Chapter 12Chapter 12

StoichiometryStoichiometry

What is Stoichiometry?What is Stoichiometry?Stoichiometry is at the heart of the production of many Stoichiometry is at the heart of the production of many

things you use in your daily life. Soap, tires, fertilizer, things you use in your daily life. Soap, tires, fertilizer, gasoline, deodorant, and chocolate bars are just a gasoline, deodorant, and chocolate bars are just a few commodities you use that are chemically few commodities you use that are chemically engineered, or produced through chemical reactions. engineered, or produced through chemical reactions. Chemically engineered commodities all rely on Chemically engineered commodities all rely on stoichiometry for their production. stoichiometry for their production.

But what is stoichiometry? Stoichiometry is the But what is stoichiometry? Stoichiometry is the calculation of quantities in chemical equations. Given calculation of quantities in chemical equations. Given a chemical reaction, stoichiometry tells us what a chemical reaction, stoichiometry tells us what quantity of each reactant we need in order to get quantity of each reactant we need in order to get enough of our desired product. Because of its real-enough of our desired product. Because of its real-life applications in chemical engineering as well as life applications in chemical engineering as well as research, stoichiometry is one of the most important research, stoichiometry is one of the most important and fundamental topics in chemistry.and fundamental topics in chemistry.

Sect. 12.1 What is Stoichiometry?Sect. 12.1 What is Stoichiometry? Mole-Mass Relationships in Mole-Mass Relationships in

Chemical ReactionsChemical Reactions Chemical reactions STOP when one Chemical reactions STOP when one

of the reactants is used up.of the reactants is used up.

StoichiometryStoichiometry – study of quantitative – study of quantitative relationships between amounts of relationships between amounts of reactants and products in a chemical reactants and products in a chemical reactionreaction Based on law of conservation of Based on law of conservation of

mass- that matter is neither mass- that matter is neither created nor destroyed in a chemical created nor destroyed in a chemical reaction.reaction.

Chemical bonds in reactants break and Chemical bonds in reactants break and new chemical bonds form to produce new chemical bonds form to produce products.products.

Mass of the reactants equals the mass of Mass of the reactants equals the mass of the products.the products.

Stoichiometry and the balanced Stoichiometry and the balanced chemical equationchemical equation

Coefficients of a balanced Coefficients of a balanced equation tell how many equation tell how many representative particles and representative particles and also how many moles of each also how many moles of each substance are involved in the substance are involved in the chemical reactionchemical reaction

4 Fe + 3O4 Fe + 3O22 2 Fe 2 Fe22OO33

Describe this reaction in terms of Describe this reaction in terms of representative particles, moles, representative particles, moles, and gramsand grams

Interpreting Chemical Equations in terms of Interpreting Chemical Equations in terms of representative particles, moles, and mass representative particles, moles, and mass

Iron + Oxygen Iron + Oxygen Iron(III) oxide Iron(III) oxide

Balanced Balanced 44 Fe + Fe + 33OO22 22 Fe Fe22OO33

RepresentativeRepresentative

particles particles

44 atoms Fe + atoms Fe + 33 molecules O molecules O22 22 formula units Fe formula units Fe22 O O33

MolesMoles 44 moles Fe + moles Fe + 33 moles O moles O22 2 2 moles Femoles Fe22 O O33

MassMass 223.4 223.4 g Fe +g Fe + 96.096.0 g O g O22 319.4319.4 g Fe g Fe22 O O33

Show Masses Show Masses are equalare equal

319.4 g reactants 319.4 g reactants 319.4 g products 319.4 g products

law of conservation of mass law of conservation of mass

The mass of the reactants equals The mass of the reactants equals the mass of products as predicted the mass of products as predicted by the law of conservation of by the law of conservation of mass. mass.

Mole ratioMole ratio – ratio between the number of – ratio between the number of moles of any 2 substances in a moles of any 2 substances in a balancedbalanced chemical equationchemical equation

Mole ratios are the key to calculations Mole ratios are the key to calculations based on a chemical equation.based on a chemical equation.

