Question

How do you know when copper (II) sulphate (hydrated) water evaporates?

Answer 1

Considering $CuS{{O}_{4}}.x{{H}_{2}}O$ as the formula for the hydrated copper (II) sulphate where, x is the water of crystallisation. When this water evaporates, the formula will turn into $CuS{{O}_{4}}$ i.e. anhydrous copper sulphate.

Complete step by step answer:
Let us see what changes take place when the water of crystallisation evaporates from the hydrated copper (II) sulphate;
- As we know, when the water of crystallisation evaporates hydrated copper (II) sulphate i.e. $CuS{{O}_{4}}.x{{H}_{2}}O$ will turn into anhydrous copper sulphate i.e. $CuS{{O}_{4}}$. The difference we can observe between both these compounds is that the one will have water of crystallisation whereas the other won’t have.
- Another remarkable difference we can observe, experimentally is the difference in the colours. If we consider copper (II) pentahydrate i.e. $CuS{{O}_{4}}.5{{H}_{2}}O$ ; its colour will be blue due to the presence of water molecules in it. After this water is removed by experimental basis, anhydrous copper (II) sulphate i.e. $CuS{{O}_{4}}$ will be white in colour.
Therefore, it means that when the water of crystallisation is removed from the molecule of copper (II) pentahydrate, its blue colour will turn into white forming copper (II) sulphate.

Note: Do note that this reaction and the results are reversible i.e. if we add water into copper (II) sulphate then the blue colour can be reformed forming the hydrated molecules of copper (II) sulphate.
- The general equation can be explained as;
${{\left( CuS{{O}_{4}}.x{{H}_{2}}O \right)}_{bright-blue}}\xrightarrow{Heat}{{\left( CuS{{O}_{4}} \right)}_{white}}+x{{H}_{2}}O$