Question

Two oxides of a metal contain 30.0% and 27.6 % of oxygen respectively. If the formula of the first oxide is $M_3{O}_4$, the formula of the second oxide, is

• A. $M_2{O}_3$
• B. $M_4{O}_3$
• C. $M_3{O}_4$
• D. $M_3{O}_2$

Answer 1

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

Formula of first oxide = $M_3{O}_4$
let mass of the metal be = x
percentage of metal in $M_3{O}_4=\frac{3x}{3x+64}\times 100$
but % age = (100-27.6) = 72.4 %
so, $\frac{3x}{3x+64}\times 100$
or x = 56.
in 2nd oxide,
oxygen = 30%....so metal = 70%
so, ratio :--
M : O
70/56 : 30/16
1.25 : 1.875
2 : 3
so, 2nd oxide =$M_2{O}_3$