Question

$0.2$ g of an organic compound containing phosphorous gave $1.877$ g of ammonium phosphomolybdate((NH₄)₃PMo₁₂O₄₀) by usual analysis. Calculate the percentage of the phosphorous in the organic compound.

• A. $12.3$%
• B. `$13.6$%
• C. $15.5$%
• D. $16.9$%

Answer 1

The correct option is C $15.5$%

The mass of the compd = 0.2 g

Mass of Mg2P2O7 = 1.877 g

Since 1 mole of $(NH4{)}_{3}PM{o}_{12}{O}_{40}$ has 1 g atoms of P, or

1876 g of $(N{H}_4{)}_{3}PM{o}_{12}O40$ = 31 g of P

So, 2.47 g of $(NH4{)}_{3}PM{o}_{12}{O}_{40}$ contains P = 31/1876 x 2.47 g

So % P present in the compound = $\frac{31}{1876}\times 0.877\times \frac{100}{0.2}$

= 15.5 %

C is correct answer