1. Hydrate Calculations

2. Hydrates Hydrates are ionic compounds that have crystalline structures involving H 2 O In MgSO 4. 7H 2 O, there are 7 water molecules present for every 1 MgSO 4. Scientists do not always know how many H 2 O molecules are present, but this can be determined through calculations

4. Anhydrous Compounds Anhydrous Compounds do not have water molecules incorporated Instead they have the ability to absorb water from the local environment.

5. Applications Hydrates can be used in Fire Retardation Anhydrates can be used as dessicants (ie. Keep the atmosphere dry) Anhydrates can indicate the presence of water.

6. Example 1: Calculate the percentage of water in Na 2 S 2 O 3. 5H 2 O % H 2 O = x 100 = x 100 = 36.30 % Mass of water Mass of hydrate 5(18.02 g/mol) 248.22 g/mol

7. Example 2: Calculate the mass of water in 215 g of Na 2 S 2 O 3. 5H 2 O Solution 1: Use the percentage 36.30% (from the previous example). In Na 2 S 2 O 3. 5H 2 O, there is 36.30% of water. In 215 g, there is x 215 g = 78.0 g of water 36.30 100

8. Solution 2: = m( H 2 O) = 78.0 g M(H 2 O) M (Total) m(H 2 O) M(Total) 5(18.02 g/mol) 248.22 g/mol m(H 2 O) 215 g

9. Example 3: A hydrate of barium chloride (BaCl 2. x H 2 O) has a mass of 1.500g. When heated to drive off the water, the residue (BaCl 2 ) has a mass of 1.279 g Step 1: Calculate the percentage of water in the hydrate. BaCl 2. x H 2 O (s)  BaCl 2 (s) + x H 2 O 1.500 g1.279g 1.500g – 1.279g = 0.221 g % H 2 O = x 100 = x 100 = 14.7% Mass of water Mass of hydrate 0.221 g 1.500 g

10. Step 2: Calculate the formula of the hydrate Mole Calculation: Using n=m/M BaCl 2 H 2 O n =n= = 0.006140 mol = 0.0123 mol 1.279g 208.3 g/mol 0.221 g 18.02 g/mol

11. Step 3: Ratio Calculation: Divide by the smallest number of moles to figure out the ratio between the atoms BaCl 2 H 2 O= =1= 2 Therefore the formula is BaCl 2. 2H 2 O 0.006140 mol 0.0123 mol 0.006140 mol

12. Example 4: A hydrate of barium hydroxide (Ba(OH) 2. x H 2 O) has a mass of 50.0g. When heated to drive off the water, the residue (Ba(OH) 2 ) has a mass of 27.2 g Step 1: Calculate the percentage of water in the hydrate. Ba(OH) 2. x H 2 O (s)  Ba(OH) 2 (s) + x H 2 O 50.0 g 27.2g 50.0g – 27.2g = 22.8 g % H 2 O = x 100 = x 100 = 45.6% Mass of water Mass of hydrate 22.8 g 50.0 g

13. Step 2: Calculate the formula of the hydrate Mole Calculation: Using n=m/M Ba(OH) 2 H 2 O n =n= = 0.159 mol = 1.27 mol 27.2g 171.3 g/mol 22.8 g 18.02 g/mol

14. Step 3: Ratio Calculation: Divide by the smallest number of moles to figure out the ratio between the atoms Ba(OH) 2 H 2 O= =1= 8 Therefore the formula is Ba(OH) 2. 8H 2 O 0.159 mol 1.27 mol 0.159 mol