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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.

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    $\begingroup$ Since you are not using units, you made a mistake in your calculation. X is the atomic mass of the metal. In the first calculation; it is 24 amu, in the second calculation, it is 72 amu. Since there are 2 metal atoms, each atom has a mass of 36 amu. The closest atom with a mass of 36 amu is chlorine, but that is not a metal; therefore $\ce {M2O3}$ is not the formula. $\endgroup$ – LDC3 Oct 29 '14 at 2:47
  • $\begingroup$ But the question is still valid. Can we assume that the metallic oxide is $MO$ without testing the other possibilities? $\endgroup$ – Gerard Oct 29 '14 at 2:52
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    $\begingroup$ In real world problems you may well have other constraints such as knowing stoichiometry. However, if you truly know only what is stated in the problem, you'd be wise to consider other stoichiometries. In any case it certainly isn't wrong to consider more possibilities and rule them out ... so long as you get the numbers right. $\endgroup$ – ZSG Oct 29 '14 at 4:52
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Your question is a good one, and no, you can't assume that the metal is $\ce{MO}$ necessarily. However, you can find the equivalent weight of the metal without any assumptions about its formula.

The equivalent weight of a metal combining with oxygen is the mass that reacts with 8g of oxygen. We know that the compound contains 40% oxygen, so it must contain 60% metal. Suppose we had 100g of this compound; that would be 40g O and 60g M.

$40g\ \ce{O}:60g\ \ce{M}$ (Divide by 5)
$8 g\ \ce{O}:12g\ \ce{M}$

The equivalent weight of the metal must be 12g.

If you had been asked to identify the metal you would have needed the formula (or would have needed to test the likely formulas to see if they produced reasonable results.)

Given an equivalent weight for M of 12g:
$\ce{M2O-> 12g/mol, carbon}$ (not a metal)
$\ce{MO-> 24g/mol, magnesium}$
$\ce{M2O3-> 36g/mol, chlorine}$ (not a metal)
$\ce{MO2->48g/mol, titanium}$

Both magnesium and titanium fit the information that was given, $\ce{MgO}$ and $\ce{TiO2}$ are common oxides of those metals.

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40% O2 So, 40/16 = 2.5 60% M So, 60/24 (given for metal M) =2.5 Therefore, metal M: oxygen= 2.5: 2.5 =1:1 So, the formula is MO

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    $\begingroup$ For homework questions we have a policy where you do not simply post the answer without explanation but rather that you give hints that lead the asker towards their own solution. We are not a homework completion service. $\endgroup$ – bon Aug 17 '16 at 7:41

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