Question

$0.5$ g of an organic substance containing phosphorus was heated with conc. $HN{O}_3$ in the carius tube. The phosphoric acid thus, formed was precipitated with magnesia mixture, $\left(MgN{H}_{4}P{O}_4\right)$ which on ignition gave a residue of $1.0$ g of magnesium pyrophosphate, $\left(M{g}_2{P}_2{O}_7\right)$. The percentage of phosphorous in the organic compound is:

• A. $55.85\%$
• B. $29.72\%$
• C. $19.81\%$
• D. $20.5\%$

Answer 1

The correct option is A $55.85\%$

$P+HN{O}_3\to H_{3}P{O}_4\underset{(N{H}_{4}Cl+MgC{l}_2)}{⟶}\underset{\underset{\underset{Magnesiumpyrrophosphate}{M{g}_2{P}_2{O}_7}}{\downarrow \Delta }}{MgN{H}_4\left(P{O}_4\right)}$
Molecular weight of $M{g}_2{P}_2{O}_7=(24\times 2+31\times 2+16\times 7)=222$ g/mol.
$\therefore \%$ of $P=\frac{62}{222}\times \frac{WeightofM{g}_2{P}_2{O}_7}{Weightofcompound}\times 100$ $=\frac{62}{222}\times \frac{1.0}{0.5}\times 100=55.85\%$