# How many atoms of oxygen are in one formula unit of Al_2(SO_4)3?

$A {l}_{2} {\left(S {O}_{4}\right)}_{3}$ formally derives from an aluminum trication, $A {l}^{3 +}$, and a sulfate dianion, $S {O}_{4}^{2 -}$. Now ionic substances form to balance charges, and the only way to achieve neutrality for these net charges is to take 6 positive charges, i.e. $\left(2 \times A {l}^{3 +}\right)$, and 6 negative charges $\left(3 \times S {O}_{4}^{2 -}\right)$.
If you count up all the atoms that carry these charges, there are $\left(2 \times A l + 3 \times S + 3 \times 4 \times O\right)$. Is this clear?
How would I form a salt from $C {a}^{2 +}$ and phosphate anion, $P {O}_{4}^{3 -}$? What composition would give electrical neutrality?