1. Mass relationships in chemical reactions: Stoichiometry
Chapter 10
Mass relationships in chemical reactions:
Stoichiometry

2. To Review How many grams in a mole of CaSO4 Al2(SO4)3
Ca 1 x 40.1 g/mol = g
S 1 x = 32.1
O 4 x = 64.0
136.2 g/mol
Al2(SO4)3
Al 2 x 27.0 g/mol =
S 3 x =
O 12 x =
312.3 g/mol

3. 2Al(s) + 3Br2(l) 1Al2Br6(s)
Stoichiometric Coefficients- the coefficients (number in front of the chemical formula) in a balanced chemical equation
Stoichiometry- the relationship between the quantities of chemical reactants and products
Stoichiometric factor- a mole ratio relating moles of one substance to another substance involved in the same chemical reaction

4. 2 H2O2(liq) ---> 2 H2O(l) + 1 O2(g)
Create two stoichiometric factors that relate the amount of the reactant H2O2 to the product O2
2 H2O2(liq) ---> 2 H2O(l) O2(g)
2 mol H2O OR mol O2
1 mol O molH2O2

5. 2 H2O2(l) ---> 2 H2O(l) + O2(g)
Create two stoichiometric factors that relate the amount of the product H2O to the reactant H2O2
2 H2O2(l) ---> 2 H2O(l) + O2(g)
2 mol H2O2 OR mol H2O
2 mol H2O mol H2O2

6. 8 mol CO2 OR 2 mol C4H10 2 C4H10 + 13O2 -----> 10H2O + 8CO2
Create two stoichiometric factors that relate the amount of the product CO2 to the reactant C4H10
2 C4H O > 10H2O CO2
8 mol CO OR mol C4H10
2 mol C4H mol CO2

7. Stoichiometric Calculations
The most common types of equation stoichiometry problems are called 3 step problems
Given the mass of a reactant or product determine the equivalent mass of a different reactant or product
These are called mass to mass or gram to gram conversions
Consider: 2 H2O > 2 H O2
How many g of H2 are produced by decomposing 1.00 g H2O?

8. Stoichiometric Calculations
All calculations are done using conversion factors
Mass H2O
Mol H2O
Mol H2
Mass H2
1 mol
MM in g
mol H2
mol H2O
MM in g
1 mol
Each arrow represents a conversion factor
Align units first, then place numbers into the
conversion factor

9. Stoichiometric Calculations
Step 1 - Convert g of given reactant or product to moles
Use molar mass of the given
Step 2 - Convert from moles of the given to moles of the unknown
Use stoich factor as conversion factor
Step 3 - Convert from moles to g of the unknown
Use molar mass of the unknown as conversion factor

10. Problem 1a. If 37.6 g of water is decomposed to hydrogen and oxygen, how many grams of hydrogen will be produced?
Write the balanced chemical equation
2 H2O > 2 H O2

11. Write the information that is given and what's asked for above the equation
Mass 37.6 g ? g
Equation 2H2O > 2H O2

12. Write the molar mass of the substances.
Mass 37.6 g ? g
Equation 2 H2O > 2H O2
Molar Masses
(g/mol)
Remember the Pathway:
g H2O 
mol H2O 
mol H2 
g H2

13. Step 1 - Convert the given mass of water to moles of water
2 H2O > 2 H O2
1 mol H2O = 18.0 g H2O
37.6 g H2O x 1 mol H2O
18.0 g H2O

14. Step 2 - Convert moles of water to moles of hydrogen using stoichiometric factor
2 H2O > 2 H O2
37.6 g H2O x 1 mol H2O x 2 mol H2
18.0 g H2O 2 mol H2O

15. Step 3 - Convert moles of hydrogen to mass of hydrogen
2 H2O > 2 H O2
1 mol H2 = g H2
37.6g H2O x 1 mol H2O x 2 mol H2 x 2.0g H2 = 4.17g H2
18.0g H2O mol H2O 1 mol H2

16. Problem 1b. At room temperature and pressure, aluminum reacts with oxygen to give aluminum oxide. If you react 161 g of Al, what mass of O2 is needed for complete reaction?
Write the chemical equation:
Al(s) + O2(g) > Al2O3

17. You need a balanced chemical equation:
Al(s) O2(g) > Al2O3 2 4 3 2

18. Record given mass and what is asked for in the problem
Mass 161 g ? g
Equation 4Al(s) + 3O2(g) ---->2Al2O3
Molar Masses
(g/mole)

