exercises: 3.64, 3.96, 3.99, and 3.103

CO2

CO2

CO2

CO2

CO2

CHEMISTRY AT WORK | and the Greenhouse EffectCO2

Coal and petroleum provide the fuels that we use to generateelectricity and power our industrial machinery. These fuels

are composed primarily of hydrocarbons and other carbon-containing substances. As we have seen, the combustion of1.00 g of produces 3.03 g of Similarly, a gallon(3.78 L) of gasoline ( and approximatecomposition ) produces about 8 kg (18 lb) of Com-bustion of such fuels releases about 20 billion tons of intothe atmosphere annually.

Much is absorbed into oceans or used by plants in pho-tosynthesis. Nevertheless, we are now generating muchfaster than it is being absorbed. Chemists have monitored at-mospheric concentrations since 1958. Analysis of airtrapped in ice cores taken from Antarctica and Greenland makesit possible to determine the atmospheric levels of duringthe past 160,000 years. These measurements reveal that the level

CO2

CO2

CO2

CO2

CO2

CO2.C8H18

density = 0.70 g>mLCO2.C4H10

1850 1870 1890 1910 1930 1950 1970 1990 2003280

290

300

310

320

330

340

350

360

370

380

Date

CO

2 co

ncen

trat

ion

(ppm

)

« Figure 3.14 Increasingconcentration of atmospheric

The world-wide concentrationof has increased from about290 ppm to over 370 ppm over thepast 150 years. The concentration inppm is the number of molecules of

per million molecules ofair. Data before 1958 came fromanalyses of air trapped in bubblesof glacial ice.

(106)CO2

CO2

CO2 .

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SAMPLE EXERCISE 3.17 | Calculating Amounts of Reactants and ProductsSolid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide.The lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium car-bonate and liquid water. How many grams of carbon dioxide can be absorbed by 1.00g of lithium hydroxide?

SolutionAnalyze: We are given a verbal description of a reaction and asked to calculate thenumber of grams of one reactant that reacts with 1.00 g of another.Plan: The verbal description of the reaction can be used to write a balanced equation:

We are given the grams of LiOH and asked to calculate grams of We can ac-complish this task by using the following sequence of conversions:

The conversion from grams of LiOH to moles of LiOH requires the molar mass ofLiOH The conversion of moles of LiOH tomoles of is based on the balanced chemical equation: 2 mol To convert the number of moles of to grams, we must use the molar mass of

Solve:

Check: Notice that and is slightly less than 1. Thus,the magnitude of the answer is reasonable based on the amount of starting LiOH; thesignificant figures and units are appropriate, too.

PRACTICE EXERCISEPropane, is a common fuel used for cooking and home heating. What mass of

is consumed in the combustion of 1.00 g of propane?Answer: 3.64 g

3.7 Limiting ReactantsSuppose you wish to make several sandwiches using one slice of cheese andtwo slices of bread for each sandwich. Using and

the recipe for making a sandwich can be represented like achemical equation:

If you have 10 slices of bread and 7 slices of cheese, you will be able to make onlyfive sandwiches before you run out of bread. You will have 2 slices of cheese leftover. The amount of available bread limits the number of sandwiches.

An analogous situation occurs in chemical reactions when one of the reac-tants is used up before the others. The reaction stops as soon as any one of thereactants is totally consumed, leaving the excess reactants as leftovers. Sup-pose, for example, that we have a mixture of and whichreact to form water:

Because the number of moles of needed to react withall the is

Moles O2 = (10 mol H2 )¢ 1 mol O2

2 mol H2 ≤ = 5 mol O2

H2

O22 mol H2 � 1 mol O2,

2 H2(g) + O2(g) ¡ 2 H2O(g)

7 mol O2,10 mol H2

2 Bd + Ch ¡ Bd2Ch

Bd2Ch = sandwich, Ch = cheese,Bd = bread,

O2

C3H8,

44>4824 * 2 = 48,23.95 L 24,

(1.00 g LiOH )a 1 mol LiOH

23.95 g LiOH b ¢ 1 mol CO2

2 mol LiOH ≤ ¢44.01 g CO2

1 mol CO2 ≤ = 0.919 g CO2

12.01 + 2(16.00) = 44.01 g>mol.CO2 :CO2

LiOH � 1 mol CO2.CO2

(6.94 + 16.00 + 1.01 = 23.95 g>mol).

Grams LiOH ¡ moles LiOH ¡ moles CO2 ¡ grams CO2.

CO2.

2 LiOH(s) + CO2(g) ¡ Li2CO3(s) + H2O(l)

Analogies dealing with stoichiometry,limiting reactants and percent yield. LilianaHaim, Eduardo Corton, Santiago Kocmur,and Lydia Galagovsky. “LearningStoichiometry with Hamburger Sandwiches.”J. Chem. Educ., Vol. 80, 2003, 1021–1022.

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3.7 | Limiting Reactants 105

10 H2O and 2 O210 H2 and 7 O2

After reactionBefore reaction

Á Figure 3.15 Example illustrating a limiting reactant. Because the is completely consumed, it isthe limiting reagent in this case. Because there is a stoichiometric excess of some is left over at the end ofthe reaction. The amount of formed is related directly to the amount of consumed.H2H2O

O2 ,H2

Because was available at the start of the reaction, will still be present when all the is consumed. The example we

have considered is depicted on a molecular level in Figure 3.15 Á.The reactant that is completely consumed in a reaction is called either the

limiting reactant or limiting reagent because it determines, or limits, the amountof product formed. The other reactants are sometimes called either excess reac-tants or excess reagents. In our example, is the limiting reactant, which meansthat once all the has been consumed, the reaction stops. is the excess re-actant, and some is left over when the reaction stops.

There are no restrictions on the starting amounts of the reactants in any re-action. Indeed, many reactions are carried out using an excess of one reagent.The quantities of reactants consumed and the quantities of products formed,however, are restricted by the quantity of the limiting reactant. When a com-bustion reaction takes place in the open air, oxygen is plentiful and is thereforethe excess reactant. You may have had the unfortunate experience of runningout of gasoline while driving. The car stops because you’ve run out of the limit-ing reactant in the combustion reaction, the fuel.

Before we leave our present example, let’s summarize the data in a tabularform:

Initial quantities: 10 mol 7 mol 0 molChange (reaction):

Final quantities: 0 mol 2 mol 10 mol

The initial amounts of the reactants are what we started with (10 mol and ). The second line in the table (change) summarizes the amountsof the reactants consumed and the amount of the product formed in the reac-tion. These quantities are restricted by the quantity of the limiting reactant anddepend on the coefficients in the balanced equation. The mole ratio

conforms to the ratio of the coefficients in the balancedequation, The changes are negative for the reactants because they areconsumed during the reaction and positive for the product because it is formedduring the reaction. Finally, the quantities in the third line of the table (finalquantities) depend on the initial quantities and their changes, and these entriesare found by adding the entries for the initial quantity and change for each col-umn. There is none of the limiting reactant left at the end of the reaction.All that remains is and 10 mol H2O.2 mol O2

(H2)

2 : 1 : 2.10 : 5 : 10=H2 : O2 : H2O

7 mol O2

H2

+10 mol-5 mol-10 mol

2 H2(g) + O2(g) ¡ 2 H2O(g)

O2H2

H2

H22 mol O2=7 mol O2 - 5 mol O27 mol O2

ANIMATIONLimiting Reagent

ACTIVITYLimiting Reagents

Students often combine the amounts ofproduct calculated from each reactant whenworking with limiting reagents. The reagentthat leads to the smallest amount of productis the limiting reagent. The amount ofproduct formed in the reaction is thesmallest calculated amount, not the sum.

Always convert the amounts of reactantsinto molar quantities and use the reactionstoichiometry to determine the limitingreagent.

Ernest F. Silversmith, “Limiting and ExcessReagents, Theoretical Yield,” J. Chem. Educ.,Vol. 62, 1985, 61.

Zoltan Toth, “Limiting Reactant; AnAlternative Analogy,” J. Chem.Educ., Vol. 76,1999, 934.

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SAMPLE EXERCISE 3.18 | Calculating the Amount of Product Formed from a Limiting ReactantThe most important commercial process for converting from the air into nitrogen-containing compounds is based on the reac-tion of and to form ammonia

How many moles of can be formed from 3.0 mol of and 6.0 mol of

SolutionAnalyze: We are asked to calculate the number of moles of product, given the quantities of each reactant, and avail-able in a reaction. Thus, this is a limiting reactant problem.Plan: If we assume that one reactant is completely consumed, we can calculate how much of the second reactant is needed in thereaction. By comparing the calculated quantity with the available amount, we can determine which reactant is limiting. We thenproceed with the calculation, using the quantity of the limiting reactant.

