1. April 7, 2014 Today: Stoichiometry and % Yield

2. Percent Yield Remember, stoichiometry is used to tell you how much product you can form from X amount of product without doing the reaction Percent yield tells you how much product you actually got in the lab compared to how much you could have got

3. Theoretical yield The maximum amount of product that can be formed from a given amount of reactant. – This is a value you calculate on paper In other words: (Write your own definition of Theoretical yield here)

4. Actual yield The measured amount of a product obtained from a reaction – This is a measurement of a product formed in an actual chemical reaction In other words: (Write your own definition of Actual yield here)

5. Percent Yield Formula Actual yield x 100 = Theoretical yield

6. Example 1 Mg(s) + 2H 2 O(g)  Mg(OH) 2 (s) + H 2 (g) A.If 16.2 g Mg are heated with excess H 2 O how many grams of hydrogen gas could theoretically be formed? Known: mass Mg Unknown: mass H 2 Plan: g Mg  mol Mg  mol H 2  g H 2 Relationships: molar mass Mg = 24.31 g/mol, molar mass H 2 = 2.02 g/mol, mole ratio = 1 mol Mg : 1 mol H 2 16.2 g Mg | 1 mol Mg | 1 mol H 2 | 2.02 g H 2 = 16.2 x 2.02 = 1.35 g 24.31 g 1 mol Mg 1 mol H 2 24.31

7. Example 1 B. If only 0.905 g H 2 are actually formed, what is the percent yield of this reaction? Actual yield x 100 = % yield Theoretical yield 0.905 g H 2 x 100 = 67.0% 1.35 g H 2 (This means that you produced or collected only 67% of the product that it was possible to form with the amount of reactant you started with.)

8. Example 2 CO(g) + 2H 2 (g)  CH 3 OH(l) If 11.0 g H 2 reacts with CO to produce 68.4 g CH 3 OH, what is the percentage yield of CH 3 OH? We need to determine the theoretical yield. Known: mass H 2 Unknown: mass CH 3 OH Plan: g H 2  mol Mg  mol H 2  g H 2 Relationships: molar mass H 2 = 2.02 g/mol molar mass CH 3 OH = 32.04 g/mol mole ratio = 2 mol H 2 : 1 mol CH 3 OH

9. Example 2 CO(g) + 2H 2 (g)  CH 3 OH(l) If 11.0 g H 2 reacts with CO to produce 68.4 g CH 3 OH, what is the percentage yield of CH 3 OH? Known: mass H 2 Unknown: mass CH 3 OH Plan: g H 2  mol Mg  mol H 2  g H 2 Relationships: molar mass H 2 = 2.02 g/mol, molar mass CH 3 OH = 32.04 g/mol, mole ratio = 2 mol H 2 : 1 mol CH 3 OH 11.0 g H 2 | 1 mol H 2 | 1 mol CH 3 OH | 32.04 g CH 3 OH = 11.0 x 32.04 = 87.2 2.02 H 2 2 mol H 2 1 mol CH 3 OH 2.02 x 2 g

10. Example 2 CO(g) + 2H 2 (g)  CH 3 OH(l) If 11.0 g H 2 reacts with CO to produce 68.4 g CH 3 OH, what is the percentage yield of CH 3 OH? Actual yield x 100 = % yield Theoretical yield 68.4 g CH 3 OH x 100 = 78.4% 87.2 g CH 3 OH