Hydrates
The strong dipole of water, which causes it to be an excelent solvent for ionic substances, also causes water molecules to attach themselves to ions in solution. Such ions are called Hydrated ions . When some solutions of hydrated ions are evaporated, the water molecules are so strongly attracted to the ions in solution that they remain attached as crystallization occurs. Water molecules are incorporated into the crystal structure. Such compounds are called hydrates . The number of moles of water present per mole of anhydrous salt is usually some simple number.
The water is not actually part of the chemical substance and this is reflected in the way the formula is written. When all hydrating water is removed by heating, the material is said to be anhydrous .
Here is the example: CuSO
4
.
5 H
2
O
This formula means that for every one mole of CuSO
4
, there are also five moles of water. The dot IS NOT a multiplication sign. There is a 1:5 ratio of moles of CuSO
4
to moles of H
2
O.
Here is the name: copper (II) sulfate pentahydrate. Notice penta-, meaning five and hydrate, meaning water. That means that when you hear "pentahydrate," you have to know to write the dot and then the 5
H
2
O.
Practice Problems
1) Suppose you heard "trihydrate." What would you write? ____________________
2) Suppose you heard "octahydrate." What would you write? ____________________
3) Name this substance: MgSO
4
.
7 H
2
O ______________________
4) Write the formula for: barium chloride dihydrate _________________

For an example calculation we will determine the number of water molecules associated with a hydrated cobalt (II) sulfate molecule, CoSO
4
• x H
2
O, with x being the unknown number of water molecules. In the laboratory you weigh out 1.0230g of the hydrate and after heating until all of the water has evaporated you are left with 0.6030g of the anhydrous compound, which is just CoSO
4 without any water molecules attached. The first step we want to take is to determine the mass of the water evaporated from the hydrated compound. Simply take the mass of the hydrated compound (1.0230g) and subtract the mass of the anhydrous compound left behind (0.6030g). The mass of the water evaporated is 0.4200g. We want to know how many moles of water there are to one mole of the anhydrous material (CoSO
4
). The next step is to convert these masses to moles.
0.4200g = x mole of H
2
O

0.02331 moles of H
2
O
18.02 g 1 mole H
2
O
0.6030g = x mole CoSO
4

0.003890 moles of CoSO
4
155.0 g 1 mole CoSO
4
The value of x can then be determined from the mole ratio. moles of H
2
O__ = 0.02331

6 moles H
2
O/ CoSO
4 moles of CoSO
4
0.003890
This tells us there is a 6:1 H
2
O:CoSO
4 ratio and so the formula for the cobalt (II) sulfate hydrate is
CoSO
4
• 6 H
2
O and its name is cobalt(II) sulfate hexahydrate.
Try this one:
We would like to find the correct formula for MgSO
4
.
xH
2
O, so we put some of the hydrate in to an evaporating dish and heat it up. An empty evaporating dish has a mass of 12.770 grams. The dish and hydrate have a mass of 13.454 grams. After heating, the dish and anhydrous salt have a mass of 13.010 grams. Follow the steps in the example problem above, but first subtract out the mass of the dish.
The final formula is: MgSO
4
.
___H
2
O

Answer:
Mass of hydrate = 13.454 - 12.770 = .684 grams
Mass of anhydrous salt = 13.010 - 12.770 = .240 grams
Mass of water = 13.454 - 13.010 = .444 grams
Moles of anhydrous salt = .240 grams MgSO
4 x
1 mol MgSO
4 =
.00199 moles MgSO
4
1 120.367 g MgSO
4
Moles of water = .444 grams H
2
O x 1 mol H
2
O
=
.0246 moles H
2
O
1 18.0148 g H
2
O
Ratio of moles of water to moles of anhydrous salt = .0246/.00199 = 12
Therefore the formula is MgSO
4
.
12H
2
O