1. Unit VIII Results 4 th Period: 4 th Period: A 3 A 3 B 7 B 7 C 4 C 4 D 1 D 1 F 5 F 5 Range= 19-102 Range= 19-102 Mean= 75 Mean= 75 S x = 20.3 S x = 20.3 Median= 79 Median= 79

2. Scores not what you’d like them to be? Go check out General Chemistry Online! Your BIGGEST problem was timing! Your BIGGEST problem was timing! For 49 minutes, 22 problems, you should be spending 2 minutes per problem so that you have enough time to go back and review!!! For 49 minutes, 22 problems, you should be spending 2 minutes per problem so that you have enough time to go back and review!!! The only way to get faster is Kill and Drill. The only way to get faster is Kill and Drill. What does that mean? Work and rework problems until you can do 5 problems in 10 minutes three times in a row. What does that mean? Work and rework problems until you can do 5 problems in 10 minutes three times in a row. Once you run out of sample lecture, book, and study guide problems, go to the internet. Look for professors who have posted sample problems (with solutions!). Here’s a good starting place. Once you run out of sample lecture, book, and study guide problems, go to the internet. Look for professors who have posted sample problems (with solutions!). Here’s a good starting place. Homepage:http://antoine.frostburg.edu/chem/senese/101/index.shtml Homepage:http://antoine.frostburg.edu/chem/senese/101/index.shtmlhttp://antoine.frostburg.edu/chem/senese/101/index.shtml Sig Figs and Dimensional Analysis Tutorial: http://antoine.frostburg.edu/chem/senese/101/measurement/index.shtml Sig Figs and Dimensional Analysis Tutorial: http://antoine.frostburg.edu/chem/senese/101/measurement/index.shtml http://antoine.frostburg.edu/chem/senese/101/measurement/index.shtml Moles/Stoichiometry Unit: http://antoine.frostburg.edu/chem/senese/101/moles/index.shtml Moles/Stoichiometry Unit: http://antoine.frostburg.edu/chem/senese/101/moles/index.shtml http://antoine.frostburg.edu/chem/senese/101/moles/index.shtml

3. Stoichiometry

4. Objectives Solve problems based on mass-mass relationships in chemical reactions. Solve problems based on mass-mass relationships in chemical reactions. Calculate percentage yield in a chemical reaction Calculate percentage yield in a chemical reaction

5. Mass-Mass Relationships Recall that Molecular/Formula Mass is determined by the sum of the atomic masses of each element in a compound. Recall that Molecular/Formula Mass is determined by the sum of the atomic masses of each element in a compound. Recall that a balanced chemical equation tells you the molar ratio in which reactants and products interact. Recall that a balanced chemical equation tells you the molar ratio in which reactants and products interact. You can now use dimensional analysis to make mass-mass conversions!! You can now use dimensional analysis to make mass-mass conversions!!

6. EXAMPLE 1: Mass-Mass Conversion 2Na + 2H 2 O  2NaOH + H 2 2Na + 2H 2 O  2NaOH + H 2 How many grams of H 2 can be produced from the reaction of 11.5 grams of sodium with an excess of water? How many grams of H 2 can be produced from the reaction of 11.5 grams of sodium with an excess of water? ? g H 2 = 11.5 g Na * 1 mol Na * 1 mol H 2 * 2.016 g H 2 ? g H 2 = 11.5 g Na * 1 mol Na * 1 mol H 2 * 2.016 g H 2 22.99 g Na 2 mol Na 1 mol H 2 22.99 g Na 2 mol Na 1 mol H 2 Ans: 0.504 g Ans: 0.504 g

7. EXAMPLE 2: Mass-Mass Conversion N 2 + 3H 2  2NH 3 N 2 + 3H 2  2NH 3 Nitrogen reacts with 2.OO grams of hydrogen. How many grams of ammonia are produced? Nitrogen reacts with 2.OO grams of hydrogen. How many grams of ammonia are produced? ? g NH 3 = 2.00 g H 2 * 1 mol H 2 * 2 mol NH 3 * 17.03 g NH 3 2.016 g H 2 3 mol H 2 1 mol NH 3 2.016 g H 2 3 mol H 2 1 mol NH 3 Ans: 11.3 g Ans: 11.3 g

