Need help with calculating the water of crystallization

A hydrate of calcium sulfate , CaSO4 XH2O contains 20.9%water. Calculate the water of crystallization (X) and write the formula of the hydrate. Please HELP!

The game plan with these is to figure out the number of moles of H2O there are per mole of CaSO4. Start by assuming any quantity of CaSO4 X H2O.

Assume 100.0 g of CaSO4 XH2O.

20.9 % of that = 20.9 g
That is the mass of water.
1 mole of water = 18.0 g
20.9 g x 1 mol/18.0 g = 1.16 mol H2O

100 % - 20.9 % = 79.1 % CaSO4
79.1 g is the mass of CaSO4
1 mole of CaSO4 = 40.1 + 32.1 + 4(16.0) = 136.2 g
79.1 g x 1 mol / 136.2 g = 0.581 mol CaSO4

Now calculate the ratio of mol H2O to mol CaSO4
1.16 / 0.581 = 2.00
So X = 2

mw of CaSO4 is 136.14 g/mol (anhydrous)
100 g CaSO4 XH2O have 20.9 g of water(mw is 18) or 1.16 mol; and 79.1 g CaSO4 or 0.581 mol, so the mol relation is approx 1:2 that means x=2
CaSO4 *2H2O

Let % = grams Grams H2O = 20.9 moles = 20.9 / 18 = 1.16
CaSO$ = 100 - 20.9 = 79.1% = g Moles = 79.1 / 136 = 0.581
Ratio is 1.16 to 0 581 = 2 to1
Formula = CaSO4.2H2O

CaSO4 weighs 136

H2O weighs 18

xH2O weighs 18x

.209 = 18x/(136 +18x)

136 +18x ~ 5*18x
136 = (90-18) x
136 = 72 x
x = 2


CaSO4*2H2O