H2,N2NH3,

H2 ?N2NH3

N2(g) + 3 H2(g) ¡ 2 NH3(g)

(NH3):H2N2

N2

Solve: The number of moles of needed for complete consumptionof 3.0 mol of isN2

H2

Moles H2 = (3.0 mol N2 )¢ 3 mol H2

1 mol N2 ≤ = 9.0 mol H2

Because only is available,we will run out of before the isgone, and will be the limiting re-actant. We use the quantity of thelimiting reactant, to calculate thequantity of produced:NH3

H2,

H2

N2H2

6.0 mol H2

Moles NH3 = (6.0 mol H2 )¢2 mol NH3

3 mol H2 ≤ = 4.0 mol NH3

Comment: The table on the rightsummarizes this example:

Initial quantities: 3.0 mol 6.0 mol 0 molChange (reaction):

Final quantities: 1.0 mol 0 mol 4.0 mol

+4.0 mol-6.0 mol-2.0 mol

N2(g) + 3 H2(g) ¡ 2 NH3(g)

Notice that we can calculate not only the number of moles of formed but also the number of moles of each of the reactantsremaining after the reaction. Notice also that although the number of moles of present at the beginning of the reactionis greater than the number of moles of present, the is nevertheless the limiting reactant because of its larger coefficient inthe balanced equation.Check: The summarizing table shows that the mole ratio of reactants used and product formed conforms to the coefficients in thebalanced equation, Also, because is the limiting reactant, it is completely consumed in the reaction, leaving 0 mol atthe end. Because has two significant figures, our answer has two significant figures.

PRACTICE EXERCISEConsider the reaction A mixture of 1.50 mol of Al and 3.00 mol of is allowed to react. (a) Which is the limiting reactant? (b) How many moles of are formed? (c) How many moles of the excess reactant remainat the end of the reaction?Answers: (a) Al, (b) 1.50 mol, (c) 0.75 mol Cl2

AlCl3

Cl22 Al(s) + 3 Cl2(g) ¡ 2 AlCl3(s).

2.0 mol H2

H21 : 3 : 2.

H2N2

H2

NH3

Romeu C. Rocha-Filho, “Electron Results andReaction Yields,” J. Chem. Educ., Vol. 64,1987, 248.

A. H. Kalantar, “Limiting Reagent ProblemsMade Simple for Students,” J. Chem. Educ.,Vol. 62, 1985, 106.

SAMPLE EXERCISE 3.19 | Calculating the Amount of Product Formed from aLimiting Reactant

Consider the following reaction:

Suppose a solution containing 3.50 g of is mixed with a solution containing6.40 g of How many grams of can be formed?

SolutionAnalyze: We are asked to calculate the amount of a product, given the amounts oftwo reactants, so this is a limiting reactant problem.Plan: We must first identify the limiting reagent. To do so, we can calculate the num-ber of moles of each reactant and compare their ratio with that required by the bal-anced equation. We then use the quantity of the limiting reagent to calculate the massof that forms.Solve: From the balanced equation, we have the following stoichiometric relations:

2 mol Na3PO4 � 3 mol Ba(NO3)2 � 1 mol Ba3(PO4)2

Ba3(PO4)2

Ba3(PO4)2Ba(NO3)2 .Na3PO4

2 Na3PO4(aq) + 3 Ba(NO3)2(aq) ¡ Ba3(PO4)2(s) + 6 NaNO3(aq)

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3.7 | Limiting Reactants 107

Using the molar mass of each substance, we can calculate the number of moles ofeach reactant:

Thus, there are slightly more moles of than moles of The coeffi-cients in the balanced equation indicate, however, that the reaction requires

for each [That is, 1.5 times more moles of are needed than moles of ] Thus, there is insufficient to complete-ly consume the That means that is the limiting reagent. We there-fore use the quantity of to calculate the quantity of product formed. We canbegin this calculation with the grams of but we can save a step by startingwith the moles of that were calculated previously in the exercise:

Check: The magnitude of the answer seems reasonable: Starting with the numbersin the two conversion factors on the right, we have The units are correct, and the number of significant figures (three) corresponds tothe number in the quantity of Comment: The quantity of the limiting reagent, can also be used to deter-mine the quantity of formed (4.16 g) and the quantity of used(2.67 g). The number of grams of the excess reagent, remaining at the end ofthe reaction equals the starting amount minus the amount consumed in the reaction,

PRACTICE EXERCISEA strip of zinc metal having a mass of 2.00 g is placed in an aqueous solution con-taining 2.50 g of silver nitrate, causing the following reaction to occur:

(a) Which reactant is limiting? (b) How many grams of Ag will form? (c) How manygrams of will form? (d) How many grams of the excess reactant will be leftat the end of the reaction?Answers: (a) (b) 1.59 g, (c) 1.39 g, (d) 1.52 g Zn

Theoretical YieldsThe quantity of product that is calculated to form when all of the limiting reactantreacts is called the theoretical yield. The amount of product actually obtained ina reaction is called the actual yield. The actual yield is almost always less than (andcan never be greater than) the theoretical yield. There are many reasons for thisdifference. Part of the reactants may not react, for example, or they may react in away different from that desired (side reactions). In addition, it is not always pos-sible to recover all of the product from the reaction mixture. The percent yield ofa reaction relates the actual yield to the theoretical (calculated) yield:

[3.14]

In the experiment described in Sample Exercise 3.19, for example, we cal-culated that 4.92 g of should form when 3.50 g of is mixedwith 6.40 g of The 4.92 g is the theoretical yield of in thereaction. If the actual yield turned out to be 4.70 g, the percent yield would be

4.70 g

4.92 g* 100% = 95.5%

Ba3(PO4)2Ba(NO3)2 .Na3PO4Ba3(PO4)2

Percent yield =actual yield

theoretical yield* 100%

AgNO3,

Zn(NO3)2

Zn(s) + 2 AgNO3(aq) ¡ 2 Ag(s) + Zn(NO3)2(aq)

3.50 g - 2.67 g = 0.82 g.

Na3PO4,Na3PO4NaNO3

Ba(NO3)2 ,Ba(NO3)2 .

200 * 0.025 = 5.600>3 = 200;

= 4.92 g Ba3(PO4)2

Grams Ba3(PO4)2 = (0.0245 mol Ba(NO3)2 )¢1 mol Ba3(PO4)2

3 mol Ba(NO3)2 ≤ ¢ 602 g Ba3(PO4)2

1 mol Ba3(PO4)2 ≤

Ba(NO3)2

Ba(NO3)2 ,Ba(NO3)2

Ba(NO3)2Na3PO4.Ba(NO3)2Na3PO4.

Ba(NO3)22 mol Na3PO4.3 mol Ba(NO3)2

Na3PO4.Ba(NO3)2

Moles Ba(NO3)2 = (6.40 g Ba(NO3)2 )¢ 1 mol Ba(NO3)2

261 g Ba(NO3)2 ≤ = 0.0245 mol Ba(NO3)2

Moles Na3PO4 = (3.50 g Na3PO4 )¢ 1 mol Na3PO4

164 g Na3PO4 ≤ = 0.0213 mol Na3PO4

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SAMPLE EXERCISE 3.20 | Calculating the Theoretical Yield and Percent Yieldfor a Reaction

Adipic acid, is used to produce nylon. The acid is made commercially bya controlled reaction between cyclohexane and

(a) Assume that you carry out this reaction starting with 25.0 g of cyclohexane andthat cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid?

(b) If you obtain 33.5 g of adipic acid from your reaction, what is the percentyield of adipic acid?

SolutionAnalyze: We are given a chemical equation and the quantity of the limiting reactant(25.0 g of ). We are asked first to calculate the theoretical yield of a product

and then to calculate its percent yield if only 33.5 g of the substance isactually obtained.Plan:

(a) The theoretical yield, which is the calculated quantity of adipic acid formed inthe reaction, can be calculated using the following sequence of conversions:

(b) The percent yield is calculated by comparing the actual yield (33.5 g) to thetheoretical yield using Equation 3.14.Solve:

(a)

(b)

Check: Our answer in (a) has the appropriate magnitude, units, and significant fig-ures. In (b) the answer is less than 100% as necessary.

Percent yield =actual yield

theoretical yield* 100% =

33.5 g

43.5 g* 100% = 77.0%

= 43.5 g H2C6H8O4

* ¢2 mol H2C6H8O4

2 mol C6H12 ≤ ¢146.0 g H2C6H8O4

1 mol H2C6H8O4 ≤

Grams H2C6H8O4 = (25.0 g C6H12 )¢ 1 mol C6H12

84.0 g C6H12 ≤

g C6H12 ¡ mol C6H12 ¡ mol H2C6H8O4 ¡ g H2C6H8O4.

