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

• A. Fe\(_2\)O
• B. Fe\(_2\)O\(_3\)
• C. Fe\(_3\)O\(_4\)
• D. FeO

Answer: (C)

Verified step by step guidance

1. Identify the elements present in the compound: iron (Fe) and oxygen (O). The problem states the compound contains 22.27% oxygen by mass, so the remaining percentage (100% - 22.27% = 77.73%) is iron by mass.
2. Assume a 100 g sample of the compound to simplify calculations. This means you have 22.27 g of oxygen and 77.73 g of iron.
3. Convert the masses of each element to moles by dividing by their respective molar masses: use 55.85 g/mol for Fe and 16.00 g/mol for O. Calculate moles of Fe as \(\frac{77.73}{55.85}\) and moles of O as \(\frac{22.27}{16.00}\).
4. Determine the simplest whole number mole ratio of Fe to O by dividing both mole values by the smaller of the two mole amounts. This will give you the ratio of atoms in the empirical formula.
5. Use the mole ratio to write the empirical formula by assigning the ratio numbers as subscripts to Fe and O. If necessary, multiply the ratio by a small whole number to get whole number subscripts, which will give you the empirical formula of the compound.