Question

0.72 gm of an oxide of a metal $M$ on reduction with $H_2$ gave 0.64 g of the metal. The atomic weight of the metal is 64. The empirical formula of the compound is:

• A. $MO$
• B. $M_{2}O$
• C. $M{O}_2$
• D. $M_2{O}_3$

Answer 1

Answer: (B)

$M_{2}O$

According to the question, $0.72g$ of the oxide contains $0.64g$ metal

Since the atomic mass of the metal is $64$, moles of metal$=\frac{0.64}{64}=0.01moles$, and

moles of oxygen$=\frac{0.72-0.64}{16}=0.005moles$

Hence, the ratio is $0.01:0.005=1:0.5=2:1$

Hence empirical formula of the compound is $M_{2}O$

Hence, option $B$ is the correct answer.