0.12g of an organic compound gave 0.22 ${\mathrm{Mg}}_2{\mathrm{P}}_2{\mathrm{O}}_7$ by the usual analysis. The percentage of phosphorus in the compound is

It is given to find out the percentage of phosphorus in the compound if

0.12g of an organic compound gave 0.22 g of magnesium pyrophosphate.

Explanation
1 mol i.e. 222g of Mg2P2O7 contains phosphorous content = 62 g
Therefore, 1 g of Mg2P2O7 contains phosphorous content = $\frac{62}{222}$g
Therefore, 0.22 g Mg2P2O7 contains phosphorous content = $\frac{62}{222}\times 0.22$ g
Thus, this is the amount of phosphorous present in 0.12 g of the organic compound.
Therefore, the percentage of phosphorous in the compound is = $$\begin{gathered} \frac{62}{222}\times 0.22\times \frac{1}{0.12}\times 100 \\ =51.16\% =51.20\% \left(\mathrm{approx}\right) \end{gathered}$$