Question

1.6 g of an organic compound gave 2.6 g of magnesium pyrophosphate. The percentage of phosphorus in the compound is:

• A. 45.38%
• B. 54.38%
• C. 37.76%
• D. 19.02%

Answer 1

Answer: (A)

45.38%

Mass of Phosphorous = 31u

Since 1 mole of ${Mg}_2{P}_2{O}_7$ has 2g atoms of $P$

Mass of Phosphorous in ${Mg}_2{P}_2{O}_7$ = 62u

Mass of ${Mg}_2{P}_2{O}_7$ = 222.5u

Weight of ${Mg}_2{P}_2{O}_7$ = 2.6g

Weight of organic compound = 1.6g

$\text{Percentage of phsphorous}=\frac{\text{Mass of phsphorous}}{\text{Mass of}{Mg}_2{P}_2{O}_7}\times \frac{\text{Weight of}{Mg}_2{P}_2{O}_7}{\text{Weight of organic compound}}\times 100$

$\text{Percentage of phsphorous}=\frac{62}{222.5}\times \frac{2.6}{1.6}\times 100=45.38\%$

Hence, $A$ is correct answer.