Question
14, 2 g of samples of hydrated sodium carbonate Na2C03.xH2O ( x is for the unknown value)was heated until no further change of mass was recorded on heating all the water of crystallisation is evaporated as follows
Na2CO3.x H2O ( x is for the unknown value which we must also find) .
Calculate the number of moles of water of crystallization in the sodium carbonate if 5, 3 g solid remained
Na2CO3.x H2O ( x is for the unknown value which we must also find) .
Calculate the number of moles of water of crystallization in the sodium carbonate if 5, 3 g solid remained
Answers
GPT-4o mini
To find the number of moles of water of crystallization (x) in the hydrated sodium carbonate \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\), we can follow these steps:
1. **Determine the mass of water lost**:
- Initial mass of the hydrated sample: 14.2 g
- Final mass after heating (anhydrous sodium carbonate): 5.3 g
- Mass of water lost = Initial mass - Final mass
\[
\text{Mass of water lost} = 14.2 \, \text{g} - 5.3 \, \text{g} = 8.9 \, \text{g}
\]
2. **Calculate the number of moles of anhydrous sodium carbonate \(\text{Na}_2\text{CO}_3\)**:
- Molar mass of \(\text{Na}_2\text{CO}_3\) = \( (2 \times 23.0) + (12.0) + (3 \times 16.0) \)
\[
= 46.0 + 12.0 + 48.0 = 106.0 \, \text{g/mol}
\]
- Moles of anhydrous sodium carbonate:
\[
\text{Moles of } \text{Na}_2\text{CO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{5.3 \, \text{g}}{106.0 \, \text{g/mol}} \approx 0.050 \, \text{mol}
\]
3. **Calculate the number of moles of water lost**:
- Molar mass of water \( \text{H}_2\text{O} = (2 \times 1.0) + (16.0) = 18.0 \, \text{g/mol} \)
- Moles of water lost:
\[
\text{Moles of water} = \frac{8.9 \, \text{g}}{18.0 \, \text{g/mol}} \approx 0.494 \, \text{mol}
\]
4. **Determine x (number of moles of water of crystallization)**:
- Assume the number of moles of sodium carbonate corresponds to x moles of water:
\[
\text{Each mole of } \text{Na}_2\text{CO}_3 \text{ contains x moles of water}
\]
- Therefore, the number of moles of water of crystallization per mole of sodium carbonate is:
\[
x = \frac{0.494 \, \text{mol of water}}{0.050 \, \text{mol of Na}_2\text{CO}_3} \approx 9.88 \approx 10
\]
Thus, the number of moles of water of crystallization is approximately **10**. Hence, x is approximately 10, and the formula of hydrated sodium carbonate is \(\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}\).
1. **Determine the mass of water lost**:
- Initial mass of the hydrated sample: 14.2 g
- Final mass after heating (anhydrous sodium carbonate): 5.3 g
- Mass of water lost = Initial mass - Final mass
\[
\text{Mass of water lost} = 14.2 \, \text{g} - 5.3 \, \text{g} = 8.9 \, \text{g}
\]
2. **Calculate the number of moles of anhydrous sodium carbonate \(\text{Na}_2\text{CO}_3\)**:
- Molar mass of \(\text{Na}_2\text{CO}_3\) = \( (2 \times 23.0) + (12.0) + (3 \times 16.0) \)
\[
= 46.0 + 12.0 + 48.0 = 106.0 \, \text{g/mol}
\]
- Moles of anhydrous sodium carbonate:
\[
\text{Moles of } \text{Na}_2\text{CO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{5.3 \, \text{g}}{106.0 \, \text{g/mol}} \approx 0.050 \, \text{mol}
\]
3. **Calculate the number of moles of water lost**:
- Molar mass of water \( \text{H}_2\text{O} = (2 \times 1.0) + (16.0) = 18.0 \, \text{g/mol} \)
- Moles of water lost:
\[
\text{Moles of water} = \frac{8.9 \, \text{g}}{18.0 \, \text{g/mol}} \approx 0.494 \, \text{mol}
\]
4. **Determine x (number of moles of water of crystallization)**:
- Assume the number of moles of sodium carbonate corresponds to x moles of water:
\[
\text{Each mole of } \text{Na}_2\text{CO}_3 \text{ contains x moles of water}
\]
- Therefore, the number of moles of water of crystallization per mole of sodium carbonate is:
\[
x = \frac{0.494 \, \text{mol of water}}{0.050 \, \text{mol of Na}_2\text{CO}_3} \approx 9.88 \approx 10
\]
Thus, the number of moles of water of crystallization is approximately **10**. Hence, x is approximately 10, and the formula of hydrated sodium carbonate is \(\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}\).