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In the Carius...

Question

In the Carius method for the estimation of phosphorus, 0.31 g of an organic compound gave 0.222 g of $M g_{2} P_{2} O_{7}$ . The percentage of phosphorus in the organic compound is .

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Solution

222 g of $M g_{2} P_{2} O_{7}$ has 62 g of Phoshphorus
So,
0.222 g of $M g_{2} P_{2} O_{7}$ = $\frac{0.222 \times 62}{222}$ g of Phosphorus= 0.062 g of phosphorus
So,
the percentage of phosphorus = $\frac{0.062}{0.31} \times 100$ = 20%

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