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

Copyright 2012 John Wiley & Sons, Inc Chapter Outline 9.1 A Short Review 9.2 Introduction to Stoichiometry 9.3 Mole-Mole Calculations 9.4 Mole-Mass Calculations 9.5 Mass-Mass Calculations 9.6 Limiting Reactant and Yield Calculations Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Molar Mass Molar Mass – sum of atomic masses of all atoms in 1 mole of an element or compound ; the units are g/mol. 6.022x1023 molecules 6.022x1023 formula units 6.022x1023 atoms 6.022x1023 ions 1 mole = Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Molar Mass What is the molar mass of Al(ClO3)3? Al 1(26.98 g) 3Cl 3(35.45 g) 9O 9(16.00 g) Al(ClO3)3 277.33 g/mol atomic mass Al 26.98 Cl 35.45 O 16.00 Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Molar Mass Calculate the mass of 2.5 moles of aluminum chlorate. Plan 2.5 mol Al(ClO3)3  g Al(ClO3)3 1 mol Al(ClO3)3 = 277.33 g Al(ClO3)3 Calculate Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Molar Mass Calculate the moles of 3.52g of aluminum chlorate. Plan 3.52 g Al(ClO3)3  mol Al(ClO3)3 1 mol Al(ClO3)3 = 277.33 g Al(ClO3)3 Calculate Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Molar Mass Calculate the number of formula units contained in 12.4 g aluminum chlorate. Plan 12.4 g Al(ClO3)3  formula units Al(ClO3)3 1 mol Al(ClO3)3 = 277.33 g = 6.022x1023 formula units Calculate Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Your Turn! What is the mass of 3.61 moles of CaCl2? 3.61 g 272 g 2.17 × 1024 g 401 g atomic mass Ca 40.08 Cl 35.45 Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc You Turn! How many moles of HCl are contained in 18.2 g HCl? 1.00 mol 0.500 mol 0.250 mol 0.125 mol atomic mass H 1.01 Cl 35.45 Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Your Turn! What is the mass of 1.60×1023 molecules of HCl? 9.69 g 137 g 0.729 g 36.5 g atomic mass H 1.01 Cl 35.45 Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Stoichiometry Stoichiometry deals with the quantitative relationships between the reactants and products in a balanced chemical equation. 1N2(g) + 3I2(s)  2NI3(s) 1 mol N2 + 3 mol I2  2 mol NI3 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, Inc

Copyright 2012 John Wiley & Sons, Inc Your Turn! Which of these statements is not true about the reaction? 1N2(g) + 3I2(s)  2NI3(s) 1 mole of nitrogen is needed for every 3 moles of iodine 1 gram of nitrogen is needed for every 3 grams of iodine Both statements are true Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Using the mole ratio Calculate the number of moles of NI3 that can be made from 5.50 mol N2 in the reaction: 1N2(g) + 3I2(s)  2NI3(s) Plan 5.50 mol N2  mol NI3 Set-Up Calculate Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Using the mole ratio Calculate the number of moles of I2 needed to react with 5.50 mol N2 in the reaction: 1N2(g) + 3I2(s)  2NI3(s) Plan 5.50 mol N2  mol I2 Set-Up Calculate Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Your Turn! How many moles of HF will be produced by the complete reaction of 1.42 moles of H2 in the following equation? H2 + F2  2HF 0.710 1.42 2.00 2.84 Copyright 2012 John Wiley & Sons, Inc

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

Copyright 2012 John Wiley & Sons, Inc Stoichiometry Copyright 2012 John Wiley & Sons, Inc

Mole-Mole Calculations How many moles of Al are needed to make 0.0935 mol of H2? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g) Plan 0.0935 mol H2  mol Al Set-Up Calculate Copyright 2012 John Wiley & Sons, Inc

Mole-Mole Calculations How many moles of HCl are needed to make 0.0935 mol of H2? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g) Plan 0.0935 mol H2  mol HCl Set-Up Calculate Copyright 2012 John Wiley & Sons, Inc

Your Turn! How many moles of H2 are made by the reaction of 1.5 mol HCl with excess aluminum? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g) 0.75 mol 3.0 mol 6.0 mol 4.5 mol Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Your Turn! How many moles of carbon dioxide are produced when 3.00 moles of oxygen react completely in the following equation? C3H8 + 5O2  3CO2 + 9H2O 5.00 mol 3.00 mol 1.80 mol 1.50 mol Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Your Turn! How many moles of C3H8 are consumed when 1.81x1023 molecules of CO2 are produced in the following equation?  C3H8 + 5O2  3CO2 + 4H2O 0.100 0.897 6.03 × 1022 5.43 × 1023 Copyright 2012 John Wiley & Sons, Inc

Mole-Mass Calculations What mass of H2 (2.02 g/mol) is made by the reaction of 3.0 mol HCl with excess aluminum? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g) Plan 3.0 mol HCl mol H2  g H2 Calculate Copyright 2012 John Wiley & Sons, Inc

Mole-Mass Calculations How many moles of HCl are needed to completely consume 2.00 g Al (26.98g/mol)? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g) Plan 2.00 g Al  mol Al  mol HCl Calculate Copyright 2012 John Wiley & Sons, Inc

Mole-Mass Calculations What mass of Al(NO3)3 (213g/mol) is needed to react with .093 mol Na2CO3? 3Na2CO3(aq) + 2Al(NO3)3(aq)  Al2(CO3)3(s) + 6NaNO3(aq) Plan 0.093 mol Na2CO3  mol Al(NO3)3  g Al(NO3)3 Calculate Copyright 2012 John Wiley & Sons, Inc

