Two oxides of a metal contain 27.3% and 30% oxygen by mass respectively. If the formula of the first oxide is MO, then the formula of the second oxide is:-

1. MO

2. $M{O}_2$

3. $M_2{O}_3$

4. $M_{2}O$

Step 1:

% of oxygen in first metal oxide = 27.3 %

$\therefore$ % of metal in first metal oxide = 100 - 27.3 = 72.4%

$\therefore (3M+4\times 16)72.4=3M$

$2.172+46.336=3M$

$46.336=0.828M$

M = 55.56 g/mol $\approx$ 56 g/mol

Step 2:

% of oxygen in second metal oxide = 30%

$\therefore$ % of metal in first metal oxide = 100 - 30 = 70%

Moles of metal in 100 g second metal oxide = $\frac{70}{55.95}=1.25$

Moles of oxygen in 100 g second metal oxide = $\frac{3}{16}=1.875$

Step 3:

Ratio of moles of metal to moles of oxygen = 1.25 : 1.875

= 1 : 1.5

= 2 : 3

Therefore, the formula of metal oxide will be $M_2{O}_3$

Hence, option (3) is correct.