Stoichiometry

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Stoichiometry

BIG

BIG Idea Idea _{Mass relationships}_{Mass relationships} in chemical reactions confirm the in chemical reactions confirm the law of conservation of mass. law of conservation of mass. 11.1

11.1 _{Defining Stoichiometry}_{Defining Stoichiometry} MAIN

MAIN Idea Idea _{The amount of each}_{The amount of each} reactant present at the start of a reactant present at the start of a chemical reaction determines how chemical reaction determines how much product can form.

much product can form. 11.2

11.2 _{Stoichiometric}_{Stoichiometric}

Calculations

MAIN

MAIN Idea Idea _{The solution to every}_{The solution to every} stoichiometric problem requires a stoichiometric problem requires a balanced chemical equation. balanced chemical equation. 11.3

11.3 _{Limiting Reactants}_{Limiting Reactants} MAIN

MAIN Idea Idea _{A chemical reaction}_{A chemical reaction} stops when one of the reactants is stops when one of the reactants is used up.

used up. 11.4

11.4 _{Percent Yield}_{Percent Yield} MAIN

MAIN Idea Idea _{Percent yield is a}_{Percent yield is a} measure of the efficiency of a measure of the efficiency of a chemical reaction.

chemical reaction.

ChemFacts

•

• Green plants make their own Green plants make their own foodfood

through photosynthesis.

•

• Photosynthesis occurs withinPhotosynthesis occurs within

structures called chloroplasts in the

cells of plants.

•

• The balanced chemical equation forThe balanced chemical equation for

the photosynthesis is:

6C

6COO22++ 6 6HH22OO → → C C66HH1212OO66++ 6 6OO22 •

• On a summer day, one acre of cornOn a summer day, one acre of corn

produces enough oxygen (a product

of photosynthesis) to meet the

respiratory needs of 130 people.

Carbon dioxide and water

Chloroplast

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Start Up Activities

Start-Up A

ctiv

ities

L

A

What evidence can you

observe

that a reaction is taking place?

During a chemical reaction, reactants are consumed as During a chemical reaction, reactants are consumed as new products are formed. Often, there are several telltale new products are formed. Often, there are several telltale signs that a chemical reaction is

signs that a chemical reaction is taking place.taking place. Procedure

Procedure

1.

1. Read and complete the lab safety form.Read and complete the lab safety form. 2.

2. Use aUse a10-mL graduated cylinder 10-mL graduated cylinder to measure to measure out 5.0 mL

out 5.0 mL 0.010.01M M potassium permanganate potassium permanganate (KMn

(KMnOO₄₄). ). Add the solution to Add the solution to aa100-mL beaker.100-mL beaker. 3.

3. Clean and dry Clean and dry the graduated cylinder, and then usethe graduated cylinder, and then use it to measure 5.0 mL

it to measure 5.0 mL0.010.01M M sodium hydrogen sodium hydrogen

sulfite solution

sulfite solution (NaHS (NaHSOO₃₃). Slowly add this solution). Slowly add this solution to the beaker while stirring with a

to the beaker while stirring with a stirring rod.stirring rod. Record your

Record your observations.observations. 4.

4. Repeat Step 3 until the KMnRepeat Step 3 until the KMnOO₄₄ solution in the solution in the beakerbeaker turns colorless. Stop

turns colorless. Stop adding the adding the NaHS NaHS OO₃₃ solution as solution as soon as you obtain a

soon as you obtain a colorless solution. Record yourcolorless solution. Record your observations.

observations. Analysis

Analysis

1.

1. IdentifyIdentifythe evidence you observed that the evidence you observed that a chemicala chemical reaction was occurring.

reaction was occurring. 2.

2. ExplainExplainwhy slowly adding the NaHSwhy slowly adding the NaHSOO₃₃ solution while solution while stirring is a better experimental technique than adding stirring is a better experimental technique than adding 5.0 mL of the solution

5.0 mL of the solution all at once.all at once. Inquiry

Inquiry Would anything more have happened if youWould anything more have happened if you continued to add NaHS

continued to add NaHSOO₃₃ solution to the solution to the beaker? Explain.beaker? Explain.

Steps in Stoichiometric Steps in Stoichiometric Calculations

Calculations Make the followingMake the following Foldable to help you summarize the Foldable to help you summarize the steps in solving a stoichiometric steps in solving a stoichiometric problem.

problem.

Visit

Visitglencoe.comglencoe.comto:to:

▶

▶ study the entire chapter onlinestudy the entire chapter online

explore explore

U

_NCH

NCH

Lab

▶ ▶ ▶

▶ take Self-Check Quizzestake Self-Check Quizzes

STEP

STEP 11 _{Fold a sheet}_{Fold a sheet}

of paper in half of paper in half lengthwise. lengthwise.

STEP

STEP 22 _{Fold in half}_{Fold in half}

widthwise and then in half again. widthwise and then in half again.

STEP

STEP 33 _{Unfold and}_{Unfold and}

cut along the folds of cut along the folds of the top flap to make the top flap to make four tabs.

four tabs.

