How much sulphur is present in an organic compound if 0.53g of the compound gave 1.158g of $B a S O_{4}$ upon analysis?

Open in App

Solution

Molar mass of $B a S O_{4}$ = 233 g
Suppose the mass of organic compound = m grams
Let the mass of barium sulphate formed = 1.158 grams
We know that 32 grams of sulphur is present in 1 mol of $B a S O_{4}$
Therefore, 233 g $B a S O_{4}$ contains 32 g sulphur
$\Rightarrow$ 1.158 g of BaSO4 contains $(32 \times 1.158)$ 233 grams of sulphur
Percentage of sulphur = $(32 \times 1.158 \times 100) / (233 \times 0.53) = 30 %$

Suggest Corrections

Similar questions

0.53 g

1.158 g

B a S O_{4}

S_{8}

(v) C a C O_{3}

x^{2} + 10 x + 20

Join BYJU'S Learning Program