(H2C6H8O4)C6H12

2 C6H12(l) + 5 O2(g) ¡ 2 H2C6H8O4(l) + 2 H2O(g)

O2 :(C6H12)H2C6H8O4,

without knowing much about the concept being tested.)Thus, you should not jump to the conclusion that becauseone of the choices looks correct, it must be so.

If a multiple-choice question involves a calculation,perform the calculation, quickly double-check your work,and only then compare your answer with the choices. Ifyou find a match, you have probably found the correctanswer. Keep in mind, though, that your instructor hasanticipated the most common errors one can make insolving a given problem and has probably listed the in-correct answers resulting from those errors. Thus, alwaysdouble-check your reasoning and make sure to use di-mensional analysis to arrive at the correct answer, withthe correct units.

In multiple-choice questions that don’t involve calcu-lations, one way to proceed if you are not sure of the cor-rect choice is to eliminate all the choices you know forsure to be incorrect. Additionally, the reasoning you usedin eliminating incorrect choices will help you in reasoningabout which is the one correct choice.

STRATEGIES IN CHEMISTRY | How to Take a Test

At about this time in your study of chemistry, you are likelyto face your first hour-long examination. The best way to

prepare for the exam is to study and do homework diligentlyand to make sure you get help from the instructor on any mate-rial that is unclear or confusing. (See the advice for learning andstudying chemistry presented in the preface of the book.) Wepresent here some general guidelines for taking tests.

Depending on the nature of your course, the exam couldconsist of a variety of different types of questions. Let’s con-sider some of the more common types and how they can bestbe addressed.

1. Multiple-choice questionsIn large-enrollment courses, the most common kind oftesting device is the multiple-choice question. You aregiven the problem and usually presented with four or fivepossible answers from which you must select the correctone. The first thing to realize is that the instructor haswritten the question so that all of the answer choices ap-pear at first glance to be correct. (There would be littlepoint in offering choices you could tell were wrong even

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Summary and Key Terms 109

S U M M A RY A N D K E Y T E R M S

Introduction and Section 3.1 The study of thequantitative relationships between chemical formulasand chemical equations is known as stoichiometry.One of the important concepts of stoichiometry is thelaw of conservation of mass, which states that the totalmass of the products of a chemical reaction is the sameas the total mass of the reactants. The same numbers ofatoms of each type are present before and after a chem-ical reaction. A balanced chemical equation shows equalnumbers of atoms of each element on each side of theequation. Equations are balanced by placing coefficientsin front of the chemical formulas for the reactants andproducts of a reaction, not by changing the subscripts inchemical formulas.

Section 3.2 Among the reaction types described in thischapter are (1) combination reactions, in which two reac-tants combine to form one product; (2) decomposition re-actions, in which a single reactant forms two or more

products; and (3) combustion reactions in oxygen, inwhich a hydrocarbon or related compound reacts with

to form and

Section 3.3 Much quantitative information can be deter-mined from chemical formulas and balanced chemicalequations by using atomic weights. The formula weight ofa compound equals the sum of the atomic weights of theatoms in its formula. If the formula is a molecular formula,the formula weight is also called the molecular weight.Atomic weights and formula weights can be used to deter-mine the elemental composition of a compound.Section 3.4 A mole of any substance is Avogadro’snumber of formula units of that substance.The mass of a mole of atoms, molecules, or ions (themolar mass) equals the formula weight of that materialexpressed in grams. The mass of one molecule of for example, is 18 amu, so the mass of 1 mol of is18 g. That is, the molar mass of is 18 g>mol.H2O

H2OH2O,

16.02 * 10232

H2O.CO2O2

2. Calculations in which you must show your workYour instructor may present you with a numerical problemin which you are to show your work in arriving at a solution.In questions of this kind, you may receive partial credit evenif you don’t arrive at the correct answer, depending onwhether the instructor can follow your line of reasoning. It isimportant, therefore, to be as neat and organized as you canbe, given the pressures of exam taking. It is helpful in ap-proaching such questions to take a few moments to thinkabout the direction you are going to take in solving the prob-lem, and you may even want to write a few words or a dia-gram on the test paper to indicate your plan of attack. Thenwrite out your calculations as neatly as you can. Show theunits for every number you write down, and use dimension-al analysis as much as you can, showing how units cancel.

3. Questions requiring drawingsSometimes a test question will require you to draw achemical structure, a diagram related to chemical bond-ing, or a figure showing some kind of chemical process.Questions of this kind will come later in the course, but it

is useful to talk about them here. (You should review thisbox before each exam you take, to remind yourself ofgood exam-taking practices.) Be sure to label your draw-ing as completely as possible.

4. Other types of questionsOther exam questions you might encounter include true-false questions and ones in which you are given a list andasked to indicate which members of the list match somecriterion given in the question. Often students answersuch questions incorrectly because, in their haste, theymisunderstand the nature of the question. Whatever thequestion form, ask yourself this: What is the instructortesting here? What material am I supposed to know thatthis question covers?Finally, if you find that you simply don’t see how to ar-

rive at a reasoned response to a question, don’t linger over thequestion. Put a check next to it and go on to the next one. Iftime permits, you can come back to the unanswered ques-tions, but lingering over a question when nothing is coming tomind is wasting time you may need to finish the exam.

PRACTICE EXERCISEImagine that you are working on ways to improve the process by which iron ore con-taining is converted into iron. In your tests you carry out the following reac-tion on a small scale:

(a) If you start with 150 g of as the limiting reagent, what is the theoretical yield ofFe? (b) If the actual yield of Fe in your test was 87.9 g, what was the percent yield?Answers: (a) 105 g Fe, (b) 83.7%

Fe2O3

Fe2O3(s) + 3 CO(g) ¡ 2 Fe(s) + 3 CO2(g)

Fe2O3

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Section 3.5 The empirical formula of any substancecan be determined from its percent composition by calcu-lating the relative number of moles of each atom in 100 gof the substance. If the substance is molecular in nature,its molecular formula can be determined from the empir-ical formula if the molecular weight is also known.

Sections 3.6 and 3.7 The mole concept can be used tocalculate the relative quantities of reactants and productsinvolved in chemical reactions. The coefficients in a bal-anced equation give the relative numbers of moles of the re-actants and products. To calculate the number of grams of a

product from the number of grams of a reactant, therefore,first convert grams of reactant to moles of reactant. We thenuse the coefficients in the balanced equation to convert thenumber of moles of reactant to moles of product. Finally,we convert moles of product to grams of product.

A limiting reactant is completely consumed in a reac-tion. When it is used up, the reaction stops, thus limiting thequantities of products formed. The theoretical yield of a re-action is the quantity of product calculated to form when allof the limiting reagent reacts. The actual yield of a reaction isalways less than the theoretical yield. The percent yieldcompares the actual and theoretical yields.

V I S U A L I Z I N G C O N C E P T S

3.3 The following diagram represents the collection of ele-ments formed by decomposition reaction. (a) If the bluespheres represent N atoms and the red ones represent Oatoms, what was the empirical formula of the originalcompound? (b) Could you draw a diagram representingthe molecules of the compound that had been decom-posed? Why or why not? [Section 3.2]

3.4 The following diagram represents the collection of and molecules formed by complete combustion ofa hydrocarbon. What is the empirical formula of thehydrocarbon? [Section 3.2]

H2OCO2

3.5 Glycine, an amino acid used by organisms to make pro-teins, is represented by the molecular model below. (a) Write its molecular formula. (b) Determine its molec-ular mass. (c) Calculate the percent nitrogen by mass inglycine. [Section 3.3 and 3.5]

3.6 The following diagram represents a high-temperaturereaction between and Based on this reaction,how many moles of each product can be obtained start-ing with 4.0 mol [Section 3.6]CH4 ?

H2O.CH4

CQCQ

CQ

CQ

CQ

CQ

3.1 The reaction between reactant A (blue spheres) and reac-tant B (red spheres) is shown in the following diagram:

Based on this diagram, which equation best describesthe reaction? [Section 3.1](a) (b)(c) (d)

3.2 Under appropriate experimental conditions, and COundergo a combination reaction to form Thedrawing below represents a sample of Make a cor-responding drawing of the CO needed to react com-pletely with the How did you arrive at the numberof CO molecules in your drawing? [Section 3.2]

H2.

H2.CH3OH.

