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%

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l dont know l am s sorry