Question

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

• A. $M{O}_2$
• B. $M_{2}O$
• C. $M_2{O}_3$
• D. $M_3{O}_2$

Answer 1

Answer: (C)

In the first oxide,
Oxygen = 27.6 parts,
Oxygen = 27.6 parts,
As the formula of the first oxide is $M_3{O}_4$, this means 72.4 parts by mass of metal = 3 atoms of metal and 4 atoms of oxygen = 27.6 parts by mass.
In the second oxide,
Oxygen = 30.0 parts by mass and metal = 100 - 30 = 70 parts by mass.
But, 72.4 parts by mass of metal = 3 atoms of metal
$\therefore 70$ parts by mass of met $=\frac{3}{72.4}\times 70$ atoms of metal
$=2.90$ atoms of metal
Also, $27.6$ parts by mass of oxygen $=4$ atoms of oxygen
$\therefore 30$ parts by mass of oxygen
$=\frac{4}{27.6}\times 30$ atoms of oxygen $=4.35$ atoms of oxygen
Hence, ratio of $M:O$ in the second oxide $=2.90:4.35=1:1.5=2:3$
$\therefore$ Formula of the metal oxide is $M_2{O}_3.$