Question

Consider compound Na₂CO₃⦁10H₂O. This compound is a hydrate.What is the percent of water in this compound?
*

Answer 1

To calculate the percent of water in the hydrate \( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \), we first need to determine the molar mass of the entire compound, including both the sodium carbonate and the water.

1. **Calculate the molar mass of \( \text{Na}_2\text{CO}_3 \)**:
- Sodium (Na): 22.99 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol

\[
\text{Molar mass of } \text{Na}_2\text{CO}_3 = (2 \times 22.99) + (1 \times 12.01) + (3 \times 16.00)
\]
\[
= 45.98 + 12.01 + 48.00 = 105.99 \, \text{g/mol}
\]

2. **Calculate the molar mass of the water in the hydrate**:
- Water (H₂O): \( 2 \times 1.01 + 16.00 = 18.02 \, \text{g/mol} \)

For \( 10 \text{H}_2\text{O} \):
\[
10 \times 18.02 = 180.20 \, \text{g/mol}
\]

3. **Total molar mass of \( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \)**:
\[
\text{Total molar mass} = 105.99 + 180.20 = 286.19 \, \text{g/mol}
\]

4. **Calculate the percent of water in the hydrate**:
\[
\text{Percent of water} = \left( \frac{\text{mass of water}}{\text{total mass of hydrate}} \right) \times 100
\]
\[
= \left( \frac{180.20}{286.19} \right) \times 100 \approx 62.91\%
\]

Thus, the percent of water in the compound \( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \) is approximately **62.91%**.