Question

The phosphorus present in an organic compound of mass 0.2 g is oxidized to phosphoric acid by heating with fuming nitric acid. The phosphoric acid so obtained is precipitated as$MgN{H}_{4}P{O}_4$, which on ignition is converted into $M{g}_2{P}_2{O}_7$. If the mass of $M{g}_2{P}_2{O}_7$ is 0.1 g, the percentage composition of phosphorus present in the sample of the organic compound is:

• A. 12.96
• B. 13.96
• C. 14.96
• D. 11.96

Answer 1

The correct option is B 13.96

The mass of the organic compound taken be$=Wg$

Mass of $M{g}_2{P}_2{O}_7$ obtained $=W1g$

From stoichimetry,

$M{g}_2{P}_2{O}_7=2P$

$1mol$ $2mol$.

$1\times$ molecular mass of $M{g}_2{P}_2{O}_7=2\times$ atomic mass of $P$

$222g2\times 31g=62g$

Then, mass of phosphorous in $W1g$ of $M{g}_2{P}_2{O}_7=W1\times \frac{62}{222}g$

Then, percentage of phosphorous in the compound = $W1\times \frac{62}{222}\times \frac{100}{W}$

Here, $W=0.2g$ and $W1=0.1g$