Two oxides of a metal contain $27.6 % a n d 30.0 %$ of Oxygen, respectively. If the formula of the first be $M_{3}$ $O_{4}$ . Find that of the second.

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Solution

The correct option is D

First oxide Second oxide

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

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

Formula = $M_{3}$ $O_{4}$ Formula = ?

Let the atomic weight the metal be x.

Therefore, percentage by weight of the metal in the compound $M_{3}$ $O_{4}$ = $\frac{3 \times x \times 100}{3 x + 64} = 72.4 (g i v e n)$

Hence, x = 55.97 = 56.

Thus, having known the atomic weights of the metal and oxygen, empirical formula of the second oxide can be worked out as under

Thus, the formula of the second oxide is $M_{2}$ $O_{3}$ .

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