Question #84161

Mar 24, 2015

Think about balancing the positive charge of the cation and the negative charge of the anion.

You get a +3 charge from the aluminium ion, $A {l}^{\textcolor{red}{3 +}}$, and a -2 charge from the sulfate ion,$S {O}_{4}^{\textcolor{b l u e}{- 2}}$. Notice that you can use 2 aluminium ions, for a total positive charge of +6, and 3 sulfate ions, for a total negative charge of -6, and achieve neutrality.

So, you need $2 A {l}^{\textcolor{red}{3 +}}$ and $3 S {O}_{4}^{\textcolor{b l u e}{2 -}}$, which will make your formula

$A {l}_{\textcolor{b l u e}{2}} {\left(S {O}_{4}\right)}_{\textcolor{red}{3}}$

The subscripts are equal to the original charges of the ions. This comes about from the cross-over rule for ionic compounds, which you can read about here:

http://socratic.org/chemistry/ionic-bonds-and-formulas/writing-ionic-formulas/cross-over-rule-for-ionic-formulas