Question

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

Answer 1

`1.58 * 10^(23)`

Explanation:

You know that magnesium chloride has a molar mass of `"95.211 g mol"^(-1)`, which means that `1` mole of magnesium chloride has a mass of `"95.211 g"`.

You also know that in order for your sample to contain `1` mole of magnesium chloride, it must contain `6.022 * 10^(23)` formula units of magnesium chloride `->` this is given by Avogadro's constant.

So if `1` mole of magnesium chloride has a mass of `"95.211 g"` and contains `6.022 * 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 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.