Ratio and proportion could also be used to solve this type of problem.

159.6 g/mol anhydrous

10.0 g anhydrous

---------------------- = --------------------

177.6 g/mol hydrate

x g hydrate

(177.6 g/mol hydrate)(10.0 g anhydrous)

---------------------------------------- = 11.1 g

159.6 g/mol anhydrous

Notice in this example that you needed a larger weight of the hydrated form to

prepare a solution of equal strength. This is because a hydrated molecule

weighs more than its anhydrous counterpart. The key is that the solution

using a hydrate instead of the prescribed form will still react in the same

manner in solution if it is prepared properly. Therefore, care must be taken

when performing the calculation to ensure the proper amount of hydrate is

used.

b. **Example 2**. How much Fe2(SO4)3.4H2O is needed to prepare 500.0 mL of a

200.0 mg/dL Fe2(SO4)3 solution?

Solution. Read the problem carefully and determine the unknown quantity.

Grams of Fe2(SO4)3.4H2O.

Calculate the amount of the anhydrous form needed.

1 dL

200.0 mg

500.0 mL X --------X ----------= 1000 mg

100 mL

1 dL

Calculate the gram molecular weight of each substance. (See Appendix C.)

Fe2(SO4)3

Fe

55.8 X 2 = 111.6

S

32.1 X 3 =

96.3

O

16.0 X 12 = + 192.0

399.9 g/mol

Fe2(SO4)3.4H2O

Fe

55.8 X 2 = 111.6

S

32.1 X 3 =

96.3

O

16.0 X 12 = 192.0

H

1.0 X 8 =

8.0

O

16.0 X 4 = + 64.0

471.9 g/mol

MD0837

2-13

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