What is the water of crystallization for barium chloride?

Many ionic compounds crystallize out of solution with water molecules inserted in regular positions throughout their crystal lattice structures. A compound with inserted water molecules is called a hydrate.

The number of water molecules present per formula unit of a compound is indicated in the formula by adding a "dot", followed by the number of water molecules. For example, the formula for hydrated copper sulfate would be written as: `~CuSO_4` `*`  `~5H_2O` .

The number of water molecules present in a barium chloride hydrate can be calculated by dividing the moles of water (`~H_2O` ) in the hydrate by the moles of the salt (`~BaCl_2` ) in the hydrate.

The number of moles of water and salt can be experimentally determined as follows:

Determine the mass of the barium chloride salt and the mass of the water molecules:

Step 1: Heat a sample of barium chloride hydrate. Let's use a 1.250 g sample of barium chloride hydrate. This would be the mass of the barium chloride salt and the attached water molecules.

Step 2: Heating causes the water molecules to detach and evaporate. So, if we weigh our sample after heating, we will have the mass of just the barium chloride salt. After heating, our sample weighs 1.060 g `~BaCl_2` .

Step 3: If we subtract the mass of our sample after heating from the mass of our sample before heating, we will have the mass of just the water molecules:

    1.250 g - 1.060 g = 0.190 g `~H_2O`

Determine the perecent composition of water and salt in the hydrate.

Step 1: Divide the mass of the water molecules by the mass of the entire hydrate and multiply by 100 to get the percent of water.

    (0.190 g water/1.250 g hydrate) x 100 = 15.2% `~H_2O`

Step 2: Subtract the percentage of water from 100% to get the percent of the barium chloride salt. 

    100% - 15.2% = 84.8% `~BaCl_2`

Convert percentages to grams.

If we assume a 100 g sample, then there will be 15.2 g `~H_2O` and 84.8 g `~BaCl_2` .

Convert the grams of water and barium chloride salt to moles by dividing each by its molar mass.

mol `~H_2O` : 15.2 g/18.0 g/mol = 0.844 mol 

mol `~BaCl_2` : 84.8 g/208.234 g/mol = 0.407 mol

Divide the moles of water by the moles of barium chloride salt to get the number of water molecules in the hydrate.

0.844 mol/ 0.407 mol = 2 `~H_2O` molecules 

So, the formula for the barium chloride hydrate is:

   `~BaCl_2` `*` `~2H_2O`