WATER of CRYSTALLISATION CALCULATIONS
hydrated and anhydrous salts
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WATER OF CRYSTALLISATION  method of
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Keywords: Quantitative chemistry
calculations How to determine the water of
crystallisation in a salt like compound, hence determine the full formula of the
hydrated salt. How to calculate the % water in a hydrated salt i.e. the
percentage of water of crystallisation in a salt. How to deduce the number of
molecules of water of crystallisation in a salt  fully worked out example
calculations of water of crystallisation. Online practice exam chemistry
CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic
starter chemical calculations for A level AS/A2/IB courses. These revision notes
and practice questions on how to do water of crystallisation chemical
calculations and worked examples should prove useful for the new AQA, Edexcel
and OCR GCSE (9–1) chemistry science courses.
Spotted any careless error?
EMAIL query ? comment or request a type of
GCSE calculation not covered?

See also
14.1
% purity of a product and assay calculations
14.2a
% reaction yield and theoretical yield calculations
and why you can't actually get 100% yield in practice
14.2b atom economy calculations
*
14.3 dilution of solutions calculations
14.5
how
much of a reactant is needed? calculation of quantities required, limiting
reactant quantities
Chemical &
Pharmaceutical Industry Economics & Sustainability, Life Cycle
Assessment, Recycling
14.4 Water of crystallisation
in a crystallised salt

Example 14.4.1:
Reminders on calculating formula mass e.g. with
MgSO_{4}.7H_{2}O

Relative atomic masses:
Mg = 24, S = 32, O = 16 and H = 1

You need to add together the formula mass
of MgSO_{4} plus the relative mass of seven water molecules.

Relative formula mass of water = (2 x 1)
+ 16 = 18

Relative formula mass of MgSO_{4}
= 24 + 32 + (4 x 16) = 120

Relative mass of seven water molecules =
7 x 18 = 126

Relative formula mass of crystals =
MgSO_{4} + (7 x H_{2}O) = 120 + 126 = 246

Example 14.4.2 How to calculate the
theoretical % of water in a hydrated salt

eg magnesium sulphate MgSO_{4}.7H_{2}O
'hydrated' salt crystals

Relative atomic masses:
Mg = 24, S = 32, O = 16 and H = 1

Relative formula mass of crystals =
24 + 32 + (4 x 16) + {7 x (1 + 1 + 16)} = 246

Relative mass of seven water molecules =
7 x 18 = 126

so % water = 126
x 100 / 246 = 51.2%

Example 14.4.3 Determination and
calculation of salt formula containing 'water of
crystallisation'.

Some salts,
when crystallised from aqueous solution, incorporate water molecules
into the structure. This is known as 'water of crystallisation', and the
'hydrated' form of the compound.

e.g. magnesium sulphate MgSO_{4}.7H_{2}O.
The formula can be determined by a simple experiment (see the copper
sulphate example below).

A known mass of the hydrated salt is gently
heated in a crucible until no further water is driven off and the weight
remains constant despite further heating.

The % water of
crystallisation and the formula and formula mass of the salt are calculated as follows:

Suppose 6.25g of blue
hydrated copper(II) sulphate, CuSO_{4}.xH_{2}O, (x
unknown) was
gently heated in a crucible until the mass remaining was a constant 4.00g.

When the mass on
subsequent weighings stays constant, you know all the water of
crystallisation has driven off by the heat.

This
is the white anhydrous copper(II) sulphate.

The mass of anhydrous
salt = 4.00g, mass of water (of crystallisation) driven off =
6.254.00 = 2.25g

The % water of
crystallisation in the crystals is 2.25 x 100 / 6.25 = 36%

[ A_{r} values:
Cu=64, S=32, O=16, H=1 ]

The mass ratio of CuSO_{4}
: H_{2}O is 4.00 : 2.25 (or 64% : 36%, doesn't matter which
mass ratio you use)

To convert from mass
ratio to mole ratio, you divide by the masses molecular/formula mass
M_{r} of each
'species'

M_{r} CuSO_{4} = 64
+ 32 + (4x18) = 160 and M_{r} H_{2}O = 1+1+16 =
18

The mole ratio of CuSO_{4}
: H_{2}O is 4.00/160 : 2.25/18

which is 0.025 : 0.125
or 1 : 5, so the formula of the hydrated salt is CuSO_{4}.5H_{2}O

The formula mass M_{r}
can then be calculated as follows:

There are some
More
sophisticated A Level problems involving titrations to determine the water of
crystallisation in a salt like compound (See Q30)
See also
14.1
% purity of a product and assay calculations
14.2a
% reaction yield and theoretical yield calculations
and why you can't actually get 100% yield in practice
14.2b atom economy calculations
*
14.3 dilution of solutions calculations
14.5
how
much of a reactant is needed? calculation of quantities required, limiting
reactant quantities
Chemical &
Pharmaceutical Industry Economics & Sustainability, Life Cycle
Assessment, Recycling
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Above is typical periodic table used in GCSE sciencechemistry specifications in
doing chemical calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
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