Stoichiometry

INTRODUCTION TO STOICHIOMETRYSection 1

•Much knowledge of chemistry based on quantitative analysis of substances in chemical reactions •Composition stoichiometry ! deals with mass

relationships of elements in compounds • Reaction stoichiometry ! deals with mass relationships

between reactants and products in a chemical reaction

• Based on chemical equations and law of conservation of matter

• All calculations start with balanced chemical equation

• This gives numbers of moles of reactants and products

REACTION-STOICHIOMETRY PROBLEMS

•Can be classified according to information given in problem and info you are expected to find, the unknown •May both be reactants, may both be products, or

both •Masses usually expressed in grams • Solved by using ratios to convert given quantities by

following methods:

PROBLEM TYPE 1:

•Given and unknown quantities are amounts in moles

•General plan:

Amt of given (mol) ! amt of unknown (mol)

PROBLEM TYPE 2:

•Given is in moles and unknown is a mass in grams

Amt given (mol) ! amt unknown (mol) ! amt unknown (g)

PROBLEM TYPE 3:

•Given is mass (g) and unknown is in moles

Mass given (g) ! amt given (mol) ! amount unknown (mol)

PROBLEM TYPE 4:

•Given is mass (g) and unknown is mass (g)

Mass given ! amt given (mol) ! amt unknown (mol) ! mass unknown

MOLE RATIO

• Solving any reaction-stoichiometry problem requires use of mole ratio to convert from moles to grams

•Mole ratio ! conversion factor that relates amounts in moles of any two substances involved in a chemical reaction

•Get directly from balanced equation

EXAMPLE: 2AL2O3(L) ! 4AL(S) + 3O2(G)

• Moles ratios:

2AL2O3(L) ! 4AL(S) + 3O2(G)

• Use mole ratios to convert from amount in moles of one substance to amount in moles of another • Ex. 13.0 mol Al2O3

MOLAR MASS

•Mass of one mole of a substance • It is the conversion factor that relates mass of

substance to amount in moles • To solve stoichiometry problems, you need to

determine molar mass of substances

•Molar masses of substances: • Al2O3 = 101.96 g/mol •O2 = 32.00 g/mol • Al = 26.98 g/mol

•Can use these as conversion factors

• Find number of grams of Al equal to 26.0 mol of Al

IDEAL STOICHIOMETRIC CALCULATIONSSection 2

IDEAL STOICHIOMETRIC CALCULATIONS

• Equations are very important because you get the mole ratio directly from it • First thing to solving these types of problems is

balancing the equation •Chemical equations help make predictions about

reactions without having to run experiments (and waste resources) in lab

•Calculations in this book are theoretical

• They tell amounts of reactants and products under ideal conditions (where all reactants completely converted into products)

• Hardly ever happens in real life

CONVERSIONS OF QUANTITIES IN MOLES

• If asked for moles of product made from specific number of moles of reactant:

• Plan requires 1 conversion factor • Mole ratio

SAMPLE PROBLEM

1. ANALYZE

• Given: • Amount of CO2 = 20 mol

• Unknown: • Amount of LiOH in moles

2. PLAN

• Amount CO2 (mol) ! amount LiOH (mol)

• Requires a mole to mole ratio between CO2 and LiOH

3. COMPUTE

• How many moles of sodium will react with water to produce 4.0 mol of hydrogen in the following reaction?

2Na(s) + 2H2O(l) ! 2NaOH(aq) + H2(g)

• 8.0 mol Na

• How many moles of lithium chloride will be formed by the reaction of chlorine with 0.046 mol of lithium bromide in the following reaction?

2LiBr(aq) + Cl2(g) ! 2LiCl(aq) + Br2(l)

• 0.046 mol LiCl

CONVERSIONS OF AMOUNTS IN MOLES TO MASS

• Usually asked to find mass in grams of product formed

PRACTICE PROBLEM

• In photosynthesis, plants use energy from the sun to produce glucose, C6H12O6, and oxygen from the reaction of carbon dioxide and water. What mass, in grams, is produced when 3.00 mol of water react with carbon dioxide?

1. ANALYZE

• Given: • Amount H2O = 3.00 mol

• Unknown: • Mass of glucose produced (in g)

2. PLAN

• Chemical equation is

• Need 2 conversion factors: • Mole ratio of CO2 to H2O

• Molar mass of CO2

3. COMPUTE

PRACTICE PROBLEM 1

• Phosphorous burns in air to produce a phosphorous oxide in the following reaction.

4P(s) + 5O2(g) ! P4O10(s) What mass of phosphorous will be needed to produce 3.25 mol P4O10?

• 403 g

PRACTICE PROBLEM 2

• Hydrogen peroxide breaks down, releasing oxygen, in the following reaction.

2H2O2(aq) ! 2H2O(l) + O2(g)

What mass of oxygen is produced when 1.840 mol of H2O2 decompose?

