Often we know what we need and how much we need but we don't have the reagent in the right form so back to the scratching around and calculating how much we need. This problem comes up when we need X grams of a reagent and all we have is the reagent in the hydrated form. So part of that reagent is water. We need to figure what fraction of it is really the reagent we need and then calculate how much to weigh out.
If we need 40g of CaCl2 but all we have is CaCl2·5H20 on the shelf, weighing out 40g of that will not give us 40g of CaCl2 as part of the 40g is water.
Since the MW of CaCl2·5H20 = 201
and the MW of CaCl2 = 111
then the hydrated form must be 201/111 times as heavy.
The way we work the problem is just as a simple proportion problem:
MW of hydrated should be the same proportion as g of hydrated
MW of anhydrous g of anhydrous
Therefore, if we need 40g of CaCl2 and only have CaCl2·5H20 on hand:
201 /111= Xg/40g
[201/111] x 40g = X
72.4g = X
We need to weigh out 72.4g of the hydrated form.
Problems:
Before doing each of these problems, think! Think about whether you will need to be weighing more or less.
6-1. How much CuS04·8H20 will you need to weigh out to substitute for 120g of CuS04?
6-2. How much BaS04 will you need to weigh out to substitute for 5g of BaS04·4H20?
6-3. You need to make 500 mL of 4N CaCl2. You only have CaCl·5H20 on the shelf. How much will you need to weigh out?
6-4. How much CaCl2·10H20 do you need to weigh out to make 200 mL of a 7% (W/V) CaCl2 solution?
6.5. You want to make 1 liter of an 8.8% (W/V) Na2S04 solution. You have Na2S04·10H20 on hand. How much will you weigh out?
Check answers