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?


1 Answer 1


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


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