Question

Two metallic oxides contain 27.6 % and 30 % oxygen respectively. If the formula of the first oxide is $X_3{O}_4$, what will be the formula of the second oxide?

• A. $XO$
• B. $X{O}_2$
• C. $X_2{O}_3$
• D. $X_2{O}_5$

Answer 1

The correct option is C $X_2{O}_3$

Let the mass of each oxide be 100 g.
For I oxide:
Oxygen = 27.6 % (given)
Metal = 72.4 % (given)
Molecular formula $=X_3{O}_4$ (given)
Let atomic mass of $X$ be $‘a^{\prime }\text{g}$.
So, $\frac{3a}{3a+64}\times 100=72.4;$
solving, we get ${}^{\prime }{a}^{\prime }=56\text{g}$
For II oxide:
Oxygen is 30 g, so metal will be 70 g.
So, the ratio of moles of $X:O$ will be $\frac{70}{56}:\frac{30}{16}=1.25:1.875$ or $2:3$
Hence, the compound is $X_2{O}_3$.