Question

A metalic element X with RAM x form two oxide p and q . Oxide p contain 70% of X and q contain 72.4% of X. If the first oxide has the formula X2O3 what is the formula of the second oxide?

Answer 1

Formula for first oxide = X₂O₃

Mass of metal X = x

% of metal in X₂O₃ = $\frac{2x}{2x+48}$ × 100

But as given % = 70%

x = 56

In second oxide

Metal given 72.4% , so oxygen will be 27.6 %

So the ratio is :

Metal : oxide

$\frac{72.4}{56}$ = $\frac{27.6}{16}$ =1.29:1.72 = 3:4

So formula for second oxide = X₃O₄ 4