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.
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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
$$M(\ce{Fe3O4})=\pu{55.845 g\cdot mol^{-1}}\times 3 + \pu{15.97 g\cdot mol^{-1}}\times 4=\pu{231.43 g}.$$


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Before I start explaining it, I would say that the data provided in the question and the correct answer corresponds to the compound $\ce{Fe3O4}$, whose molar mass is $\pu{231.43 g}$. But, I am not proceeding with this known fact rather I'm considering only the given data and the molar mass of iron, $\pu{55.845 g}$. 


In $\pu{1 mol}$ of the compound, $\ce{Fe3X4}$, there are $\pu{3 mol}$ of $\ce{Fe}$ and $\pu{4 mol}$ of the unknown element $\ce{X}$.
Let molar mass of the compound be $M$ grams.
There is $3\times\pu{55.845 g}$ of $\ce{Fe}$ in $M$ grams of the compound.


Given, in $\pu{3.26 g}$ of the compound, $\pu{2.36 g}$ of $\ce{Fe}$ is present.
Therefore, $\pu{1 g}$ of $\ce{Fe}$ is present in $\pu{(3.26 / 2.36) g}$ of the compound.
Therefore, $3\times\pu{55.845 g}$ of $\ce{Fe}$ is present in
$(3.26 / 2.36) \times 3 \times \pu{55.845 g}$ of the compound.


Therefore, $M = (3.26 / 2.36) \times 3 \times 55.845 = 231.43$, the molar mass is $\pu{231.43g/mol}$.


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