Chapter 9
Stoichiometry

Introduction to Stoichiometry

Composition stoichiometry - definition
Reaction stoichiometry - definition

Reaction-Stoichiometry Problems

use ratios from balanced equations
mole-mole problems
amount given is in moles, amount unknown and to find in moles
mole-mass problems
amount given is in moles, amount unknown and to find in grams
mass-mole problems
amount given is in grams, amount unknown and to find in moles
mass-mass problems
amount given is in grams, amount unknown and to find in grams

Mole Ratio

mole ratio - definition

for the reaction 2 Al₂O₃ ---> 4 Al + 3 O₂

2 moles of aluminum oxide produces 4 moles of aluminum and 3 moles of oxygen OR

2 moles of aluminum oxide produces 4 moles of aluminum i.e. 2 molAl₂O₃/ 4 mol Al OR

2 moles of aluminum oxide produces 3 moles of oxygen
i.e. 2 mol Al₂O₃/ 3 mol O₂

4 moles of aluminum is produced with 3 moles of oxygen
i.e. 4 mol Al / 3 mol O₂

Example: How many moles of aluminum can be produced from 13.0 mol of aluminum oxide according to the balanced chemical equation:

2 Al₂O₃ ---> 4 Al + 3 O₂

Solution: relationship between aluminum and aluminum oxide is 4 mol Al / 2 mol Al₂O₃. We can ignore the oxygen.

Aluminum is in numerator because you are asked how many moles of aluminum can be produced i.e. it is the unknown for which you are solving; also, you need the aluminum oxide to cancel out because you are solving for moles of aluminum

13.0 mol Al₂O₃ x 4 mol Al / 2 mol Al₂O₃ = 26.0 mol Al

N.B. mole ratios do not limit significant figures

Example - alternate method

Homework: 9.1

Molar Mass

molar mass - definition
used as conversion factor in changing moles to grams and grams to moles
use periodic table -- round to second place to right of decimal point.
Example: convert 26 mol of aluminum to grams

Solution: molar mass is 26.98 g Al / 1 mol Al

26.0 mol Al x 26.98 g Al / 1 mol Al = 701 g Al

Homework: 9.2

Ideal Stoichiometric Calculations

must have balanced chemical equation to do stoichiometry
problems we will be doing are for ideal conditions,
rarely exist in the laboratory.
assumes all reactants are completely converted to products which rarely happens.
problems will indicate the maximum amount of a substance that can be produced from a given amount of a reactant.
Conversions of Quantities in Moles (Mole-Mole Problems)
revolves around balanced equation
coefficients are used to created a mole ratio
start calculation with given, multiply by mole ratio so that units (substances) cancel

e.g. Sample Problem 9-1 page 281
How many moles of lithium hydroxide are required to react with 20 mol of CO₂ according to the following balanced equation

CO₂ + 2 LiOH ---> Li₂CO₃ + H₂O

given
20 mol CO₂
x mol LiOH

Solution: ratio is either 1 mol CO₂ / 2 mol LiOH or
2 mol LiOH / 1 mol CO₂ --- from the coefficients in balanced equation

start calculation with given information 20 mol CO₂

20 mol CO₂ x 2 mol LiOH / 1 mol CO₂ = 40 mol LiOH

Alternate: equation method

Sample Problem 9-1 page 281

Homework: 9.3

Conversions of Amounts in Moles to Mass (Mole-mass problems)

Need two conversion factors
moles to moles as above then moles to grams using molar mass

Sample Problem 9-2
Given: mass in g of glucose (C₆H₁₂O₆) = ?
3.00 mol water
6 H₂O + 6 CO₂ ---> C₆H₁₂O₆ + 6 O₂

Solution: convert mole of water into mole of glucose then convert mole of glucose into gram of glucose

convert moles of water to moles of glucose
conversion factors:
6 mol H₂O / 1 mol C₆H₁₂O₆ or
1 mol C₆H₁₂O₆ /6 mol H₂O

3.00 mol H₂O x 1 mol C₆H₁₂O₆ /6 mol H₂O = 0.5 mol C₆H₁₂O₆

convert moles of glucose to grams
conversion factors:
C: 6 X 12.01g = 72.06 g
H: 12 x 1.01 g = 12.12 g
O: 6 x 16.00 g = 96.00 g
molar mass = 180.18g

