Question

A 240g sample of hydrated sodium sulphide cointains 162g of water of crystallization.Whqt is the correct formula of this compound

Answer 1

To find the correct formula of the hydrated sodium sulfide, we first need to determine the moles of sodium sulfide (\( \text{Na}_2\text{S} \)) and the moles of water of crystallization (\( \text{H}_2\text{O} \)) in the sample.

1. **Calculate the mass of sodium sulfide**:
\[
\text{Mass of } \text{Na}_2\text{S} = \text{Total mass} - \text{Mass of water} = 240 \, \text{g} - 162 \, \text{g} = 78 \, \text{g}
\]

2. **Calculate the moles of sodium sulfide (\( \text{Na}_2\text{S} \))**:
- The molar mass of sodium sulfide (\( \text{Na}_2\text{S} \)) is:
\[
\text{Na}: 22.99 \, \text{g/mol} \times 2 + \text{S}: 32.07 \, \text{g/mol} = 22.99 \times 2 + 32.07 = 77.05 \, \text{g/mol}
\]
- Moles of sodium sulfide:
\[
\text{Moles of } \text{Na}_2\text{S} = \frac{78 \, \text{g}}{77.05 \, \text{g/mol}} \approx 1.01 \, \text{mol}
\]

3. **Calculate the moles of water (\( \text{H}_2\text{O} \))**:
- The molar mass of water is approximately:
\[
\text{H}_2\text{O} = 2 \times 1.01 \, \text{g/mol} + 16.00 \, \text{g/mol} = 18.02 \, \text{g/mol}
\]
- Moles of water:
\[
\text{Moles of } \text{H}_2\text{O} = \frac{162 \, \text{g}}{18.02 \, \text{g/mol}} \approx 8.99 \, \text{mol}
\]

4. **Determine the ratio of moles of water to moles of sodium sulfide**:
- The ratio is:
\[
\frac{\text{Moles of } \text{H}_2\text{O}}{\text{Moles of } \text{Na}_2\text{S}} \approx \frac{8.99}{1.01} \approx 8.90
\]

Since this is very close to 9, we can round it to 9.

5. **Write the formula**:
Based on this ratio, the formula of the hydrated sodium sulfide can be written as:
\[
\text{Na}_2\text{S} \cdot 9 \text{H}_2\text{O}
\]

Thus, the correct formula of the compound is:
\[
\text{Na}_2\text{S} \cdot 9\text{H}_2\text{O}
\]