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

Aug 21, 2016

$\text{55.5 g}$

#### Explanation:

What you need to do here is use Avogadro's number to calculate how many moles of calcium chloride, ${\text{CaCl}}_{2}$, are equivalent to that many formula units of this salt.

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 $6.022 \cdot {10}^{23}$ 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 ${\text{111 g mol}}^{- 1}$, which means that every mole of calcium chloride has a mass of $\text{111 g}$.

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.