# How many atoms, and of what type, are present in a 222*g mass of "calcium chloride"?

Jun 4, 2016

$\text{Moles of calcium chloride}$ $=$ $\frac{222 \cdot g}{110.98 \cdot g \cdot m o {l}^{-} 1}$ $=$ ??*mol

#### Explanation:

From the above, clearly we have $2$ $m o l$ of calcium chloride.

What does this mean? It means that I have $2 \times {N}_{A}$ individual $C a C {l}_{2}$ units, where ${N}_{A}$ $=$ $\text{Avogadro's number}$ $=$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

Here I use ${N}_{A}$ as I would any other collective number: dozen; bakers' dozen; score; Botany Bay dozen; gross; etc.

The reason why we use such an absurdly large number is that ${N}_{A}$ ""^1H atoms have a mass of $1 \cdot g$ precisely. ${N}_{A} , \text{ Avogadro's number}$, is thus the link between the micro world of atoms and molecules, that which we cannot perceive, with the macro world of grams, and kilos, and litres, that which we can measure conveniently in a laboratory.

So, if I have $2$ $m o l$ $C a C {l}_{2}$, how many moles of calcium and chlorine atoms do I have?