Question

Two oxides of a metals contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is M3O4, find that of the second .

Answer 1

Answer:

One oxide of a metal contains 27.6 % amount of oxygen where as the oxide of the element contains 30.0 % amount of oxygen.

The formula given for the first oxide is : $\tt{M_3O_4}$

Let us understand about the first metal oxide firstly.

It contains 27.6 parts by mass of oxygen and contains 100 - 27.6 = 72.4 parts of the metal by mass.

Now, as its formula is $\tt{M_3O_4}$, it contains 3 atoms of the metal and 4 atoms of oxygen.

Similarly, in the second oxide of the metal.

It contains 30.0 parts by mass of oxygen and contains 100 - 30.0 = 70.0 parts of the metal by mass.

Calculating the number of atoms of metal :

$\tt{\longrightarrow \dfrac{3}{72.4}\times 70 =2.90\:atoms.}$

Calculating the number of atoms of oxygen :

$\tt{\longrightarrow \dfrac{4}{27.4}\times 30 =4.37\:atoms.}$

Now, Finding the ratio of metal to oxygen = M : O = 2.90 : 4.37 = 1 : 1.5 = 2 : 3.

Hence, the formula of the second oxide : $\bf{M_2O_3.}$