Question

In a gravimetric determination of phosphorus of an organic compound of mass 0.248g was strongly heated in a carius tube with concentrated HNO3, phosphoric acid so produced was precipitated as MgNH4PO4 which on ignition yielded 0.44g of Mg2P2O7. Find the percentage of phosphorus in compound

Answer 1

Answer:

The % of P is 0.5×100= 50. (Ans)

Explanation:

We know the molecular mass of Mg2P2O7 is 222 gm

Now,

222g of Mg2P2O7 has 62g of P

1 g of Mg2P2O7 has 62/222 g of p

0.44 g of Mg2P2O7

has( 62/222) × 0.444  gm of

P = 0.124 g of P

Now,  

0.248 g of organic compound has

(0.124/0.248 ) = 0.5 g of P.

Hence, The % of P is 0.5×100= 50. (Ans)