A 1.543 gram sample containing sulfate ion was treated with barium chloride reagent, and 0.2243 grams of barium sulfate was isolated. calculate the percentage of sulfate ion in the sample.

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Chemistry

Mallory Hensley

13 December, 22:41

A 1.543 gram sample containing sulfate ion was treated with barium chloride reagent, and 0.2243 grams of barium sulfate was isolated. calculate the percentage of sulfate ion in the sample.

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Jovany Love · Answer 1

Chemical reaction: SO₄²⁻ + Ba²⁺ → BaSO₄.

m (sample) = 1,543 g.

m (BaSO₄) = 0,2243 g.

n (BaSO₄) = m (BaSO₄) : M (BaSO₄).

n (BaSO₄) = 0,2243 g : 233,4 g/mol.

n (BaSO₄) = 0,00096 mol.

n (BaSO₄) = n (SO₄²⁻).

ω (SO₄²⁻) = m (SO₄²⁻) : m (sample).

ω (SO₄²⁻) = 0,00096 mol · 96 g/mol : 1,543 g.

ω (SO₄²⁻) = 0,059 = 5,9%.