H2

A + B2 ¡ AB22 A + B4 ¡ 2 AB2

A2 + 4 B ¡ 2 AB2A2 + B ¡ A2B

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CQ

Exercises 111

3.7 Nitrogen and hydrogen react to form ammo-nia Consider the mixture of and shown inthe accompanying diagram. The blue spheres representN, and the white ones represent H. Draw a representa-tion of the product mixture, assuming that the reactiongoes to completion. How did you arrive at your repre-sentation? What is the limiting reactant in this case?[Section 3.7]

H2N2(NH3).(H2)(N2) 3.8 Nitrogen monoxide and oxygen react to form nitrogen

dioxide. Consider the mixture of NO and shown in theaccompanying diagram. The blue spheres represent N, andthe red ones represent O. (a) Draw a representation of theproduct mixture, assuming that the reaction goes to com-pletion. What is the limiting reactant in this case? (b) Howmany molecules would you draw as products if thereaction had a percent yield of 75%? [Section 3.7]

NO2

O2

E X E R C I S E S

(d)(e)(f)

(g)3.13 Write balanced chemical equations to correspond to each

of the following descriptions: (a) Solid calcium carbide,reacts with water to form an aqueous solution of

calcium hydroxide and acetylene gas, (b) Whensolid potassium chlorate is heated, it decomposes to formsolid potassium chloride and oxygen gas. (c) Solid zincmetal reacts with sulfuric acid to form hydrogen gas andan aqueous solution of zinc sulfate. (d) When liquid phos-phorus trichloride is added to water, it reacts to formaqueous phosphorous acid, and aqueous hy-drochloric acid. (e) When hydrogen sulfide gas is passedover solid hot iron(III) hydroxide, the resultant reactionproduces solid iron(III) sulfide and gaseous water.

3.14 Convert these descriptions into balanced equations: (a) When sulfur trioxide gas reacts with water, a solu-tion of sulfuric acid forms. (b) Boron sulfide, re-acts violently with water to form dissolved boric acid,

and hydrogen sulfide gas. (c) Phosphine,combusts in oxygen gas to form gaseous water

and solid tetraphosphorus decoxide. (d) When solidmercury(II) nitrate is heated, it decomposes to formsolid mercury(II) oxide, gaseous nitrogen dioxide, andoxygen. (e) Copper metal reacts with hot concentratedsulfuric acid solution to form aqueous copper(II) sul-fate, sulfur dioxide gas, and water.

PH3(g),H3BO3,

B2S3(s),

H3PO3(aq),

C2H2.CaC2,

CH3NH2(g) + O2(g) ¡ CO2(g) + H2O(g) + N2(g)Ag2SO4(s) + NaNO3(aq)

AgNO3(aq) + Na2SO4(aq) ¡Ca(OH)2(aq) + H3PO4(aq) ¡ Ca3(PO4)2(s) + H2O(l)Ca3P2(s) + H2O(l) ¡ Ca(OH)2(aq) + PH3(g)Balancing Chemical Equations

3.9 (a) What scientific principle or law is used in the processof balancing chemical equations? (b) In balancing equa-tions, why shouldn’t subscripts in chemical formulas bechanged? (c) What are the symbols used to representgases, liquids, solids, and aqueous solutions in chemicalequations?

3.10 (a) What is the difference between adding a subscript 2to the end of the formula for CO to give and addinga coefficient in front of the formula to give 2 CO? (b) Isthe following chemical equation, as written, consistentwith the law of conservation of mass?

Why or why not?

3.11 Balance the following equations:(a)(b)(c)(d)(e)(f)(g)

3.12 Balance the following equations:(a)(b)(c) NH4NO3(s) ¡ N2(g) + O2(g) + H2O(g)

La2O3(s) + H2O(l) ¡ La(OH)3(aq)Li(s) + N2(g) ¡ Li3N(s)

MgSO4(aq) + (NH4)2SO4(aq)Mg3N2(s) + H2SO4(aq) ¡Fe(OH)3(s) + H2SO4(aq) ¡ Fe2(SO4)3(aq) + H2O(l)C5H10O2(l) + O2(g) ¡ CO2(g) + H2O(g)Al4C3(s) + H2O(l) ¡ Al(OH)3(s) + CH4(g)CH4(g) + Cl2(g) ¡ CCl4(l) + HCl(g)N2O5(g) + H2O(l) ¡ HNO3(aq)CO(g) + O2(g) ¡ CO2(g)

Mg3(PO4)2(s) + 6 H2O(l)

3 Mg(OH)2(s) + 2 H3PO4(aq) ¡

CO2

equation for the reaction. (b) When a hydrocarbonburns in air, what reactant besides the hydrocarbon isinvolved in the reaction? What products are formed?Write a balanced chemical equation for the combustionof benzene, in air.C6H6(l),

Patterns of Chemical Reactivity

3.15 (a) When the metallic element sodium combines withthe nonmetallic element bromine, how can youdetermine the chemical formula of the product? How doyou know whether the product is a solid, liquid, or gasat room temperature? Write the balanced chemical

Br2(l),

CQ

CQ

CQ

CQ

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3.16 (a) Determine the chemical formula of the productformed when the metallic element calcium combineswith the nonmetallic element oxygen, Write the bal-anced chemical equation for the reaction. (b) Whatproducts form when a compound containing C, H, andO is completely combusted in air? Write a balancedchemical equation for the combustion of acetone,

in air.

3.17 Write a balanced chemical equation for the reaction thatoccurs when (a) reacts with (b) bariumcarbonate decomposes into barium oxide and carbondioxide gas when heated; (c) the hydrocarbon styrene,

is combusted in air; (d) dimethylether,is combusted in air.

3.18 Write a balanced chemical equation for the reaction thatoccurs when (a) aluminum metal undergoes a combina-tion reaction with (b) copper(II) hydroxide de-composes into copper(II) oxide and water when heated;

O2(g);

CH3OCH3(g),C8H8(l),

Cl2(g);Mg(s)

C3H6O(l),

O2.

(c) heptane, burns in air; (d) the gasoline additiveMTBE (methyl tert-butyl ether), burns in air.

3.19 Balance the following equations, and indicate whetherthey are combination, decomposition, or combustionreactions:(a)(b)(c)(d)(e)

3.20 Balance the following equations, and indicate whetherthey are combination, decomposition, or combustionreactions:(a)(b)(c)(d)(e) K2O(s) + H2O(l) ¡ KOH(aq)

N2(g) + H2(g) ¡ NH3(g)C5H6O(l) + O2(g) ¡ CO2(g) + H2O(g)NH4NO3(s) ¡ N2O(g) + H2O(g)C3H6(g) + O2(g) ¡ CO2(g) + H2O(g)

C7H8O2(l) + O2(g) ¡ CO2(g) + H2O(g)PbCO3(s) ¡ PbO(s) + CO2(g)Li(s) + N2(g) ¡ Li3N(s)C2H4(g) + O2(g) ¡ CO2(g) + H2O(g)Al(s) + Cl2(g) ¡ AlCl3(s)

C5H12O(l),C7H16(l),

3.26 Calculate the percentage of carbon by mass in each ofthe compounds represented by the following models:

C

C

N

CC

H

H

H

(a) (b)

(d)(c)

S

O O

Isopentyl acetate(banana flavor)

(c) H3C

H3C

CH3C

H

H

C

H

H

C O C

H O

(b) HO C

O

H3CO H

H H

Vanillin(vanilla flavor)H

C

C C

C

C C

Formula Weights

3.21 Determine the formula weights of each of the followingcompounds: (a) (b) (c)(d) (e) aluminum sulfide, (f) iron(III) sul-fate, (g) disilicon hexabromide.

3.22 Determine the formula weights of each of the followingcompounds: (a) nitrous oxide, known as laughinggas and used as an anesthetic in dentistry; (b) benzoicacid, a substance used as a food preserva-tive; (c) the active ingredient in milk of mag-nesia; (d) urea, a compound used as anitrogen fertilizer; (e) isopentyl acetate, responsible for the odor of bananas.

3.23 Calculate the percentage by mass of oxygen in the fol-lowing compounds: (a) (b)(c) (d) sodium sulfate; (e) ammonium nitrate.

3.24 Calculate the percentage by mass of the indicated ele-ment in the following compounds: (a) carbon in acety-lene, a gas used in welding; (b) hydrogen inascorbic acid, also known as vitamin C;(c) hydrogen in ammonium sulfate, a sub-stance used as a nitrogen fertilizer; (d) platinum in

a chemotherapy agent called cisplatin;(e) oxygen in the female sex hormone estradiol,

(f) carbon in capsaicin, the com-pound that gives the hot taste to chili peppers.

3.25 Based on the following structural formulas, calculate thepercentage of carbon by mass present in each compound:

H C

O

H H

H H

Benzaldehyde(almond fragrance)H

C

C C

C

C C

(a)

C18H27NO3,C18H24O2 ;

PtCl2(NH3)2 ,

(NH4)2SO4,HC6H7O6,

C2H2,

Cr(NO3)3 ;CH3COOCH3 ;SO3 ;

CH3CO2C5H11,(NH2)2CO,

Mg(OH)2,HC7H5O2,

N2O,

Ca(HCO3)2 ,(NH4)3PO4,CuSO4,N2O5,

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Exercises 113

3.36 (a) What is the mass, in grams, of 0.0714 mol ofiron(III) sulfate?

