Question

The percentage of water of crystalization in
ZnSO4 7H2O is

Answer 1

To calculate the percentage of water of crystallization in zinc sulfate heptahydrate (ZnSO₄·7H₂O), we first need to find the molar masses of ZnSO₄ and H₂O.

1. **Molar mass of ZnSO₄:**
- Zinc (Zn): 65.38 g/mol
- Sulfur (S): 32.07 g/mol
- Oxygen (O): 16.00 g/mol (4 oxygen atoms)

Molar mass of ZnSO₄ = 65.38 + 32.07 + (4 × 16.00) = 65.38 + 32.07 + 64.00 = 161.45 g/mol

2. **Molar mass of 7H₂O:**
- Water (H₂O): 1.01 g/mol (2 hydrogen atoms) + 16.00 g/mol (1 oxygen atom) = 18.02 g/mol
- Molar mass of 7H₂O = 7 × 18.02 = 126.14 g/mol

3. **Total molar mass of ZnSO₄·7H₂O:**
- Total molar mass = Molar mass of ZnSO₄ + Molar mass of 7H₂O
- Total molar mass = 161.45 g/mol + 126.14 g/mol = 287.59 g/mol

4. **Percentage of water of crystallization:**
\[
\text{Percentage of } H_2O = \left( \frac{\text{Molar mass of } 7H_2O}{\text{Total molar mass}} \right) \times 100
\]
\[
\text{Percentage of } H_2O = \left( \frac{126.14}{287.59} \right) \times 100 \approx 43.9\%
\]

Therefore, the percentage of water of crystallization in ZnSO₄·7H₂O is approximately **43.9%**.