Question

0.12 gm of an organic compound containing phosphorous gave 0.22 gm of Mg₂P₂O₇ by the usual analysis. Calculate the percentage of phosphorous in compound.

Options

• 31.2%

• 41.2%

• 51.2%

• 37.2%

Answer 1

51.2%

Explanation: 

Here, the mass of the compound taken = 0.12 g
Mass of Mg₂P₃O₇ = 2 g atoms of P or (2 x 24 x 2x 31 + 16 ×7) = 222 g of Mg₂P₂O₇ = 62 g of P i.e., 222 g of Mg₂P₂O₇ contain phosphorous = 62 g
∴ 0.22 g of Mg₂P₂O₇ will contain phosphorous
\[=\frac{62}{222}\times0.22\mathrm{g}\]
But this is the amount of phosphorous percent in 0.12 g of the organic compound.
∴ Percentage of phosphorous
\[=\frac{62}{222}\times\frac{0.22}{0.12}\times100=51.20\]