Question

Two oxides of a metal contain 27.6% and 30.0% of oxygen, respectively. If the formula of the first oxide be ${M_{3}O}_4$, find that of the second oxide.

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

Answer 1

Answer: (C)

Given that, formula of first oxide$=M_3{O}_4$

Let mass of the metal = x

% of metal in $M_3{O}_4$ $=\frac{3x}{3x+64}\times 100$

But as given % age $=(100-27.6)=72.4$%

so, $\frac{3x}{3x+64}\times 100$ = 72.4

or $x=56$.

In 2nd oxide,

Given, oxygen = 30%, So metal = 70%

So, the ratio is

$M:O$=$\frac{70}{56}:\frac{30}{16}$

$⟹1.25:1.875$

$⟹2:3$

So, 2nd oxide is $M_2{O}_3$