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Chapter 9 Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena Calculations from.

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Presentation on theme: "Chapter 9 Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena Calculations from."— Presentation transcript:

1 Chapter 9 Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena Calculations from Chemical Equations Accurate measurement and calculation of the correct dosage are important in dispensing the correct medicine to patients throughout the world.

2 Chapter Outline Copyright 2012 John Wiley & Sons, Inc A Short ReviewA Short Review 9.2 Introduction to StoichiometryIntroduction to Stoichiometry 9.3 Mole-Mole CalculationsMole-Mole Calculations 9.4 Mole-Mass CalculationsMole-Mass Calculations 9.5 Mass-Mass CalculationsMass-Mass Calculations 9.6 Limiting Reactant and Yield CalculationsLimiting Reactant and Yield Calculations

3 Molar Mass Molar Mass – sum of atomic masses of all atoms in 1 mole of an element or compound ; the units are g/mol x10 23 molecules 6.022x10 23 formula units 6.022x10 23 atoms 6.022x10 23 ions Copyright 2012 John Wiley & Sons, Inc9-3 1 mole =

4 Molar Mass What is the molar mass of Al(ClO 3 ) 3 ? Al1(26.98 g) 3Cl3(35.45 g) 9O9(16.00 g) Al(ClO 3 ) g/mol Copyright 2012 John Wiley & Sons, Inc7-4 atomic mass Al26.98 Cl35.45 O16.00

5 Molar Mass Calculate the mass of 2.5 moles of aluminum chlorate. Copyright 2012 John Wiley & Sons, Inc9-5 Plan 2.5 mol Al(ClO 3 ) 3  g Al(ClO 3 ) 3 Calculate 1 mol Al(ClO 3 ) 3 = g Al(ClO 3 ) 3

6 Calculate the moles of 3.52g of aluminum chlorate. Molar Mass Copyright 2012 John Wiley & Sons, Inc9-6 Plan 3.52 g Al(ClO 3 ) 3  mol Al(ClO 3 ) 3 Calculate 1 mol Al(ClO 3 ) 3 = g Al(ClO 3 ) 3

7 Molar Mass Calculate the number of formula units contained in 12.4 g aluminum chlorate. Copyright 2012 John Wiley & Sons, Inc9-7 Plan 12.4 g Al(ClO 3 ) 3  formula units Al(ClO 3 ) 3 1 mol Al(ClO 3 ) 3 = g = 6.022x10 23 formula units Calculate

8 Your Turn! What is the mass of 3.61 moles of CaCl 2 ? a.3.61 g b.272 g c.2.17 × g d.401 g Copyright 2012 John Wiley & Sons, Inc9-8 atomic mass Ca40.08 Cl35.45

9 You Turn! How many moles of HCl are contained in 18.2 g HCl? a.1.00 mol b mol c mol d mol Copyright 2012 John Wiley & Sons, Inc9-9 atomic mass H1.01 Cl35.45

10 Your Turn! What is the mass of 1.60×10 23 molecules of HCl? a.9.69 g b.137 g c g d.36.5 g Copyright 2012 John Wiley & Sons, Inc9-10 atomic mass H1.01 Cl35.45

11 Stoichiometry Stoichiometry deals with the quantitative relationships between the reactants and products in a balanced chemical equation. 1N 2(g) + 3I 2(s)  2NI 3(s) 1 mol N mol I 2  2 mol NI 3 Mole ratios come from the coefficients in the balanced equation: The 3 other possibilities are the inverse of these ratios. Copyright 2012 John Wiley & Sons, Inc9-11

12 Your Turn! Which of these statements is not true about the reaction? 1N 2(g) + 3I 2(s)  2NI 3(s) a.1 mole of nitrogen is needed for every 3 moles of iodine b.1 gram of nitrogen is needed for every 3 grams of iodine c.Both statements are true Copyright 2012 John Wiley & Sons, Inc9-12

