% Fe in a sample

Seems so simple but is confusing me as when I take into account final weight I don't get one of the answers.


An iron ore was analysed by dissolving a $1.1324~\mathrm{g}$ sample in concentrated $\ce{HCl}$ and converting it to give $0.5394~\mathrm{g}$ of pure $\ce{Fe2O3}$. The $\mathrm{m\%}\_\ce{Fe}$ in the sample is either $18.9~\%$, $25.2~\%$, $37.7~\%$ or $50.4~\%$.


**My approach:**  

If I take the molar mass of iron as $55.85~\mathrm{g/mol}$ and oxygen as $15.99~\mathrm{g/mol}$. I have a $159.67~\mathrm{g/mol}$ weight for $\ce{Fe2O3}$. From this I calculated that amount of $\ce{Fe2O3}$ is$$\frac{0.5394~\mathrm{g}}{159.67~\mathrm{g/mol}}=3.378 \cdot 10^{-3}~\mathrm{mol}$$


The answer is then multiplied by ($2 \times 55.85~\mathrm{g/mol}$) giving $0.3773~\mathrm{g}$ iron but this is what is confusing.


Isn't the answer I got just for $0.5394~\mathrm{g}$ so shouldn't I take into account that I started off with $1.1324~\mathrm{g}$? If I do I get $33.32~\%$ so doesn't match an answer.


