# How many grams are in #3.01*10^23# formula units of #CaCl_2#?

##### 1 Answer

#### Explanation:

What you need to do here is use **Avogadro's number** to calculate how many *moles* of calcium chloride,

Once you know how many *moles* you have, you can use calcium chloride's **molar mass** to convert those to *grams*.

So, *Avogadro's number*, which essentially acts as the definition of a **mole**, tells you that **one mole** of any ionic compound contains **formula units** of that compound.

As you can see, the sample given to you contains approximately **half a mole** of calcium chloride, since

#3.01 * 10^(23)color(red)(cancel(color(black)("f. units CaCl"_2))) * "1 mole CaCl"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("f. units CaCl"_2)))) = "0.4998 moles CaCl"_2#

Now, calcium chloride has a molar mass of approximately **every mole** of calcium chloride has a mass of

Your sample will thus be equivalent to

#0.4998 color(red)(cancel(color(black)("moles CaCl"_2))) * "111 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("55.5 g")color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.