Question

Two oxides of metal contain 27.6% and 30% of oxygen respectively. If the formula of first oxide is $M_{3}O_{4}$, find that of second.

Answer 1

From the given percentages of metal oxides are 27.6% and 30%

Formula of first oxide = ${ M }_{ 3 }{ O }_{ 4 }$

Let mass of the metal = x

% of metal in ${ M }_{ 3 }{ O }_{ 4 } = \left( \frac { 3x }{ 3x+64 }  \right)\times100$

but as given, %$= (100 - 27.6) = 72.4$%

So,$\left( \frac { 3x }{ 3x + 64 }  \right) \times 100$= 72.4

(or) x = 56

In second oxide,

Oxygen = 30%

So, metal = (100 - 30) = 70%

So the ratio is M:O $= \frac { 70 }{ 56 } :\frac { 30 }{ 16 }$ = 1.25:1.8752

Therefore, formula of second oxide is ${ M }_{ 2 }{ O }_{ 3 }$.

Answer 2

Explanation:

Given that, the formula of first oxide =M

3

O4

Let mass of the metal = x

% of metal in M

3

O

4

=(3x/3x+64)×100

But as given percentage = (100-27.6) = 72.4 %

so, (3x/3x+64)×100=72.4

or x = 56.

In 2nd oxide,

oxygen = 30%....so metal = 70%

so, the ratio is

M : O

70/56 : 30/16

1.25: 1.875

2 : 3

so, 2nd oxide is M

2

O

3

Hence, the correct option is (D).