STEP

STEP 44 _{Label the tabs}_{Label the tabs}

with the steps in stoichiometric with the steps in stoichiometric calculations.

calculations. 

 Use this Foldable with Section 11.2.Use this Foldable with Section 11.2. AsAs

you read this section, summarize each step on a

you read this section, summarize each step on a tab andtab and include an example of the step.

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Section

1

1.1

Objectives Objectives Describe

Describethe types of relationshipsthe types of relationships

indicated by a balanced chemical

equation.

State

Statethe mole ratios from athe mole ratios from a

balanced chemical equation.

balanced chemical equation. Review Vocabulary

Review Vocabulary

reactant:

reactant:the starting substance in athe starting substance in a

chemical reaction chemical reaction New Vocabulary New Vocabulary stoichiometry stoichiometry mole ratio mole ratio

Defining Stoichiometry

MAIN

MAIN Idea Idea _{The amount of each reactant present at the}_{The amount of each reactant present at the}_{start of a}_{start of a}

chemical reaction determines how much product can form. chemical reaction determines how much product can form.

Real-World Reading Link

Real-World Reading Link Have you ever watched a candle burning? YouHave you ever watched a candle burning? You might have watched the candle burn out as the last of the wax was used up. might have watched the candle burn out as the last of the wax was used up. Or, maybe you used a candle snuffer to put out the flame. Either way, when the Or, maybe you used a candle snuffer to put out the flame. Either way, when the candle stopped burning, the combustion reaction ended.

candle stopped burning, the combustion reaction ended.

Particle and Mole Relationships

In doing the Launch L

In doing the Launch Lab, were you surprised when the purple color ofab, were you surprised when the purple color of

potassium permanganate disap

potassium permanganate disappeared as you added peared as you added sodium hydrogensodium hydrogen

sulfite? If you concluded that the pot

sulfite? If you concluded that the potassium permanganaassium permanganate had beente had been

used up and the reaction

used up and the reaction had stopped, you are right. Chemical reactionshad stopped, you are right. Chemical reactions

stop when one of the

stop when one of the reactants is used up. When planning the reactionreactants is used up. When planning the reaction

of potassium permanganate and sodium hydrogen sulfite, a

of potassium permanganate and sodium hydrogen sulfite, a chemistchemist

might ask, “How many grams of potassium permanganate are needed to

react completely with a known mass of so

react completely with a known mass of sodium hydrogen sulfite?” Or,dium hydrogen sulfite?” Or,

when analyzing a photosynthesis reaction, you might ask,

when analyzing a photosynthesis reaction, you might ask, “How muc“How muchh

oxygen and carbon dioxide are needed to form a known mass of sugar.”

Stoichiometry is the tool for answering these questions.

Stoichiometry is the tool for answering these questions. Stoichiometry

Stoichiometry The study of quantitative relationshipThe study of quantitative relationships between ts between thehe

amounts of reactants used and amounts of products

amounts of reactants used and amounts of products formed by a chemi-formed by a

chemi-cal reaction is chemi-called

cal reaction is calledstoichiometry.stoichiometry. Stoichiometry is based on the law Stoichiometry is based on the law

of conservation of mass. Recall from Chapter 3 that the

of conservation of mass. Recall from Chapter 3 that the law states thatlaw states that

matter is neither created nor destroyed in a chemical

matter is neither created nor destroyed in a chemical reaction. In anyreaction. In any

chemical reaction, the amount of matter present at the end

chemical reaction, the amount of matter present at the end of the reac-of the

reac-tion is the

tion is the same as the amount of matter present at the beginning.same as the amount of matter present at the beginning.

Therefore, the mass of the reactants equals the

Therefore, the mass of the reactants equals the mass of the products.mass of the products.

Note the reaction of powdered iron (Fe) with oxygen (

Note the reaction of powdered iron (Fe) with oxygen (OO22) shown in) shown in

Figure 11.1.

Figure 11.1. Although iron reacts with oxygen to form a new com- Although iron reacts with oxygen to form a new

com-pound, iron(III) oxide (F

pound, iron(III) oxide (Fee22OO33), the total ), the total mass is unchanged.mass is unchanged.

■

■ Figure 11.1 Figure 11.1 The The balanced chemi-balanced

chemi-cal equation for this reaction between cal equation for this reaction between iron and oxygen provides the iron and oxygen provides the relation-ships between amounts of

ships between amounts of reactantsreactants and products.

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V

OCABULARYOCABULARY W

WORDORDORIGINORIGIN Stoichiometry Stoichiometry comes from the Greek

comes from the Greek wordswords

stoikheion,

stoikheion, which means element, which means element,

and

andmetron,metron, which means to which means to

measure measure

Table

1 11.11.1

Relationships

Derived from a

Balanced Chemical Equation

4Fe(s)

4Fe(s) ++ 33OO₂₂(g)(g) → → 2F2Fee₂₂OO₃₃(s)(s)

iron

iron ++ oxygenoxygen → → iron(III) iron(III) oxideoxide

4

4 atoms atoms FeFe ++ 3 molecules O3 molecules O₂₂ → → 2 formula units F2 formula units Fee₂₂OO₃₃

4 mol Fe

4 mol Fe ++ 3 mol O3 mol O₂₂ → → 2 2 mol mol FFee₂₂OO₃₃

223.4 g Fe

223.4 g Fe ++ 96.00 g O96.00 g O₂₂ → → 319.4 319.4 g g FFee₂₂OO₃₃

319.4 g reactants

319.4 g reactants → → 319.4 319.4 g g productsproducts

Interactive Table

Interactive Table ExploreExplore balanced chemical equations balanced chemical equations at

atglencoe.comglencoe.com..

