Question

A 0.4550 g solid mixture containing CaSO4 is distilled in water and treated with an access of Ba(NO3)2, resulting in the precipitation of 0.6168 g of BaSO4.
What is the percentage of CaSO4 in the mixture? State the answer to 4 significant figures.Hint: Write balance equation first before calculations

Answer 1

To determine the percentage of \( \text{CaSO}_4 \) in the mixture, we need to begin with the balanced chemical equation for the reaction between \( \text{CaSO}_4 \) and \( \text{Ba(NO}_3\text{)}_2 \).

1. **Write the balanced equation**:
\[
\text{CaSO}_4 (s) + \text{Ba(NO}_3\text{)}_2 (aq) \rightarrow \text{BaSO}_4 (s) + \text{Ca(NO}_3\text{)}_2 (aq)
\]
This indicates that one mole of \( \text{CaSO}_4 \) reacts with one mole of \( \text{Ba(NO}_3\text{)}_2 \) to produce one mole of \( \text{BaSO}_4 \).

2. **Molar masses**:
- Molar mass of \( \text{CaSO}_4 \):
- \( \text{Ca} = 40.08 \, \text{g/mol} \)
- \( \text{S} = 32.07 \, \text{g/mol} \)
- \( \text{O} = 16.00 \, \text{g/mol} \) (4 O's)
- \( \text{Total} = 40.08 + 32.07 + (16.00 \times 4) = 136.14 \, \text{g/mol} \)

- Molar mass of \( \text{BaSO}_4 \):
- \( \text{Ba} = 137.33 \, \text{g/mol} \)
- \( \text{S} = 32.07 \, \text{g/mol} \)
- \( \text{O} = 16.00 \, \text{g/mol} \) (4 O's)
- \( \text{Total} = 137.33 + 32.07 + (16.00 \times 4) = 233.39 \, \text{g/mol} \)

3. **Calculate moles of \( \text{BaSO}_4 \)** formed:
\[
\text{Moles of BaSO}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.6168 \, \text{g}}{233.39 \, \text{g/mol}} \approx 0.002644 \, \text{mol}
\]

4. **Moles of \( \text{CaSO}_4 \)** that reacted:
From the balanced equation, the moles of \( \text{CaSO}_4 \) is equal to the moles of \( \text{BaSO}_4 \) formed. Thus,
\[
\text{Moles of CaSO}_4 = 0.002644 \, \text{mol}
\]

5. **Calculate mass of \( \text{CaSO}_4 \)**:
\[
\text{Mass of CaSO}_4 = \text{moles} \times \text{molar mass} = 0.002644 \, \text{mol} \times 136.14 \, \text{g/mol} \approx 0.3595 \, \text{g}
\]

6. **Calculate percentage of \( \text{CaSO}_4 \) in the mixture**:
\[
\text{Percentage of CaSO}_4 = \left( \frac{\text{mass of CaSO}_4}{\text{mass of mixture}} \right) \times 100 = \left( \frac{0.3595 \, \text{g}}{0.4550 \, \text{g}} \right) \times 100 \approx 78.9735\%
\]

7. **Final result (to 4 significant figures)**:
\[
\text{Percentage of CaSO}_4 = 78.97\%
\]

Therefore, the percentage of \( \text{CaSO}_4 \) in the mixture is **78.97%**.