In phosphorus estimation,0.31 g of an organi | vedclass

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(A) The formula for the percentage of phosphorus is: $	ext{Percentage of P} = \frac{62}{222} 	imes \frac{	ext{mass of } Mg_2P_2O_7}{	ext{mass of o

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$40$

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(A) The formula for the percentage of phosphorus is: $	ext{Percentage of P} = \frac{62}{222} 	imes \frac{	ext{mass of } Mg_2P_2O_7}{	ext{mass of organic compound}} 	imes 100$. Given: Mass of organic compound $= 0.31 \ g$,Mass of $Mg_2P_2O_7 = 0.444 \ g$,Molar mass of $Mg_2P_2O_7 = 222 \ g \ mol^{-1}$. Since $1 \ mol$ of $Mg_2P_2O_7$ contains $2 \ mol$ of $P$,the mass of $P$ in $0.444 \ g$ of $Mg_2P_2O_7$ is $\frac{2 	imes 31}{222} 	imes 0.444 = 0.124 \ g$. Percentage of $P = \frac{0.124}{0.31} 	imes 100 = 40 \%$.

In phosphorus estimation,$0.31 \ g$ of an organic compound gave $0.444 \ g$ of magnesium pyrophosphate (molar mass $= 222 \ g \ mol^{-1}$). The percentage of phosphorus in the compound is

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