A 1.345 g sample of solid residue from a hazardous waste dump is being analyzed for barium. The sample is dissolved and then sodium sulphate is added. The insoluble barium sulphate is dried, and a total of 73.8 mg of BaSO₄ is collected. What percent of the sample is barium?

Question

Quizplus · Accepted Answer

To find the percent of barium in the sample, first, we need to calculate the moles of barium sulfate (BaSO4) formed, using its molar mass (233.4 g/mol for BaSO4). The mass of BaSO4 collected is 73.8 mg, which is 0.0738 g. The moles of BaSO4 are calculated as $ \frac{0.0738 \, \text{g}}{233.4 \, \text{g/mol}} \approx 0.000316 \, \text{mol} $. Since the mole ratio of Ba to BaSO4 is 1:1, there are also approximately 0.000316 moles of Ba. The molar mass of Ba is 137.3 g/mol, so the mass of Ba is $ 0.000316 \, \text{mol} \times 137.3 \, \text{g/mol} \approx 43.4 \, \text{mg} $ or 0.0434 g. The percent of barium in the sample is then $ \frac{0.0434 \, \text{g}}{1.345 \, \text{g}} \times 100\% \approx 3.23\% $.

A 1345 G Sample of Solid Residue from a Hazardous Waste