# How to find the amount of moles given mass percentage data

A hydrated aluminium sulphate, $$\ce{Al2(SO4)3.xH2O}$$, contains 8.10% of aluminium by mass. Find the value of x.

My attempt: (Assuming 100g sample)

1. Calculate the amount of aluminium using the formula $$\mathrm{amount = \frac{mass}{molar~mass}}$$, yielding $$\mathrm{0.300~mol}~\ce{Al}$$ having a mass of $$\mathrm{8.10~g}$$.

2. Determine the amount of sulphate ($$\ce{SO4^2-}$$) by recognising a ratio between the aluminium and sulphate. 2 moles of aluminium to 3 moles of sulphate, 2:3.

3. Using the ratio, I determined the amount to be $$\mathrm{0.450~mol}~\ce{SO4^2-}$$, which has a mass of $$\mathrm{43.2~g}$$.

4. I can now calculate the mass of the anhydrous compound, aluminium sulphate ($$\ce{Al2(SO4)3}$$), to be $$\mathrm{51.3~g}$$, leaving $$\mathrm{48.7~g}$$ of anhydrous substance, or $$\ce{H2O}$$.

5. Using the figure of $$\mathrm{48.7~g}$$, I can calculate the amount of $$\ce{H2O}$$ using the formula $$\mathrm{amount = \frac{mass}{molar~mass}}$$ which produced a figure of $$\mathrm{2.71~mol}~\ce{H2O}$$.

6. By using the principle of relative amounts, I find that x=9.

The answer is incorrect. Where did I go wrong?

$m_{\ce{Al}}=8.10\ \mathrm g$
$n_{\ce{Al}}=\frac{8.10\ \mathrm g}{27\ \mathrm{g/mol}}=0.3\ \mathrm{mol}$
$n_{\ce{Al2(SO4)3}}=0.5 \times n_{\ce{Al}}=0.15\ \mathrm{mol}$
$m_{\ce{Al2(SO4)3}}=0.15\ \mathrm{mol} \times 342.15\ \mathrm{g/mol}=51.3\ \mathrm g$
$m_{\ce{H2O}}=100\ \mathrm g-51.3\ \mathrm g=48.65\ \mathrm g$
$n_{\ce{H2O}}=\frac{48.65\ \mathrm g}{18\ \mathrm{g/mol}}=2.7\ \mathrm{mol}$
$x=\frac{2.7\ \mathrm{mol}}{0.15\ \mathrm{mol}}=18$