1. Hydrated Ionic Compounds
You have learned how to find the molar mass of a compound with the help of your periodic table. And you now know how to find the empirical and molecular formulae of a compound.
Sometimes though, chemists know most of the compound’s formula, but a significant piece of information is missing.
Some ionic compounds crystallize from a water solution with water molecules incorporated into their structure. These are called hydrates.

2. Hydrates have a specific number of water molecules chemically bonded to each formula unit.
Epsom salts for example, consist of magnesium sulfate with seven waters attached.
MgSO4.7H2O
This is called Magnesium sulfate heptahydrate
Compounds that have no water molecules incorporated into them are called anhydrous. This distinguishes them from their hydrated forms.

3. Magnesium sulfate heptahydrate is calculated as follows:
The molar mass of a hydrated compound must include the mass of any water molecules that are attached to the compound.
Magnesium sulfate heptahydrate is calculated as follows: Mg S
4O 4(15.999)
7 H2O 7(18.015)
g/mol

4. Determining the Formula of a Hydrate
A hydrate of Barium hydroxide, Ba(OH)2.?H2O, is used to make barium salts and to prepare certain organic compounds.
If you have a 50.0 g sample of the hydrate and it contains 27.2 of Ba(OH)2, calculate the percent by mass of water in the hydrate.
Then find out how many water molecules are attached to each Barium hydroxide.

5. The total mass of the compound is 50. 0 g and 27
The total mass of the compound is 50.0 g and 27.2 g is from the Ba(OH)2.
50.0 – 27.2 = 22.8 g of water in the sample.
To find the percent mass of water, take the mass of water and divide it by the total, then multiply by 100%.
22.8/50.0 x 100% = 45.6%
Or to find percent mass of Ba(OH)2 take it’s mass, divide by total, and multiply by 100.
27.2/50.0 x 100% = 54.4%

6. Find the empirical formula.
moles of Ba(OH)2 = mass of Ba(OH)2 = 27.2 = 0.159
molar mass from PT
moles of H2O = mass of H2O = 22.8 = 1.27
molar mass from PT
Ba(OH)2  0.159/0.159 = 1
H2O  1.27/0.159 = 7.9 ~ 8
Ba(OH)2.8H2O