13 Using the mole ratio Calculate the number of moles of NI 3 that can be made from 5.50 mol N 2 in the reaction: 1N 2(g) + 3I 2(s)  2NI 3(s) Copyright 2012 John Wiley & Sons, Inc9-13 Plan Calculate 5.50 mol N 2  mol NI 3 Set-Up

14 Using the mole ratio Calculate the number of moles of I 2 needed to react with 5.50 mol N 2 in the reaction: 1N 2(g) + 3I 2(s)  2NI 3(s) Copyright 2012 John Wiley & Sons, Inc9-14 Plan Calculate 5.50 mol N 2  mol I 2 Set-Up

15 Your Turn! How many moles of HF will be produced by the complete reaction of 1.42 moles of H 2 in the following equation? H 2 + F 2  2HF a b.1.42 c.2.00 d.2.84 Copyright 2012 John Wiley & Sons, Inc9-15

16 Stoichiometry Problem Solving Strategy for stoichiometry problems: 1.Convert starting substance to moles. 2.Convert the moles of starting substance to moles of desired substance. 3.Convert the moles of desired substance to the units specified in the problem. Copyright 2012 John Wiley & Sons, Inc9-16

17 Stoichiometry Copyright 2012 John Wiley & Sons, Inc9-17

18 Mole-Mole Calculations How many moles of Al are needed to make mol of H 2 ? 2Al (s) + 6HCl (aq)  2AlCl 3(aq) + 3H 2(g) Copyright 2012 John Wiley & Sons, Inc9-18 Plan Calculate mol H 2  mol Al Set-Up

19 Mole-Mole Calculations How many moles of HCl are needed to make mol of H 2 ? 2Al (s) + 6HCl (aq)  2AlCl 3(aq) + 3H 2(g) Copyright 2012 John Wiley & Sons, Inc9-19 Plan Calculate mol H 2  mol HCl Set-Up

20 Your Turn! How many moles of H 2 are made by the reaction of 1.5 mol HCl with excess aluminum? 2Al (s) + 6HCl (aq)  2AlCl 3(aq) + 3H 2(g) a.0.75 mol b.3.0 mol c.6.0 mol d.4.5 mol Copyright 2012 John Wiley & Sons, Inc9-20

21 Your Turn! How many moles of carbon dioxide are produced when 3.00 moles of oxygen react completely in the following equation? C 3 H 8 + 5O 2  3CO 2 + 9H 2 O a.5.00 mol b.3.00 mol c.1.80 mol d.1.50 mol Copyright 2012 John Wiley & Sons, Inc9-21

22 Your Turn! How many moles of C 3 H 8 are consumed when 1.81x10 23 molecules of CO 2 are produced in the following equation? C 3 H 8 + 5O 2  3CO 2 + 4H 2 O a b c.6.03 × d.5.43 × Copyright 2012 John Wiley & Sons, Inc9-22

23 Mole-Mass Calculations What mass of H 2 (2.02 g/mol) is made by the reaction of 3.0 mol HCl with excess aluminum? 2Al (s) + 6HCl (aq)  2AlCl 3(aq) + 3H 2(g) Copyright 2012 John Wiley & Sons, Inc9-23 Plan Calculate 3.0 mol HCl  mol H 2  g H 2

24 Mole-Mass Calculations How many moles of HCl are needed to completely consume 2.00 g Al (26.98g/mol)? 2Al (s) + 6HCl (aq)  2AlCl 3(aq) + 3H 2(g) Copyright 2012 John Wiley & Sons, Inc9-24 Plan Calculate 2.00 g Al  mol Al  mol HCl

25 Mole-Mass Calculations What mass of Al(NO 3 ) 3 (213g/mol) is needed to react with.093 mol Na 2 CO 3 ? 3Na 2 CO 3(aq) + 2Al(NO 3 ) 3(aq)  Al 2 (CO 3 ) 3(s) + 6NaNO 3(aq) Copyright 2012 John Wiley & Sons, Inc9-25 Plan Calculate mol Na 2 CO 3  mol Al(NO 3 ) 3  g Al(NO 3 ) 3

