Question

0.33 g of an organic compound containing phosphorous gave 0.397 g of Mg₂P₂O₇ by the analysis. Calculate the percentage of P in the compound.

Answer 1

Weight of organic compound = 0.33g ;

Weight of Mg₂P₂O₇ = 0.397g

222 g of Mg₂P₂O₇ contains 62 g of phosphorous.

0.397 g of Mg₂P₂O₇ will contain \(\frac{62}{222}\) x 0.397 g of P

0.33 g of organic compound contains \(\frac{62}{222}\) x \(\frac{0.397}{0.33}\) g of P

100 g of organic compound will contain \(\frac{62}{222}\) x \(\frac{0.397}{0.33}\) x 100

= \(\frac{2461.4}{73.26}\) = 33.59%

Percentage of phosphorous = 33.59 %