The balanced chemical equation for the chemical reaction shown in

Figure 11.1

Figure 11.1 is as follows. is as follows.

4Fe(s)

4Fe(s)++ 3 3OO₂₂(g)(g) → → 2F 2Fee₂₂OO₃₃(s)(s)

Y

You can interpret thou can interpret this equation in is equation in terms of representative particles byterms of representative particles by

saying that four atoms of iron react with three molecu

saying that four atoms of iron react with three molecules of oxygen toles of oxygen to

produce two formula units of iron(III) oxide. Remember that co

produce two formula units of iron(III) oxide. Remember that coeffi-

effi-cients in an equation represent not only numbers of i

cients in an equation represent not only numbers of individual particlesndividual particles

but also numbers of moles of particles. The

but also numbers of moles of particles. Therefore, you can also say thatrefore, you can also say that

four moles of iron react with

four moles of iron react with three moles of oxygen to three moles of oxygen to produce twoproduce two

moles of iron(III) oxide.

The chemical equation does not directly tell you anything about the

masses of the

masses of the reactants and products. Howeverreactants and products. However, by converting t, by converting thehe

known mole quantities to mass, the

known mole quantities to mass, the mass relationshipmass relationships become obvious.s become obvious.

Recall that moles are converted to mass by multiplying by the

Recall that moles are converted to mass by multiplying by the molarmolar

mass. The masses of the react

mass. The masses of the reactants are as follows.ants are as follows.

4 4 mol mol FeFe××

_

55.85 g Fe55.85 g Fe 1 mol Fe 1 mol Fe == 223.4 g Fe 223.4 g Fe 3 3 mol mol OO22×× 32.00 g O 32.00 g O22

_

1 mol O 1 mol O22 = = 96.00 g O 96.00 g O₂₂

The total mass of the reactants is: (223.4 g

The total mass of the reactants is: (223.4 g++ 96.00 g) 96.00 g) == 319.4 g 319.4 g

Similarly

Similarly, the , the mass of the mass of the product is calculated as follows:product is calculated as follows:

2 2 mol mol FFee22OO33×× 159.7 g F 159.7 g Fee22OO33

__

1 mol F 1 mol Fee22OO33 = = 319.4 g 319.4 g

Note that the mass of the

Note that the mass of the reactants equals the mass of the product.reactants equals the mass of the product.

mass

mass of of reactantsreactants== mass of products mass of products

319.4

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EXAMPLE

Problem 11.1

Interpret

Interpreting ing Chemical EquationsChemical Equations The combustion of propane (The combustion of propane (CC33HH88) provides) provides energy for heating homes, cooking food, and soldering metal parts. Interpret the

energy for heating homes, cooking food, and soldering metal parts. Interpret the

equation for the combustion of propane in terms of representative particles,

moles, and mass. Show that the law of conservation of mass is observed.

1

_Analyze

_the

_Problem

The coefficients in the balanced chemical equation shown below represent both moles

and representative particles, in this case molecules. Therefore, the equation can be

interpreted in terms of molecules and moles. The law of conservation of mass will be

verified if the masses of the reactants and products are equal.

Known Known C C33HH88(g) + 5(g) + 5OO22(g)(g) → → 3C 3COO22(g) + 4(g) + 4HH22O(g)O(g) Unknown Unknown

Equation interpreted in terms of molecules Equation interpreted in terms of molecules ==_?_? Equation interpreted in terms of moles

Equation interpreted in terms of moles ==?? Equation interpreted in terms of mass Equation interpreted in terms of mass ==??

2

_Solve

_{for the}

_Unknown

The coefficients in the chemical equation indicate the number of molecules.

1 molecule C

1 molecule C33HH88++ 5 molecules O 5 molecules O → →22 3 molecules CO 3 molecules CO22++ 4 molecules H 4 molecules H22OO

The coefficients in the chemical equation also indicate the number of moles.

1 mol C

1 mol C33HH88++ 5 mol O 5 mol O → →22 3 mol CO 3 mol CO22++ 4 mol H 4 mol H22OO

To verify that mass is conserved, first convert moles of reactant and product to mass

by multiplying by a conversion factor—the molar mass—that relates grams to moles.

moles of reactant or product

moles of reactant or product ××

___

grams reactant or productgrams reactant or product 1 mol reactant or product

1 mol reactant or product == grams of reactant or product grams of reactant or product

1 mol C 1 mol C33HH88×× 44.09 g C 44.09 g C33HH88

__

1 mol C 1 mol C33HH88 =

= 44.09 g C 44.09 g C₃₃HH₈₈ Calculate the mass of the reactant CCalculate the mass of the reactant C33HH88..

