The sulfur content of an ore is determined gravimetrically by reacting the ore with concentrated nitric acid and potassium chlorate,converting all sulfur to sulfate.The excess nitrate and chlorate is removed by reaction with concentrated hydrochloric acid and the sulfate is precipitated using barium cation. Ba²⁺ (aq) + SO₄^2- (aq) $\rightarrow$ BaSO₄ (s)
Analysis of 10.1830 grams of a sulfur containing ore yielded 13.0221 grams of BaSO₄.What is the percent by mass sulfur in the ore? (BaSO₄ = 233.43 g/mol)

Question

The sulfur content of an ore is determined gravimetrically by reacting the ore with concentrated nitric acid and potassium chlorate,converting all sulfur to sulfate.The excess nitrate and chlorate is removed by reaction with concentrated hydrochloric acid and the sulfate is precipitated using barium cation. Ba²⁺ (aq) + SO₄^2- (aq)

\rightarrow

BaSO₄ (s)
Analysis of 10.1830 grams of a sulfur containing ore yielded 13.0221 grams of BaSO₄.What is the percent by mass sulfur in the ore? (BaSO₄ = 233.43 g/mol)

Quizplus · Accepted Answer

First, calculate the moles of BaSO4 produced: 13.0221 g / 233.43 g/mol = 0.0558 mol. Since the molar ratio of sulfur to BaSO4 is 1:1, there are also 0.0558 mol of sulfur in the ore. The mass of sulfur is 0.0558 mol * 32.07 g/mol (molar mass of sulfur) = 1.789 g. The percent by mass of sulfur in the ore is then (1.789 g / 10.1830 g) * 100% = 17.57%.

The Sulfur Content of an Ore Is Determined Gravimetrically by Reacting