Question

Find the empirical formula of 62.2% Fe, 35.6% O, 2.2% H

Answer 1

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:

`\begin{gathered} 62.2gFe\cdot(molFe)/(55.8g)=1.11molFe \\ 35.6gO\cdot(molO)/(32g)=1.11molO \\ 2.2gH\cdot(molH)/(1g)=2.2molH \end{gathered}`

Now, divide every result by the smallest result of them, it means divide each result by 1.11:

`\begin{gathered} (1.11molFe)/(1.11mol)=1 \\ (1.11molO)/(1.11mol)=1 \\ (2.2molH)/(1.11mol)=2 \end{gathered}`

The obtained quotients will be the subscripts for each element in the empirical formula. It means that the empirical formula:

`FeOH_2`