Question

Two oxides of a metal contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is $M_3O_4$, then second one is

• A. $MO_2$
• B. $M_2O$
• C. $M_2O_3$
• D. $M_3O_2$

Answer 1

Answer: (C)

$M_2O_3$

Answer 2

In the first oxide, Oxygen = 27.6 parts, Oxygen = 27.6 parts, As the formula of the first oxide is $M_3O_4$, 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 $=\frac{3}{72.4}\times70$ 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 $M_2O_3.$