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.

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

GPT-4o mini · Answer

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%**.