Year 11 Chemistry: Chapter 15:~ Calculations in Chemistry: Stoichiometry
 Notes from ‘Chapter 4: Relative atomic mass and mole’ will help
Reacting Quantities
A chemist working for a drug company needs to be precise when weighing the reactants
required to prepare a particular compound. Chemists involved in the preparation and
analysis of a wide range of compounds rely heavily on calculations of masses of reactants
and products. Calculations such as these are based on balanced, full chemical equations
for each reaction. All particles that existed at the start of a chemical reaction must
be accounted for after the reaction is complete.
2H2(g) + O2(g)  2H2O(g)
This equation gives the following information:
 It identifies the reactants and products.
 It identifies the states of the reactants and products
 Because it is balanced, it gives the ratio in which the substances react.
The coefficients in a balanced equation therefore give us the ratio of the amounts (in
mol) of reactants and products. This mole ratio will be used in calculating masses of
reactants and products.
15.1 Calculations based on amount of a reactant or product
Mass-Mass Stoichiometry
In these problems, the mass of one of the reactants or products is known and is used to
determine the mass of one of the other reactants or products involved in the reaction.
Step 1: Write a balanced equation for the reaction.
Step 2: Calculate the amount (in mol) of the substance with the known mass.
m
n = M
Step 3: Use the mole ratio from the equation to calculate the amount (in mol)
of the required substance.
Step 4: Calculate the mass required
m = nM
microgram
g
milligram
mg
gram
g
kilogram
kg
tonne
t
Example:
Iron rusts slowly when exposed to air and water. The reaction involved in rusting can be
represented by the equation below. What mass of iron oxide is produced when 120g of
iron completely in air?
Step 1: Write a balanced equation
4Fe(s) + 3O2(g) 
2Fe2O3(s)
Step 2: Calculate the amount (in mol) of the substance with the known mass.
m
n = M
M(Fe2O3) =
Step 3: Mole ratio
n (unknown)
n (know)
=
Step 4: Calculate the mass required
n = m x M
Example:
A phosphorus manufacturer is to extract 1.00 tonne of phosphorus per day by the
process given by the equation.
2Ca3(PO4)2(s) + 6SiO2(s) + 10C(s)  P4(s) + 10CO(g) + 6CaSiO3(s)
Calculate the mass required daily of:
a) calcium phosphate
b) silicon dioxide
a) Calcium phosphate
Step 1: Write the balanced equation
Step 2: Calculate mole of known substance
Step 3: mole ratio
Step 4: Calculate the mass required
b) Silicon Dioxide
Step 1: Write the balanced equation
Step 2: Calculate mole of known substance
Step 3: mole ratio
Step 4: Calculate the mass required
Stoichiometry involving solutions
Many chemical reactions occur between reactants in solution. The amount of a reactant
present in a solution can be determined from the volume of the solution used and its
concentration, by using the relationship developed in Chapter 11.
n(mol) = c (mol L-1) x V (L)
Example: Precipitation reactions occur when two solutions containing soluble reactants
are mixed together to produce an insoluble product. This product is collected as a
precipitate. What mass of barium sulfate is produce when 100 mL of a 1.52M solutions
of barium nitrate reacts completely with a sodium sulfate solution.
Step 1: Write a balanced equation
Step 2: Calculate mole of known solution
Step 3: mole ratio
Step 4: Calculate mass of unknown solution
Example: What volume of 0.100M sulfuric acid reacts completely with 17.8 mL of
0.152M potassium hydroxide.
Step 1: Write a balanced equation
Step 2: Calculate mole of known solution
Step 3: mole ratio
Step 4: Calculate volume of unknown solution
Questions: Week 6
Year 11 Chemistry: Chapter 15:~ Calculations in Chemistry: Stoichiometry
15.2 Calculations based on amount of two reactants
In the previous examples the quantity of one substance was given and used to calculate
the quantity of other reactants or products. The previous calculations assumed that any
other reactants were present in sufficient quantities to react completely with all of the
given substance.
In the following problems, fixed quantities of two different reactants are mixed. It is
likely that one of the reactants will be used up before the other. The reaction stops
when one reactant is used up (limiting reagent), even though some of the other
substance is unreacted. The other reactant is said to be in excess (excess reagent).
2H2(g) + O2(g)  2H2O(g)
Container A: holds 8 hydrogen molecules and 6 oxygen molecules, they react
Container B: 8 water molecules and two oxygen molecules left.
2H2(g)
+
8 mol
8 mol
0 mol
O2(g)

2H2O(g)
Initial amount
React’n change according to eq’n
Final amounts
After the reaction:
 no hydrogen gas remains, since this is the limiting reagent.
 The amount of water vapour formed is limited by the amount of hydrogen gas
available.
 2 mol of oxygen gas remains unreacted. Oxygen gas is in excess in this example.
Example: A gaseous mixture of 25.0 g of hydrogen gas and 100.0 g of oxygen gas are
mixed and ignited. The water produced is collected and weighed. What is the expected
mass of water produced?
Step 1: Write a balanced chemical equation
Step 2: Calculate the amount of each reactant.
m
n (H2) = M
n(O2) =
m
M
M(H2) =
M(O2) =
Step 3: Identify the reagent in excess, using mole ratio. Use the reactant with the
smallest mole.
Step 4: Calculate the amount of the limiting reagent to determine the amount (in mol)
of the required product.
Step 5: Calculate mass of product
Example: 2.50 g of aluminium is mixed with 5.00 g of iodine and allowed to react.
a) What mass of aluminium iodide would be produced?
b) What is the mass of the reactant in excess?
Step 1: Write a balanced chemical equation
Step 2: Calculate the amount of each reactant
Step 3: Identify reactant in excess.
Step 4: use the limiting reagent to determine the quantity required to produce.
Step 5: Calculate the mass.
a) aluminium iodide
b) mass of aluminium in excess
Questions: 9, 10, 11, 34, 35, 36 & 39
Volumetric Analysis - Titrations
There are many situations when it is essential to know the exact amount of acid or base
in a substance. The concentration of solutions of acids and bases can be determined
accurately by a technique called volumetric analysis. This involves reacting the
solution of unknown concentration with a solution of accurately known concentration
(standard solution).
 Volumetric flask:
 Pipette:
 Aliquot:
 Burette:
 Titre:
 Equivalence point: is where the acid and base are neutralised.
 End point: indicator changes colour close to the equivalence point.
Since both acid and base solutions are often colourless, there needs to be some way to
determine when the reaction is complete. This is done by adding an indicator that
changes colour at (or very close) to the equivalence point of the titration (end point).
Rinsing equipment: Experiment 54 as example
Pipette: water then, sample from volumetric flask
Burette: water then, solution that is being placed in the burette
Conical flask: distilled water
Volumetric flask: distilled water.
Questions: 12, 13, 40, 41 & 43.
Questions: 1, 2, 3, 4, 5, 6, 7, 8 (should already be completed), 15, 16, 17, 19, 21, 22, 24,
26, 28, 29, 30, 31, 33, & 45.