Introduction
To determine the empirical formula of an oxide of iron, we need to know the percentage composition of its elements. In this case, we know that the oxide contains 69.9% iron and 30.1% dioxygen by mass.

Calculating the empirical formula
To calculate the empirical formula, we need to convert the mass percentages to mole fractions. We can do this by assuming a fixed mass for the sample, such as 100 g. This means that the sample contains 69.9 g of iron and 30.1 g of dioxygen.

Calculating the moles
Next, we need to calculate the moles of each element in the sample. To do this, we divide the mass of each element by its molar mass. The molar mass of iron is 55.845 g/mol, and the molar mass of dioxygen is 32.00 g/mol.

Therefore, the moles of iron in the sample are:

69.9 g / 55.845 g/mol = 1.251 mol

The moles of dioxygen in the sample are:

30.1 g / 32.00 g/mol = 0.9406 mol

Determining the empirical formula
The next step is to determine the empirical formula of the oxide. We do this by dividing the number of moles of each element by the smallest number of moles.

In this case, the smallest number of moles is 0.9406, which corresponds to dioxygen. Dividing the moles of iron by 0.9406 gives us:

1.251 mol / 0.9406 mol = 1.329

Rounding to the nearest whole number, we get:

Iron: 1
Dioxygen: 1.329 ≈ 1.33

Therefore, the empirical formula of the oxide of iron is:

FeO1.33

Conclusion
The empirical formula of the oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass, is FeO1.33. This means that the ratio of iron to dioxygen in the oxide is approximately 1:1.33.