5 mol O 5 mol O22×× 32.00 g O 32.00 g O22

_

1 mol O 1 mol O22 =

= 160.0 g O 160.0 g O₂₂ Calculate the mass of the reactant OCalculate the mass of the reactant O22..

3 mol C 3 mol COO22×× 44.01 g C 44.01 g COO22

_

1 mol C 1 mol COO22 =

= 132.0 g C 132.0 g COO₂₂ Calculate the mass of the product COCalculate the mass of the product CO22..

4 mol H 4 mol H22OO×× 18.02 g H 18.02 g H22OO

_

1 mol H 1 mol H22OO =

= 72.08 g H 72.08 g H₂₂OO Calculate the mass of the product HCalculate the mass of the product H22OO

44.09 g C

44.09 g C33HH88++ 160.0 g O 160.0 g O22==204.1 g reactants204.1 g reactants Add the masses of the reactan Add the masses of the reactants.ts.

132.0 g C

132.0 g COO22++ 72.08 g H 72.08 g H22OO==204.1 g products204.1 g products Add the masses of the products Add the masses of the products.. 204.1 g reactants

204.1 g reactants 204.1 g products204.1 g products The law of conservation of mass is observe The law of conservation of mass is observed.d.

Rounding Rounding page 952 page 952 Math Handbook Math Handbook

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Mole ratios

Mole ratios YYou have read that the coeou have read that the coefficients in a chemical fficients in a chemical equa-

equa-tion indicate the

tion indicate the relationshiprelationships between moles s between moles of reactants and products.of reactants and products.

Y

You can use ou can use the relationships between cothe relationships between coefficients to efficients to derive conversionderive conversion

factors called mole ratios. A

factors called mole ratios. A mole ratiomole ratio is a ratio between the numbers is a ratio between the numbers

of moles of any two of the

of moles of any two of the substances in a balanced chemical substances in a balanced chemical equation.equation.

For example, consider the reaction shown in

For example, consider the reaction shown inFigure 11.2.Figure 11.2. In this In this

reaction, potassium (K) reacts with bromine (B

reaction, potassium (K) reacts with bromine (Brr22) to form potassium) to form potassium

bromide (KBr). The product of the reaction,

bromide (KBr). The product of the reaction, the ionic salt pthe ionic salt potassiumotassium

bromide, is prescribed by veterinarians as

bromide, is prescribed by veterinarians as an antiepileptic medicationan antiepileptic medication

for dogs and cats.

2K(s)

2K(s)++BBrr₂₂(l)(l) → → 2KBr(s) 2KBr(s)

What mole ratios can be written for this reaction? Starting with the

reactant potassium, you can write a mole ratio t

reactant potassium, you can write a mole ratio that relates the moles ofhat relates the moles of

potassium to each of the other two substances in the equation. Thus,

one mole ratio relates the moles of pot

one mole ratio relates the moles of potassium used to the moles assium used to the moles of bro-of

bro-mine used. The other mole ratio relates the moles of potassium used to

mine used. The other mole ratio relates the moles of potassium used to

the moles of potassium bromide formed.

the moles of potassium bromide formed. 2 mol K

2 mol K

_

1 mol B

1 mol Brr22 andand

2 mol K 2 mol K

_

2 mol KBr 2 mol KBr Tw

Two other mole ratios show how o other mole ratios show how the moles of bromine relate to thethe moles of bromine relate to the

moles of the other

moles of the other two substances in the two substances in the equation—potassium andequation—potassium and

potassium bromide. potassium bromide. 1 mol B 1 mol Brr22

_

2 mol K

2 mol K andand

1 mol B 1 mol Brr22

_

2 mol KBr 2 mol KBr Similarly

Similarly, two ratios relate the moles of , two ratios relate the moles of potassium bromide to the molespotassium bromide to the moles

of potassium and bromine.

2 mol KBr

2 mol KBr _and_and 2 mol KBr2 mol KBr

■

■ Figure 11.2 Figure 11.2 Potassium Potassium metal andmetal and

liquid bromine react vigorously to form liquid bromine react vigorously to form the ionic compound

the ionic compound potassium bromide.potassium bromide. Bromine is one of the two elements that Bromine is one of the two elements that are liquids at room

are liquids at room temperature (mercurytemperature (mercury is the other). Potassium is a highly

is the other). Potassium is a highly reactivereactive metal.

metal.

PRACTICE

Problems

Extra PracticeExtra Practice Pages 982–983Pages 982–983andandglencoe.comglencoe.com

1.

