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Answer 1 · 2016-09-13T21:34:09

$sf(A_r=11)$

Since the metal is trivalent we can say that the formula of the oxide will be $sf(M_2O_3)$ .

This is because $sf(2M^(3+))$ ions has the same charge as $sf(3O^(2-))$ ions.

The percentage by mass of $sf(M=100-68=32)$

So the ratio of $sf(M:O)$ by mass is:

$sf(32:68)$

To get the ratio by moles we divide by the mass of 1 mole which is the $sf(A_r)$ of the element in grams:

Ratio by moles $sf(M:O)rArr$

$sf(32/A_r:68/16=2:3)$

$:.$ $sf(32/A_r:4.25=2:3)$

$:.$ $sf((32/A_r)xx3=4.25xx2)$

$:.$ $sf(96/A_r=8.5)$

$sf(A_r=96/8.5=11)$ to 2sig fig.