# Question #bd6ed

Sep 13, 2016

$\textsf{{A}_{r} = 11}$

#### Explanation:

Since the metal is trivalent we can say that the formula of the oxide will be $\textsf{{M}_{2} {O}_{3}}$.

This is because $\textsf{2 {M}^{3 +}}$ ions has the same charge as $\textsf{3 {O}^{2 -}}$ ions.

The percentage by mass of $\textsf{M = 100 - 68 = 32}$

So the ratio of $\textsf{M : O}$ by mass is:

$\textsf{32 : 68}$

To get the ratio by moles we divide by the mass of 1 mole which is the $\textsf{{A}_{r}}$ of the element in grams:

Ratio by moles $\textsf{M : O} \Rightarrow$

$\textsf{\frac{32}{A} _ r : \frac{68}{16} = 2 : 3}$

$\therefore$$\textsf{\frac{32}{A} _ r : 4.25 = 2 : 3}$

$\therefore$$\textsf{\left(\frac{32}{A} _ r\right) \times 3 = 4.25 \times 2}$

$\therefore$$\textsf{\frac{96}{A} _ r = 8.5}$

$\textsf{{A}_{r} = \frac{96}{8.5} = 11}$ to 2sig fig.