Water of Crystallisation Calculations

Water of crystallisation calculations involve finding the value of n in a hydrated salt formula such as CuSO4.nH2O. The method uses mass data from a heating experiment to determine how many water molecules are bonded to each formula unit of the salt.

The method

1. Find the mass of water lost by subtracting the mass of anhydrous salt from the mass of hydrated salt.
2. Calculate the moles of anhydrous salt: moles = mass / Mr.
3. Calculate the moles of water: moles = mass of water / 18.
4. Find the simplest ratio of moles of salt to moles of water.
5. This ratio gives the value of n.

The formula

mass of water = mass of hydrated salt - mass of anhydrous salt

moles of anhydrous salt = mass / Mr

moles of water = mass of water / 18

n = moles of water / moles of anhydrous salt

Worked examples

Example 1: Copper sulfate

Question: 6.25 g of hydrated copper sulfate (CuSO4.nH2O) was heated until all the water was removed. The anhydrous salt weighed 4.00 g. Find the value of n. (Ar: Cu = 64, S = 32, O = 16, H = 1)

Working:

• Mass of water = 6.25 - 4.00 = 2.25 g
• Mr of CuSO4 = 64 + 32 + (4 x 16) = 160
• Moles of CuSO4 = 4.00 / 160 = 0.025 mol
• Moles of H2O = 2.25 / 18 = 0.125 mol
• Ratio = 0.125 / 0.025 = 5
• n = 5

Answer: CuSO4.5H2O

Examiner note: This is the classic example. The ratio must be a whole number. If you get 4.9 or 5.1, round to 5.

Example 2: Magnesium sulfate

Question: 12.3 g of hydrated magnesium sulfate (MgSO4.nH2O) gave 6.0 g of anhydrous magnesium sulfate after heating. Find the value of n. (Ar: Mg = 24, S = 32, O = 16, H = 1)

Working:

• Mass of water = 12.3 - 6.0 = 6.3 g
• Mr of MgSO4 = 24 + 32 + (4 x 16) = 120
• Moles of MgSO4 = 6.0 / 120 = 0.05 mol
• Moles of H2O = 6.3 / 18 = 0.35 mol
• Ratio = 0.35 / 0.05 = 7
• n = 7

Answer: MgSO4.7H2O

Examiner note: Magnesium sulfate heptahydrate (Epsom salt) is a common exam example.

Example 3: Iron(II) sulfate

Question: When 5.56 g of FeSO4.nH2O was heated, 3.04 g of anhydrous FeSO4 remained. Find n. (Ar: Fe = 56, S = 32, O = 16, H = 1)

Working:

• Mass of water = 5.56 - 3.04 = 2.52 g
• Mr of FeSO4 = 56 + 32 + (4 x 16) = 152
• Moles of FeSO4 = 3.04 / 152 = 0.02 mol
• Moles of H2O = 2.52 / 18 = 0.14 mol
• Ratio = 0.14 / 0.02 = 7
• n = 7

Answer: FeSO4.7H2O

Examiner note: Show all working clearly. The examiner needs to see both mole calculations and the ratio step.

Example 4: Sodium carbonate

Question: 14.3 g of Na2CO3.nH2O was heated until constant mass. The residue weighed 5.3 g. Find n. (Ar: Na = 23, C = 12, O = 16, H = 1)

Working:

• Mass of water = 14.3 - 5.3 = 9.0 g
• Mr of Na2CO3 = (2 x 23) + 12 + (3 x 16) = 106
• Moles of Na2CO3 = 5.3 / 106 = 0.05 mol
• Moles of H2O = 9.0 / 18 = 0.5 mol
• Ratio = 0.5 / 0.05 = 10
• n = 10

Answer: Na2CO3.10H2O

Examiner note: Washing soda (Na2CO3.10H2O) has 10 water molecules. The high water content means the mass loss is large.

Example 5: Finding percentage of water in a hydrated salt

Question: Calculate the percentage by mass of water in CuSO4.5H2O. (Ar: Cu = 64, S = 32, O = 16, H = 1)

Working:

• Mr of CuSO4.5H2O = 160 + (5 x 18) = 160 + 90 = 250
• Mass of water = 5 x 18 = 90
• Percentage of water = (90 / 250) x 100 = 36%

Answer: 36%

Examiner note: This is the reverse problem: given the formula, find the percentage of water. Use Mr of the whole hydrated formula as the denominator.

Common mistakes

1. Using the mass of the hydrated salt as the mass of anhydrous salt. The anhydrous mass is always the smaller number (after heating). The mass of water is found by subtraction.
2. Using Mr of the hydrated salt instead of the anhydrous salt. When calculating moles of anhydrous salt, use the Mr of the anhydrous form only.
3. Not rounding n to a whole number. The value of n must be a whole number. Values like 4.9 or 7.1 should be rounded, as slight discrepancies come from experimental error.
4. Forgetting that Mr of water is 18. This is used every time. Mr of H2O = (2 x 1) + 16 = 18.

When this appears

Water of crystallisation questions are Extended tier content and appear on Paper 4. They combine moles from mass and ratio calculations. See the stoichiometry topic for context.