1. Interpret the following balanced chemical equations in terms of particles, moles, andInterpret the following balanced chemical equations in terms of particles, moles, and

mass. Show that the law of conservation of mass is observed.

a.

a. NN22(g)(g) ++ 3 3HH22(g)(g) → → 2N 2NHH33((g)g)

b.

b. HCl(aq)HCl(aq) ++ KOH(aq) KOH(aq) → → KCl(aq) KCl(aq) ++ H H₂₂O(l)O(l) c.

c. 2Mg(s)2Mg(s)++ O O₂₂(g)(g) → → 2MgO(s) 2MgO(s) 2. Challenge

2. Challenge For each of the following, balance the chemical equation; interpret theFor each of the following, balance the chemical equation; interpret the

equation in terms of particles, moles, and mass; and show that the law of conservation

of mass is observed.

a.

a. ___Na(s) ___Na(s) ++ ___ ___HH₂₂O(l)O(l) → → ___NaOH(aq) ___NaOH(aq) ++ ___ ___HH₂₂((g)g) b.

b. ___Zn(s) ___Zn(s) ++ ___HN ___HNOO₃₃(aq)(aq) → → ___Zn(N ___Zn(NOO₃₃))₂₂(aq)(aq) ++ ___ ___NN₂₂O(g)O(g) ++ ___ ___HH₂₂O(l)O(l)

Personal Tutor

Personal Tutor For an online tutorialFor an online tutorial

on ratios, visit

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Section

1

1 .

.

1

1 Assessment

Assessment

The decomposition of p

The decomposition of potassium chlorate (KClotassium chlorate (KClOO33) is sometimes) is sometimes

used to obtain small amounts of oxygen in

used to obtain small amounts of oxygen in the laboratory.the laboratory.

2KCl

2KClOO33(s)(s) → → 2KCl(s) 2KCl(s)++ 3 3OO22(g)(g)

The mole ratios that can b

The mole ratios that can be written for this e written for this reaction are as follows.reaction are as follows. 2 mol KCl

2 mol KClOO33

_

2 mol KCl

2 mol KCl andand

2 mol KCl 2 mol KClOO33

_

3 mol O 3 mol O22 2 mol KCl 2 mol KCl

_

2 mol KCl

2 mol KClOO33 andand

2 mol KCl 2 mol KCl

_

3 mol O 3 mol O22 3 mol O 3 mol O22

_

2 mol KCl

2 mol KClOO33 andand

3 mol O 3 mol O22

_

2 mol KCl 2 mol KCl

Note that the number of mole

Note that the number of mole ratios you can write for a chemicalratios you can write for a chemical

reaction involving a total of

reaction involving a total ofnn substances is ( substances is (nn)()(nn–1). Thus, for reactions–1). Thus, for reactions

involving four and five substances, you can write 12 and 20 moles ratios,

respectively.

Four

Four substances: substances: (4)(3)(4)(3)== 12 mole ratios 12 mole ratios

Five

Five substances: substances: (5)(4)(5)(4)== 20 mole ratios 20 mole ratios

V

OCABULARYOCABULARY A

ACADEMICCADEMICVOCABULARYVOCABULARY Derive

Derive

to obtain from a specified source

The researcher was able to derive

the meaning of the illustration

from ancient t

from ancient texts.exts.

PRACTICE

Problems

Extra PracticeExtra Practice Page 983 aPage 983 andndglencoe.comglencoe.com

3.

3. Determine all possible mole ratios for the following balanced chemicalDetermine all possible mole ratios for the following balanced chemical

equations.

a.

a. 4Al(s)4Al(s)++₃₃_O_O₂₂_(g)_(g) → → 2A 2All₂₂OO₃₃(s)(s) b.

b. 3Fe(s)3Fe(s)++ 4 4HH₂₂O(l)O(l) → → F Fee₃₃OO₄₄(s)(s) ++ 4 4HH₂₂((g)g) c.

c. 2HgO(s)2HgO(s) → → 2Hg(l) 2Hg(l)++ O O₂₂((g)g) 4. Challenge

4. Challenge Balance the following equations, and determine theBalance the following equations, and determine the

possible mole ratios.

a.

a. ZnO(s)ZnO(s) ++ HCl(aq) HCl(aq) → → ZnC ZnCll₂₂(aq)(aq) ++ H H₂₂O(l)O(l) b.

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Section

Objectives Objectives List

Listthe sequence of steps used inthe sequence of steps used in

solving stoichiometric

solving stoichiometric problems.problems.

Solve

Solvestoichiometric problems.stoichiometric problems. Review Vocabulary

Review Vocabulary

chemical reaction:

chemical reaction:a process ina process in

which the atoms of one or more

substances are rearranged to form

different substances

Stoichiometric Calculations

MAIN

MAIN Idea Idea _{The solution to}_{The solution to}_{every stoichiometric problem requir}_{every stoichiometric problem requir}_{es a}_{es a}

balanced chemical equation. balanced chemical equation. Real-World Reading Link

Real-World Reading Link Baking requires accurate measurements. That isBaking requires accurate measurements. That is why it is

why it is necessary to follow a recipe when baking cookies from scratch. If younecessary to follow a recipe when baking cookies from scratch. If you need to make more cookies than a recipe yields, what must you do?

need to make more cookies than a recipe yields, what must you do?

