Question

(ii) 0.16g of an organic substance was heated in carius tube and the sulphuric
acid formed was precipitated as BaSO4 with BaCl2.The mass of dry BaSO4
was 0.35g. Calculate the percentage of sulphur in the organic compound.

please be fast
and send me solution​

Answer 1

Answer : The percentage of sulfur in the organic compound is, 30 %

Solution : Given,

Mass of organic compound = 0.16 g

Mass of $BaSO_4$ = 0.35 g

Molar mass of $BaSO_4$ = 233 g/mole

Molar mass of sulfur = 32 g/mole

First we have to calculate the mass of sulfur.

As, 233 g of $BaSO_4$ contains = 32 g of sulfur

So, 0.35 g of $BaSO_4$ contains = $\frac{32g}{233g}\times 0.35g=0.048g$ of sulfur

Now we have to calculate the percentage of sulfur in organic compound.

$\%\text{ of sulfur}=\frac{\text{Mass of sulfur}}{\text{Mass of organic compound}}\times 100=\frac{0.048g}{0.16g}\times 100=30\%$

Therefore, the percentage of sulfur in the organic compound is, 30 %