Unit 6
Stoichiometry
Chapter 9
Stoichiometry
CHAPTER 9
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.
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
4 mol Al
2 mol Al2O3
3 mol O2
4 mol Al
3 mol O2
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) = 18.02 g/mol.
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?
This is what
we’re given:
Use mole ratio as
a conversion factor
12 mol NH3 x
3 mol H2
2 mol NH3
= 18 mol H2
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:
Given
Conversion factor
20 mol CO2 x 2 mol LiOH = 40 mol LiOH
1 mol CO2
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.
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.
Given
2.00 mol Mg x
1st C.F.
Mole Ratio
2nd C.F.
Molar Mass
of the Unknown
2 mol MgO x 40.3 g MgO
= 80.6 g MgO
2 mol Mg
1 mol MgO
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.
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?
Given
125 g O2
1st C.F.
Molar Mass
of the Given
x 1 mol O2
32 g O2
2nd C.F.
Mole Ratio
x
2 mol HgO
1 mol O2
= 7.81 mol HgO
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:
Given
1st C.F.
Mole Ratio
2nd C.F.
Molar Mass
of the Unknown
1
180.2
mol
C
H
O
g C6H12O6
6
12
6
x
x
3.00 mol H2O
= 90.1 g
6 mol H2O
1 mol C6H12O6
C6H12O6
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?
st C.F.
1
Solution:
Molar Mass
of the Given
Given
1
mol NH3 x
x
824 g NH3
17.0 g NH3
2nd C.F.
Mole Ratio
4 mol NO
4 mol NH3
= 48.5 mol NO
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.
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?
1st
Given
C.F.
Molar Mass
of the Given
1
mol O2
x
45 g O2
32 g O2
3rd C.F.
Molar Mass of
2nd C.F.
Mole Ratio the Unknown
x 2 mol HgO x 216.6 g HgO = 609.2 g
1 mol O2
1 mol HgO HgO
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
30.0 g HF with Sn?
Solution: 1st C.F.
3rd C.F.
Molar Mass 2nd C.F. Molar Mass of
Given of the Given Mole Ratio the Unknown
30.0 g HF x 1 mol HF x 1 mol SnF2 x 156.7 g SnF2 = 117.5 g
SnF2
20.0 g HF 2 mol HF
1 mol SnF2
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
Limiting Reactants
• 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.
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.
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
Limiting Reactants (continued)
Example:
Silicon dioxide reacts with hydrogen fluoride according
to the following equation:
The limiting
SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)
reactant
If 6.0 mol HF is added to 4.5 mol SiO2, which makes the
is the limiting reactant? Mole ratio C.F.
least product.
Set up 2
6 mol HF x
equations,
one for each
given:
4.5 mol SiO2 x
1 mol SiF4 = 1.5 mol SiF44
4 mol HF Limiting Reactant is HF
1 mol SiF4 = 4.5 mol SiF4
1 mol SiO2
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)
a. If 1 mol CS2 reacts with 1 mol O2, identify the limiting reactant.
1 mol CS2 x 2 mol SO2 = 2 mol SO2
Limiting
1 mol CS2
Reactant is O2
1 mol O2 x 2 mol SO2 = 0.67 mol SO2
Mol of excess
3 mol O2
reactant that
b. How many moles of excess reactant remain?
Work
0.67 mol SO2 x 1 mol CS2 = 0.33 mol CS2 was used up
backwards from
2 mol SO2
Moles
actual mol of
remaining
product formed 1 mol CS2 - 0.33 mol CS2 = 0.67 mol CS2
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
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.
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?
Set up a mass to mass conversion
This is the
Theoretical Yield
36.8 g C6H6 x 1 mol C6H6 x 1 mol C6H5Cl x 112.5 g C6H5Cl = 53.1 g
78.0 g C6H6 1 mol C6H6
1 mol C6H5Cl C6H5Cl
Actual
38.8 g C6H5Cl 100
Percent
100
x
x
= 73.1%
Now use
=
=
Theoretical
53.1 g C6H5Cl
the formula Yield
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?
Set up a mass to mass conversion
This is the
Theoretical Yield
75.0 g CO x 1 mol CO x 1 mol CH3OH x 32.0 g CH3OH = 85.7 g
28.0 g CO
1 mol CO
1 mol CH3OH
CH3OH
Multiply theoretical
yield by percentage
to get actual yield
85.7 g CH3OH x 80% =
68.6 g CH3OH