Question
If a metallic oxide has $40\,\%$ oxygen, find the equivalent weight of the metal.
This amounts to finding the atomic weight of the metal and the charge on the cation (in effect identifying the metal itself).
I didn't know how to do this, so I looked up the solution. It goes as follows.
Solution
Assume the oxide has the formula $\ce{MO}$. Let the atomic weight of $\ce{M}$ be $x$ Then $$\frac{16}{x + 16} = \frac{40}{100}$$ Simplifying and solving for $x$ yields $$x = 24$$ Thus the metal is magnesium and the equivalent weight is $12$.
My problem is, how can we assume that the metallic oxide has the formula $\ce{MO}$? For all we know, the metallic oxide may be $\ce{M2O}$ or $\ce{M2O3}$. In fact, if we assume the latter formula we get
$$\frac{48}{2x + 48} = \frac{40}{100}$$
giving
$$x = 72$$
which I believe is Hafnium.