# Question #45a3b

Dec 17, 2014

The molecule $A l {\left(P {O}_{4}\right)}_{3}$ contains the elements Aluminum, Phosphorus and Oxygen.

For this molecule the Aluminum has one (1) atom.
The Phosphate ion $\left(P {O}_{4}\right)$ has three units which means the subscript 3 outside of the parenthesis gets distributed to each value inside thus there are three (3) Phosphorus atoms and twelve (12) Oxygen atoms.

Al = 1
P = 3
O = 16

This means that the molecule $A l {\left(P {O}_{4}\right)}_{3}$ contains a total of 16 atoms (1 +3+12 = 16).

Technically however, there is no molecule of $A l {\left(P {O}_{4}\right)}_{3}$ since Aluminum has an ionic charge of +3, $A {l}^{+} 3$ and Phosphate has a charge of -3, $P {O}_{4}^{-} 3$. This means that the actual molecule is $A l P {O}_{4}$.

Therefore the atom count is

Al = 1
P = 1
O = 4

For a total of 6 atoms in the molecule.