1. Hydrate Calculations

2. Example 1: Calculate the percentage of water in Na 2 S 2 O 3. 5H 2 O % H2O = x 100 = x 100 = 36.30 % Mass of water Mass of hydrate 5(18.02 g/mol) 248.22 g/mol

3. 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

4. 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

5. Example 2: 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

6. 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

7. 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