Question

How much of sulphur is present in an organic compound if 0.53g of compound gave 1.158g of baso4 on analysis

Answer 1

we know, mass of organic compound=0.53 grams

Mass of barium sulphate formed=1.158 grams

32 grams of sulphur is present in 1 mol of BaSO4.

Molar mass of BaSO4=233 grams

therefore, 233 grams of BaSO4 contains 32 grams of sulphur

Therefore, 1.158 grams of BaSO4 contains
=(32*1.158)/233

Therefore, % of sulphur in it
=(32*1.158)/233*0.53 ×100
=30%
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