(b) How many moles of ammonium ions are in 8.776 gof ammonium carbonate?

(c) What is the mass, in grams, of moleculesof aspirin,

(d) What is the molar mass of diazepam (Valium®) if0.05570 mol weighs 15.86 g?

3.37 The molecular formula of allicin, the compound respon-sible for the characteristic smell of garlic, is (a) What is the molar mass of allicin? (b) How manymoles of allicin are present in 5.00 mg of this substance?(c) How many molecules of allicin are in 5.00 mg of thissubstance? (d) How many S atoms are present in5.00 mg of allicin?

3.38 The molecular formula of aspartame, the artificialsweetener marketed as NutraSweet®, is (a) What is the molar mass of aspartame? (b) How manymoles of aspartame are present in 1.00 mg of aspartame?(c) How many molecules of aspartame are present in1.00 mg of aspartame? (d) How many hydrogen atomsare present in 1.00 mg of aspartame?

3.39 A sample of glucose, contains atoms of carbon. (a) How many atoms of hydrogen doesit contain? (b) How many molecules of glucose does itcontain? (c) How many moles of glucose does it con-tain? (d) What is the mass of this sample in grams?

3.40 A sample of the male sex hormone testosterone,contains atoms of hydrogen.

(a) How many atoms of carbon does it contain? (b) Howmany molecules of testosterone does it contain? (c) Howmany moles of testosterone does it contain? (d) What isthe mass of this sample in grams?

3.41 The allowable concentration level of vinyl chloride,in the atmosphere in a chemical plant is

How many moles of vinyl chloride in eachliter does this represent? How many molecules per liter?

3.42 At least of tetrahydrocannabinol (THC), the activeingredient in marijuana, is required to produce intoxica-tion. The molecular formula of THC is Howmany moles of THC does this represent? Howmany molecules?

25 mgC21H30O2.

25 mg

2.0 * 10-6 g>L.C2H3Cl,

7.08 * 1020C19H28O2,

1.250 * 1021C6H12O6,

C14H18N2O5.

C6H10OS2.

C9H8O4 ?6.52 * 1021

Avogadro’s Number and the Mole

3.27 (a) What is Avogadro’s number, and how is it related tothe mole? (b) What is the relationship between the for-mula weight of a substance and its molar mass?

3.28 (a) What is the mass, in grams, of a mole of (b) Howmany carbon atoms are present in a mole of

3.29 Without doing any detailed calculations (but using a peri-odic table to give atomic weights), rank the followingsamples in order of increasing number of atoms: 0.50 mol

23 g Na, molecules.3.30 Without doing any detailed calculations (but using a pe-

riodic table to give atomic weights), rank the followingsamples in order of increasing number of atoms:

molecules of 2.0 mol 32 g 3.31 What is the mass, in kilograms, of an Avogadro’s num-

ber of Olympic shot-put balls if each one has a mass of16 lb? How does this compare with the mass of Earth,

3.32 If Avogadro’s number of pennies is divided equallyamong the 292 million men, women, and children in theUnited States, how many dollars would each receive?How does this compare with the national debt of theUnited States, which was $7.0 trillion at the time of thewriting of this text?

3.33 Calculate the following quantities:(a) mass, in grams, of (b) moles of in 5.35 g of this substance(c) number of molecules in 0.0305 mol (d) number of C atoms in 0.585 mol

3.34 Calculate the following quantities:(a) mass, in grams, of (b) number of moles of in 48.3 g of this substance(c) number of molecules in (d) number of O atoms in

3.35 (a) What is the mass, in grams, of ofammonium phosphate?

(b) How many moles of chloride ions are in 0.2550 g ofaluminum chloride?

(c) What is the mass, in grams, of moleculesof caffeine,

(d) What is the molar mass of cholesterol if 0.00105 molweighs 0.406 g?

C8H10N4O2 ?7.70 * 1020

2.50 * 10-3 mol4.88 * 10-3 mol Al(NO3)3

0.05752 mol HCHO2

NH4Cl1.906 * 10-2 mol BaI2

C4H10

CH3OHMg(NO3)2

0.773 mol CaH2

5.98 * 1024 kg?

O2.CH4,H2O2,3.0 * 1023

6.0 * 1023 N2H2O,

12C?

12C?

3.46 Determine the empirical formulas of the compoundswith the following compositions by mass:(a) 55.3% K, 14.6% P, and 30.1% O(b) 24.5% Na, 14.9% Si, and 60.6% F(c) 62.1% C, 5.21% H, 12.1% N, and 20.7% O

3.47 What is the molecular formula of each of the followingcompounds?(a) empirical formula (b) empirical formula

3.48 What is the molecular formula of each of the followingcompounds?(a) empirical formula (b) empirical formula molar mass = 88 g>molC2H4O,

HCO2, molar mass = 90.0 g>mol

NH2Cl, molar mass = 51.5 g>molCH2, molar mass = 84 g>mol

Empirical Formulas

3.43 Give the empirical formula of each of the followingcompounds if a sample contains (a) 0.0130 mol C,0.0390 mol H, and 0.0065 mol O; (b) 11.66 g iron and 5.01g oxygen; (c) 40.0% C, 6.7% H, and 53.3% O by mass.

3.44 Determine the empirical formula of each of the follow-ing compounds if a sample contains (a) 0.104 mol K,0.052 mol C, and 0.156 mol O; (b) 5.28 g Sn and 3.37 g F;(c) 87.5% N and 12.5% H by mass.

3.45 Determine the empirical formulas of the compoundswith the following compositions by mass:(a) 10.4% C, 27.8% S, and 61.7% Cl(b) 21.7% C, 9.6% O, and 68.7% F(c) 32.79% Na, 13.02% Al, and 54.19% F

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3.49 Determine the empirical and molecular formulas ofeach of the following substances:(a) Styrene, a compound substance used to make Styro-

foam® cups and insulation, contains 92.3% C and7.7% H by mass and has a molar mass of

(b) Caffeine, a stimulant found in coffee, contains 49.5%C, 5.15% H, 28.9% N, and 16.5% O by mass and hasa molar mass about

(c) Monosodium glutamate (MSG), a flavor enhancer incertain foods, contains 35.51% C, 4.77% H, 37.85%O, 8.29% N, and 13.60% Na, and has a molar massof

3.50 Determine the empirical and molecular formulas ofeach of the following substances:(a) Ibuprofen, a headache remedy, contains 75.69% C,

8.80% H, and 15.51% O by mass, and has a molarmass about

(b) Cadaverine, a foul smelling substance produced bythe action of bacteria on meat, contains 58.55% C,13.81% H, and 27.40% N by mass; its molar mass is

(c) Epinephrine (adrenaline), a hormone secreted intothe bloodstream in times of danger or stress,contains 59.0% C, 7.1% H, 26.2% O, and 7.7% N bymass; its MW is about 180 amu.

3.51 (a) Combustion analysis of toluene, a common organicsolvent, gives 5.86 mg of and 1.37 mg of If thecompound contains only carbon and hydrogen, what is itsempirical formula? (b) Menthol, the substance we cansmell in mentholated cough drops, is composed of C, H,

H2O.CO2

102.2 g>mol.

206 g>mol.

169 g>mol.

195 g>mol.

104 g>mol.

and O. A 0.1005-g sample of menthol is combusted, pro-ducing 0.2829 g of and 0.1159 g of What is theempirical formula for menthol? If the compound has amolar mass of what is its molecular formula?

3.52 (a) The characteristic odor of pineapple is due to ethylbutyrate, a compound containing carbon, hydrogen,and oxygen. Combustion of 2.78 mg of ethyl butyrateproduces 6.32 mg of and 2.58 mg of What isthe empirical formula of the compound? (b) Nicotine, acomponent of tobacco, is composed of C, H, and N. A5.250-mg sample of nicotine was combusted, producing14.242 mg of and 4.083 mg of What is the em-pirical formula for nicotine? If the substance has a molarmass of what is its molecular formula?

3.53 Washing soda, a compound used to prepare hard waterfor washing laundry, is a hydrate, which means that acertain number of water molecules are included in thesolid structure. Its formula can be written as

where x is the number of moles of per mole of When a 2.558-g sample of washingsoda is heated at 25°C, all the water of hydration is lost,leaving 0.948 g of What is the value of x?

3.54 Epsom salts, a strong laxative used in veterinary medi-cine, is a hydrate, which means that a certain number ofwater molecules are included in the solid structure. Theformula for Epsom salts can be written as where x indicates the number of moles of per moleof When 5.061 g of this hydrate is heated to250°C, all the water of hydration is lost, leaving 2.472 g of

What is the value of x?MgSO4.

