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

##### 1 Answer

#### Explanation:

You know that magnesium chloride has a **molar mass** of **mole** of magnesium chloride has a mass of

You also know that in order for your sample to contain **mole** of magnesium chloride, it must contain **formula units** of magnesium chloride **Avogadro's constant**.

So if **mole** of magnesium chloride has a mass of

#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 color(red)(cancel(color(black)("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.