Question #ac390

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, #"CaCl"_2 * color(red)(6)"H"_2"O"#.

Notice that one formula unit of this hydrate contains

• one mole of anhydrous calcium chloride, #1 xx "CaCl"_2#
• six moles of water of crystallization, #color(red)(6) xx "H"_2"O"#

This means that for every mole of calcium chloride hexahydrate, you get #1# mole of anhydrous calcium chloride and #color(red)(6)# 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

#(1 color(red)(cancel(color(black)("mole CaCl"_2))) * 110.98 "g"/color(red)(cancel(color(black)("mole CaCl"_2))))/(1color(red)(cancel(color(black)("mole CaCl"_2))) * 110.98"g"/color(red)(cancel(color(black)("mole CaCl"_2))) + color(red)(6)color(red)(cancel(color(black)("moles H"_2"O"))) * 18.015"g"/(1color(red)(cancel(color(black)("mole H"_2"O"))))) xx 100#

which gets you

#"% CaCl"_2 = 50.67%#

This means that for every #"100 g"# of calcium chloride hexahydrate you get #"50.67 g"# of anhydrous calcium chloride.

Since you want to get #"0.2 g"# of the anhydrous salt, you will need

#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.