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CHEM 101 LECTURE NOTES Fall 2001 Tuesday, September 11, 2001 Chapter 3 Dr. Joy Heising S 572-580 First exam coming up 1 week from this Thursday will cover Chapters 1-3 and at least part of Chapter 4. - read textbook - attend lecture - do homework/practice problems - ask questions in recitation DO NOT procrastinate! help session tomorrow night o Wed. Sept 12, HELD 100, 7-9pm review session o Wed. Sept 19, HELD 200, 7-9pm come with textbook, calculators, and questions! 1 CHEM 101 LECTURE NOTES Fall 2001 Tuesday, September 11, 2001 Chapter 3 Dr. Joy Heising S 572-580 Calculations involving balanced equations. Consider the following cooking analogy: chicken breast + veggies + oil + chicken stir fry rice à we can use the recipe to write a balanced equation: 4 chicken 8 3 4 8 chicken breasts + cups veggies + tbsp oil + cups rice à servings stir fry But what if we only have 5 dinner guests and we want to cut back on the amount of food we make? How much chicken do we need? How many cups of vegetables? We set up a ratio that relates the ingredients to one another. 5 servings x 4 chicken breasts 8 servings = 2.5 chicken breasts 5 servings x 8 cups veggies 8 servings = 5 cups veggies 2 CHEM 101 LECTURE NOTES Fall 2001 Tuesday, September 11, 2001 Chapter 3 Dr. Joy Heising S 572-580 We use balanced equations to set up mole ratios to do the same kind of calculations. Ex. if we want to make 0.750 moles of AgCl, how much CaCl2 do we need? How much AgNO3? CaCl2 + 2AgNO3 à Ca(NO3)2 + 2AgCl 0.750 mol AgCl x 1 mol CaCl2 2 mol AgCl = 0.750 mol AgCl x 2 mol AgNO3 = 2 mol AgCl 0.375 mol CaCl2 0.750 mol AgNO3 How much Ca(NO3)2 will we also make? 0.750 mol AgCl x 1 mol Ca(NO3)2 = 0.375 mol AgNO3 2 mol AgCl This example involves moles. But what about grams? 3 CHEM 101 LECTURE NOTES Fall 2001 Tuesday, September 11, 2001 Chapter 3 CaCl2 + 2AgNO3 à Dr. Joy Heising S 572-580 Ca(NO3)2 + 2AgCl 10.0 g CaCl2 yields how many grams of AgCl? Strategy: see next slide Molar mass CaCl2 = Molar mass AgCl = 40.08 + 2(35.45) = 110.98 g/mol 107.9 + 35.45 = 143.4 g/mol 10.0g CaCl2 x 1 mol CaCl2 x 2 mol AgCl x 143.4g AgCl = 25.8 g AgCl 110.0g CaCl2 1 mol CaCl2 1 mol AgCl 4 CHEM 101 LECTURE NOTES Fall 2001 Tuesday, September 11, 2001 Chapter 3 Dr. Joy Heising S 572-580 Problem solving strategy: 1. what do we want to know? # g AgCl 2. what do we know? #g CaCl2 3. how can we get from #1 to #2? g CaCl2 g AgCl ? How can you get AgCl from CaCl2? Use the balanced equation to create a mole ratio 10.0 g CaCl2 mol CaCl2 à mol AgCl ?g AgCl ¦ how can I do this kind of problem when the quantities that are given and/or requested are in grams? Convert to moles! Calculate the molar masses of AgCl and CaCl2 Why? to relate grams to moles 10.0 g CaCl2 à mol CaCl2 mol AgCl à ?g AgCl 5 CHEM 101 LECTURE NOTES Fall 2001 Tuesday, September 11, 2001 Chapter 3 Dr. Joy Heising S 572-580 In these calculations, we have been assuming that we have plenty of the ‘other reactants’ - plenty of AgNO3for CaCl2 but what if I have limited quantities of both reactants? Which quantity determines the amount of product? Limiting reactant – a substance that limits the amount of product(s) that can be formed. 2 oz. Cherry syrup + 1 can coke = 1 cherry coke 8 oz. Cherry syrup 6-pack coke how many cherry cokes? 8 oz. Syrup 6 cokes x x 1 cherry coke = 2 oz. cherry syrup 1 cherry coke 1 coke = 4 cherry cokes 6 cherry cokes only have enough to make 4 cherry cokes – syrup is limiting reactant. 