A $10.0 g$ sample consisting of $CuSO X_{4} \cdot 5 H X_{2} O$ mixed with a non-volatile inert solid loses $1.80 g$ on strong heating, all of the loss being the water of crystallisation (Cu = 63.5, S = 32, O = 16, H = 1). What is the mass percentage of $CuSO X_{4} \cdot 5 H X_{2} O$ in the sample? Give one decimal place.

Numerical answer type, answer in %, tolerance 0.1.

Answer: 49.9 %

Water lost

= 1.80/18 = 0.100

mol. Each formula unit holds

5

water, so hydrate

= 0.0200

mol.

M = 249.5

, mass

= 0.0200 \times 249.5 = 4.99

g. Percentage

= 4.99/10.0 \times 100 = 49.9%