Using Stoichiometry

What tools are needed to perform stoichiometric calculations? All

stoi-chiometric calculations begin with a

chiometric calculations begin with a balanced chemical equation. Mobalanced chemical equation. Molele

ratios based on the balanced chemical

ratios based on the balanced chemical equatioequation are needed, as well asn are needed, as well as

mass-to-mole conversions.

Stoichiometric mole-to-mole conversion

Stoichiometric mole-to-mole conversion The vigorous reac-The vigorous

reac-tion between potassium and water is shown in

tion between potassium and water is shown in Figure 11.3.Figure 11.3. The balanced The balanced

chemical equation is as follows.

2K(s)

2K(s)++ 2 2HH₂₂O(l)O(l) → → 2KOH(aq) 2KOH(aq)++ H H₂₂(g)(g)

From the balanced equation, you know that two moles of potassium

yields one mole of hydrogen. But how much hydrogen is produced if

only 0.0400 mol of potassium is used? To answer this question, identify

the given, or known, substance and the substance that you need to

determine. The given substance is 0.0400 mol of potassium. The

unknown is the number of moles of hydrogen. Because the given

sub-stance is in moles and the

stance is in moles and the unknown substance to be determined is alsounknown substance to be determined is also

in moles, this problem involves a mole-to-mole conversion.

To solve the problem, you need to know how the unknown moles of

hydrogen are related to the known moles of potassium. In Section 11.1,

you learned to derive mole ratios from the b

you learned to derive mole ratios from the balanced chemical equatioalanced chemical equation.n.

Mole ratios are used as conversion factors to convert the known number

of moles of one substance to the unknown number of moles of another

substance in the same reaction. Several mole ratios can b

substance in the same reaction. Several mole ratios can be written frome written from





Incorporate information

from this section into

your Foldable.

11.2

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As shown below, the correct mole ratio, 1 mol H

As shown below, the correct mole ratio, 1 mol H22 to 2 mol K, has to 2 mol K, has

moles of unknown in the numerator and moles of known in the

denom-inator. Using this mole ratio converts the moles of potassium to the

inator. Using this mole ratio converts the moles of potassium to the

unknown number of moles of hydrogen.

moles

moles of of knownknown ××

__

moles of unknownmoles of unknown

moles of known

moles of known == moles of unknown moles of unknown

0.0400

0.0400 mol mol KK××

_

1 mol H1 mol H22

2 mol K

2 mol K == 0.0200 mol H 0.0200 mol H22

The following Example Problems show mole-to-mole, mole-to-mass,

and mass-to-mass stoichiometry problems. The process used to solve

these problems is outlined in the Problem-Solving Strategy below.

Problem-Solving Strategy

Mastering Stoichiometry

The flowchart below outlines the steps used to solve mole-to-mole,

mole-to-mass, and mass-to-mass stoichiometric problems.

Step 1

Start with a balanced equation. Start with a balanced equation.

Interpret the equation in terms of moles. Interpret the equation in terms of moles.

1.

1. Complete Step 1 by writing the balanced chemical equationComplete Step 1 by writing the balanced chemical equation

for the reaction.

2.

2. To determine where to start your calculations, note the unitTo determine where to start your calculations, note the unit

of the given substance.

•

• If mass (in grams) of the given substance is the startingIf mass (in grams) of the given substance is the starting

unit, begin your calculations with Step 2.

•

• If amount (in moles) of the given substance is theIf amount (in moles) of the given substance is the

starting unit, skip Step 2 and begin your calculations

with Step 3.

3.

3. The end point of the calculation depends onThe end point of the calculation depends on

the desired unit of the unknown substance.

•

• If the answer must be in moles, stop afterIf the answer must be in moles, stop after

completing Step 3.

•

• If the answer must be in grams, stop afterIf the answer must be in grams, stop after

completing Step 4.

Apply the Strategy

Apply

Applythe Problem-Solving Strategy to Example Problems 11.2, 11.3,the Problem-Solving Strategy to Example Problems 11.2, 11.3,

and 11.4.

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EXAMPLE

Problem 11.2

Mole-to-Mole

Mole-to-Mole StoichiometryStoichiometry One disadvantage of burningOne disadvantage of burning

propane (

propane (CC33HH88) is that carbon dioxide (C) is that carbon dioxide (COO22) is one of the products.) is one of the products.

The released carbon dioxide increases the concentration of C

The released carbon dioxide increases the concentration of COO22

in the atmosphere. How many moles of C

in the atmosphere. How many moles of COO22 is produced when is produced when

10.0 mol of C

10.0 mol of C33HH88 is burned in excess oxygen in a gas grill? is burned in excess oxygen in a gas grill?

1

_Analyze

_the

_Problem

You are given moles of the reactant, C

You are given moles of the reactant, C33HH88 and must find the moles of and must find the moles of

the product, C

the product, COO22. First write the balanced chemical equation, then. First write the balanced chemical equation, then

convert from moles of C

convert from moles of C33HH88 to moles of C to moles of COO22. The correct mole ratio. The correct mole ratio

has moles of unknown substance in the numerator and moles of

known substance in the denominator.

