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0.5 g 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 $({MgNH}_{4} {PO}_{4})$ which on ignition gave a residue of 1.0 g of magnesium pyrophosphate. The percentage of phosphorous in the organic compound is?

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answer is 55.85.

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Detailed Solution

$P + H N O_{3} \to H_{3} P O_{4} \underset{(N H_{4} C l + M g C l_{2})}{\to}$

$M g N H_{4} P O_{4} \underset{Δ}{\to} M g_{2} P_{2} O_{7}$

M.Wt of $M g_{2} P_{2} O_{7} = 222$

% of $P = \frac{62}{222} \times \frac{w t o f M g_{2} P_{2} O_{7}}{W t o f O . C} \times 100 = \frac{62}{222} \times \frac{1.0}{0.5} \times 100 = 55.85 %$

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