Determining the degree of hydration experimentally

A hydrated compound loses water of crystallisation when it is heated. As it loses water of crystallisation, it loses mass. When it has lost all of its water of crystallisation it is anhydrous.

We can use difference in the mass between the hydrated and anhydrous compound to calculate the mass of water of crystallisation removed by heating.

Heat to constant mass to ensure all of the water of crystallisation is removed.

Heating to constant mass involves heating the same for several minutes, weighing it and repeating this until two consecutive mass measurements are the same.

Example

The following measurements were taken when a sample of hydrated aluminium nitrate Al(NO3)3.n H2O was heated to a constant mass in an oven.

Mass of evaporating basin = 52.67 g

Mass of evaporating basin + hydrated salt = 56.42 g

Mass of evaporating basin and contents after heating to constant mass = 54.80 g

Use the figures above to calculate the degree of hydration in Al(NO3)3·n H2O.

Mass of anhydrous salt = 54.80 – 52.67 = 2.13 g

Mass of water lost = 56.42 – 54.80 = 1.62 g

Now do an empirical formula calculation using these masses (higher tier):

Al(NO3)3H2O
Write the masses2.13 g1.62 g
Write the Mr values21318
Divide masses by Mr to find the moles2.13 ÷ 213 = 0.011.62 ÷ 18 = 0.09
Divide by the smallest number of moles0.01 ÷ 0.01 = 10.09 ÷ 0.01 = 9
Write the formulaAl(NO3)3 . 9H2O

Prescribed Practical C5 - Determine the mass of water present in hydrated crystals