Question

what is the empirical formula of a compound that contains only iron and oxygen and is 22.27% oxygen?

Explanation:

Step1: Calculate the mass percentage of iron

Since the compound contains only iron and oxygen, and the oxygen percentage is \(22.27\%\), the mass percentage of iron is \(100\% - 22.27\%=77.73\%\)

Step2: Assume a 100 - g sample

Assume we have a \(100\) - g sample of the compound. Then the mass of oxygen (\(m_O\)) is \(22.27\) g and the mass of iron (\(m_{Fe}\)) is \(77.73\) g.

Step3: Calculate the number of moles of each element

The molar mass of oxygen (\(M_O\)) is \(16\space g/mol\), and the molar mass of iron (\(M_{Fe}\)) is \(55.85\space g/mol\).
The number of moles of oxygen (\(n_O\)): \(n_O=\frac{m_O}{M_O}=\frac{22.27\space g}{16\space g/mol}\approx1.392\space mol\)
The number of moles of iron (\(n_{Fe}\)): \(n_{Fe}=\frac{m_{Fe}}{M_{Fe}}=\frac{77.73\space g}{55.85\space g/mol}\approx1.392\space mol\)

Step4: Find the mole ratio

Divide each number of moles by the smaller number of moles (in this case, both are approximately \(1.392\space mol\))
The mole ratio of \(Fe:O=\frac{1.392}{1.392}:\frac{1.392}{1.392}= 1:1\)

Answer:

\(FeO\)