Barium can be analyzed by precipitating it as BaS04 and weighing the precipitate. When a 0.713-g sample of a barium compound was treated with excess H2S04, 0.5331 g of BaSO4 formed. What is the percentage of barium in the compound?

Molecular mass of BaSO4 233.39 g/mol

Molecular mass of Ba = 137.327 g/mol

31.37%

Explanation:

For this case, you should consider the following reaction:

Ba⁺²₍aq₎ + H₂SO₄ ₍aq₎ → BaSO₄ ₍s₎ + H₂O

For which you obtain the precipitate of BaSO₄

In order to obtain the mas of Barium on the precipitate, you may use the following formula:

gBa = M₍BaSO₄₎x (M₍Ba₎/M₍BaSO₄₎)

Where:

gBa = mass of Barium

M₍BaSO₄ ₎ = mass of BaSO₄ from the precipitate

M₍Ba₎ = mass of Barium from the original sample

M₍BaSO₄₎ = mass of BaSO₄ from the precipitate

gBa = (0.5331) x (137.327/233.39) = 0.3136 g

Then we ontain the percentage of Barium multiplying by 100:

% Ba on the original sample = 31.36%