Use ratio and proportion to determine the amount of hydrate needed.
399.9 g/mol anhydrous
1000 mg anhydrous
----------------------= --------------------
471.9 g/mol hydrate
x mg hydrate
(1000 mg anhydrous) (471.9 g/mol hydrate)
x mg hydrate = ------------------------------------------
399.9 g/mol anhydrous
x mg hydrate = 1180 mg
c. Example 3. How many grams of CaCl2.2H2O are needed to make 1.00 liter
of a 20.0 mg/dL CaCl2 solution?
Solution. Read the problem carefully and select the unknown quantity.
Grams of CaCl2.2H2O.
Calculate the amount of the anhydrous form needed.
Calculate the amount of the anhydrous form needed
10 dL
20.0 mg
1.00 L X --------X -------- = 200 mg
1L
1 dL
Calculate the gram molecular weight of each substance. (See Appendix C.)
CaCl2
Ca
40.1 X 1 = 40.1
Cl
35.5 X 2 = + 71.0
111.1 g/mol
CaCl2.2H2O
Ca 40.1 X 1 = 40.1
Cl 35.5 X 2 = 71.0
H
1.0 X 4 =
4.0
O 16.0 X 2 = + 32.0
147.1 g/mol
MD0837
2-14