• 29.44 g

CONVERSIONS OF MASS TO AMOUNTS IN MOLES

• In this type of problem you are starting with mass of some substance • Plan:

SAMPLE PROBLEM

1. ANALYZE

• Given: • Mass of NH3 = 824 g

• Unknown: • a. amount of NO produced (in mol) • b. amount of H2O produced (in mol)

2. PLAN

• First write balanced equation

• Mole ratios needed:

3. COMPUTE

MASS-MASS CALCULATIONS

• Since you can never measure moles directly, mass-mass calculations are more common

SAMPLE PROBLEM

1. ANALYZE

• Given: • Amount of HF = 30.00 g

• Unknown: • Mass of SnF2 produced in grams

2. PLAN

3. COMPUTE

PRACTICE PROBLEM 1

• Calculate the mass of silver bromide produced from 22.5 g of silver nitrate in the following reaction:

2AgNO3 + MgBr2 ! 2AgBr + Mg(NO3)2

• 24.9 g AgBr

PRACTICE PROBLEM 2

• What mass of acetylene, C2H2, will be produced from the reaction of 90. g of calcium carbide, CaC2, with water in the following reaction?

CaC2(s) + 2H2O(l) ! C2H2(s) + Ca(OH)2(aq)

• 37 g

LIMITING REACTANTS AND PERCENT YIELDSection 3

• In experiments, a reaction is rarely ever done with exact amounts of reactants

• Usually one or more reactants is in excess (too much) • Once one of the reactants is used up, the reaction stops • Substance used up first is the limiting reactant

• Limiting reactant ! the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction

• Excess reactant ! the substance that is not used up completely in reaction

• Consider the following reaction

C(s) + O2(g) ! CO2(g)

• According to equation, 1 mol C reacts with 1 mol oxygen to form 1 mol carbon dioxide

• Supposed you could mix 5 mol C with 10 mol O2

C(S) + O2(G) ! CO2(G)

• There is more oxygen (10 mol) than needed to react with 5 mol C

• C is limiting reactant – limits amount of CO2 formed

• Oxygen is excess reactant

• 5 mol O2 left over at the end

SAMPLE PROBLEM

1. ANALYZE

• Given: • Amount of HF = 2.0 mol • Amount of SiO2 = 4.5 mol

• Unknown: • Limiting reactant

2. PLAN

• The given amount of either reactant is used to calculate required amount of other reactant

• Calculated amount compared with amount you actually have

• Limiting reactant can be identified

• Under ideal conditions, 2.0 mol HF requires 0.50 mol SiO2 for complete reaction

• Because 4.5 mol SiO2 is available, that is more than is required so….

• HF is limiting reactant

PRACTICE PROBLEM 1

• Aluminum oxidizes according to the following equation:

4Al + 3O2 ! 2Al2O3

• Powdered Al (0.048 mol) is placed into a container containing 0.030 mol O2. What is the limiting reactant?

• O2

PRACTICE PROBLEM 2• Heating zinc sulfide in the presence of oxygen yields the

following:

ZnS + O2 ! ZnO + SO2

• If 1.72 mol ZnS is heated in the presence of 3.04 mol O2, which reactant will be used up?

• ZnS

PRACTICE PROBLEM 3

• Use the following equation for the oxidation of aluminum in the following problems:

4Al + 3O2 ! 2Al2O3

4AL + 3O2 ! 2AL2O3

• Which reactant is limiting if 0.32 mol Al and 0.26 mol O2 are available? • Al • How many moles of Al2O3 are formed from the reaction of 6.38 x 10-3 mol of

O2 and 9.15 x 10-3 mol of Al?

• 4.25 x 10-3 mol Al2O3

• If 3.17 g Al and 2.55 g of O2 are available, which reactant is limiting?

• O2

PRACTICE PROBLEM 4

ZrSiO4 + 2Cl2 ! ZrCl4 + SiO2 + O2

• What mass of ZrCl4 can be produced if 862 g of ZrSiO4 and 950 g of Cl2 are available?

• 1.10 x 103 g

PERCENT YIELD

• Amounts of products calculated in problems represent theoretical yields

• Theoretical yield ! maximum amount of product that can be made from a given amount of reactant

• In lab, amount of product is usually less than theoretical yield

WHY?

• Some reactant may be used competing in side reactions that reduce amount of product

• Once product is formed, usually collected in impure form • Some product lost during purification • Actual yield ! measured amount of product gotten from reaction

• Chemists usually interested in efficiency of reaction • Expressed by comparing actual and theoretical yields

• Percent yield ! ratio of actual yield to theoretical yield multiplied by 100

SAMPLE PROBLEM

1. ANALYZE

• Given: • Mass C6H6 = 36.8 g

• Mass of Cl2 = excess

• Actual yield of C6H5Cl = 38.8 g

• Unknown: • Percent yield of C6H5Cl

2. PLAN

• First do mass-mass calculation to find theoretical yield

• Then percent yield can be found

3. COMPUTE

PRACTICE PROBLEM

• Calculate the percent yield in each of the following cases • T Yield = 50.0 g, A Yield = 41.9 g • 83.8% • T yield = 290 kg, A Yield = 270 kg • 93% • T Yield = 0.00192 g, A yield = 0.00089 g • 46%