# Question #485ae

##### 1 Answer

#### Answer:

#### Explanation:

You should know that in order to get from *grams* to *atoms*, you need to go through **moles** first.

So you will need to use the **molar mass** of barium chloride to convert the sample to moles, **Avogadro's constant** to convert the number of moles of barium chloride to **formula units**, and the chemical formula of the salt to convert the number of formula units to a number of **atoms**.

Since every formula unit of barium chloride contains

,onebarium cation#1 xx "Ba"^(2+)# ,twochloride anions#2 xx "Cl"^(-)#

you can say that **every formula unit** of barium chloride contains **atoms**, i.e.

This means that you will have

#2.08 color(red)(cancel(color(black)("g"))) * overbrace((1 color(red)(cancel(color(black)("mole BaCl"_2))))/(208.23color(red)(cancel(color(black)("g")))))^(color(blue)("molar mass")) * overbrace((6.022 * 10^(23)color(red)(cancel(color(black)("formula units BaCl"_2))))/(1color(red)(cancel(color(black)("mole BaCl"_2)))))^(color(blue)("Avogadro's constant")) * "3 atoms"/(1color(red)(cancel(color(black)("formula unit BaCl"_2)))#

# = color(darkgreen)(ul(color(black)(1.80 * 10^(22)color(white)(.)"atoms")))#

The answer is rounded to three **sig figs**, the number of sig figs you have for the mass of barium chloride.