How to calculate the empirical formula of a hydrate

I have CrxSyOz.nH2O, total mass 50 g, mass of salt alone after heating is 27,38 g
I got Cr2S3O12 after calculations but still cant get *n* of the cristalwater
the answer is 18 though but I want to know hoe to get that, whats the process ?

You start with 50.0g of the hydrated salt. The mass of salt alone = 27.38g. Therefore you have evaporated 50.0-27.38 = 22.62 g H2O

Molar mass Cr2S3O12 = 392.1820 g/mol
Molar mass H2O = 18.0153g/mol

Divide each compound mass by respective molar mass:
Cr2S3O12 = 27.38/392.1820 = 0.0698
H2O = 22.62/18.0153 = 1.2556

Divide by smaller:
Cr2S3O12 = 1
H2O = 1.2556/0.0698 = 18
Ratio: salt : water = 1:18

There you have it!
Cr2(SO4)3.18H2O is the formula of the compound

Cr2S3O12
The formula above can be simplified to Cr2(SO4)3 as shown below.
Cr = +3
SO4 = -2
Cr2(SO4)3


The letter is the number of moles of water per mole of chromium sulfate.

Number of moles = mass ÷ mass of 1 mole
Mass of 1 mole of Cr2S3O12 = 2 * 52 + 3 * 32.1 + 12 * 16 = 392.3 g
Mass of 1 mole of water = 18 g

Total mass = 50 g
Mass of salt alone after heating is 27.38 g
Moles of salt = 27.38 ÷ 392.3 = 0.06979

Mass of H2O = 50 – 27.38 = 22.62 g
Mole of H2O = 22.68 ÷ 18 = 1.25667

Mole ratio = Moles of H2O / Moles of Cr2S3O12
Mole ratio = 1.25667 / 0.06979 = 18
This is the value of n

Cr2(SO4)3 * 18 H2O is the formula of hydrated chromium sulfate