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Q. $0.5 \,gm$ of an organic substance containing phosphorous was heated with conc. $HNO _3$ in the Carius tube. The phosphoric acid thus formed was precipitated with magnesia mixture( $\left.MgNH _4 PO _4\right)$ which on ignition gave a residue of $1.0$ gm of magnesium pyrophosphate $\left( Mg _2 P _2 O _7\right)$. The percentage of phosphorous in the organic compound is:

Organic Chemistry – Some Basic Principles and Techniques

Solution:

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Mw of $Mg _2 P _2 O _7=24 \times 2+31 \times 2+16 \times 7=222$
Percentage of $P =\frac{62}{222} \times \frac{\text { Weight of } Mg _2 P _2 O _7}{\text { Weight of compound }} \times 100$
$=\frac{62}{222}=\frac{1.0}{0.5} \times 100=55.85 \%$