1. Hydrates (11.5)
Hydrate- compound with a specific amount of water molecules bound to it

2. Hydrates are often used as drying agents or desiccators because they can absorb moisture
Silica gel- used in electronic equipment packaging, cat litter

3. Chemicals that do not contain water in their structures are called anhydrous (without water)
Hydrates are easy to spot because their formulas show the water molecules
Anhydrous cobalt (ii) chloride Cobalt (II) chloride hexahydrate CoCl2 CoCl2 . 6H2O

4. What is a Hydrate?
Any salt that has water chemically bonded to the ions in the crystal structure is a hydrate or hydrated crystal.
Copper(II) sulfate is a hydrate.
Hydrated copper(II) sulfate is deep blue in color.

5. What does the Chemical Formula of A Hydrate Look Like?
BaCl2•2H2O
FeSO4•6H2O
Na2CO3•10H2O
CuSO4•5H2O

6. How are Names of Hydrates Written?
BaCl2•2H2O
barium chloride dihydrate
FeSO4•6H2O
iron(II) sulfate hexahydrate
Na2CO3•10H2O
sodium carbonate decahydrate
CuSO4•5H2O
copper(II) sulfate pentahydrate
BaCl2•2H2O
FeSO4•6H2O
Na2CO3•10H2O
CuSO4•5H2O

7. What prefixes are used? 0.5 hemi 7 hepta 1 mono 8 octa 2 di 9 nona 3
tri 10
deca 4
tetra 11
undeca 5
penta 12
dodeca 6
hexa 13
triskaideca

9. To determine hydrate formula
Determine moles of water and of anhydrous compound
Find x = moles of water
moles of anhydrous compound
Plug in x
Formula for compound X H2O

10. The water molecule is highly polar

11. Water’s polar ends are attracted to the + and – elements of the crystal

12. The # of water molecules associated with the compound depends on the size of the crystal

13. The chemical formula for a hydrate includes the
formula unit for the ionic compound and the number
of complexed water molecules separated by a dot:
a) CaSO4 • 2 H2O
calcium sulfate dihydrate
b) CoCl2 • 6 H2O
cobalt II chloride hexahydrate

14. Analysis of an unknown hydrate:
There is no simple way to predict the number of
water molecules found in a hydrate compound.
It must be determined empirically by heating the
compound and evaporating the water

15. How can the water be removed?
Heat the crystal. The water is loosely bound, and will come away as water vapor.
Put the crystal in contact with or near a desiccant, maybe in a desiccator.

16. What is the compound called after the water has been removed?
Anhydride (noun)
The light blue powder is the anhydride.
Anhydrous (adjective)
Anhydrous copper(II) sulfate is left in the test tube after heating.

17. Example: A 14 gram sample of hydrated cobalt II
chloride is heated thoroughly to remove all the water.
The resulting anhydrous cobalt II chloride weighs
7.65 g. What is the chemical formula of the hydrated
cobalt II chloride
To know the chemical formula, you must determine
the number of moles of water present per mole of
cobalt II chloride
Approach: a) the 7.65 g is only anhydrous cobalt II
chloride
b) the difference between 7.65 g and
the original 14 g is the amount of
water present in the original sample

18. Approach: a) the 7.65 g is only anhydrous cobalt II
chloride
b) the difference between 7.65 g and
the original 14 g is the amount of
water present in the original sample
7.65 g 1 mole = mol CoCl2
130 g
= 6.35 g H2O 1 mole = mol H2O
18 g
0.353 mol H2O = 6 mol H2O  CoCl2 • 6 H2O
mol CoCl mol CoCl2

19. CuSO4 x 5H2O 2.50 = CuSO4 + H2O CuSO4= 1.59 g
A mass of 2.50 g blue hydrated copper sulfate (CuSO4 x XH2O) is place in a crucible and heated. After heating, 1.59 g of copper sulfate remains. What is the formula for the hydrate?
2.50 = CuSO4 + H2O CuSO4= 1.59 g
2.50 = H2O H2O = 0.91 g
CuSO4 = x 16 = g/mol
CuSO4 = 1.59 / = mol
H2O = 2 x = 18 g/mol
H2O = 0.91 / 18 = mol
X = / = 5
CuSO4 x 5H2O

20. What is the mole ratio of the hydrate CuSO4 x ___H2O, if the anhydrous mass of CuSO4 is 25.42g and mass of water is 14.3 g?
CuSO4 = x 16 = g/mol
CuSO4 = / = mol
H2O = 2 x = 18 g/mol
H2O = 14.3 / 18 = 0.79 mol
X = 0.79 / = 4.96 ~ 5
CuSO4 x 5H2O
1:5

21. What is the mole ratio of hydrate Na2CO3 x ____ H2O, if the original mass of the hydrated Na2CO3 was g and the mass of the anhydrous Na2CO3 is 16.19g?
Hydrate = anhydrous + H2O
43.69 = H2O
H2O = g Na2CO3 = g
Na2CO3 = 2 x x 16 = g/mol
Na2CO3 = / = 0.15 mol
H2O = 2 x = 18 g/mol
H2O = / 18 = 1.5 mol
X = 1.5 / 0.15 = 10
Na2CO3 x 10 H2O