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

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