MgSO4.H2OMgSO4 #xH2O,

Na2CO3.

Na2CO3.H2ONa2CO3 #xH2O,

60 ; 5 g>mol,

H2O.CO2

H2O.CO2

156 g>mol,

H2O.CO2

(b) How many grams of are needed to form7.50 g of

(c) How many grams of form when 7.50 g ofare produced?

3.59 Several brands of antacids use to react withstomach acid, which contains primarily HCl:

(a) Balance this equation.(b) Calculate the number of grams of HCl that can react

with 0.500 g of (c) Calculate the number of grams of and the

number of grams of formed when 0.500 g ofreaacts.

(d) Show that your calculations in parts (b) and (c) areconsistent with the law of conservation of mass.

3.60 An iron ore sample contains together with other sub-stances. Reaction of the ore with CO produces iron metal:

(a) Balance this equation.(b) Calculate the number of grams of CO that can react

with 0.150 kg of (c) Calculate the number of grams of Fe and the

number of grams of formed when 0.150 kg ofreacts.

(d) Show that your calculations in parts (b) and (c) areconsistent with the law of conservation of mass.

Fe2O3

CO2

Fe2O3.

Fe2O3(s) + CO(g) ¡ Fe(s) + CO2(g)

Fe2O3

Al(OH)3

H2OAlCl3

Al(OH)3.

Al(OH)3(s) + HCl(aq) ¡ AlCl3(aq) + H2O(l)

Al(OH)3

C2H5OHCO2

C2H5OH?C6H12O6Calculations Based on Chemical Equations

3.55 Why is it essential to use balanced chemical equationswhen determining the quantity of a product formedfrom a given quantity of a reactant?

3.56 What parts of balanced chemical equations give infor-mation about the relative numbers of moles of reactantsand products involved in a reaction?

3.57 Hydrofluoric acid, cannot be stored in glass bot-tles because compounds called silicates in the glass areattacked by the Sodium silicate forexample, reacts as follows:

(a) How many moles of HF are needed to react with0.300 mol of

(b) How many grams of NaF form when 0.500 mol ofHF reacts with excess

(c) How many grams of can react with 0.800 gof HF?

3.58 The fermentation of glucose produces ethylalcohol and

(a) How many moles of are produced when0.400 mol of reacts in this fashion?C6H12O6

CO2

C6H12O6(aq) ¡ 2 C2H5OH(aq) + 2 CO2(g)

CO2 :(C2H5OH)(C6H12O6)

Na2SiO3

Na2SiO3 ?

Na2SiO3 ?

H2SiF6(aq) + 2 NaF(aq) + 3 H2O(l)

Na2SiO3(s) + 8 HF(aq) ¡

(Na2SiO3),HF(aq).

HF(aq),

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Exercises 115

3.61 Aluminum sulfide reacts with water to form aluminumhydroxide and hydrogen sulfide. (a) Write the balancedchemical equation for this reaction. (b) How manygrams of aluminum hydroxide are obtained from 6.75 gof aluminum sulfide?

3.62 Calcium hydride reacts with water to form calcium hy-droxide and hydrogen gas. (a) Write a balanced chemicalequation for the reaction. (b) How many grams of calci-um hydride are needed to form 8.500 g of hydrogen?

3.63 Automotive air bags inflate when sodium azide, rapidly decomposes to its component elements:

(a) How many moles of are produced by the decom-position of 1.50 mol of

(b) How many grams of are required to form10.0 g of nitrogen gas?

(c) How many grams of are required to produceof nitrogen gas if the gas has a density of

3.64 The complete combustion of octane, a compo-nent of gasoline, proceeds as follows:

(a) How many moles of are needed to burn 1.25 molof

(b) How many grams of are needed to burn 10.0 g of

(c) Octane has a density of at 20°C. Howmany grams of are required to burn 1.00 gal ofC8H18 ?

O2

0.692 g>mLC8H18 ?

O2

C8H18 ?O2

2 C8H18(l) + 25 O2(g) ¡ 16 CO2(g) + 18 H2O(g)

C8H18,1.25 g>L?10.0 ft3

NaN3

NaN3

NaN3 ?N2

2 NaN3(s) ¡ 2 Na(s) + 3 N2(g)

NaN3,

3.65 A piece of aluminum foil 1.00 cm square and 0.550 mmthick is allowed to react with bromine to form alu-minum bromide as shown in the accompanying photo.

(a) How many moles of aluminum were used? (Thedensity of aluminum is ) (b) How manygrams of aluminum bromide form, assuming that thealuminum reacts completely?

3.66 Detonation of nitroglycerin proceeds as follows:

(a) If a sample containing 2.00 mL of nitroglycerinis detonated, how many total

moles of gas are produced? (b) If each mole of gas occu-pies 55 L under the conditions of the explosion, howmany liters of gas are produced? (c) How many gramsof are produced in the detonation?N2

(density = 1.592 g>mL)

12 CO2(g) + 6 N2(g) + O2(g) + 10 H2O(g)

4 C3H5N3O9(l) ¡

2.699 g>cm3.

Which reagent is the limiting reactant when 0.500 moland 0.500 mol are allowed to react?

How many moles of can form under theseconditions? How many moles of the excess reactant re-main after the completion of the reaction?

3.73 The fizz produced when an Alka-Seltzer® tablet is dis-solved in water is due to the reaction between sodiumbicarbonate and citric acid

In a certain experiment 1.00 g of sodium bicarbonateand 1.00 g of citric acid are allowed to react. (a) Whichis the limiting reactant? (b) How many grams of car-bon dioxide form? (c) How many grams of the excessreactant remain after the limiting reactant is complete-ly consumed?

3 CO2(g) + 3 H2O(l) + Na3C6H5O7(aq)

3 NaHCO3(aq) + H3C6H5O7(aq) ¡

(H3C6H5O7) :(NaHCO3)

Al2(SO4)3

H2SO4Al(OH)3

Limiting Reactants; Theoretical Yields

3.67 (a) Define the terms limiting reactant and excess reactant.(b) Why are the amounts of products formed in a reactiondetermined only by the amount of the limiting reactant?

3.68 (a) Define the terms theoretical yield, actual yield, andpercent yield. (b) Why is the actual yield in a reaction al-most always less than the theoretical yield?

3.69 A manufacturer of bicycles has 4815 wheels, 2305frames, and 2255 handlebars. (a) How many bicyclescan be manufactured using these parts? (b) How manyparts of each kind are left over? (c) Which part limits theproduction of bicycles?

3.70 A bottling plant has 121,515 bottles with a capacity of355 mL, 122,500 caps, and 40,875 L of beverage. (a) Howmany bottles can be filled and capped? (b) How muchof each item is left over? (c) Which component limits theproduction?

3.71 Sodium hydroxide reacts with carbon dioxide as follows:

Which reagent is the limiting reactant when 1.85 molNaOH and 1.00 mol are allowed to react? Howmany moles of can be produced? How manymoles of the excess reactant remain after the completionof the reaction?

3.72 Aluminum hydroxide reacts with sulfuric acid as follows:

Al2(SO4)3(aq) + 6 H2O(l)2 Al(OH)3(s) + 3 H2SO4(aq) ¡

Na2CO3

CO2

2 NaOH(s) + CO2(g) ¡ Na2CO3(s) + H2O(l)

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3.74 One of the steps in the commercial process for convertingammonia to nitric acid is the conversion of to NO:

In a certain experiment, 1.50 g of reacts with 2.75 gof (a) Which is the limiting reactant? (b) How manygrams of NO and of form? (c) How many grams ofthe excess reactant remain after the limiting reactant iscompletely consumed? (d) Show that your calculationsin parts (b) and (c) are consistent with the law of conser-vation of mass.

3.75 Solutions of sodium carbonate and silver nitrate react toform solid silver carbonate and a solution of sodium ni-trate. A solution containing 3.50 g of sodium carbonate ismixed with one containing 5.00 g of silver nitrate. Howmany grams of sodium carbonate, silver nitrate, silvercarbonate, and sodium nitrate are present after the reac-tion is complete?

3.76 Solutions of sulfuric acid and lead(II) acetate react toform solid lead(II) sulfate and a solution of acetic acid. If7.50 g of sulfuric acid and 7.50 g of lead(II) acetate aremixed, calculate the number of grams of sulfuric acid,lead(II) acetate, lead(II) sulfate, and acetic acid presentin the mixture after the reaction is complete.

3.77 When benzene reacts with bromine bro-mobenzene is obtained:

C6H6 + Br2 ¡ C6H5Br + HBr

(C6H5Br)(Br2),(C6H6)

H2OO2.

