To find the empirical formula of a compound, we follow these steps:

Mass of Fe = 20.2 g
Mass of S = 11.5 g
Mass of O = 23.0 g
Mass of $\text{H}_2\text{O}$ = 45.3 g

Molar mass of Fe = 56 g/mol
Molar mass of S = 32 g/mol
Molar mass of O = 16 g/mol
Molar mass of $\text{H}_2\text{O} = (2 \times 1) + 16 = 18 \text{ g/mol}$

Moles of Fe $= \frac{20.2 \text{ g}}{56 \text{ g/mol}} = 0.3607 \text{ mol}$
Moles of S $= \frac{11.5 \text{ g}}{32 \text{ g/mol}} = 0.3594 \text{ mol}$
Moles of O $= \frac{23.0 \text{ g}}{16 \text{ g/mol}} = 1.4375 \text{ mol}$
Moles of $\text{H}_2\text{O} = \frac{45.3 \text{ g}}{18 \text{ g/mol}} = 2.5167 \text{ mol}$

The smallest number of moles among Fe, S, and O is approximately 0.36. We will use the moles of S (0.3594 mol) as the smallest for the anhydrous part.

Ratio for Fe: $\frac{0.3607}{0.3594} \approx 1.00$
Ratio for S: $\frac{0.3594}{0.3594} = 1.00$
Ratio for O: $\frac{1.4375}{0.3594} \approx 4.00$

This gives the anhydrous part of the formula as $\text{FeSO}_4$.

Now, we find the ratio of water molecules to the anhydrous salt. We use the moles of $\text{FeSO}_4$ (which is 0.3594 mol, based on the 1:1 ratio of Fe and S).

Ratio for $\text{H}_2\text{O}$: $\frac{2.5167}{0.3594} \approx 7.00$

The ratio of $\text{FeSO}_4$ to $\text{H}_2\text{O}$ is 1:7.

The formula of the hydrated salt is $\text{FeSO}_4 \cdot 7\text{H}_2\text{O}$.