# Explain how 4"Ca"_3("PO"_4)_2 contains 52 total atoms?

## Ca=? P=? O=? (I'm getting 22 total and not quite understanding what I've done wrong)

Feb 19, 2018

Here's how I do it.

#### Explanation:

Assume you have one formula unit "Ca"_3("PO"_4)_2.

The subscripts represent the number of atoms or groups immediately before them.

Thus,

${\text{Ca}}_{3}$ means that you have three $\text{Ca}$ atoms.

${\left({\text{PO}}_{4}\right)}_{2}$ means that you have two ${\text{PO}}_{4}$ groups.

One ${\text{PO}}_{4}$ group contains one $\text{P}$ atom and four $\text{O}$ atoms.

∴ Two ${\text{PO}}_{4}$ group contain two $\text{P}$ atoms and eight $\text{O}$ atoms.

Let's count the number of atoms in one formula unit.

$\text{Ca" color(white)(m)= color(white)(l)"3 atoms}$
$\text{P"color(white)(mll) =color(white)(l) "2 atoms}$
$\underline{\text{O" color(white)(mll)= color(white)(l)"8 atoms}}$
$\text{Total = 13 atoms}$

If 1 formula unit contains 13 atoms, then
$\textcolor{w h i t e}{l l}$ 4 formula units contains 4 × 13 = 52 atoms.