Known

moles C

moles C33HH88== 10.0 mol C 10.0 mol C33HH88

Unknown Unknown moles CO

moles CO22== ? mol CO? mol CO22

2

_Solve

_{for the}

_Unknown

Write the balanced chemical equation for the combustion of C

Write the balanced chemical equation for the combustion of C33HH88..

Use the correct mole ratio to convert moles of known (

Use the correct mole ratio to convert moles of known (CC33HH88) to) to

moles of unknown (C

moles of unknown (COO22).).

10.0 mol

10.0 mol ? mol? mol C

C33HH88(g)(g) ++ 5 5OO22(g)(g) → → 3C 3COO₂₂(g)(g) ++₄₄_H_H₂₂_O(g)_O(g) Mole ratio:

Mole ratio:

_

3 mol C3 mol COO22

1 mol C 1 mol C33HH88 10.0 mol C 10.0 mol C33HH88×× 3 mol C 3 mol COO22

_

1 mol C 1 mol C33HH88 = = 30.0 mol CO30.0 mol CO₂₂ Burning 10.0 moles of C

Burning 10.0 moles of C33HH88 produces 30.0 moles C produces 30.0 moles COO22..

3

_Evaluate

_the

_Answer

Because the given number of moles has three significant figures, the

answer also has three figures. The balanced chemical equation

indi-cates that 1 mol of C

cates that 1 mol of C33HH88 produces 3 mol of C produces 3 mol of COO22. Thus, 10.0 mol of C. Thus, 10.0 mol of C33HH88

produces three times as many moles of C

produces three times as many moles of COO , or 30.0 mol., or 30.0 mol.

Real-Wo

rld

Chemistry

Outdoor Cooking

Gas Grills

Gas Grills Using outdoor grills isUsing outdoor grills is a popular way to cook. Gas grills a popular way to cook. Gas grills burn either natural gas or propane burn either natural gas or propane that is mixed with air. The initial that is mixed with air. The initial spark is provided by a grill starter. spark is provided by a grill starter. Propane is more commonly used Propane is more commonly used for fuel because it can be

for fuel because it can be suppliedsupplied in liquid form in a portable tank. in liquid form in a portable tank.

Ratios Ratios page 964 page 964 Math Handbook Math Handbook

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EXAMPLE

Problem 11.3

Mole-to-Mass

Mole-to-Mass StoichiometrStoichiometryy Determine the mass of sodium chloride (NaCl),Determine the mass of sodium chloride (NaCl),

commonly called table salt, produced when 1.25 mol of chlorine gas (C

commonly called table salt, produced when 1.25 mol of chlorine gas (Cll22) reacts) reacts

vigorously with excess sodium.

vigorously with excess sodium. 1

1

_Analyze

_the

_Problem

You are given the moles of the reactant, C

You are given the moles of the reactant, Cll22, and must determine the mass of the product,, and must determine the mass of the product,

NaCl. You must convert from moles of C

NaCl. You must convert from moles of Cll22 to moles of NaCl using the mole ratio from the to moles of NaCl using the mole ratio from the

equation. Then, you need to convert moles of NaCl to grams of NaCl using the molar mass

as the conversion factor.

Known

Known UnknownUnknown moles of chlorine

moles of chlorine == 1.25 mol C 1.25 mol Cll₂₂ mass of sodium chloridemass of sodium chloride==? g NaCl? g NaCl

2

_Solve

_{for the}

_Unknown

1.25

1.25 molmol ? g? g 2Na(s)

2Na(s)++_C_C_l_l₂₂_(g)_(g) → → 2NaCl(s) 2NaCl(s) _{Write the balanced chemica}_{Write the balanced chemica}_{l equation, and}_{l equation, and}

identify the known and the unknown values.

identify the known and the unknown values. Mole ratio:

Mole ratio:

_

2 mol NaCl2 mol NaCl

1 mol C

1 mol Cll22

1.25 mol C

_

ll22×× 2 mol NaCl2 mol NaCl_{1 mol C}_{1 mol C}_l_l

2 2

=

= 2.50 mol NaCl 2.50 mol NaCl Multiply moles of CMultiply moles of Cll22 by the mole ratio to get moles of NaCl. by the mole ratio to get moles of NaCl.

2.50 mol NaCl

2.50 mol NaCl ××

_

58.44 g NaCl58.44 g NaCl 1 mol NaCl

1 mol NaCl ==146 g NaCl146 g NaCl Multiply moles of NaCl by the molar mass to Multiply moles of NaCl by the molar mass to get grams of NaCl.get grams of NaCl.

3

_Evaluate

_the

_Answer

Because the given number of moles has three significant figures, the mass of NaCl also

has three. To quickly assess whether the calculated mass value for NaCl is correct,

perform the calculations in reverse: divide the mass of NaCl by the molar mass of NaCl,

Calculations with Calculations with Significant Figures Significant Figures pages 952–953 pages 952–953 Math Handbook Math Handbook

Stoichiometric mole-to-mass conversion

Stoichiometric mole-to-mass conversion Now, suppose youNow, suppose you

know the number of moles of a reactant or product in a reaction and

you want to calculate the mass of another product or reactant. This is an

example of a mole-to-mass conversion.

