Given:

- Mass of chromium = 104.0 g

- Mass of oxygen = 48.0 g

- Molar mass of chromium = 51.996 g/mol

- Molar mass of oxygen = 15.999 g/mol

Let's calculate the moles of chromium and oxygen in the compound:

- Moles of chromium = 104.0 g / 51.996 g/mol = 2.00 mol

- Moles of oxygen = 48.0 g / 15.999 g/mol = 3.00 mol

To find the simplest whole number ratio of chromium to oxygen, we divide both moles by the smallest number of moles:

- Moles of chromium (simplified) = 2.00 mol / 2.00 mol = 1

- Moles of oxygen (simplified) = 3.00 mol / 2.00 mol = 1.5

Since we cannot have half moles in a chemical formula, we multiply both simplified moles by 2 to get whole numbers:

- Moles of chromium (final) = 1 * 2 = 2

- Moles of oxygen (final) = 1.5 * 2 = 3

Therefore, the most likely empirical formula for this compound is Cr2O3.