Determining the degree of hydration experimentally
A hydratedMeans that the solid crystals contain water of crystallisation. compound loses water of crystallisationWater that is chemically bonded into a crystal structure. 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 anhydrousContains no water of crystallisation..
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路n H2O.
Mass of anhydrous salt = 54.80 鈥 52.67 = 2.13 g
Mass of water lost = 56.42 鈥 54.80 = 1.62 g
Now do an empirical formula calculation using these masses (higher tier):
| Al(NO3)3 | H2O | |
| Write the masses | 2.13 g | 1.62 g |
| Write the Mr values | 213 | 18 |
| Divide masses by Mr to find the moles | 2.13 梅 213 = 0.01 | 1.62 梅 18 = 0.09 |
| Divide by the smallest number of moles | 0.01 梅 0.01 = 1 | 0.09 梅 0.01 = 9 |
| Write the formula | Al(NO3)3 . 9H2O |
| Write the masses | |
| Al(NO3)3 | 2.13 g |
| H2O | 1.62 g |
| Write the Mr values | |
| Al(NO3)3 | 213 |
| H2O | 18 |
| Divide masses by Mr to find the moles | |
| Al(NO3)3 | 2.13 梅 213 = 0.01 |
| H2O | 1.62 梅 18 = 0.09 |
| Divide by the smallest number of moles | |
| Al(NO3)3 | 0.01 梅 0.01 = 1 |
| H2O | 0.09 梅 0.01 = 9 |
| Write the formula | |
| Al(NO3)3 | Al(NO3)3 . 9H2O |