Chapter 3: Stoichiometry Chem 6A Michael J. Sailor, UC San Diego 1 Announcements: Thursday (Sep 29) quiz: Bring student ID or we cannot accept your quiz! No notes, no calculators Covers chapters 1 and 2 Need to know your name, PID, and section # Chem 6A Michael J. Sailor, UC San Diego 2
Cover page of Thursday s quiz Chem 6A 2010 (Sailor) Name: Student ID Number: Section Number: QUIZ 1 18 1 2 H 2 13 14 15 16 17 He 1.0079 4.0026 3 4 5 6 7 8 9 10 Li Be B C N O F Ne 6.941 9.01218 10.811 12.011 14.0067 15.9994 18.9984 20.1797 11 12 13 14 15 16 17 18 Na Mg 3 4 5 6 7 8 9 10 11 12 Al Si P S Cl Ar 22.9898 24.305 26.9815 28.0855 30.9738 32.066 35.4527 39.948 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.0983 40.078 44.9559 47.88 50.9415 51.9961 54.9381 55.847 58.9332 58.69 63.546 65.39 69.723 72.61 74.9216 78.96 79.904 83.8 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.4578 87.62 88.9059 91.224 92.9064 95.94 98.9063 101.07 102.906 106.42 107.868 112.411 114.82 118.71 121.75 127.6 126.905 131.29 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.905 137.327 138.906 178.49 180.948 183.85 186.207 190.2 192.22 195.08 196.967 200.59 204.383 207.2 208.98 208.982 209.987 222.018 87 88 89 104 105 106 107 108 109 Fr Ra Ac 223.02 226.025 227.028 - - - - - - 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140.12 140.91 144.24 146.92 150.35 151.96 157.25 158.92 162.5 164.93 167.26 168.93 173.04 174.97 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232.038 231.04 238.03 237.05 239.05 241.06 247.07 249.08 251.08 254.09 257.1 258.1 255 262.1 Some useful constants and relationships: Ideal gas constant: 0.08206 L. atm. mol -1. K -1 = 8.31451 J. mol -1. K -1 Avogadro constant: 6.022 x 10 23 mole -1 Planck's constant = h = 6.6261 x 10-34 J. s speed of light: 3.00 x 10 8 m/s 101.325 J = 1 L. atm 1J = 1kg. m 2 /s 2 1 atm = 760 Torr 1 ev = 1.6022 x 10-19 J E = -RHh/n 2 RH = 3.29 x 10 15 Hz C2 = second radiation constant = 1.44 x 10-2 K. m Emitted power (W) = (constant)t 4 Surface area (m 2 ) max = 1 5 C 2 E = hc " Chem 6A Michael J. Sailor, UC San Diego 3 The Periodic Table of the Elements QUIZ THURS Oct 20 (front page) 1 18 1 2 2 13 14 15 16 17 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 3 4 5 6 7 8 9 10 11 12 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 Chem 6A Michael J. Sailor, UC San Diego 4
The Periodic Table of the Elements QUIZ THURS Oct 20 (back page) Provide the names of the following elements: Element H Li Fr Pt Ag Sb Etc 20 elements total Name hydrogen lithium francium platinum silver antimony Chem 6A Michael J. Sailor, UC San Diego 5 Problem: Mass-to-mass calculations Since the bronze age (4000 b.c.), copper metal has been produced by smelting, in which Cu 2 O ore is reduced with excess carbon (charcoal). How much copper can be produced from smelting of 1.00 kg of pure Cu 2 O? a) 222 g b) 888 g c) 444 g d) 2252 g e) none of the above Chem 6A Michael J. Sailor, UC San Diego 6
Solution: Mass-to-mass calculations (1) The balanced equation for the reaction is: 2Cu 2 O + C 4Cu + CO 2 (2) Find out how many moles of Cu are there, then convert to grams Cu: 1000 g Cu 2 O Mol Cu 2 O 4 mol Cu 63.546 g Cu 143.1 g Cu 2 O 2 mol Cu 2 O 1 mol Cu = 888 g Cu Chem 6A Michael J. Sailor, UC San Diego 7 Problem: Limiting reactant In the Haber process, hydrogen (H 2 ) reacts with nitrogen (N 2 ) in a reactor to make ammonia. The reactor is initially charged with 2 mol of H 2 and 3 mol of N 2 and the reaction is allowed to go to completion. Fill out the table below indicating the amounts of reactants and products present after the reaction is complete. 3 H 2 + N 2 2 NH 3 Moles H 2 Moles N 2 Moles NH 3 Chem 6A Michael J. Sailor, UC San Diego 8
