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 0.6168 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 start by calculating the number of moles of BaSO4 that were precipitated.

1. Calculate the molar mass of BaSO4:

• Barium (Ba): 137.33 g/mol
• Sulfur (S): 32.07 g/mol
• Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol

Total molar mass of BaSO4: \[ 137.33 + 32.07 + 64.00 = 233.40 \text{ g/mol} \]

2. Calculate the number of moles of BaSO4 precipitated:

\[ \text{Number of moles of BaSO4} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.6168 \text{ g}}{233.40 \text{ g/mol}} = 0.00264 \text{ mol} \]

3. Since the precipitation reaction is: \[ \text{CaSO4} + \text{Ba(NO3)2} \rightarrow \text{BaSO4} + \text{Ca(NO3)2} \] The moles of CaSO4 that reacted will equal the moles of BaSO4 formed. Therefore, we have: \[ \text{Number of moles of CaSO4} = 0.00264 \text{ mol} \]

4. Calculate the mass of CaSO4 that corresponds to the moles we calculated:

First, we calculate the molar mass of CaSO4:

• Calcium (Ca): 40.08 g/mol
• Sulfur (S): 32.07 g/mol
• Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol

Total molar mass of CaSO4: \[ 40.08 + 32.07 + 64.00 = 136.15 \text{ g/mol} \]

Now, we can calculate the mass of CaSO4: \[ \text{Mass of CaSO4} = \text{Number of moles} \times \text{Molar mass} = 0.00264 \text{ mol} \times 136.15 \text{ g/mol} = 0.3594 \text{ g} \]

5. Calculate the percentage of CaSO4 in the original mixture: \[ \text{Percentage of CaSO4} = \left( \frac{\text{mass of CaSO4}}{\text{mass of mixture}} \right) \times 100 = \left( \frac{0.3594 \text{ g}}{0.4550 \text{ g}} \right) \times 100 = 78.93% \]

Therefore, the percentage of CaSO4 in the mixture, to four significant figures, is:

\[ \boxed{78.93%} \]