Question

A 3.060 g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample.The resultant reaction produced a precipitate of barium sulfate, which collected by filtration, washed, dried, weight?

If 0.2745 g of barium sulfate was obtained, what was the mass percentage of barium in the sample?

Answer 1

The mass percentage of barium was 5.265 %.

Explanation:

For `"BaSO"_4`,

`M_r = "137.33 + 32.06 + 64.00 = 233.39"`

∴ `0.2745 color(red)(cancel(color(black)("g BaSO"_4))) × "137.33 g Ba"/(233.99 color(red)(cancel(color(black)("g BaSO"_4)))) = "0.1611 g Ba"`

`"% Ba" = ("0.1611" color(red)(cancel(color(black)("g"))))/(3.060 color(red)(cancel(color(black)("g")))) × 100 % = 5.265 %`