8. EXAMPLE 3: Mass-Mass Conversion O 2 + C  CO 2 O 2 + C  CO 2 How many grams of oxygen are required to burn 85.6 grams of carbon? How many grams of oxygen are required to burn 85.6 grams of carbon? ? g O 2 = 85.6 g C * 1 mol C * 1 mol O 2 * 32.00 g O 2 12.01 g C 1 mol C 1 mol O 2 12.01 g C 1 mol C 1 mol O 2 Ans: 228 g Ans: 228 g

9. Additional Examples: Mass-Mass Conversion 4. The action of carbon monoxide on iron(III) oxide (ferric oxide) can be represented by the equation, Fe2O3 (c) + 3CO(g) ----> 2Fe (c) + 3CO (g). What would be the amount of carbon monoxide used if 18.7 grams of iron were produced? Ans: 14.1g. 4. The action of carbon monoxide on iron(III) oxide (ferric oxide) can be represented by the equation, Fe2O3 (c) + 3CO(g) ----> 2Fe (c) + 3CO (g). What would be the amount of carbon monoxide used if 18.7 grams of iron were produced? Ans: 14.1g. 5. How many grams of hydrochloric acid are required to react with 75.1 grams of calcium hydroxide? Remember the rules for parentheses for calcium hydroxide. Ans: 74.6g. 5. How many grams of hydrochloric acid are required to react with 75.1 grams of calcium hydroxide? Remember the rules for parentheses for calcium hydroxide. Ans: 74.6g. 6. How many grams of hydrogen gas are produced when 5.62 grams of aluminum react with hydrochloric acid? Hint: hydrochloric acid is hydrogen chloride, hydrogen gas is diatomic. Ans: 0.631g. 6. How many grams of hydrogen gas are produced when 5.62 grams of aluminum react with hydrochloric acid? Hint: hydrochloric acid is hydrogen chloride, hydrogen gas is diatomic. Ans: 0.631g.

10. Percent Yield The problems we’ve been working assume that a reaction goes to completion. The problems we’ve been working assume that a reaction goes to completion. i.e. Whatever you start out with, you completely use up in the reaction. i.e. Whatever you start out with, you completely use up in the reaction. In reality, that doesn’t happen. Why not? In reality, that doesn’t happen. Why not? Some reactions go “backwards” Some reactions go “backwards” Some reactions form unintended side reactions Some reactions form unintended side reactions Impurities in reactants Impurities in reactants Mechanical losses (i.e. experimenter error, like dropping a crucible and losing your sample!) Mechanical losses (i.e. experimenter error, like dropping a crucible and losing your sample!) This means that there is a theoretical yield (what we’ve been calculating) and an actual yield (what you actual get when performing the reaction) This means that there is a theoretical yield (what we’ve been calculating) and an actual yield (what you actual get when performing the reaction)

11. Calculating Percent Yield Actual Yield/Theoretical Yield * 100 Actual Yield/Theoretical Yield * 100 Example 1: Example 1: 2Na + 2H 2 O  2NaOH + H 2 2Na + 2H 2 O  2NaOH + H 2 How many grams of H 2 can theoretically be produced from the reaction of 11.5 grams of sodium with an excess of water? How many grams of H 2 can theoretically be produced from the reaction of 11.5 grams of sodium with an excess of water? ? g H 2 = 11.5 g Na * 1 mol Na * 1 mol H 2 * 2.016 g H 2 ? g H 2 = 11.5 g Na * 1 mol Na * 1 mol H 2 * 2.016 g H 2 22.99 g Na 2 mol Na 1 mol H 2 22.99 g Na 2 mol Na 1 mol H 2 Ans: 0.504 g theoretical yield Ans: 0.504 g theoretical yield If Matt actually obtains 0.475 g H 2 when running this reaction, what is his percent yield? If Matt actually obtains 0.475 g H 2 when running this reaction, what is his percent yield? 0.475/0.504*100 = 94% 0.475/0.504*100 = 94%