Mole-Mass Calculations How many moles of Al2(CO3)3 are made by the reaction of 3.45g Na2CO3 (105.99 g/mol) with excess Al(NO3)3? 3Na2CO3(aq) + 2Al(NO3)3(aq)  Al2(CO3)3(s) + 6NaNO3(aq) Plan 3.45g Na2CO3  mol Na2CO3  g Al2(CO3)3 Calculate Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Your Turn! How many moles of oxygen are consumed when 38.0g of aluminum oxide are produced in the following equation? 4Al(s) + 3O2(g)  2Al2O3(s)  0.248 0.559 1.50 3.00 atomic mass Al 26.98 O 16.00 Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Your Turn! What mass of HCl is produced when 1.81x1024 molecules of H2 react completely in the following equation? H2(g) + Cl2(g)  2HCl(g) 54.7g 72.9g 109g 219g atomic mass H 1.01 Cl 35.45 Copyright 2012 John Wiley & Sons, Inc

Mass-Mass Calculations Now we will put it all together. What mass of Br2 (159.80 g/mol) is needed to completely consume 7.00 g Al (26.98 g/mol)? 2Al(s) + 3Br2(l)  2AlBr3(s) 7.00 g Al mol Al  mol Br2  g Br2 Copyright 2012 John Wiley & Sons, Inc

Mass-Mass Calculations What mass of Br2 (159.80 g/mol) is needed to completely consume 7.00 g Al (26.98 g/mol)? 2Al(s) + 3Br2(l)  2AlBr3(s) Plan 7.00 g Al mol Al  mol Br2  g Br2 Calculate Copyright 2012 John Wiley & Sons, Inc

Mass-Mass Calculations What mass of Fe2S3 (207.91g/mol) can be made from the reaction of 9.34 g FeCl3 (162.20 g/mol) with excess Na2S? 2FeCl3(aq) + 3Na2S(aq)  Fe2S3(s) + 6NaCl(aq) Plan 9.34 g FeCl3  mol FeCl3  mol Fe2S3  g Fe2S3 Calculate Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Your Turn! What mass of oxygen is consumed when 54.0g of water is produced in the following equation?  2H2 + O2  2H2O 0.167 g 0.667 g 1.50 g 47.9 g atomic mass H 1.01 O 16.00 Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Your Turn! What mass of H2O is produced when 12.0g of HCl react completely in the following equation?  6HCl + Fe2O3  2FeCl3 + 3H2O 2.97 g 39.4 g 27.4 g 110. g atomic mass H 1.01 O 16.00 Cl 35.45 Copyright 2012 John Wiley & Sons, Inc

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

Copyright 2012 John Wiley & Sons, Inc 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

Copyright 2012 John Wiley & Sons, Inc 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

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

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

Limiting Reactant Calculation Calculate the number of moles of water that can be made by the reaction of 1.51 mol H2 with 0.932 mol O2. 2H2(g) + O2(g)  2H2O(g) Calculate the theoretical yield of H2O assuming H2 is the limiting reactant and that O2 is the excess reactant. Calculate the theoretical yield of H2O assuming that O2 is the limiting reactant and that H2 is the excess reactant. Copyright 2012 John Wiley & Sons, Inc

Limiting Reactant Calculation continued Assuming that H2 is limiting and O2 is excess: Assuming that O2 is limiting and H2 is excess: So what is the maximum yield of H2O? Copyright 2012 John Wiley & Sons, Inc

Limiting Reactant Calculation continued How much H2 and O2 remain when the reaction stops? H2: Limiting Reactant – None remains. It was used up in the reaction. O2: Excess Reactant – Calculate the amount of O2 used in the reaction with H2. Then subtract that from the original amount. Copyright 2012 John Wiley & Sons, Inc

Limiting Reactant Calculation 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) + 3CuSO4(aq)  Al2(SO4)3(aq) + 3Cu(s) Plan 15 g Al  mol Al  mol Cu  g Cu 25 g CuSO4  mol CuSO4  mol Cu  g Cu Compare answers. The smaller number is the right answer. Copyright 2012 John Wiley & Sons, Inc

Limiting Reactant Calculation continued 2Al(s) + 3CuSO4(aq)  Al2(SO4)3(aq) + 3Cu(s) 1. Assume Al is limiting and CuSO4 is in excess. 53.0 g Cu 2. Assume CuSO4 is limiting and Al is in excess. 9.96 g Cu 3. Compare answers. CuSO4 is the limiting reagent. The theoretical yield of Cu is 9.96 g. Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc 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. True False Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc 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 + 2AgNO3  Cu(NO3)2 + 2Ag Cu AgNO3 Cu(NO3)2 Ag Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc Your Turn! What is the mass of silver (107.87 g/mol) produced by the reaction of 3.00 moles of copper with 3.00 moles of silver nitrate?   Cu + 2AgNO3  Cu(NO3)2 + 2Ag 162g 216g 324g 647g Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc 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, Inc

Copyright 2012 John Wiley & Sons, Inc Percent Yield Calculate the % yield of PCl3 that results from reacting 5.00 g P with excess Cl2 if only 17.2 g of PCl3 were recovered. 2P + 3Cl2 2PCl3 Compute the expected yield of PCl3 from 5.00 g P with excess Cl2. 22.2g PCl3 Compute the % Yield. Copyright 2012 John Wiley & Sons, Inc

Copyright 2012 John Wiley & Sons, Inc 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, Inc