IGCSE Chemistry 0620: Hydrated Copper(II) Sulfate and Reversible Dehydration

Blue hydrated copper(II) sulfate crystals, $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ , contain water of crystallisation.

A 25.0 g sample of these hydrated crystals contains 9.00 g of water of crystallisation; the remaining 16.0 g is anhydrous copper(II) sulfate.

(a) A second sample of the same hydrated crystals has a mass of 60.0 g. Assuming this sample has the same composition as the first, calculate the mass of water of crystallisation it contains, and hence the mass of anhydrous copper(II) sulfate that would remain if it were heated to constant mass. [2]

(b) State the colour change observed as the 60.0 g sample is heated strongly until no further change occurs, and name the white solid formed. [2]

(c) Write a balanced symbol equation, including state symbols and the $\rightleftharpoons$ symbol, to represent this reversible reaction. [2]

(d) A few drops of water are then added to the white solid formed in (b). Describe what would be observed, and state what this shows about the reaction. [2]

Part (a): Scaling the composition to a new mass

In the first sample, 9.00 g of the 25.0 g of hydrated crystals is water of crystallisation. This is a fixed proportion of the mass, no matter how large the sample is:

$\text{fraction of water} = \frac{9.00}{25.0} = 0.36$

For the 60.0 g sample, the mass of water of crystallisation is:

$60.0 \times 0.36 = 21.6\text{ g}$

Since the rest of the sample is anhydrous copper(II) sulfate, subtract this from the total mass (this is the “hence” step):

$60.0 - 21.6 = 38.4\text{ g}$

So the sample contains 21.6 g of water of crystallisation and would leave 38.4 g of anhydrous copper(II) sulfate.

Part (b): Colour change on heating

Hydrated copper(II) sulfate crystals are blue. As the sample is heated strongly, the water of crystallisation is driven off as steam, and the crystals turn white.

The white solid remaining is anhydrous copper(II) sulfate.

Part (c): Writing the reversible symbol equation

The equation must show:

The correct formulae for hydrated and anhydrous copper(II) sulfate and water.
Correct balancing (5 water molecules per formula unit).
State symbols showing solid crystals releasing water.
The $\rightleftharpoons$ symbol, since the change can go in either direction.

$\text{CuSO}_4 \cdot 5\text{H}_2\text{O(s)} \rightleftharpoons \text{CuSO}_4\text{(s)} + 5\text{H}_2\text{O(g)}$

Part (d): Adding water back

When a few drops of water are added to the white anhydrous copper(II) sulfate, the solid turns blue again (and the mixture may feel slightly warm), reforming the original hydrated crystals.

This shows that the reaction is reversible: the same chemical system can be made to go in either direction (losing water on heating, or regaining it when water is added) exactly as represented by the $\rightleftharpoons$ symbol in part (c).

Final answers

(a) Water of crystallisation $=$ 21.6 g; anhydrous copper(II) sulfate $=$ 38.4 g
(b) Blue $\rightarrow$ white; the white solid is anhydrous copper(II) sulfate
(c) $\text{CuSO}_4 \cdot 5\text{H}_2\text{O(s)} \rightleftharpoons \text{CuSO}_4\text{(s)} + 5\text{H}_2\text{O(g)}$
(d) The solid turns blue again, showing the reaction is reversible