1. Stoichiometry Chapter 9
Unit 6
Stoichiometry
Chapter 9

2. Stoichiometry
Chapter 9

3. Chapter 9 – Section 1: Introduction to Stoichiometry
What is Stoichiometry?
Stoichiometry is the branch of chemistry that deals with the mass relationships of elements in a chemical reaction.
Stoichiometry calculations always start with a balanced chemical equation.

4. Chapter 9 – Section 1: Introduction to Stoichiometry
Mole Ratio
A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction
Example: 2Al2O3(l) → 4Al(s) + 3O2(g)
Mole Ratios: 2 mol Al2O3 2 mol Al2O3 4 mol Al
4 mol Al mol O mol O2

5. Chapter 9 – Section 1: Introduction to Stoichiometry
Molar Mass (Review)
Molar Mass is the mass of one mole of a pure substance.
Molar mass units are g/mol.
The molar mass of an element is the same number as its atomic mass, only the units are different.
Example: Molar mass of water (H2O).
(2 mol H x 1.01) + (1 mol O x 16.00) = g/mol.

6. Mole to Mole Conversions
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Mole to Mole Conversions
Using the mole ratio, you can convert from moles of one substance to moles of any other substance in a chemical reaction.
Example:
For the reaction N2 + 3H2 → 2NH3, how many moles of H2 are required to produce 12 moles of NH3?
Use mole ratio as a conversion factor
This is what
we’re given:
3 mol H2
12 mol NH3 x =
18 mol H2
2 mol NH3

7. Mole/Mole Conversions Sample Problem 1
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Mole/Mole Conversions Sample Problem 1
In a spacecraft, the CO2 exhaled by astronauts can be removed by the following reaction with LiOH:
CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O(l)
How many moles of lithium hydroxide are required to react with 20 mol CO2, the average amount exhaled by a person each day?
Solution:
Conversion factor
Given
2 mol LiOH
20 mol CO2 x =
40 mol LiOH
1 mol CO2

8. Mole to Mass Conversions
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Mole to Mass Conversions
To convert from moles of one substance to grams of another, you need 2 conversion factors: 1. mole ratio. 2. molar mass of the unknown.
To set up your conversion factor, always put the units you have on the bottom and the units you need on the top.

9. Mole to Mass Conversions (continued)
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Mole to Mass Conversions (continued)
Example:
Given the equation: 2Mg(s) + O2(g)→ 2MgO(s),
Calculate the mass in grams of magnesium oxide which is produced from 2.00 mol of magnesium.
2nd C.F.
Molar Mass
of the Unknown
1st C.F.
Mole Ratio
Given 2
mol MgO
40.3
g MgO
2.00 mol Mg x x
80.6 g MgO = 2
mol Mg 1
mol MgO

10. Mass to Mole Conversions
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Mass to Mole Conversions
To convert from grams of one substance to moles of another, you need 2 conversion factors: 1. molar mass of the given. 2. mole ratio.
Mass to mole conversion factors are the inverse of mole to mass conversion factors.

11. Mass to Mole Conversions (continued)
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Mass to Mole Conversions (continued)
Example:
Given the equation: 2HgO(s) → 2Hg(s) + O2(g),
How many moles of HgO are needed to produce 125g of O2?
1st C.F.
Molar Mass of the Given
2nd C.F.
Mole Ratio
Given 1
mol O2 2
mol HgO
125 g O2 x x =
7.81 mol HgO 32
g O2 1
mol O2

12. Mole/Mass Conversions Sample Problem 1
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Mole/Mass Conversions Sample Problem 1
The balanced equation for photosynthesis is as follows:
6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g)
What mass, in grams, of glucose (C6H12O6) is produced when 3.00 mol of water react with carbon dioxide?
Solution:
2nd C.F.
Molar Mass
of the Unknown
1st C.F.
Mole Ratio
Given 1
mol C6H12O6
180.2
g C6H12O6
3.00 mol H2O x x
90.1 g C6H12O6 = 6
mol H2O 1
mol C6H12O6

13. Mole/Mass Conversions Sample Problem 2
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Mole/Mass Conversions Sample Problem 2
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
If the reaction is run using 824 g NH3 and excess oxygen, how many moles of NO are formed?
Solution:
1st C.F.
Molar Mass of the Given
2nd C.F.
Mole Ratio
Given 1
mol NH3 4
mol NO
824 g NH3 x x =
48.5 mol NO
17.0
g NH3 4
mol NH3

14. Mass to Mass Conversions
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Mass to Mass Conversions
To convert from grams of one substance to grams of another, you need 3 conversion factors: 1. molar mass of the given. 2. mole ratio. 3. molar mass of the unknown.

