# Question 33cb2

Jul 26, 2016

$2 \cdot {10}^{24} \text{atoms of Ca}$

#### Explanation:

The first thing to do here is figure out how many formula units of calcium chloride, ${\text{CaCl}}_{2}$, you have in $3$ moles of this ionic compound.

To do that, you must use the definition of a mole, which as you know is simply a very, very large collections of atoms. More specifically, in order to have one mole of an ionic compound, you need to have $6.022 \cdot {10}^{23}$ formula units of said compound.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = 6.022 * 10^(23)"f units} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ Avogadro's number

Now, your sample will contain a total of

3 color(red)(cancel(color(black)("moles CaCl"_2))) * (6.022 * 10^(23)"f units CaCl"_2)/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = 1.8 * 10^(24)"f units CaCl"_2

As you know, a formula unit of calcium chloride is made up of

• one atom of calcium, $1 \times \text{Ca}$
• two atoms of chlorine, $2 \times \text{Cl}$

Since every formula unit of calcium chloride contains one atom of calcium, it follows that your answer will be

1.8 * 10^(24) color(red)(cancel(color(black)("f units CaCl"_2))) * "1 atom Ca"/(1color(red)(cancel(color(black)("f unit CaCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2 * 10^(24)"atoms of Ca")color(white)(a/a)|)))#

The answer is rounded to one sig fig, the number of sig figs you have for the number of moles of calcium chloride.