180.18 g / mole or mole / 180.18 g

0.5 mole glucose X 180.18 g glucose/ mole glucose = 90.09 g

Alternate method

Sample Problem 9-3 page 283
Given: 3.00 mol water
mass in grams of carbon dioxide = ?
6 H₂O + 6 CO₂ ---> C₆H₁₂O₆ + 6 O₂

Conversion Factors:
ratio of CO₂ to water is 6:6
molar mass of CO₂: C: 1 x 12.01g = 12.01 g
O: 2 x 16.00g = 32.00g
molar mass = 44.01g

3 mol water x 6 mol CO₂/6 mol water x 44.01 g CO₂/mol CO₂

132 g CO₂

Alternate method

Conversion of Mass to Amounts in Moles (Mass-Mole Problems)

Conversion of Mass to Amounts in Moles (Mass-Mole Problems)

Sample Problem 9-4 page 285

Given:
824 g of NH₃
moles of NO = ?
moles of water = ?

4NH₃ + 5O₂ ---> 4NO + 6H₂O
Conversion Factors Part A:
ratio of NO to NH₃ is 4:4
ratio of water to ammonia is 6:4
molar mass of ammonia: N: 1 x 14.01g = 14.01g
H: 3 x 1.01 g = 3.03g
molar mass = 17.04g

824 g NH₃ x ( 1 mol NH₃/17.04 g NH₃) x (4 mol NO/4 mole ammonia) =

48.4 mol NO
Conversion Factors Part B:
824 g of ammonia
moles of water = ?
molar ratio of ammonia to water is 4:6

824 g NH₃ x ( 1 mol NH₃/17.04 g NH₃) x (6 mol water/4 mol NH₃) =

72.6 mol water

Alternate method

Homework: 9.4

Mass-Mass Calculations
mass in grams --> amount in moles --> amount in moles --> mass in grams

Sample Problem 9-5 page 286

Given Information:
Sn + 2 HF ---> SnF₂ + H₂
30.00 g HF
mass of SnF₂ produced = ?

Conversion Factors:
molar mass of HF 20.01 g/mole
molar mass of SnF₂: 156.71 g/mole
molar ratio: SnF₂ to HF is 1:2

30.00 g HF x ( 1 mol HF/20.01 gHF) x (1 mol SnF₂/2 mol HF) x

(156.71 g SnF₂ / 1 mole SnF₂) = 117.5 g SnF₂

alternate method

Homework: 9.5

Limiting Reactants and Percent Yield

Deals with forcing a reaction to produce a minimum of a product by adding an excess of one reactant. When the reactant not in excess is used up the reaction stops.

Limiting reactant - definition
Excess reactant - definition

Sample problem 9-6 page 289
Given: 2.0 mol HF
4.5 mol SiO₂
SiO₂ + 4 HF ---> SiF4 + 2 H₂O
limiting reactant = ?

molar ratio for SiO₂ to HF is 1:4 from the balanced equation
2.0 mol HF x (1 mol SiO₂/4 mol HF) = 0.5 mol SiO₂

Tells us that 2 mol of HF requires 0.5 mol SiO₂ to react completely.Since we are told that we have 4.5 mol SiO₂, SiO₂ is present in excess. That means HF is the limiting reactant.

Alternate method

Homework: 9.6

Percent Yield
theoretical yield - definition
actual yield - definition
percent yield - definition

Sample problem 9-8 page 293

Given: 36.8 g C₆H₆
excess of Cl₂
yield of C₆H₅Cl = 38.8 g

% yield of C₆H₅Cl = ?

C₆H₆ + Cl₂ ---> C₆H₅Cl + HCl

molar ratio of C₆H₅Cl to benzene is 1:1
molar mass of benzene = 78.12g/mole
molar mass of C₆H₅Cl = 112.56 g/mole

36.8 g benzene x (mole benzene/78.12g benzene) x (1 mole C₆H₅Cl / 1 mole benzene) x (112.56 g C₆H₅Cl /mole C₆H₅Cl = 53.0 g C₆H₅Cl

(actual yield / theoretical yield) x 100 = percentage yield
(38.8 g / 53.0 g) x 100 = 73.2 % yield

alternate method

Homework: 9.7

END OF NOTES

Limiting reactant is 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 is the substance that is not used up completely in a reaction.

Theoretical yield is the maximum amount of product that can be produced from a given amount of reactant.

Actual yield is the measured amount of a product obtained form a reaction.

Percent yield is the ratio of the actual yield to the theoretical yield multiplied by 100.