Water of crystallisation

Last time

Before we start: last time, in Gas volumes at r.t.p., you learned these points. Here they are again on one card. Keep them in mind, because today builds on them.

Blue crystals that go white

Blue copper sulfate crystals look dry, but heat them and steam rises, and they turn into a white powder. The crystals contained water, locked into their structure. That water is called water of crystallisation, and we can calculate exactly how much there is.

Hydrated and anhydrous

Water of crystallisation is water that is part of the structure of a crystal, in a fixed proportion. A salt that contains it is hydrated. The formula is written with a dot, such as CuSO₄·5H₂O. A salt without it is anhydrous. Anhydrous copper sulfate is a white powder. The five in front of the H₂O means five moles of water for every one mole of CuSO₄.

A reversible change

Heating the blue hydrated crystals drives off the water, and leaves the white anhydrous powder. Adding water to the white powder turns it blue again, and gives out heat. This is a reversible reaction, and it is used as a chemical test for water. Cobalt(II) chloride paper does it in reverse colours: blue when anhydrous, pink with water.

Mr of a hydrated salt

To find the Mr of a hydrated salt, add the Mr of the salt to the Mr of all its water. For CuSO₄·5H₂O, the salt is sixty-four plus thirty-two plus sixty-four, which is one hundred and sixty. Five waters are five times eighteen, which is ninety. The total is two hundred and fifty.

Percentage of water

The percentage of water is the mass of the water divided by the Mr of the hydrated salt, times one hundred. In CuSO₄·5H₂O, that is ninety over two hundred and fifty, times one hundred. It is thirty-six per cent water.

Finding x by experiment

Weigh an empty crucible, then add the hydrated salt, and weigh again. The difference is the mass of the hydrated salt. Heat gently, let it cool, and weigh again. Then heat again, and repeat until the mass no longer changes. The mass lost is the mass of the water driven off. The mass left is the anhydrous salt.

The calculation

First, find the mass of the water: hydrated mass minus anhydrous mass. Second, change the water mass into moles, dividing by eighteen. Third, change the anhydrous mass into moles, dividing by its Mr. Fourth, divide the moles of water by the moles of salt. That gives x.