NH3

4 NH3(g) + 5 O2(g) ¡ 4 NO(g) + 6 H2O(g)

NH3

(a) What is the theoretical yield of bromobenzene in thisreaction when 30.0 g of benzene reacts with 65.0 g ofbromine? (b) If the actual yield of bromobenzene was56.7 g, what was the percentage yield?

3.78 When ethane reacts with chlorine the mainproduct is but other products containing Cl,such as are also obtained in small quantities.The formation of these other products reduces the yieldof (a) Calculate the theoretical yield of when 125 g of reacts with 255 g of assumingthat and react only to form and HCl.(b) Calculate the percent yield of if the reactionproduces 206 g of

3.79 Lithium and nitrogen react to produce lithium nitride:

If 5.00 g of each reactant undergoes a reaction with a88.5% yield, how many grams of are obtained fromthe reaction?

3.80 When hydrogen sulfide gas is bubbled into a solution ofsodium hydroxide, the reaction forms sodium sulfideand water. How many grams of sodium sulfide areformed if 1.50 g of hydrogen sulfide is bubbled into a so-lution containing 2.00 g of sodium hydroxide, assumingthat the sodium sulfide is made in 92.0% yield?

Li3N

6 Li(s) + N2(g) ¡ 2 Li3N(s)

C2H5Cl.C2H5Cl

C2H5ClCl2C2H6

Cl2 ,C2H6

C2H5ClC2H5Cl.

C2H4Cl2 ,C2H5Cl;

(Cl2),(C2H6)

(a) Calculate the mass in grams of a quantum dotconsisting of 10,000 atoms of silicon.

(b) Assuming that the silicon in the dot has a density ofcalculate its volume.

(c) Assuming that the dot has the shape of a cube,calculate the length of each edge of the cube.

3.86 Serotonin is a compound that conducts nerve impulses inthe brain. It contains 68.2 mass percent C, 6.86 mass percentH, 15.9 mass percent N, and 9.08 mass percent O. Its molarmass is Determine its molecular formula.

3.87 The koala dines exclusively on eucalyptus leaves. Its di-gestive system detoxifies the eucalyptus oil, a poison toother animals. The chief constituent in eucalyptus oil isa substance called eucalyptol, which contains 77.87% C,11.76% H, and the remainder O. (a) What is the empiri-cal formula for this substance? (b) A mass spectrum ofeucalyptol shows a peak at about 154 amu. What is themolecular formula of the substance?

3.88 Vanillin, the dominant flavoring in vanilla, contains C,H, and O. When 1.05 g of this substance is completelycombusted, 2.43 g of and 0.50 g of are pro-duced. What is the empirical formula of vanillin?

[3.89] An organic compound was found to contain only C, H,and Cl. When a 1.50-g sample of the compound wascompletely combusted in air, 3.52 g of was formed.In a separate experiment the chlorine in a 1.00-g sampleof the compound was converted to 1.27 g of AgCl. De-termine the empirical formula of the compound.

[3.90] An oxybromate compound, where x is un-known, is analyzed and found to contain 52.92% Br.What is the value of x?

KBrOx ,

CO2

H2OCO2

176 g>mol.

2.3 g>cm3,

Additional Exercises

3.81 Write the balanced chemical equation for (a) the com-plete combustion of butyric acid, a com-pound produced when butter becomes rancid; (b) thedecomposition of solid nickel(II) hydroxide into solidnickel(II) oxide and water vapor; (c) the combination re-action between zinc metal and chlorine gas.

3.82 The effectiveness of nitrogen fertilizers depends on boththeir ability to deliver nitrogen to plants and the amountof nitrogen they can deliver. Four common nitrogen-containing fertilizers are ammonia, ammonium nitrate,ammonium sulfate, and urea Rank thesefertilizers in terms of the mass percentage nitrogen theycontain.

3.83 (a) Diamond is a natural form of pure carbon. Howmany moles of carbon are in a 1.25-carat diamond

How many atoms are in this dia-mond? (b) The molecular formula of acetylsalicylic acid(aspirin), one of the most common pain relievers, is

How many moles of are in a0.500-g tablet of aspirin? How many molecules of

are in this tablet?3.84 (a) One molecule of the antibiotic known as penicillin G

has a mass of What is the molar mass ofpenicillin G? (b) Hemoglobin, the oxygen-carrying pro-tein in red blood cells, has four iron atoms per moleculeand contains 0.340% iron by mass. Calculate the molarmass of hemoglobin.

3.85 Very small crystals composed of 1000 to 100,000 atoms,called quantum dots, are being investigated for use inelectronic devices.

5.342 * 10-21 g.

HC9H7O4

HC9H7O4HC9H7O4.

(1 carat = 0.200 g)?

[(NH2)2CO].

HC4H7O2(l),

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Exercises 117

[3.91] An element X forms an iodide and a chlorideThe iodide is quantitatively converted to the

chloride when it is heated in a stream of chlorine:

If 0.5000 g of is treated, 0.2360 g of is obtained.(a) Calculate the atomic weight of the element X.(b) Identify the element X.

3.92 If 1.5 mol of each of the following compounds is com-pletely combusted in oxygen, which one will produce thelargest number of moles of Which will producethe least? Explain.

3.93 A method used by the Environmental ProtectionAgency (EPA) for determining the concentration ofozone in air is to pass the air sample through a “bub-bler” containing sodium iodide, which removes theozone according to the following equation:

(a) How many moles of sodium iodide are needed to re-move (b) How many grams of sodi-um iodide are needed to remove 0.550 mg of

3.94 A chemical plant uses electrical energy to decomposeaqueous solutions of NaCl to give and NaOH:

If the plant produces (1500 metric tons)of daily, estimate the quantities of and NaOHproduced.

3.95 The fat stored in the hump of a camel is a source of bothenergy and water. Calculate the mass of producedby metabolism of 1.0 kg of fat, assuming the fat consistsentirely of tristearin a typical animal fat,and assuming that during metabolism, tristearin reactswith to form only and

[3.96] When hydrocarbons are burned in a limited amount ofair, both CO and form. When 0.450 g of a particu-lar hydrocarbon was burned in air, 0.467 g of CO,0.733 g of and 0.450 g of were formed.H2OCO2,

CO2

H2O.CO2O2

(C57H110O6),

H2O

H2Cl2

1.5 * 106 kg

2 NaOH(aq) + H2(g) + Cl2(g)2 NaCl(aq) + 2 H2O(l) ¡

H2,Cl2 ,

O3 ?O3 ?3.8 * 10-5 mol

O2(g) + I2(s) + 2 NaOH(aq)O3(g) + 2 NaI(aq) + H2O(l) ¡

CH3CH2COCH3 C3H8,C2H5OH,H2O?

XCl3XI3

2 XI3 + 3 Cl2 ¡ 2 XCl3 + 3 I2

(XCl3).(XI3) (a) What is the empirical formula of the compound?

(b) How many grams of were used in the reaction?(c) How many grams would have been required forcomplete combustion?

3.97 A mixture of and reacts in a closed contain-er to form ammonia, The reaction ceases beforeeither reactant has been totally consumed. At this stage2.0 mol 2.0 mol and 2.0 mol are present.How many moles of and were present originally?

[3.98] A mixture containing and KClwas heated, producing and gases accord-ing to the following equations:

The KCl does not react under the conditions of the reac-tion. If 100.0 g of the mixture produces 1.80 g of 13.20 g of and 4.00 g of what was the composi-tion of the original mixture? (Assume complete decom-position of the mixture.)

3.99 When a mixture of 10.0 g of acetylene and 10.0 gof oxygen is ignited, the resultant combustion reac-tion produces and (a) Write the balancedchemical equation for this reaction. (b) Which is the lim-iting reactant? (c) How many grams of and are present after the reaction is complete?

3.100 Aspirin is produced from salicylic acidand acetic anhydride

(a) How much salicylic acid is required to produceof aspirin, assuming that all of the sali-

cylic acid is converted to aspirin? (b) How muchsalicylic acid would be required if only 80% of the sali-cylic acid is converted to aspirin? (c) What is the theo-retical yield of aspirin if 185 kg of salicylic acid isallowed to react with 125 kg of acetic anhydride? (d) Ifthe situation described in part (c) produces 182 kg ofaspirin, what is the percentage yield?

1.5 * 102 kg

C7H6O3 + C4H6O3 ¡ C9H8O4 + HC2H3O2

(C4H6O3):(C7H6O3)(C9H8O4)

H2OCO2,O2,C2H2,

H2O.CO2

(O2)(C2H2)

O2,CO2,H2O,

K2CO3(s) ¡ K2O(s) + CO2(g)

2 KHCO3(s) ¡ K2O(s) + H2O(g) + 2 CO2(g)

2 KClO3(s) ¡ 2 KCl(s) + 3 O2(g)

H2OO2,CO2,KHCO3,K2CO3,KClO3,

H2N2

NH3H2,N2,

NH3(g).H2(g)N2(g)

O2

Assume that the gasoline is composed of octane,whose density is

3.104 In 1865 a chemist reported that he had reacted aweighed amount of pure silver with nitric acid andhad recovered all the silver as pure silver nitrate. Themass ratio of silver to silver nitrate was found to be0.634985. Using only this ratio and the presently ac-cepted values for the atomic weights of silver and oxy-gen, calculate the atomic weight of nitrogen. Comparethis calculated atomic weight with the currently ac-cepted value.

