Question

Two oxides of a metal 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

In the first oxide,

% of oxygen = 27.6

% of metal = 100 – 27.6 = 72.4 parts by mass

Formula of first oxide = M₃O₄

So, 72.4 parts by mass of metal = 3 atoms of metal and 4 atoms of oxygen = 27.6 parts by mass of oxygen.

In the second oxide,

Oxygen = 30.0 parts by mass

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 metal = \(\frac{3}{72.4}\)x 70

= 2.90 atoms of metal

\(\because\) 276 parts by mass oxygen = 4 atoms oxygen

\(\therefore\) 30 parts by mass of oxygen = \(\frac{4}{27.6}\) x 30

= 4.35 atoms of oxygen

Ratio of M : O in the second oxide

= 2.90 : 4.35 = 1 : 1.5 or 2 : 3

= Formula of second oxide = M₂O₃