Respuesta :
The first step is to take the given percents as if they were masses.
It means, we are going to use 62.2g of Fe, 35.6g of O and 2.2g of H.
Convert the given mass to moles using the corresponding molecular mass:
[tex]\begin{gathered} 62.2gFe\cdot\frac{molFe}{55.8g}=1.11molFe \\ 35.6gO\cdot\frac{molO}{32g}=1.11molO \\ 2.2gH\cdot\frac{molH}{1g}=2.2molH \end{gathered}[/tex]Now, divide every result by the smallest result of them, it means divide each result by 1.11:
[tex]\begin{gathered} \frac{1.11molFe}{1.11mol}=1 \\ \frac{1.11molO}{1.11mol}=1 \\ \frac{2.2molH}{1.11mol}=2 \end{gathered}[/tex]The obtained quotients will be the subscripts for each element in the empirical formula. It means that the empirical formula:
[tex]FeOH_2[/tex]