3.105 A particular coal contains 2.5% sulfur by mass. Whenthis coal is burned, the sulfur is converted into sulfurdioxide gas. The sulfur dioxide reacts with calciumoxide to form solid calcium sulfite. (a) Write the bal-anced chemical equation for the reaction. (b) If thecoal is burned in a power plant that uses 2000 tons ofcoal per day, what is the daily production of calciumsulfite?

0.69 g>mL.C8H18(l),Integrative Exercises

(These exercises require skills from earlier chapters as well asskills from the present chapter.)

3.101 Consider a sample of calcium carbonate in the form ofa cube measuring 2.005 in. on each edge. If the samplehas a density of how many oxygen atomsdoes it contain?

3.102 (a) You are given a cube of silver metal that measures1.000 cm on each edge. The density of silver is

How many atoms are in this cube? (b) Be-cause atoms are spherical, they cannot occupy all of thespace of the cube. The silver atoms pack in the solid insuch a way that 74% of the volume of the solid is actual-ly filled with the silver atoms. Calculate the volume of asingle silver atom. (c) Using the volume of a silver atomand the formula for the volume of a sphere, calculate theradius in angstroms of a silver atom.

3.103 If an automobile travels 225 mi with a gas mileage ofhow many kilograms of are produced?CO220.5 mi>gal,

0.49 g>cm3.

2.71 g>cm3,

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3.106 Copper is an excellent electrical conductor widely used inmaking electric circuits. In producing a printed circuitboard for the electronics industry, a layer of copper is lam-inated on a plastic board. A circuit pattern is then printedon the board using a chemically resistant polymer. Theboard is then exposed to a chemical bath that reacts withthe exposed copper, leaving the desired copper circuit,which has been protected by the overlaying polymer. Fi-nally, a solvent removes the polymer. One reaction used toremove the exposed copper from the circuit board is

A plant needs to produce 5000 circuit boards, eachwith a surface area measuring Theboards are covered with a 0.65-mm layer of copper. Insubsequent processing, 85% of the copper is removed.Copper has a density of Calculate themasses of and needed to produceNH3Cu(NH3)4Cl2

8.96 g>cm3.

2.0 in. * 3.0 in.

2 Cu(NH3)4Cl(aq)

Cu(s) + Cu(NH3)4Cl2(aq) + 4 NH3(aq) ¡

the circuit boards, assuming that the reaction usedgives a 97% yield.

3.107 Hydrogen cyanide, HCN, is a poisonous gas. The lethaldose is approximately 300 mg HCN per kilogram of airwhen inhaled. (a) Calculate the amount of HCN thatgives the lethal dose in a small laboratory room measur-ing The density of air at 26°C is

(b) If the HCN is formed by reaction ofNaCN with an acid such as what mass of NaCNgives the lethal dose in the room?

(c) HCN forms when synthetic fibers containing Orlon®

or Acrilan® burn. Acrilan® has an empirical formula ofso HCN is 50.9% of the formula by mass. A

rug measures and contains 30 oz of Acrilan®

fibers per square yard of carpet. If the rug burns, will alethal dose of HCN be generated in the room? Assumethat the yield of HCN from the fibers is 20% and that thecarpet is 50% consumed.

12 * 15 ftCH2CHCN,

2 NaCN(s) + H2SO4(aq) ¡ Na2SO4(aq) + 2 HCN(g)

H2SO4,0.00118 g>cm3.

12 * 15 * 8.0 ft.

e M E D I A E X E R C I S E S

These exercises make use of the interactive objects available on-line in OneKey or the Companion Website, and on your Acceler-ator CD. Access to these resources comes in your MediaPak.

3.108 (a) Balance the three reactions available in theBalancing Equations activity (3.1). (b) If, in the case ofthe reduction of you were to multiply each coeffi-cient in the balanced equation by the same number,would the reaction still be balanced? (c) Would it still bebalanced if you were to square each coefficient?(d) What would the coefficients be in the balanced equa-tion if the reduction reaction produced carbon monoxideinstead of carbon dioxide?

3.109 Calculate the percentage composition of each compoundin Exercise 3.22. Use the Molecular Weight and WeightPercent activity (3.3) to check your answers.

3.110 Consider the reaction of zinc metal with hydrochloricacid shown in the animation Limiting Reagent (3.7).(a) What would the limiting reagent have been if 100 mgof Zn had been combined with of HCl?2.0 * 10-3 mol

Fe2O3

(b) What volume of hydrogen gas would the reactionhave produced? (c) For the purposes of the limitingreagent experiment shown in the movie, why is itimportant that the hydrogen gas have a low solubility inwater? How would the apparent yield of a reaction beaffected if the evolved gas were very soluble in (or reac-tive with) water?

3.111 Write the balanced equation and predict the masses ofproducts and remaining reactant for each of the follow-ing combinations. Use the Limiting Reagents (3.7) ac-tivity to check your answers. (a) 50 g and 55 g

to form and (b) 150 g and125 g to form FeS and NaCl; (c) 96 g and62 g to form and

3.112 (a) In the reaction between and NaOH to produceand NaCl, what mass of sodium hydroxide

would be required to completely consume 50 g ofiron(III) chloride? (b) Use the Limiting Reagents activity(3.7) to combine 50 g with as close as you can get tothe stoichiometric amount of sodium hydroxide, makingsure that you add enough to completely consume the

How much sodium hydroxide is left over?FeCl3 .

FeCl3

Fe(OH)3

FeCl3

NaNO3.CaCO3Na2CO3

Ca(NO3)2Na2SFeCl2KNO3 ;PbCrO4K2CrO4

Pb(NO3)2

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eLaboratory Exercises 119

e L A B O R ATO RY E X E R C I S E S

These exercises allow students to apply the concepts and skillsfor this chapter in the simulated laboratory of Virtual ChemLab.Worksheets for each of these exercises are found in the VirtualChemLab Workbook in your MediaPak.

3.113 (VCL 3-1) Counting AtomsIt is a common problem for students to be asked to calcu-late the number of atoms in a pure elemental material orcompound, given the mass of the sample. In this problemyou will go into the virtual laboratory, retrieve a sample ofgold (Au), use an analytical balance to measure its mass,and then calculate the number of atoms in the sample.

3.114 (VCL 3-2) Counting AtomsIn this problem you will go into the virtual laboratory, re-trieve a sample of lead (Pb) and of uranium (U), use an an-alytical balance to measure the mass of each sample, andthen calculate the number of atoms in each sample.

3.115 (VCL 3-3) Counting AtomsIn this problem you will go into the virtual laboratory;retrieve samples of erbium (Er), sodium (Na), tungsten(W), and a fourth sample of your choice; use an analyti-cal balance to measure the mass of each sample; andthen calculate the number of atoms in each sample.

3.116 (VCL 3-4) Counting MoleculesIn this problem you will go into the virtual laboratoryand retrieve a bottle of NaCl, weigh out approximately1 g of the sample on an analytical balance, and then cal-culate the number of atoms of Na and Cl in the sample.

3.117 (VCL 3-5) Counting MoleculesIn this problem you will go into the virtual laboratoryand retrieve a bottle of sugar, weigh out approximately1 g of the sample on an analytical balance, and then cal-culate the number of atoms of C, H, and O in the sam-ple. You will also repeat this with a bottle of

3.118 (VCL 3-6) Counting Protons, Neutrons, and ElectronsIn this problem you will enter the virtual laboratory andweigh out an isotopically pure sample of scandium

calculate the number of moles in the sample,and then calculate the number of protons, neutrons, andelectrons.

3.119 (VCL 3-7) Counting Protons, Neutrons, and ElectronsIn this problem you will enter the virtual laboratory andweigh out an isotopically pure sample of bismuth

calculate the number of moles in the sample,and then calculate the number of protons, neutrons, andelectrons in

3.120 (VCL 3-8) Writing Balanced Precipitation ReactionsIt is an important skill that students be able to write bal-anced net ionic reactions for reactions that are observedin the laboratory. In this problem you will go into thevirtual laboratory and perform a series of precipitationreactions using and After observing thereactions, you will write the net ionic equations repre-senting these reactions and then balance them.

Sb3+.Ag+, Pb2+,

209Bi5+.

(209Bi),

(45Sc),

NH4Cl.

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