# 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?

• (3.26-2.36)/(2.36/55.845/3*4)*4+55.845*3 Commented Jan 30, 2018 at 5:48
• @WeijunZhou thank you but what 55.845 represent? Commented Jan 30, 2018 at 5:51
• @napi15 The molar mass of iron. Commented Jan 30, 2018 at 5:52

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

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 $\pu{231.43g/mol}$.

• The correct unit for (relative) molar mass is $\pu{g/mol}$, you are a bit inconsistent with that. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. Commented Jan 30, 2018 at 7:53
• Yes I knew since begining it was the oxygene , but I must follow the flow of the question Commented Jan 30, 2018 at 16:10