Water of Crystallisation – IGCSE Chemistry Definition and Key Facts

Water of crystallisation is tested in the Cambridge 0620 syllabus under experimental techniques and stoichiometry. Calculations involving water of crystallisation — finding x in a formula like CuSO₄.xH₂O — appear regularly on Paper 4, typically for 3–4 marks.

The 0620 definition

Water of crystallisation is water that is chemically combined in definite proportions within the crystal structure of a substance. A salt that contains water of crystallisation is described as hydrated. A salt that has lost its water of crystallisation is anhydrous.

The dot notation

The formula uses a dot (.) to show the water:

CuSO₄.5H₂O — copper(II) sulfate pentahydrate

The 5 means there are 5 molecules of water for every one formula unit of CuSO₄.

Common hydrated salts

Hydrated salt	Formula	Colour	Anhydrous form	Colour
Copper(II) sulfate	CuSO₄.5H₂O	Blue crystals	CuSO₄	White powder
Cobalt(II) chloride	CoCl₂.6H₂O	Pink crystals	CoCl₂	Blue
Iron(II) sulfate	FeSO₄.7H₂O	Green crystals	FeSO₄	White
Sodium carbonate	Na₂CO₃.10H₂O	White crystals	Na₂CO₃	White powder
Magnesium sulfate	MgSO₄.7H₂O	Colourless crystals	MgSO₄	White powder

The copper(II) sulfate and cobalt(II) chloride colour changes are used as tests for water.

Tests for water

Test	Observation if water present
Add to anhydrous copper(II) sulfate	White powder turns blue
Add to anhydrous cobalt(II) chloride paper	Blue paper turns pink

These tests confirm the presence of water but do not prove it is pure water. To confirm pure water, check that it boils at 100 °C and freezes at 0 °C.

Heating a hydrated salt

When CuSO₄.5H₂O is heated:

CuSO₄.5H₂O → CuSO₄ + 5H₂O

Observations:

• Blue crystals turn to a white powder
• Water droplets (steam) appear on cool parts of the test tube
• Mass decreases (water is lost)

The process is reversible: adding water to anhydrous CuSO₄ regenerates the blue hydrated form and the mixture warms (exothermic).

Calculating x in CuSO₄.xH₂O

This is a classic Paper 4 calculation:

Question: 6.25 g of hydrated copper(II) sulfate was heated until constant mass. The anhydrous residue weighed 4.00 g. Find x.

Step 1: Mass of water lost = 6.25 - 4.00 = 2.25 g

Step 2: Moles of CuSO₄ = 4.00 / 160 = 0.025 mol

Step 3: Moles of H₂O = 2.25 / 18 = 0.125 mol

Step 4: Ratio = 0.125 / 0.025 = 5

Therefore x = 5 and the formula is CuSO₄.5H₂O

”Heated until constant mass”

The phrase “heated until constant mass” means the salt was heated, cooled, weighed, heated again, cooled, and weighed again — repeated until two successive masses were the same. This ensures all the water of crystallisation has been driven off. Examiners may ask why this procedure is necessary.

Effect on Mr calculations

When calculating the relative formula mass of a hydrated salt, the water must be included:

Mr of CuSO₄.5H₂O = 64 + 32 + 64 + 5(18) = 160 + 90 = 250

Mr of the anhydrous form CuSO₄ = 160

Percentage of water = (90 / 250) x 100 = 36%

Worked exam question

3.86 g of hydrated magnesium sulfate, MgSO₄.xH₂O, was heated to constant mass. The residue weighed 1.88 g. Calculate x. (Ar: Mg=24, S=32, O=16, H=1) [4]

Mass of water = 3.86 - 1.88 = 1.98 g [1]. Moles of MgSO₄ = 1.88 / 120 = 0.01567 mol [1]. Moles of H₂O = 1.98 / 18 = 0.110 mol [1]. x = 0.110 / 0.01567 = 7, so the formula is MgSO₄.7H₂O [1].

Common exam mistakes

1. Forgetting to include the water in Mr calculations when the question gives a hydrated formula.
2. Not subtracting the residue mass from the original mass to find the mass of water lost.
3. Using the Mr of the hydrated salt instead of the anhydrous salt when calculating moles of the salt residue.
4. Not heating to constant mass — if some water remains, the calculated x will be too low.