26 How many moles of Al 2 (CO 3 ) 3 are made by the reaction of 3.45g Na 2 CO 3 ( g/mol) with excess Al(NO 3 ) 3 ? 3Na 2 CO 3(aq) + 2Al(NO 3 ) 3(aq)  Al 2 (CO 3 ) 3(s) + 6NaNO 3(aq) Copyright 2012 John Wiley & Sons, Inc9-26 Plan Calculate 3.45g Na 2 CO 3  mol Na 2 CO 3  g Al 2 (CO 3 ) 3 Mole-Mass Calculations

27 Your Turn! How many moles of oxygen are consumed when 38.0g of aluminum oxide are produced in the following equation? 4Al (s) + 3O 2(g)  2Al 2 O 3(s) a b c.1.50 d.3.00 Copyright 2012 John Wiley & Sons, Inc9-27 atomic mass Al26.98 O16.00

28 Your Turn! What mass of HCl is produced when 1.81x10 24 molecules of H 2 react completely in the following equation? H 2(g) + Cl 2(g)  2HCl (g) a.54.7g b.72.9g c.109g d.219g Copyright 2012 John Wiley & Sons, Inc9-28 atomic mass H1.01 Cl35.45

29 Mass-Mass Calculations Now we will put it all together. What mass of Br 2 ( g/mol) is needed to completely consume 7.00 g Al (26.98 g/mol)? 2Al (s) + 3Br 2(l)  2AlBr 3(s) Copyright 2012 John Wiley & Sons, Inc g Al  mol Al  mol Br 2  g Br 2

30 Mass-Mass Calculations What mass of Br 2 ( g/mol) is needed to completely consume 7.00 g Al (26.98 g/mol)? 2Al (s) + 3Br 2(l)  2AlBr 3(s) Copyright 2012 John Wiley & Sons, Inc9-30 Plan Calculate 7.00 g Al  mol Al  mol Br 2  g Br 2

31 Mass-Mass Calculations What mass of Fe 2 S 3 (207.91g/mol) can be made from the reaction of 9.34 g FeCl 3 ( g/mol) with excess Na 2 S? 2FeCl 3(aq) + 3Na 2 S (aq)  Fe 2 S 3(s) + 6NaCl (aq) Copyright 2012 John Wiley & Sons, Inc9-31 Plan Calculate 9.34 g FeCl 3  mol FeCl 3  mol Fe 2 S 3  g Fe 2 S 3

32 Your Turn! What mass of oxygen is consumed when 54.0g of water is produced in the following equation? 2H 2 + O 2  2H 2 O a g b g c.1.50 g d.47.9 g Copyright 2012 John Wiley & Sons, Inc9-32 atomic mass H1.01 O16.00

33 Your Turn! What mass of H 2 O is produced when 12.0g of HCl react completely in the following equation? 6HCl + Fe 2 O 3  2FeCl 3 + 3H 2 O a.2.97 g b.39.4 g c.27.4 g d.110. g Copyright 2012 John Wiley & Sons, Inc9-33 atomic mass H1.01 O16.00 Cl35.45

34 Determine the number of that can be made given these quantities of reactants and the reaction equation: Limiting Reactant Copyright 2012 John Wiley & Sons, Inc  + 

35 Limiting Reactant The limiting reactant is the reactant that limits the amount of product that can be made. The reaction stops when the limiting reactant is used up. What was the limiting reactant in the reaction: The small blue balls. Copyright 2012 John Wiley & Sons, Inc 

36 Excess Reactant The excess reactant is the reactant that remains when the reaction stops. There is always left over excess reactant. What was the excess reactant in the reaction: The excess reactant was the larger blue ball. Copyright 2012 John Wiley & Sons, Inc 

37 Limiting reactant Copyright 2012 John Wiley & Sons, Inc9-37 Figure 9.2 The number of bicycles that can be built from these parts is determined by the “limiting reactant” (the pedal assemblies).

