Question

Two oxides of a metal contain 27.6% and 30.0% of oxygen, respectively. If the formula of the first be M3O4. Then that of the second is .

• A. $M_3$$O_5$
• B. $M_2$$O_3$
• C. $M_3$$O_3$
• D. $M_3$O${}_2$

Answer 1

The correct option is

B

$M_2$$O_3$

In the first oxide, oxygen=27.6,metal=100−27.6=72.4 parts by mass.

In the second oxide, oxygen=30.0 parts by mass and metal =100−30=70 parts by mass.

First oxide Second oxide

Oxygen = $27.6\%$ Oxygen = $30.0\%$

Metal = $72.4\%$ Metal = $70.0\%$

Given that, formula of first oxide = $M_3$$O_4$
Let mass of the metal = x
% of metal in $M_3$$O_4$ = ( 3x/3x+64 )×100= 72.4
x = 56.
In 2nd oxide,
oxygen = 30%, metal = 70%
so, the ratio is
M : O
70/56 : 30/16
1.25 : 1.875
2 : 3
so, 2nd oxide is M₂O_3.