Question

The phosphorus present in an organic compound of mass 30 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 percentage composition of phosphorus present in the sample of the organic compound is 50 percent, find the approximate mass of $M{g}_2{P}_2{O}_7$ formed.

• A. 17 g
• B. 25 g
• C. 10 g
• D. None of these

Answer 1

The correct option is A 17 g

The mass of the organic compound taken be = W gMass of $M{g}_2{P}_2{O}_7$obtained = W1 g
From stoichiometry, $M{g}_2{P}_2{O}_7$= 2P1 mol 2 mol1 x molecular mass of $M{g}_2{P}_2{O}_7$= 2 x atomic mass of P/222 g 2 x 31 g = 62 g
Then, mass of phosphorous in W1 g of $M{g}_2{P}_2{O}_7$ = W1×6/2222g
Then, percentage of phosphorous in the compound = W1×6/2222×100W=50
Here, W1=30 g from above equation weight of W can easily be calculated.