# What is the mass of 1.2*10^18 formula units of calcium chloride?

Mar 20, 2018

$2.2 \cdot {10}^{- 4}$ $\text{g}$

#### Explanation:

The thing to keep in mind about formula units is that you need $6.022 \cdot {10}^{23}$ of them to have exactly $1$ mole of an ionic compound $\to$ think Avogadro's constant here.

In this case, you know that $6.022 \cdot {10}^{23}$ formula units of calcium chloride are needed in order to have exactly $1$ mole of calcium chloride.

Moreover, you know that calcium chloride has a molar mass of ${\text{110.98 g mol}}^{- 1}$, which means that $1$ mole of calcium chloride has a mass of $\text{110.98 g}$.

So if $1$ mole of calcium chloride contains $6.022 \cdot {10}^{23}$ formula units and has a mass of $\text{110.98 g}$, you can say that $6.022 \cdot {10}^{23}$ formula units of calcium chloride have a mass of $\text{110.98 g}$.

This means that your sample has a mass of

$1.2 \cdot {10}^{8} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{f. units CaCl"_2))) * "110.98 g"/(6.022 * 10^(23)color(red)(cancel(color(black)("f. units CaCl"_2)))) = color(darkgreen)(ul(color(black)(2.2 * 10^(-4) quad "g}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the number of formula units.