19. 3 Step Calculations 4Al(s) + 3O2(g) ---->2Al2O3
g Al  mol Al  mol O2  g O2
161 g Al ? g O2
4Al(s) + 3O2(g) ---->2Al2O3
27.0 g/mol g/mol
161 g Al
x 1 mol Al
27.0 g Al
x 3 mol O2
4 mol Al
x g O2
1 mol O2
= 143 g O2

20. What mass of MgO can be produced from igniting 1.5 g of Mg in oxygen?
Mass g ? g
Equation 2 Mg(s) + O2(g) ---> 2 MgO
Molar Masses
(g/mole)
1.5 g Mg x
1 mol Mg x
24.3 g Mg
2 mol MgO x
2 mol Mg
40.3 g MgO
1 mol MgO
= 2.49 g MgO

21. Theoretical vs Actual Yield
Theoretical yield = the amount of product that could be produced if the reaction is perfect
The amount that results from the 3 step process
This never happens
You usually get less than the theoretical yield
Actual yield = The amount of product you recover when you run the experiment

22. Limiting and Excess Reactant
In chem reactions, reactants are never present in the exact ratio to ensure they are completely used up in the reaction
You usually run out of one reactant, and the other is in excess (left over)
Limiting reactant = the one completely used up
Excess reactant = the one with some left over

23. A Chemical Reaction Reactants Products
All of is used up = Limiting Reactant
Some of is left over = Excess Reactant

24. Limiting Reactant Problem
Determine the limiting and excess reactant if 1.40 g of N2 reacts with 1.00 g of H2 to produce ammonia, NH3.
1.40 g g
N H2  2NH3
28.O g/mol g/mol
1.40 g N2
x 1 mol N2
28.0 g N2
x 3 mol H2
1 mol N2
x 2.0 g H2
1 mol H2
= 0.30 g H2
There is more than 0.30 g of H2 present, therefore
H2 will be left over (is in excess) and N2 will be the
limiting reactant

25. Percent Yield
A means of quantifying how efficient a chemical reaction is
Actual Yield
Theoretical Yield
% Yield =
x 100
% Yields are usually below 100%
less product is recover than is possible
% Yields can be above 100%
What’s better, a % yield of 90% or 110% ?

26. Solving Percent Yield Probs
1 – Determine the limiting reactant
2 – Calculate the theoretical yield from the given amount of the limiting reactant
3 – Calculate the % yield (the actual yield needs to be given to you in the problem).
Actual Yield
Theoretical Yield
% Yield =
x 100

27. Percent Yield Practice Prob #1
A student reacts 6.00 g of N2 reacts with g of H2
to produce ammonia, NH3. if 2.59 g of NH3 is recovered
from the reaction, what is the student’s % yield?
N H2  2NH3
Step 1 Determine the limiting reactant.
6.00 g N2
x 1 mol N2
28.0 g N2
x 3 mol H2
2 mol N2
x 2.0 g H2
1 mol H2
= g H2
Therefore, g H2 is the limiting reactant

28. Percent Yield Practice Prob #1
Step 2 Deter the theoretical yield from limiting reactant
0.500 g H2
x 1 mol H2
2.0 g H2
x 2 mol NH3
3 mol H2
x 17.0 g NH3
1 mol NH3
= 2.83 g NH3
Actual yield = 2.59 g NH3 (given in the problem)
Theoretical yield = 2.83 g NH3
% yield = actual yield x 100
theroretical yield
= g x = 91.5%
2.83 g

29. Lab Exercise A student reacts 2.40 g of NaHCO3 with an excess of HCL
according to the reaction:
2.40 g X g
NaHCO3 + HCl ---> NaCl H2O + CO2
84.0 g/mol g/mol
Determine the theoretical yield of NaCl (3 step process)
g NaHCO3  mol NaHCO3  mol NaCl  g NaCl
2.40 g NaHCO3
x 1 mol NaHCO3
84.0 g NaHCO3
x 1 mol NaCl
1 mol NaHCO3
x g NaCl =
1 mol NaCl
= 1.67 g NaCl

30. Lab Exercise
The reaction from your lab is conducted with a 2.40 g sample of NaHCO3. In the lab, the reaction produces 1.57 g of NaCl. What is the percent yield?
x 1 mol NaHCO3
84.0 g/mol
x 1 mol NaCl
1 mol NaHCO3
x g NaCl =
1 mol NaCl
1.67 g
NaCl
2.40 g NaHCO3
% Yield = actual yield x 100
theoretical yield
= g x 100
1.67 g
= 94.0%