Question

A 0.4550 g solid mixture containing CaSO4 is dissolved in water and treated with an excess of Ba(NO3)2, resulting in the precipitation of 0.6158 g of BaSO4 What is the percentage of CaSO4 in the mixture? state the answer to 4 s.f.

Answer 1

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:

\[
\text{Percentage of CaSO}_4 \text{ in the mixture} = 79.03\%
\]