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?
The mass of aluminium sulfate has to be taken along with the water of crystallization.    

Now molar masses
$$
\begin{array}{c| c}
\text{compound} &amp; \text{ mass (g)} \\
\hline
\ce{Al2(SO4)} &amp; 342 \\
\ce{6H2O} &amp; 108 \\
\ce{Al2(SO4)3\cdot 6H2O} &amp; 450\\
\end{array}
$$  

After that it's simply ratio and proportion find out the moles of hydrated salt then find out the moles of water finally get the mass of water. 

$$\frac{13.5}{450} \times 6 \times 18 = 3.42$$