38 Limiting Reactant Calculations Technique for solving limiting reactant problems: 1.Convert reactant 1 to moles or mass of product 2.Convert reactant 2 to moles or mass of product 3.Compare answers. The smaller answer is the maximum theoretical yield. Copyright 2012 John Wiley & Sons, Inc9-38

39 Calculate the number of moles of water that can be made by the reaction of 1.51 mol H 2 with mol O 2. 2H 2(g) + O 2(g)  2H 2 O (g) 1.Calculate the theoretical yield of H 2 O assuming H 2 is the limiting reactant and that O 2 is the excess reactant. 2.Calculate the theoretical yield of H 2 O assuming that O 2 is the limiting reactant and that H 2 is the excess reactant. Copyright 2012 John Wiley & Sons, Inc9-39 Limiting Reactant Calculation

40 Assuming that H 2 is limiting and O 2 is excess: So what is the maximum yield of H 2 O? Assuming that O 2 is limiting and H 2 is excess: Copyright 2012 John Wiley & Sons, Inc9-40 Limiting Reactant Calculation continued

41 How much H 2 and O 2 remain when the reaction stops? H 2 : Limiting Reactant – None remains. It was used up in the reaction. O 2 : Excess Reactant – Calculate the amount of O 2 used in the reaction with H 2. Then subtract that from the original amount. Copyright 2012 John Wiley & Sons, Inc9-41

42 Calculate the mass of copper that can be made from the combination of 15.0 g aluminum with 25.0 g copper(II) sulfate. 2Al (s) + 3CuSO 4(aq)  Al 2 (SO 4 ) 3(aq) + 3Cu (s) Copyright 2012 John Wiley & Sons, Inc9-42 Limiting Reactant Calculation Plan 15 g Al  mol Al  mol Cu  g Cu 25 g CuSO 4  mol CuSO 4  mol Cu  g Cu Compare answers. The smaller number is the right answer.

43 2Al (s) + 3CuSO 4(aq)  Al 2 (SO 4 ) 3(aq) + 3Cu (s) 1. Assume Al is limiting and CuSO 4 is in excess. 2. Assume CuSO 4 is limiting and Al is in excess g Cu 9.96 g Cu 3. Compare answers. CuSO 4 is the limiting reagent. The theoretical yield of Cu is 9.96 g. Copyright 2012 John Wiley & Sons, Inc9-43 Limiting Reactant Calculation continued

44 Your Turn! True/False: You can compare the quantities of reactant when you work a limiting reactant problem. The reactant you have the least of is the limiting reactant. a.True b.False Copyright 2012 John Wiley & Sons, Inc9-44

45 Your Turn! Which is the limiting reactant when 3.00 moles of copper are reacted with 3.00 moles of silver nitrate in the following equation? Cu + 2AgNO 3  Cu(NO 3 ) 2 + 2Ag a.Cu b.AgNO 3 c.Cu(NO 3 ) 2 d.Ag Copyright 2012 John Wiley & Sons, Inc9-45

46 Your Turn! What is the mass of silver ( g/mol) produced by the reaction of 3.00 moles of copper with 3.00 moles of silver nitrate? Cu + 2AgNO 3  Cu(NO 3 ) 2 + 2Ag a.162g b.216g c.324g d.647g Copyright 2012 John Wiley & Sons, Inc9-46

47 Percent Yield To determine the efficiency of a process for making a compound, chemists compute the percent yield of the reaction. The theoretical yield is the result calculated using stoichiometry. The actual yield of a chemical reaction is the experimental result, which is often less than the theoretical yield due to experimental losses and errors along the way. Copyright 2012 John Wiley & Sons, Inc9-47

48 Percent Yield Calculate the % yield of PCl 3 that results from reacting 5.00 g P with excess Cl 2 if only 17.2 g of PCl 3 were recovered. 2P + 3Cl 2  2PCl 3 Compute the expected yield of PCl 3 from 5.00 g P with excess Cl g PCl 3 Compute the % Yield. Copyright 2012 John Wiley & Sons, Inc9-48

49 Your Turn! In a reaction to produce ammonia, the theoretical yield is 420. g. What is the percent yield if the actual yield is 350. g? A. 83.3% B. 20.0% C. 16.7% D. 120.% Copyright 2012 John Wiley & Sons, Inc9-49


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