6 CHEM 101 LECTURE NOTES Fall 2001 Tuesday, September 11, 2001 Chapter 3 Dr. Joy Heising S 572-580 What is the maximum mass of sulfur dioxide which can be obtained from the reaction of 95.6 g of carbon disulfide with 110. g of oxygen? Strategy: see next slide Sulfur dioxide Carbon disulfide Oxygen SO2 CS2 O2 molar mass molar mass molar mass = = = 64.07 g/mol 76.14 g/mol 32.00 g/mol Write equation: CS2 + O2 à SO2 CS2 + O2 à SO2 + CO2 CS2 + 3O2 à 2SO2 + CO2 à mol à carb. disulf. a. 95.6 g carbon disulfide need one more product – CO2 next balance mol sulf. diox. à ?g sulfur dioxide 95.6 g CS2 x 1 mol CS2 x 2 mol SO2 x 64.07g SO2 = 161 g SO2 76.14 g CS2 1 mol CS2 1 mol SO2 b. 110. g oxygen 110. g O2 c. à mol oxygen à mol sulf. diox. à ?g sulfur dioxide x 1 mol O2 x 2 mol SO2 x 64.07g SO2 = 147 g SO2 32.00g O2 3 mol O2 1 mol SO2 147g is the maximum mass (there isn’t enough O2 to make 161 g SO2 ). 7 CHEM 101 LECTURE NOTES Fall 2001 Tuesday, September 11, 2001 Chapter 3 Dr. Joy Heising S 572-580 Problem solving strategy: 1. don’t panic. 2. what do we want to know? ‘maximum mass’ sulfur dioxide 3. what do we know? #g carbon disulfide, #g oxygen 4. how can we get from #1 to #2? a. 95.6 g………………………………………………………? g carbon disulfide sulfur dioxide b. 110. g……………………………………………………….? g oxygen sulfur dioxide c. compare results of a and b and choose the smaller number. Writing a master plan for steps a and b, then: a. 95.6 g à mol à mol à carbon disulfide carb. disulf. sulf. diox. b. 110. g oxygen à mol à mol à oxygen sulf. diox. ?g sulfur dioxide ?g sulfur dioxide ? How can I get to sulfur dioxide from carbon disulfide? From oxygen? How did I do it before? Used the balanced equation to create a mole ratio ? How am I going to get a balanced equation? 1. write the formulas for the names 2. figure out on which ‘side’ of the equation each compound belongs 3. make sure it obeys the law of conservation of matter (balance it). 8 CHEM 101 LECTURE NOTES Fall 2001 Tuesday, September 11, 2001 Chapter 3 Dr. Joy Heising S 572-580 But what if we had only gotten 100. g of SO 2 with those amounts of CS2 and O2? - reaction did not go to completion - several competing reactions involving CS2 and/or O2 were going on at the same time. Theoretical yield – the amount of product we should get when a reactant is completely consumed according to our chemical equation Actual yield – the amount of product actually measured The actual yield is always equal to or lower than the theoretical yield Percent yield actual yield theoretical yield x 100 the percent yield of SO 2 is 100. g SO2 146.8 g SO2 x 100 = 68.1 % 9 CHEM 101 LECTURE NOTES Fall 2001 Tuesday, September 11, 2001 Chapter 3 Dr. Joy Heising S 572-580 We have done a lot of calculations between moles and grams. However, a lot of reactions happen in water. Solution – a homogeneous mixture of two or more substances (ex. NaCl in H2O) Solute – a substance that is dissolved in another substance (NaCl) Solvent – the substance in which the other substance is dissolved (H2O, liquid) Aqueous solution – when solvent is H2O For a given quantity of a solution, how can we know: - amount of solute? - amount of solvent? Concentration – the amount of solute per solution 10 CHEM 101 LECTURE NOTES Fall 2001 Tuesday, September 11, 2001 Chapter 3 Dr. Joy Heising S 572-580 Percent by mass: Percent solute = mass of solute x 100 mass of solution ex. if we have 20.0 g NaOH dissolved in H2O to make 250. g of solution, what is the percent by mass of NaOH? 20.0 g NaOH = 250 g solution 8.00 % NaOH (230 g H2O) 11 CHEM 101 LECTURE NOTES Fall 2001 Tuesday, September 11, 2001 Chapter 3 Dr. Joy Heising S 572-580 TEAM QUIZ #2 5.00 g CH3OH reacts with 8.00 g O2. What is the limiting reactant? If only 5.00 g of H2O is the observed yield, what is the percent yield? 2CH3OH + 3O2 à 2CO2 + 4H2O molar mass CH3OH = 32.04 g/mol molar mass H2O = 18.02 g/mol 12