2 Al + 3 Br2 Al + 3 Br22 2 AlBr 2 AlBr33

# of possible ratios = 3 parts of the equation # of possible ratios = 3 parts of the equation x 2 other parts = 6 mole ratiosx 2 other parts = 6 mole ratios

2 mol Al2 mol Al 2 mol Al2 mol Al 3 mol Br3 mol Br22

3 mol Br3 mol Br22 2 mol AlBr2 mol AlBr33 2 mol Al2 mol Al

3 mol Br3 mol Br22 2 mol AlBr2 mol AlBr33 2 mol AlBr2 mol AlBr33

2 mol AlBr2 mol AlBr33 2 mol Al2 mol Al 3 mol Br3 mol Br22

12.2 Stoichiometric Calculations12.2 Stoichiometric Calculations

A. Mole to Mole conversion:A. Mole to Mole conversion: Step 1: Begin with a balanced chemical Step 1: Begin with a balanced chemical

equationequation Step 2: Identify the substance that you Step 2: Identify the substance that you

know, and the substance that you need know, and the substance that you need to determine.to determine.

Step 3: Write mole ratiosStep 3: Write mole ratios Step 4: Do the mole to mole Step 4: Do the mole to mole

conversionsconversions

Example:Example: Potassium reacts in water to produce Potassium reacts in water to produce

Potassium hydroxide and hydrogen gas.Potassium hydroxide and hydrogen gas. Determine the moles of hydrogen Determine the moles of hydrogen

produced when produced when 0.0400 moles0.0400 moles of of potassium is used. potassium is used.

Step 1Step 1: Write the balanced equation: Write the balanced equation 2K + 2H2K + 2H22O O 2KOH + H 2KOH + H22

Step 2Step 2: Identify the substance that you know, : Identify the substance that you know, and the substance you need to determine.and the substance you need to determine.

2K 2K + + 2H2H22O O 2KOH + 2KOH + HH22

0.0400 mol K0.0400 mol K ? need to ? need to KnownKnown determine determine

UnknownUnknown Step 3Step 3: Write the mole ratios needed: Write the mole ratios needed

1 mol H1 mol H22

2 mol K2 mol K

Step 4Step 4: Mole to mole conversion. We use : Mole to mole conversion. We use conversion factors to cancel units.conversion factors to cancel units.

Moles of known x Moles of known x Moles of unknownMoles of unknown = moles of = moles of Moles of known Moles of known

unknownunknown

0.0400 mol K x 0.0400 mol K x 1 mol H1 mol H2 2 = 0.0200 mol H = 0.0200 mol H22

2 mol K2 mol K

Mole to Mass conversionMole to Mass conversion

ExampleExample: What mass of NaCl produced : What mass of NaCl produced from 1.25 moles of Chlorine reacting with from 1.25 moles of Chlorine reacting with Sodium.Sodium.

Step 1Step 1: Write the balanced chemical : Write the balanced chemical equation and identify the known and equation and identify the known and unknown substances.unknown substances.

2Na + Cl2Na + Cl22 2NaCl2NaCl

1.25 mols Cl1.25 mols Cl22 ? g Unknown ? g Unknown

Step 2Step 2: Write the mole ratio that relates the : Write the mole ratio that relates the moles of the 2 substances.moles of the 2 substances.

2 mol NaCl2 mol NaCl

1 mol Cl1 mol Cl22 Step 3Step 3: Multiply the number of moles by the : Multiply the number of moles by the

mole ratio.mole ratio.

1.25 mol Cl1.25 mol Cl22 x x 2 mol NaCl 2 mol NaCl = 2.50 mol NaCl = 2.50 mol NaCl

1 mol Cl1 mol Cl22 Step 4Step 4: Multiply mol by the molar mass of the : Multiply mol by the molar mass of the

substancesubstance

2.50 mol NaCl x 2.50 mol NaCl x 58.44 g NaCl58.44 g NaCl = 146 g NaCl = 146 g NaCl

1 mol NaCl1 mol NaCl

Mass to Mass ConversionMass to Mass Conversion Example:Example: Determine the mass of water produced Determine the mass of water produced

from the decomposition of 25.0 g of from the decomposition of 25.0 g of ammonium nitrate into water and Nammonium nitrate into water and N22O gas.O gas.

