Question

Combustion of $3.10g$ of phosphorous produces a white, waxy solid with a mass of $5.50g$. What is the empirical formula for the compound produced in this reaction?

• A. $P_4{O}_7$
• B. $P_2{O}_5$
• C. $P^{}{O}^{}$
• D. $P_2{O}_3$

Answer 1

Answer: (D)

$P_2{O}_3$

$P+O_2\to P_x{O}_y$

Given, Mass of phosphorus $=3.10g$

Mass of product $=5.5g$

So, mass of oxygen present $=5.5-3.10=2.4g$

No. of moles of phosphorus present $=\frac{3.1}{31}=0.1moles$

No. of moles of oxygen present $=\frac{2.4}{16}=\frac{0.3}{2}moles$

$\Rightarrow$ Mole ratio $=x:y=P:O=0.1:\frac{0.3}{2}=2:3$

So, the empirical formula is $P_2{O}_3$