Question

A mixture of unknown salts containing a sulfate salt has been weighed to be 1.50 g. After dissolving the solution in 500 mL of water, heating it, and adding hot barium chloride (40.0 mL, 0.500 M); the precipitate has been collected after succeeding digestion, and the mass was found to be 0.325 g. What is the percentage (by mass) of sulfur in the mixture? 

Answer 1

BaCl₂ + SO₄^2- = BaSO₄ + 2Cl^-

The precipitate is BaSO₄;

M(BaSO₄) = 233 g/mole;

n(BaSO₄) = m/M = 0.325/233 = 0.0014 moles;

According to the chemical reaction n(BaCl₂) = n(SO₄^2-) = n(BaSO₄) = 0.0014 moles.

According to the task n(BaCl₂) = C(BaCl₂) * V(BaCl₂) = 0.5 * 0.04 = 0.02 moles.

Therefore, salt BaCl₂ is in excess.

n(SO₄^2-) = n(S) = 0.0014 moles;

m(SO₄^2-) = n(SO₄^2-) * M(SO₄^2-) = 0.0014 * 96 = 0.1344 g;

m(S) = n(S) * M(S) = 0.0014 * 32 = 0.0448 g;

X(SO₄^2-) = m(SO₄^2-) * 100%/m(mixture) = 0.1344 * 100%/1.5 = 9.0%;

X(S) = m(S) * 100%/m(mixture) = 0.0448 * 100%/1.5 = 3.0%.

Answer: 3.0%