# How many formula units make up 37.4 g of magnesium chloride (MgCl_2)?

Nov 10, 2016

$2.37 \cdot {10}^{23}$

#### Explanation:

The first thing to do here is to convert the mass of magnesium chloride to moles by using the compound's molar mass.

Magnesium chloride has a molar mass of ${\text{95.211 g mol}}^{- 1}$, which means that every mole of magnesium chloride has a mass of $\text{95.211 g}$.

This means that your sample will contain

37.4 color(red)(cancel(color(black)("g"))) * "1 mole MgCl"_2/(95.211 color(red)(cancel(color(black)("g")))) = "0.3928 moles MgCl"_2

Now, in order to find the number of formula units of magnesium chloride present in the sample, you must use Avogadro's constant to go from moles to formula units.

$\textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 mole" = 6.022 * 10^(23)"f. units}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ Avogadro's constant

In your case, the sample will contain

0.3928 color(red)(cancel(color(black)("moles MgCl"_2))) * (6.022 * 10^(23)"f. units MgCl"_2)/(1color(red)(cancel(color(black)("mole MgCl"_2))))

$= \textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{2.37 \cdot {10}^{23} {\text{f units MgCl}}_{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.