# Question #ac390

##### 1 Answer

#### Explanation:

The idea here is that you're dealing with a **hydrate**, which is a compound that contains *water of crystallization* in its structure.

In your case, you have *calcium chloride hexahydrate*,

Notice that **one formula unit** of this hydrate contains

,one moleofanhydrouscalcium chloride#1 xx "CaCl"_2# ,six molesof water of crystallization#color(red)(6) xx "H"_2"O"#

This means that **for every mole** of calcium chloride hexahydrate, you get **mole** of anhydrous calcium chloride and **moles** of water.

Now, you know the mass of **one mole** of anhydrous calcium chloride from the compound's **molar mass**. The same goes for water. This means that you can use that information to determine the **percent composition** of the hydrate.

So, you have

#M_("M CaCl"_2) = "110.98 g mol"^(-)#

#M_("M H"_2"O") = "18.015 g mol"^(-1)#

If you assume that you have **one mole** of the hydrate, you can say that the percent composition of the anhydrous salt will be

#color(purple)(|bar(ul(color(white)(a/a)color(black)("% CaCl"_2 = "mass of CaCl"_2/("mass of CaCl"_2*6"H"_2"O") xx 100)color(white)(a/a)|)))#

The right side of the equation will be equal to

which gets you

#"% CaCl"_2 = 50.67%#

This means that **for every**

Since you want to get

#0.2 color(red)(cancel(color(black)("g CaCl"_2))) * overbrace(("100 g CaCl"_2 * 6"H"_2"O")/(50.67color(red)(cancel(color(black)("g CaCl"_2)))))^(color(blue)(= "% CaCl"_2)) = "0.39 g CaCl"_2*6"H"_2"O"#

I'll leave the answer rounded to two **sig figs**.