Question

Two oxides of metal contain $27.6$% and $30$% oxygen respectively. If the formula of the first oxide is $M_3{O}_4$, then the formula of second oxide is :

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

Answer 1

The correct option is C $M_2{O}_3$

Let the mass of the metal be $x$.

Percentage of metal in $M_3{O}_4=\left(\frac{3x}{3x+64}\right)\times 100$

Given percentage$=100-27.6=72.4\%$

So,

$\left(\frac{3x}{3x+64}\right)\times 100=72.4$

$\left(\frac{3x}{3x+64}\right)=\frac{72.4}{100}$

$300x=217.2x+4633.6$

$x=55.96\approx 59$

So, the ratio is:

$\left(70/56\right):\left(30/16\right)$

$2:3$

So the second oxide is $M_2{O}_3$