Question

An oxide of metal m has 47% by mass of oxygen.Metal M has atomic mass of 24 the empirical formula of oxide is

Answer 1

The formulae of the oxide is :

M x O y

Let X be moles of M and y be moles of Oxygen.

Mass of oxygen is 0.47g

Mass of M is 0.53g

Moles of M = 0.53 / 24 = 0.02

Moles of Oxygen = 0.47 / 16 = 0.03

Mole ratio :

M : O = 0.02 : 0.03

= 2:3

The empirical formula of the oxide is thus :

M₂O₃

Answer 2

Answer : The empirical formula of metal oxide is $M_3O_4$.

Solution : Given,

Molar mass of metal = 24 g/mole

Molar mass of oxide = 16 g/mole

47 % by mass of oxygen means 47 g of oxygen present in 100 g of metal oxide.

So, mass of metal present in metal oxide = 100 - 47 = 53 g

Now we have to calculate the moles of metal and oxide.

Moles of Metal = $\frac{\text{ Given mass of metal}}{\text{ Molar mass of metal}}=\frac{53}{24}=2.2\approx 2$

Moles of Oxide = $\frac{\text{ Given mass of oxide}}{\text{ Molar mass of oxide}}=\frac{47}{16}=2.9\approx 3$

The mole ratio of Metal : Oxide,

$\frac{Metal}{Oxide}=\frac{2.2}{2.9}= \frac{2.2/2.2}{2.9/2.2}= \frac{1}{1.3}=\frac{1\times 3}{1.3\times 3}= \frac{3}{3.9}= \frac{3}{4}$

Therefore, the empirical formula of metal oxide is, $M_3O_4$.