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

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Solution

Assuming X as M
$\begin{matrix} F i r s t o x i d e & S e c o n d o x i d e O x y g e n = 27.6 % & O x y g e n = 30 % M e t a l = 72.4 % & M e t a l = 70 % \end{matrix}$
Formula of $1^{s t} O x i d e = M_{3} O_{4}$
Let atomic mass of metal =x
$= \frac{3 x}{3 x + 64} \times 100$
So
$= \frac{3 x}{3 x + 64} \times 100 = 72.4$
$= x = 56$
Now in second oxide, Metal and oxygen are $70 %$ and $30 %$ therefore,
Their atomic ratio is
$\begin{matrix} M & O \frac{70}{56} & : & \frac{30}{16} 1.25 & : & 1.875 1 & : & 1.5 2 & : & 3 \end{matrix}$
Therefore, formula of the compound = $M_{2} O_{3}$

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