Question

The metal salt A is blue in colour. When salt A is heated strongly over a burner, then a substance B is eliminated and a white powder C is left behind. When a few drops of a liquid D are added to powder C, it becomes blue again. What could be A, B, C and D?

Answer 1

A is copper sulphate pentahydrate, CuSO₄.5H₂O; B is water of crystallisation, 5H₂O; C is anhydrous copper sulphate, CuSO₄; D is water, H₂O.

The blue colour of copper sulphate is due to the presence of 5 water molecules. When copper sulphate crystals are heated, they lose their water molecules and turn into a grey-white powder known as anhydrous copper sulphate. The chemical équation for this process is:

\[\ce{CuSO4 * 5H2O -> CuSO4 + 5H2O}\]

After adding a few drops of water (D), the anhydrous copper sulphate rehydrates into copper sulphate pentahydrate, and due to the re-entry of water molecules into its crystal lattice, it becomes blue again. The chemical equation for this process is:

\[\ce{CuSO4 + 5H2O -> CuSO4 * 5H2O}\]