Question

Two oxides of a metal contain $27.6\%and30.0\%$ of oxygen, respectively. If the formula of the first is $M_3$$O_4$, the second oxide is.

• A. $M_3$$O_5$
• B. $M_3$$O_3$
• C. $M_3$$O_2$
  
• D. $M_2$$O_3$

Answer 1

The correct option is D $M_2$$O_3$

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

Hence, $x=55.97=56$.

Knowing the atomic weights of the metal and oxygen, the empirical formula of the second oxide can be worked out as:
$M_{\frac{70}{56}}{O}_{\frac{30}{16}}=M_{1.25}{O}_{1.875}=M_2{O}_3$

Thus, the formula of the second oxide is $M_2$$O_3$.