15. Mass to Mass Conversions (continued)
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Mass to Mass Conversions (continued)
Example:
Given the equation: 2HgO(s) → 2Hg(s) + O2(g),
How many grams of HgO are needed to produce 45g of O2?
3rd C.F.
Molar Mass of the Unknown
1st C.F.
Molar Mass of the Given
2nd C.F.
Mole Ratio
Given 1
mol O2 2
mol HgO
216.6
g HgO
45 g O2 x x x =
609.2 g HgO 32
g O2 1
mol O2 1
mol HgO

16. Mass/Mass Conversions Sample Problem 1
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Mass/Mass Conversions Sample Problem 1
Tin(II) fluoride, SnF2, is used in some toothpastes. It is made by the following reaction :
Sn(s) + 2HF(g) → SnF2(s) + H2(g)
How many grams of SnF2 are produced from the reaction of g HF with Sn?
Solution:
3rd C.F.
Molar Mass of the Unknown
1st C.F.
Molar Mass of the Given
2nd C.F.
Mole Ratio
Given 1
mol HF 1
mol SnF2
156.7
g SnF2
117.5 g SnF2
30.0 g HF x x x =
20.0
g HF 2
mol HF 1
mol SnF2

17. Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
When combining 2 or more different things to make a product, you have to stop when one of the things is used up.
For example, no matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made. 

18. Limiting Reactants (continued)
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
Limiting Reactants (continued)
The limiting reactant is the reactant that limits the amount of product formed.
The excess reactant is the substance that is not used up completely.
Once the limiting reactant is used up, a chemical reaction will stop.

19. Limiting Reactants (continued)
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
Limiting Reactants (continued)
Example: Silicon dioxide reacts with hydrogen fluoride according to the following equation: SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l) If 6.0 mol HF is added to 4.5 mol SiO2, which is the limiting reactant?
The limiting reactant makes the least product.
Mole ratio C.F.
Set up 2 equations, one for each given:
1 mol SiF4
6 mol HF
6 mol HF x =
1.5 mol SiF4
1.5 mol SiF4
4 mol HF
Limiting Reactant is HF
1 mol SiF4
4.5 mol SiO2 x =
4.5 mol SiF4
1 mol SiO2

20. Limiting Reactants Sample Problem 1
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
Limiting Reactants Sample Problem 1
According to the following reaction :
CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)
If 1 mol CS2 reacts with 1 mol O2, identify the limiting reactant.
How many moles of excess reactant remain? x
2 mol SO2 =
2 mol SO2
1 mol CS2
Limiting Reactant is O2
1 mol CS2
1 mol O2
1 mol O2 x
2 mol SO2 =
0.67 mol SO2
0.67 mol SO2
3 mol O2
Mol of excess reactant that was used up
Work backwards from actual mol of product formed
0.67 mol SO2 x
1 mol CS2 =
0.33 mol CS2
2 mol SO2
Moles remaining =
1 mol CS2
mol CS2
0.67 mol CS2

21. Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant (found with stoichiometry).
The actual yield of a product is the measured amount of that product obtained from a reaction (given in problem).
The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100.

22. Percentage Yield (continued)
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
Percentage Yield (continued)
Example: Given the following equation: C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g) When 36.8 g C6H6 react with excess Cl2, the actual yield of C6H5Cl is 38.8 g. What is the percentage yield of C6H5Cl?
This is the Theoretical Yield
Set up a mass to mass conversion
36.8 g C6H6 x 1
mol C6H6 x 1
mol C6H5Cl x
112.5
g C6H5Cl =
53.1 g C6H5Cl
g C6H6
78.0 1
mol C6H6 1
mol C6H5Cl
Actual
38.8 g C6H5Cl
Percent Yield x
100 x
100 =
73.1%
Now use the formula = =
Theoretical
53.1 g C6H5Cl

23. Percentage Yield Sample Problem 1
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
Percentage Yield Sample Problem 1
According to the following reaction : CO(g) + 2H2(g) → CH3OH(l) If the typical yield is 80%, what mass of CH3OH should be expected if 75.0 g of CO reacts with excess hydrogen gas?
This is the Theoretical Yield
Set up a mass to mass conversion
75.0 g CO x 1
mol CO x 1
mol CH3OH x
32.0
g CH3OH =
85.7 g CH3OH
28.0
g CO 1
mol CO 1
mol CH3OH
Multiply theoretical yield by percentage to get actual yield
85.7 g CH3OH x
80%
68.6 g CH3OH =