# How many formula units make up "25.0 g" of magnesium chloride?

Apr 8, 2018

$1.58 \cdot {10}^{23}$

#### Explanation:

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

You also know that in order for your sample to contain $1$ mole of magnesium chloride, it must contain $6.022 \cdot {10}^{23}$ formula units of magnesium chloride $\to$ this is given by Avogadro's constant.

So if $1$ mole of magnesium chloride has a mass of $\text{95.211 g}$ and contains $6.022 \cdot {10}^{23}$ formula units of magnesium chloride, you can say that

color(white)(overbrace(underbrace(color(black)("95.211 g"))_ (color(red)("given by the molar mass")))^(color(blue)("= 1 mole MgCl"_ 2)) color(black)(=) " "overbrace(underbrace(color(black)(6.022 * 10^(23) quad "formula units MgCl"_ 2))_ (color(red)("given by Avogadro's constant")))^(color(blue)("= 1 mole MgCl"_2))

This means that your sample will contain

$25.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g"))) * (6.022 * 10^(23) quad "form. units MgCl"_2)/(95.211color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(1.58 * 10^(23) quad "form. units MgCl}}_{2}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the mass of the sample.