Analysis of organic compound (0.36 g) containing phosphorus gave 0.66 g of `Mg_(2)P_(2)O_(7)` when treated with concentrated nitric acid followed by magnesia mixture. Calculate the amount of phosphorus present in the compound.

Options

A. `51.20%`
B. `61.20%`
C. `73.5%`
D. `68.3%`

Answer

A

Explanation

The percentage of phosphorus in an organic compound when estimated as magnesium pyrophosphate is given by `62/222 xx ("Mass of" Mg_(2) P_2O_7 "formed" (m_1)xx100)/("Mass of compound taken (m)")%`
Given that mass of compound taken (m)=0.36 g and amount of `Mg_2P_2O_7` formed `(m_1)=0.66` g
Substituting the values in the equation, we get
Percentage of phosphorous `=(62 xx m_1 xx100)/(222xxm) % (62xx0.66xx100)/(222xx0.36)=51.20%`