A chemist added an excess of sodium sulfate to a solution of a soluble barium compound to precipitate all of the barium ions as barium sulfate, BaSO4. How many grams of barium ions are in a 441-mg sample of the barium compound if a solution of the sample gave 403 mg BaSO4 precipitate? What is the mass percentage of barium compound?

259.497 mg, 58.84%

Explanation:

BaSO₄ → Ba²⁺ + SO₄²⁻

to calculate the mole of BaSO₄

mole BaSO₄ = mass given / molar mass = 403 mg / 233.38 g/mol = 1.7268 mol

comparing the mole ratio

1.7268 mol of BaSO₄ yields 1.7268 mol of Ba²⁺

403 mg BaSO₄ yields (1.7268 * 137.327) where 137.327 is the molar mass of Barium mol of Ba²⁺

441 mg BaSO₄ will yield (1.7268 * 137.327 * 441 mg) / 403 mg = 259.497 mg

mas percentage of the Barium compound = 259.497 mg / 441 mg * 100 = 58.84%