Question

The phosphorus present in an organic compound of mass 75 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 Mg2P2O7 is 50 g, the percentage composition of phosphorus present in the sample of the organic compound is:

• A. 19.5
• B. 18.5
• C. 20
• D. 18.6

Answer 1

The correct option is D 18.6

The mass of the organic compound taken be = W g

Mass of $M{g}_2{P}_2{O}_7$ obtained = W1 g
From stoichiometry, $M{g}_2{P}_2{O}_7$ = 2P
2 x molecular mass of $M{g}_2{P}_2{O}_7$ = 2 x atomic mass of PP
Then, mass of phosphorous in W1 g of $M{g}_2{P}_2{O}_7$= W1×62222g
Then, percentage of phosphorous in the compound = W1×62222×100W
Here, W=75 g and W1 = 50 g