Step 1Step 1: Write the balanced chemical : Write the balanced chemical equation and identify the known and equation and identify the known and unknown substances.unknown substances.

NHNH44NONO3 3 N N22O + HO + H22OO

25.0 g Known25.0 g Known ? Unknown? Unknown

Step 2:Step 2: Convert grams to moles. Convert grams to moles.25.0 g NH25.0 g NH44NONO33 x x 1 mol NH1 mol NH44NONO33 = 0.312 mol NH= 0.312 mol NH44NONO33

80.04 g NH 80.04 g NH44NONO33

Step 3Step 3: Determine the mole ratio.: Determine the mole ratio.

2 mol H2 mol H22O__O__

1 mol NH1 mol NH44NONO33

Step 4Step 4: Multiply mol of substance determined : Multiply mol of substance determined by the mole ratio.by the mole ratio.

0.312 mol NH0.312 mol NH44NONO33 X X 2 mol H2 mol H22O___O___ = 0.624mol H = 0.624mol H22OO

1 mol NH1 mol NH44NONO33

Step 5Step 5: Calculate the mass using molar mass.: Calculate the mass using molar mass.

0.624 mol H0.624 mol H22O X O X 18.02 g H18.02 g H22O O = 11.2 g H = 11.2 g H22OO

1 mol H1 mol H22OO

pg 363 flow chart of stepspg 363 flow chart of steps Practice Problems:Practice Problems:

pg 359 9, 10pg 359 9, 10 Pg 360 11,12Pg 360 11,12 Pg 362 13,14Pg 362 13,14

Sect. 12.4: % YieldSect. 12.4: % Yield

Calculations are done to determine the Calculations are done to determine the success of chemical reactions.success of chemical reactions.

Most reactions never succeed in Most reactions never succeed in producing the predicted amount of producing the predicted amount of product.product.

3. Not every reaction goes cleanly or 3. Not every reaction goes cleanly or completely:completely:

a. Stop before all reactants are used up.a. Stop before all reactants are used up.

b. Liquids may adhere to the containers or b. Liquids may adhere to the containers or evaporateevaporate

c.c. Solid product may be left on the filter Solid product may be left on the filter paper paper or lost in the purification process.or lost in the purification process.

d. Products other than the intended ones may d. Products other than the intended ones may form. form.

How Much Product?How Much Product?

Theoretical yieldTheoretical yield – maximum – maximum amount of product that can be amount of product that can be produced from a given amount of produced from a given amount of reactantreactant calculated value…typically mass-calculated value…typically mass-

massmass

Actual yieldActual yield – amount of product – amount of product produced when a reaction is produced when a reaction is carried outcarried out must be given in the problem must be given in the problem

(found in the experiment)(found in the experiment)

percent yieldpercent yield – ratio of actual yield – ratio of actual yield to theoretical yield expressed as a to theoretical yield expressed as a percentpercentShows efficiency of a reactionShows efficiency of a reaction% yield = % yield = actual actual x 100 x 100

theoreticaltheoretical

30 g actual yield from the experiment30 g actual yield from the experiment X 100 = X 100 =

60 g theoretical yield calculated60 g theoretical yield calculated

50% percent yield50% percent yield

Limiting Reagent (reactant)Limiting Reagent (reactant)

The reactant in a The reactant in a chemical reaction that chemical reaction that limits the quantity of a limits the quantity of a product obtained.product obtained.

Calculating limiting reagentCalculating limiting reagent

1.1. Convert atoms to moles or grams Convert atoms to moles or grams to moles.to moles.

2.2. Calculate the amount of product. Calculate the amount of product. Convert moles of reactant to Convert moles of reactant to moles of product.moles of product.

3.3. Choose the reactant that produces Choose the reactant that produces the least amount of product.the least amount of product.

Excess ReagentExcess Reagent

The reactant that is The reactant that is left over after the left over after the reaction is complete.reaction is complete.

Calculating Excess reagentCalculating Excess reagent

1. Calculate the # of moles of the excess 1. Calculate the # of moles of the excess reactant.reactant.

2. Subtract the amount determined in step 2. Subtract the amount determined in step 1 from the original molar amount present.1 from the original molar amount present.

3. Change moles to grams if needed.3. Change moles to grams if needed.