How to find the molar mass of an iron compound based on its mass and mole fraction in the compound? I'm trying to solve this question:


  We have a molecule composed of $3$ iron atoms, and $4$ atoms of another element. We are given the following information: it has $\pu{2.36 g}$ of iron for $\pu{3.26 g}$ of molecule.


I want to find the molar mass of the compound, I have tried so far: 
$$ m = \pu{3.26 g} = \pu{0.00326 kg}$$
Since it has $3$ atoms of $\ce{Fe}$ and $4$ atoms of an unknown substance, therefore:
$$3 + 4 = 7~\text{atoms},\\
  \pu{1 mol} = \pu{6.022* 10^23 atoms}\\
\frac{7}{\pu{6.022* 10^23}} = \pu{1.16 * 10^-23}$$ 

As we know: $M = m / n,$ I tried to divide $0.00326$  by $\pu{1.16 * 10^-23}$ and I obtained $\pu{2.79429 * 10^19}$, but the correct answer is $\pu{231.43 g/mol}$. 

What I have done wrong?
Let $\ce Z$ denote the unknown element.
The total amount of iron atom in the compound is,

$$n(\ce{Fe})=\frac{\pu{2.36g}}{\pu{55.845 g\cdot mol^{-1}}}=\pu{0.0423 mol},$$

The molecule comprises of 3 iron atoms and 4 other atoms, so the amount of the unknown atom is,

$$n(\ce Z)=\pu{0.0423 mol}\times \frac43=\pu{0.0564 mol},$$

The molar mass of $\ce Z$ is,

$$M(\ce Z)=\frac{\pu{3.26g}-\pu{2.36g}}{\pu{0.0564 mol}}=\pu{15.97 g\cdot mol^{-1}}$$

So $\ce Z$ is $\ce{O}$. The molar mass is
$$M(\ce{Fe3O4})=\pu{55.845 g\cdot mol^{-1}}\times 3 + \pu{15.97 g\cdot mol^{-1}}\times 4=\pu{231.43 g}.$$