To determine the percentage of CaSO₄ in the mixture, we can use the mass of BaSO₄ precipitated to find the amount of CaSO₄ that was originally present. The reaction that occurs when CaSO₄ is treated with Ba(NO₃)₂ is:
\[ \text{CaSO}_4 + \text{Ba(NO}_3\text{)}_2 \rightarrow \text{BaSO}_4 \downarrow + \text{Ca(NO}_3\text{)}_2 \]
We know the molar masses of the compounds involved:
- Molar mass of \( \text{BaSO}_4 \) = 137.33 (Ba) + 32.06 (S) + 4 × 16.00 (O) = 233.39 g/mol
- Molar mass of \( \text{CaSO}_4 \) = 40.08 (Ca) + 32.06 (S) + 4 × 16.00 (O) = 136.14 g/mol
Next, we calculate the number of moles of BaSO₄ that precipitated:
\[ \text{moles of BaSO}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.6158 \text{ g}}{233.39 \text{ g/mol}} \approx 0.00264 \text{ mol} \]
Since the stoichiometry of the reaction is 1:1 for CaSO₄ to BaSO₄, it means that the moles of CaSO₄ that reacted is also 0.00264 mol.
Now we can calculate the mass of CaSO₄ that corresponds to these moles:
\[ \text{mass of CaSO}_4 = \text{moles} \times \text{molar mass} = 0.00264 \text{ mol} \times 136.14 \text{ g/mol} \approx 0.3594 \text{ g} \]
Now we find the percentage of CaSO₄ in the original mixture:
\[ \text{percentage of CaSO}_4 = \left( \frac{\text{mass of CaSO}_4}{\text{mass of mixture}} \right) \times 100 = \left( \frac{0.3594 \text{ g}}{0.4550 \text{ g}} \right) \times 100 \approx 79.0% \]
To present the answer to four significant figures:
\[ \text{Percentage of CaSO}_4 \text{ in the mixture} = 79.03% \]