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Stoichiometric mass-to-mass conversion

Stoichiometric mass-to-mass conversion If you were prepar-If you were

prepar-ing to carry

ing to carry out a chemical reaction in out a chemical reaction in the laboratorythe laboratory, you would need, you would need

to know how much of each reactant to use in order to produce the mass

of product you required. Example Problem 11.4 demonstrates how you

can use a measured mass of

can use a measured mass of the known substance, the balanced chemi-the known substance, the balanced

chemi-cal equation, and mole ratios from the equation to find the mass of the

cal equation, and mole ratios from the equation to find the mass of the

unknown substance. The ChemLab at the end of this chapter will

pro- vide you with la

vide you with laboratory experience boratory experience in determining a moin determining a mole ratiole ratio..

EXAMPLE

Problem 11.4

Mass-to-Mass

Mass-to-Mass StoichiometrStoichiometryy Ammonium nitrate (NAmmonium nitrate (NHH44NNOO33), an important), an important

fertilizer

fertilizer, produces dinitrogen , produces dinitrogen oxide (oxide (NN22O) gas and HO) gas and H22O when it decomposes.O when it decomposes.

Determine the mass of H

Determine the mass of H22O produced from the decomposition of 25.0 g ofO produced from the decomposition of 25.0 g of

solid N

solid NHH44NNOO33..

1

_Analyze

_the

_Problem

You are given a description of the chemical reaction and the mass of the reactant.

You need to write the balanced chemical equation and convert the known mass of the

reactant to moles of the reactant. Then, use a mole ratio to relate moles of the reactant to

moles of the product. Finally, use the molar mass to convert from moles of the product to

the mass of the product.

Known

Known UnknownUnknown mass of ammonium nitrate

mass of ammonium nitrate == 25.0 g N 25.0 g NHH₄₄NNOO₃₃ mass of watermass of water ==? g H? g H₂₂OO

2

_Solve

_{for the}

_Unknown

Dimensional Analysis Dimensional Analysis page 956 page 956 Math Handbook Math Handbook

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Apply Stoichiometry

How much sodium carbonate (N

How much sodium carbonate (Naa₂₂CCOO₃₃) is pro-) is pro-duced when baking soda

duced when baking soda decomposes?decomposes? BakingBaking

soda is used in many baking recipes because

soda is used in many baking recipes because itit

makes batter rise, which results in a light and fluffy

texture. This occurs because baking soda, sodium

hydrogen carbonate (NaHC

hydrogen carbonate (NaHCOO33), decomposes upon), decomposes upon

heating to form carbon dioxide gas according to

the following equation.

2NaHC

2NaHCOO33 → → N Naa22CCOO33 ++ C COO22++ H H22OO

Procedure

1.

1. Read and complete the lab safety form.Read and complete the lab safety form. 2.

2. Create a data table to record your Create a data table to record your experimentalexperimental

data and observation.

data and observation. 3.

3. Use aUse abalancebalance to measure the mass of a to measure the mass of a clean,clean,

dry

drycrucible.crucible. Add about 3.0 g of Add about 3.0 g of sodium hydro-sodium

hydro-gen carbonate (NaHC

gen carbonate (NaHCOO33),), and measure the and measure the

combined mass of the crucible and NaHC

combined mass of the crucible and NaHCOO33..

Record both masses in your data table, and

calculate the mass of the NaHC

calculate the mass of the NaHCOO33..

4.

4. Use this starting mass of NaHCUse this starting mass of NaHCOO33 and the bal- and the

bal-5.

5. Set up aSet up a ring standring stand with a with aringring and andclay triangleclay triangle

for heating the crucible.

for heating the crucible. 6.

6. Heat the crucible with aHeat the crucible with a Bunsen Bunsen burnerburner,, slowly slowly

at first and then with a stronger flame, for

7–8 min. Record your observations during the

heating.

heating. 7.

7. Turn off the burner, and useTurn off the burner, and use crucible tongscrucible tongs to to

remove the hot crucible.

WARNING:

WARNING:Do not touch the hot crucible with yourDo not touch the hot crucible with your

hands.

8.

8. Allow the crucible to cool, and then measureAllow the crucible to cool, and then measure

the mass of the crucible and NaHC

the mass of the crucible and NaHCOO33..

Analysis

1. Describe

1. Describe what you observed during the heatingwhat you observed during the heating

of the baking soda.

of the baking soda. 2. Compare

2. Compare your calculated mass of NaHCyour calculated mass of NaHCOO33

with the actual mass you obtained from the

experiment.

experiment. 3. Calculate

3. Calculate Assume that the mass of NAssume that the mass of Naa22HCHCOO33

that you calculated in Step 4 is

that you calculated in Step 4 is the acceptedthe accepted

value for the mass of product that

value for the mass of product that will form.will form.

Calculate the error and percent error associated

with the experimentally measured mass.

with the experimentally measured mass. 4. Identify

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