Question

a 15.79 g sample of a hydrate of an iron(ii) sulfate compound was heated, without decomposing the sulfate to drive off the water. the mass was reduced to 8.63 g. what is the formula of the hydrate?

Explanation:

Step1: Find mass of water lost

Mass of hydrate = 15.79 g, mass of anhydrate (after heating) = 8.63 g.
Mass of water = \( 15.79 - 8.63 = 7.16 \) g.

Step2: Molar mass of \( \text{FeSO}_4 \)

Molar mass of Fe = 55.85 g/mol, S = 32.07 g/mol, O = 16.00 g/mol.
\( \text{Molar mass of FeSO}_4 = 55.85 + 32.07 + (4\times16.00) = 151.92 \) g/mol.

Step3: Moles of \( \text{FeSO}_4 \)

Moles of \( \text{FeSO}_4 = \frac{8.63\ \text{g}}{151.92\ \text{g/mol}} \approx 0.0568\ \text{mol} \).

Step4: Molar mass of \( \text{H}_2\text{O} \)

Molar mass of \( \text{H}_2\text{O} = (2\times1.01) + 16.00 = 18.02 \) g/mol.

Step5: Moles of \( \text{H}_2\text{O} \)

Moles of \( \text{H}_2\text{O} = \frac{7.16\ \text{g}}{18.02\ \text{g/mol}} \approx 0.397\ \text{mol} \).

Step6: Find mole ratio

Divide moles of \( \text{H}_2\text{O} \) by moles of \( \text{FeSO}_4 \):
\( \frac{0.397}{0.0568} \approx 7 \).

Answer:

The formula of the hydrate is \( \text{FeSO}_4 \cdot 7\text{H}_2\text{O} \) (Iron(II) sulfate heptahydrate).