Answer to Question #114235 in Chemistry for Arvic

Question

A student was tasked to perform gravimetric analysis of a soluble sulfate. His unknown sample weighed 0.7543 g. The sample underwent
precipitation using BaCl2 and was digested for overnight. The precipitate was then filtered off to obtain white crystalline precipitate that was collected in
an ash less filter paper. In performing constant weighing, he obtained a crucible mass that is 29.9442 g. After burning his samples inside the crucible,
the obtained mass was 30.3375 g. Compute for the mass (g) of BaSO4 from the experiment.

Answer 1

The soluble sulfate is precipitated by BaCl₂, barium sulphate is formed:

$SO_4^{2-} + Ba^{2+}\rightarrow BaSO_4\downarrow$ .

The number of the moles of the precipitate barium sulphate is equal to the number of the moles of soluble sulphate:

$n(\text{SO}_4^{2-})=n(\text{BaSO}_4^{2-})$ .

The mass of the barium sulphate is:

$m(\text{BaSO}_4) = m(\text{crucible with BaSO}_4) - m(\text{crucible})$

$m(\text{BaSO}_4) =30.3375 - 29.9442 = 0.3933$ g.

Answer: the mass (g) of BaSO₄ from the experiment is 0.3933 g