# How to calculate mass of water of crystallisation?

How do you work out the mass of water of crystallisation for 13.5 g of

$$\ce{Al2(SO4)3*6H2O}$$

The correct answer is 3.24 g of water but I'm unable to derive the answer.

What I have tried:

I worked out the mols of aluminium sulfate which is approx 0.038 mol by dividing 13.5 g/353.5.

Since the ratio of aluminium sulfate: water of crystallisation is 1:6, I multiplied the number of mols by 6 to result in 0.228 mol of water.

To work out the mass, I multiplied mols of water by molar mass to result in 4.101 g which is NOT the answer of 3.24 g .

What am I doing wrong and what is the correct method?

$$\begin{array}{c| c} \text{compound} & \text{ mass (g)} \\ \hline \ce{Al2(SO4)} & 342 \\ \ce{6H2O} & 108 \\ \ce{Al2(SO4)3\cdot 6H2O} & 450\\ \end{array}$$
$$\frac{13.5}{450} \times 6 \times 18 = 3.42$$