Question

cerium (III) iodide (CeI3) occurs as a hydrate with the composition 76.3% CeI3 and 23.7% H2O. Calculate the formula for the hydrate.

Answer 1

To determine the formula of the hydrate of cerium (III) iodide (\( \text{CeI}_3 \cdot x \text{H}_2\text{O} \)), we first need to find the molar masses of cerium (Ce), iodine (I), and water (H\(_2\)O).

1. **Molar masses**:
- Molar mass of Ce: \( 140.12 \, \text{g/mol} \)
- Molar mass of I: \( 126.90 \, \text{g/mol} \)
- Molar mass of H\(_2\)O: \( 18.02 \, \text{g/mol} \)

2. **Calculate the molar mass of \(\text{CeI}_3\)**:
\[
\text{Molar mass of CeI}_3 = 140.12 \, \text{g/mol} + 3 \times 126.90 \, \text{g/mol}
\]
\[
= 140.12 + (3 \times 126.90) = 140.12 + 380.70 = 520.82 \, \text{g/mol}
\]

3. **Determine the mass of the hydrate**:
The composition of the hydrate indicates that \( 76.3\% \) is \(\text{CeI}_3\) and \( 23.7\% \) is water. Let's assume we have \( 100 \, \text{g} \) of the hydrate:
- Mass of \(\text{CeI}_3\): \( 76.3 \, \text{g} \)
- Mass of \(\text{H}_2\text{O}\): \( 23.7 \, \text{g} \)

4. **Calculate the number of moles of \(\text{CeI}_3\)**:
\[
\text{Moles of CeI}_3 = \frac{76.3 \, \text{g}}{520.82 \, \text{g/mol}} \approx 0.146 \, \text{mol}
\]

5. **Calculate the number of moles of water**:
\[
\text{Moles of H}_2\text{O} = \frac{23.7 \, \text{g}}{18.02 \, \text{g/mol}} \approx 1.314 \, \text{mol}
\]

6. **Calculate the ratio of moles of water to moles of \(\text{CeI}_3\)**:
\[
x = \frac{\text{Moles of } H_2O}{\text{Moles of } CeI_3} = \frac{1.314 \, \text{mol}}{0.146 \, \text{mol}} \approx 9.0
\]

Since \( x \) is a whole number, we can conclude that the hydrate is \(\text{CeI}_3 \cdot 9 \text{H}_2\text{O}\).

**Final Answer**: The formula for the hydrate is \(\text{CeI}_3 \cdot 9 \text{H}_2\text{O}\).