# How many atoms are in in 13.2 mol copper?

$13.2 \times {N}_{A} \text{ copper atoms}$, where ${N}_{A} = \text{Avogadro's number}$.
${N}_{A} = \text{Avogadro's number} = 6.022140857 \left(74\right) \times {10}^{23} \cdot m o {l}^{-} 1$.
I may use ${N}_{A}$ as I would use any other collective number: $\text{dozens, Bakers' dozens, scores, gross, etc.}$ Of course, ${N}_{A}$ is an unfeasibly large number, and it happens that ${N}_{A}$ $\text{^1H" atoms}$ have a mass of precisely $1 \cdot g$. ${N}_{A}$ is thus the link between the sub-micro world of atoms, and molecules, that which we can conceive of, and theorize about, and the macro world of $\text{grams}$, and $\text{litres}$, and $\text{kilograms}$, that which we measure in a lab.
In an exam, would you have to remember ${N}_{A}$? The answer is no, as it will quoted to you as supplementary material. You do have to be able to manipulate it, and to demonstrate its significance.