Chapter 9 Notes
STOICHIOMETRY deals with mass relationships. COPOSITION STOICHIOMETRY
is the mass relationship between the elements in a compound. REACTION
STOICHIOMETRY is the mass relationship between the reactants and products of a
chemical reaction.
You have been studying composition stoichiometry in the previous chapters. Our study
of chemical reactions introduces mass relationships between the elements or compounds
found in chemical reactions. Stoichiometry in regards to a chemical reaction can be
compared to a recipe used in cooking. In a recipe, you can find a list of ingredients.
These ingredients can be compared to the reactants in a chemical reaction. The product
would be comparable to the final dish prepared from the recipe. Anytime the recipe is
adjusted due to the lack of ingredients or to produce larger amounts of a product is the
same as stoichiometry.
With a chemical reaction, it is very important that the chemical equation is balanced. A
proper balanced equation is our recipe. From this balanced chemical equation we find
MOLAR RATIOS. Molar ratios are conversion factors between the numbers of moles of
any two substances in a chemical reaction.
For the equation:
2Al2O3  4Al + 3O2
The following molar ratios can be derived.
2 moles Al2O3
4 moles Al
4 moles Al
2 moles Al2O3
4 moles Al
3 moles O2
3 moles O2
4 moles Al
2 moles Al2O3
3 moles O2
3 moles O2
2 moles Al2O3
So if you have 13 moles of Al2O3, how many moles of Al can be produced?
113 moles Al2O3 x 4 moles Al
2 moles Al2O3
=
26 moles of Al
Using the molar mass of Al, we can convert to grams.
26 moles Al
x
26.98 grams Al
1 mole Al
=
701 grams Al
There are 4 basic types of stoichiometric problems; mole to mole, gram to mole, mole to
gram, and gram to gram. However, volumes of a reactant or product can be introduced
using the conversion factors of density for liquids or solids and 22.4 liters/1 mole for
gases.
Common to all these problems is a “given” and an “unknown”. Identifying these in the
problem is very important. The stoichiometric map provided outlines the steps involved
in solving these problems.
Let’s look at each of these types of problems.
MOLE TO MOLE
In a space craft, the carbon dioxide exhaled by astronauts can be removed by the reaction
with lithium hydroxide, LiOH, according to the following chemical equation.
CO2(g) + 2LiOH(s)  Li2CO3(s) +H2O(l)
How many moles of LiOH are required to react with 20 moles of CO2, the average
amount exhaled by a person per day?
Given: 20 moles CO2
Unknown: ? moles LiOH
The molar ratio is 2 moles LiOH to 1 mole CO2. This is found in the balanced chemical
equation.
20 moles CO2
x
2 moles LiOH
1 mole CO2
=
40 moles LiOH
MOLES TO GRAMS
In photosynthesis, plants use energy from the sun to produce glucose, C6H12O6 from the
reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when
3.00 moles of water react with carbon dioxide?
6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g)
Given: 3.00 moles water
Unknown: ? grams of glucose
From the balanced chemical equation, the molar ratio is 6 moles H2O to 1 mole C6H12O6.
Using the periodic table to calculate, the molar mass of C6H12O6 is 180.18 grams.
3.00 moles H2O
x
1 mole C6H12O6 x
6 moles H2O
180.18 grams = 90.1 grams C6H12O6
1 mole C6H12O6
GRAMS TO MOLES
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of
ammonia.
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
Using 824 grams of NH3 and excess oxygen, how many moles of NO is formed?
Given: 824 grams NH3
Unknown: ? moles of NO
From the balanced chemical equation, the molar ratio is 4 moles NH3 to 4 moles NO.
Using the periodic table to calculate, the molar mass of NH3 is 17.04 grams.
824g NH3 x 1 mole NH3 x 4 moles NO = 48.4 moles NO
17.04 grams
4 moles NH3
GRAMS TO GRAMS
Tin (II) fluoride, SnF2, is used in some toothpaste. It is made by reacting tin with
hydrogen fluoride according to the following equation.
Sn(s) + 2HF(g) SnF2(s) + H2(g)
How many grams of SnF2 are produced from 30.00 grams HF?
Given: 30.00 grams HF
Unknown: ? grams of SnF2
From the balanced chemical equation, the molar ratio is 1 mole SnF2 to 2 moles HF.
Using the periodic table to calculate, the molar mass of HF is 20.01 grams and the molar
mass of SnF2 is 156.71 grams.
30.00g HF x 1 mole HF x 1 mol SnF2 x 156.71g SnF2 = 117.5 grams SnF2
20.01g HF
2 mol HF
1 mol SnF2
In most reactions, one of the reactants is completely used up in the reaction and the other
reactant will have some amount remaining. The reactant that is completely consumed in
the reaction is known as the LIMITTING REACTANT. The limiting reactant determines
how much product is produced. The other reactant will be in excess and is called an
EXCESS REACTANT. Limiting reactant problems use stoichiometric solutions.
EXAMPLE
Silicon dioxide is usually unreactive, but reacts readily with hydrogen fluoride according
to the following equation.
SiO2(s) + 4HF(g)  SiF4(g) + 2H2O(l)
If 6.0 moles of HF is added to 4.5 moles of SiO2, which is the limiting reactant?
Given: 6.0 moles HF and 4.5 moles SiO2
Unknown: ? Limiting reactant
You will have to select a product. It is not suggested that water is chosen as a product
unless it is the only product in the equation. So in this case, we will choose SiO4. We
now work a mole to mole stoichiometric problem with both reactants.
6.0 moles HF x 1 mole SiO4 = 1.5 moles SiO4
4 moles HF
4.5 moles SiO2 x 1 mole SiO4 = 4.5 moles SiO4
1 mole SiO2
Because 6.0 moles of HF produce the lesser amount of product, SiO4, it is considered to
be the limiting reactant.
Stoichiometric problems can also be used to calculate the amount of product is expected
to be produced in a reaction. This calculated amount of product can be compared to the
amount actually produced in a laboratory setting. In order to make this type of
comparison, the calculated amount of product expected to be produced is known as the
THEORETICAL YIELD. Using the following equation,
Percentage yield = experimental yield x 100
theoretical yield
EXAMPLE
Chlorobenzene, C6H5Cl, is used in the production of many important chemicals, such as
aspirin, dyes, and disinfectants. One industrial method of preparing chlorobenzene is to
react benzene, C6H6, with chlorine.
C6H6(l) + Cl2(g)  C6H5Cl(l) + HCl(g)
When 36.8 grams of benzene react with excess chlorine, the actual yield of
chlorobenzene is 38.8 grams. What is the percentage yield?
Given: 36.8 grams C6H6 and actual yield is 38.8 grams
Unknown: Percentage yield of chlorobezene
From the balanced chemical equation, the molar ratio is 1 mole of C6H6 to 1 mole of
C6H5Cl. Using the periodic table to calculate, the molar mass of benzene is 78.12 grams
and the molar mass of chlorobenzene is 112.56 grams.
36.8g C6H6 x 1 mol C6H6 x 1 mol C6H5Cl x 112.56g = 53.0 grams C6H5Cl
78.12 grams
1 mol C6H6
1 mol C6H5Cl
Percentage yield =
38.8 grams
53.0 grams
x
100
= 73.2 %