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  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
NOTE:
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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