0.12g of an organic compound containing phosphorous gave 0.22 gram Mg, `P_2O_7` by the usual analysis. Calculate the percentage of phosphorus in the compound.

Explanation

Mass of organic compound = 0.12g
Mass of `Mg_(2) P_(2)O_(7)=0.22g`
Now 1 mol of `Mg_(2)P_(2)O_(7)="2 mol of .P. atom"`
i.e, `222"g of "Mg_(2)P_(2)O_(7)=(62)/(222)"g of P"`
`0.22" g of "Mg_(2) P_(2)O_(7)=(62)/(222)" xx 0.22"g of P"`
0.12 g of the compound contain `=(62)/(222) xx 0.22"g of P"`
100g of the compound will contain `=(62)/(222) xx (0.22)/(0.12) xx 100`
=51.20%