Water of crystallisation – Calculations
Some salts contain water molecules that form part of their crystal structure. This water is called
water of crystallisation and is written in the formula of the salt.
e.g. MgSO4.H2O refers to hydrated magnesium sulphate.
Na2SO4.6H2O refers to hydrated sodium sulphate.
The % water of crystallisation and the formula of the salt are calculated as follows:
• A known mass of hydrated salt is heated gently in a crucible until it reaches constant
mass.
• The mass of anhydrous salt remaining and the mass of water lost are then calculated.
• These are converted to moles and the formula of the hydrated salt can be found from the
mole ratio.
Worked example.
6.25g of blue hydrated copper(II) sulphate, CuSO4.xH2O, (x unknown) was gently heated in a
crucible until the mass remaining was 4.00g. (This is the white anhydrous copper(II) sulphate
CuSO4).
The mass of anhydrous salt (CuSO4) = 4.00g,
Mass of water (of crystallisation) driven off = 6.25 - 4.00 = 2.25g
RFM (Mr) CuSO4 = 64 + 32 + (4x18) = 160 and RFM (Mr) H2O = 1+1+16 = 18
Moles of CuSO4 = 4 / 160 = 0.025
Moles of H2O = 2.25 / 18 = 0.125
The mole ratio of CuSO4 : H2O is 0.025 : 0.125 or 1 : 5
So the formula of the hydrated salt is CuSO4.5H2O
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Calculations from Equations: Reacting masses
Normally you will be given an equation, and asked a question which concerns only
two substances, for one of which the mass is given, and for the other the mass
needs to be calculated:
e.g. (6) What mass of substance A is needed to give x g of substance B?
What mass of substance C is produced from y g of substance D?
The steps involved in a calculation are as follows :
(a) Convert the information given to moles of one substance.
(b) Use the chemical equation to find moles of other substance
needed.
(c) Convert back from moles to mass (or concentration, volume
etc.)
e.g. What mass of oxygen is needed to burn 3.00 kg of propane, C3H8?
In this case the chemical equation is not given, so we must start by
writing it down: C3H8 + 5 O2
3CO2 + 4H2O
(a) Convert the information given to moles of one substance.
We are given the mass of propane, so it is this we must convert to moles.
Formula mass of C3H8 = 3 × 12 + 8 × 1 = 44 So molar mass = 44 g/mol
Number of moles of C3H8 in 3.00kg = = 68.2 mol
(b) Use the chemical equation to find moles of other substance
needed. 1 mol of propane reacts with 5 mol of oxygen molecules.
So 68.2 mol of propane reacts with 5 × 68.2 = 341 mol of oxygen molecules
(c) Convert back from moles to mass (or concentration, volume
etc.)
Molar mass of O2 molecules = 2 × 16 = 32 g/mol
So mass of oxygen needed = 341 mol × 32 g/mol = 10.9 kg
e.g. In the reaction Fe3O4 + 4CO → 3Fe + 4CO2 what mass of the iron oxide is
needed to form 2.50 g of carbon dioxide?
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(a) molar mass of CO2 = 12 + 2 × 16 = 44 g/mol
amount of CO2
= = 0.0568 mol
(b) From equation, 4 mol of CO2 is formed from 1 mol Fe3O4
So 0.0568 mol of CO2 is formed from = 0.0142 mol Fe3O4
(c) Molar mass of Fe3O4 = 3×56 + 4×16 = 232 g/mol mass of 0.0142
mol Fe3O4 = 0.0142 mol × 232 g/mol = 3.29g (3 s.f.)
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