Question

what is the empirical formula of a compound composed of 69.94% fe and 30.06% o by mass?

Explanation:

Step1: Assume 100g of the compound

Assume the mass of the compound is 100g. Then there are 69.94g of Fe and 30.06g of O.

Step2: Calculate the moles of each element

The molar - mass of Fe is $M_{Fe}=55.85g/mol$, and the molar - mass of O is $M_{O}=16.00g/mol$.
The number of moles of Fe, $n_{Fe}=\frac{m_{Fe}}{M_{Fe}}=\frac{69.94g}{55.85g/mol}\approx1.25mol$.
The number of moles of O, $n_{O}=\frac{m_{O}}{M_{O}}=\frac{30.06g}{16.00g/mol}\approx1.88mol$.

Step3: Find the mole - ratio

Divide each number of moles by the smaller number of moles. Here, the smaller number of moles is 1.25mol.
The mole - ratio of Fe: $r_{Fe}=\frac{1.25mol}{1.25mol}=1$.
The mole - ratio of O: $r_{O}=\frac{1.88mol}{1.25mol}\approx1.5$.
Since we need whole - number ratios, multiply both ratios by 2 to get whole numbers.
The mole - ratio of Fe: $r_{Fe}=2$.
The mole - ratio of O: $r_{O}=3$.

Answer:

$Fe_2O_3$