Water of crystallisation

Worked example

A student heated 4.76 g of hydrated magnesium sulfate, MgSO₄·xH₂O, to constant mass. The anhydrous MgSO₄ left weighed 2.33 g. Calculate x. M(MgSO₄) = 120.4 g mol⁻¹, M(H₂O) = 18.0 g mol⁻¹.

1. 1. Subtract m(hydrated salt) − m(after heating) → m(water lost)

   = 2.430 g

2. 2. Moles from mass: solve for n(H₂O)

   $n = \dfrac{m}{M}$

   = 0.1350 mol

3. 3. Moles from mass: solve for n(MgSO₄)

   $n = \dfrac{m}{M}$

   = 0.01935 mol

4. 4. Divide n(H₂O) ÷ n(MgSO₄) → x = n(H₂O) ÷ n(MgSO₄)

   = 6.976

Answer: x = 6.98

Method

Mass of water lost = mass before − mass after. Convert both the water and the anhydrous salt to moles; x is their ratio. Heating to constant mass makes sure all the water has gone.

Common mistakes

• The moles of MgSO₄ come from the anhydrous mass left after heating, not the starting mass of the hydrate.

Formulas used

• Moles from mass

  $n = \dfrac{m}{M}$

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