Question

$0.40$ g of an organic compound containing phosphorous gave $0.555$ g of $M{g}_2{P}_2{O}_7$ by usual analysis. Calculate the percentage of the phosphorous in the organic compound.

• A. $38.75$%
• B. $31.56$%
• C. $25.99$%
• D. $63.45$%

Answer 1

The correct option is A $38.75$%

The mass of the compd = 0.40 g

Mass of Mg2P2O7 = 0.55 g

Since 1 mole of $M{g}_2{P}_2{O}_7$ has 2 g atoms of P, or

222 g of $M{g}_2{P}_2{O}_7$ = 62 g of P

So, 0.55 g of $M{g}_2{P}_2{O}_7$ contains P = 62/222 x 0.55 g

So % P present in the compound = $62/222\times 0.55\times 100/0.40$

= 38.75

A is correct answer