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.
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Calculate the mass of sodium sulfide: \[ \text{Mass of } \text{Na}_2\text{S} = \text{Total mass} - \text{Mass of water} = 240 , \text{g} - 162 , \text{g} = 78 , \text{g} \]
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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: \[ \text{Moles of } \text{Na}_2\text{S} = \frac{78 , \text{g}}{77.05 , \text{g/mol}} \approx 1.01 , \text{mol} \]
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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: \[ \text{Moles of } \text{H}_2\text{O} = \frac{162 , \text{g}}{18.02 , \text{g/mol}} \approx 8.99 , \text{mol} \]
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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.
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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: \[ \text{Na}_2\text{S} \cdot 9\text{H}_2\text{O} \]