0.12 gm of an organic compound containing phosphorus gave 0.22 gm of `Mg_2P_2O_7` by the usual analysis. Calculate the percentage of phosphorus in the compound.

Explanation

Here, the mas of the compound taken = 0.13 g
Mass of `Mg_(2)P_(2)O_(7) -= 2g` atoms of P or `(2 xx 24 + 2 xx 31 + 16 xx 7) = 222g` of `Mg_(2)P_(2)O_(7) -= 62 g` of P
i.e., 222g of `Mg_(2)P_(2)O_(7)` contain phosphorus = 62 g
`:.` 0.22 g of `Mg_(2)P_(2)O_(7)` will contain phosphorus `= (62)/(222) xx 0.22 g`
But this is the amount of phosphorus present in 0.12 g of the organic compound.
`:.` Percentage of phosphorus `= (62)/(222) xx (0.22)/(0.12) xx 100 = 51.20`