Solution: Limiting reactant The reactor is initially charged with 2 mol of H 2 and 3 mol of N 2 Balanced equation: 3 H 2 + N 2 2 NH 3 If all 2 mol of H 2 reacts, how many moles of N 2 are needed? Before reaction Moles H 2 2 Moles N 2 3 Moles NH 3 0 After reaction Chem 6A Michael J. Sailor, UC San Diego 9 Solution: Limiting reactant The reactor is initially charged with 2 mol of H 2 and 3 mol of N 2 Balanced equation: 3 H 2 + N 2 2 NH 3 If all 2 mol of H 2 reacts, how many moles of N 2 are needed? = 2/3 mol N 2 needed We have 3 mole of N 2, so N 2 is in excess and H 2 is limiting Chem 6A Michael J. Sailor, UC San Diego 10
Solution: Limiting reactant The reactor is initially charged with 2 mol of H 2 and 3 mol of N 2 and the reaction is allowed to go to completion. Balanced equation: 3 H 2 + N 2 2 NH 3 Before reaction What s used? What s left? After reaction, Moles H 2 = 2-2 = 0 Moles H 2 2 2 0 Moles N 2 3 2/3 2.33 Moles NH 3 0-1.33 Moles N 2 = 3-2/3 = 7/3, or 2.33 Moles NH 3 = 0 + 2 mol H 2 2 mol NH 3 = 4/3, or 1.33 3 mol H 2 Chem 6A Michael J. Sailor, UC San Diego 11 Solution: Limiting reactant What if we chose to calculate it assuming the N 2 is limiting? Balanced equation: 3 H 2 + N 2 2 NH 3 If all 3 mol of N 2 reacts, how many moles of H 2 are needed? Before reaction Moles H 2 2 Moles N 2 3 Moles NH 3 0 After reaction Chem 6A Michael J. Sailor, UC San Diego 12
Solution: Limiting reactant If all 3 mol of N 2 reacts, how many moles of H 2 are needed? Balanced equation: 3 H 2 + N 2 2 NH 3 = 9 mol H 2 needed We only have 2 mole of H 2, so H 2 is limiting Chem 6A Michael J. Sailor, UC San Diego 13 Chapter 3: Stoichiometry (cont) Chem 6A, Section D Sept 29, 2011 Chem 6A Michael J. Sailor, UC San Diego 14
Problem: Limiting reactant Since the bronze age (4000 b.c.), copper metal has been produced by smelting, in which Cu 2 O ore is reduced with excess carbon (charcoal). How much copper can be produced from smelting of 1.00 kg of pure Cu 2 O with 25 g of charcoal? Assume the byproduct of the reaction is CO2. a) 222 g b) 888 g c) 444 g d) 2252 g e) none of the above Chem 6A Michael J. Sailor, UC San Diego 15 Solution: Limiting reactant The balanced equation for the reaction is: 2Cu 2 O + C 4Cu + CO 2 Find out how many moles of Cu 2 O we have: 1000 g Cu 2 O Mol Cu 2 O 143.1 g Cu 2 O Find out how many moles of C we have: 25 g C Mol C 12.011 g C how many moles of C are needed: = 6.99 mol Cu 2 O = 2.08 mol C 1000 g Cu 2 O Mol Cu 2 O 1 mol C 143.1 g Cu 2 O 2 mol Cu 2 O = 3.49 mol C needed Chem 6A Michael J. Sailor, UC San Diego 16
Solution: Limiting reactant We have 2.08 mol C. We need 3.49, so all of the C will be used up before all of the Cu 2 O has reacted. So C (carbon) is the limiting reactant. how many moles of Cu will be produced using this much C? 25 g C Mol C 4 mol Cu 63.546 g Cu 12.011 g C 1 mol C Mol Cu = 529 g Cu Chem 6A Michael J. Sailor, UC San Diego 17 Problem: Combustion analysis Combustion analysis is carried out on 1.621 g of a compound that contains only carbon, hydrogen, and oxygen. The masses of water and carbon dioxide produced are 1.902 g and 3.095 g, respectively. What is the empirical formula of the compound? See problem 3.34 from the book Chem 6A Michael J. Sailor, UC San Diego 18
Solution: Combustion analysis Like excess-limiting reagent problem except the oxygen is never limiting: C x H y O z + xs O 2 x CO 2 + y/2 H 2 O Figure out number of moles and mass of carbon, hydrogen: 3.095 g CO 2 1mol CO 2 1 mole C 44.01 g CO 2 1 mol CO 2 = 0.07032 moles C 0.07032 moles C 12.011 g C 1 mol C = 0.8446 g C 1.902 g H 2 O 1mol H 2 O 2 mole H 18.02 g H 2 O 1 mol H 2 O = 0.2111 moles H 0.2111 moles H 1.0079 g H 1 mol H = 0.2128 g H Chem 6A Michael J. Sailor, UC San Diego 19 Solution: Combustion analysis C x H y O z + xs O 2 x CO 2 + y/2 H 2 O Mass left over = 1.621-0.2128-0.8446 = 0.5636 g mass C x H y O z mass H mass C So 0.5636 g is the mass of O (NOT O 2 ) in the sample. The number of moles of O is: 0.5636 g O 1mol O 15.9994 g O = 0.0352 mol O Chem 6A Michael J. Sailor, UC San Diego 20
Solution: Combustion analysis C x H y O z + xs O 2 x CO 2 + y/2 H 2 O Dividing by the least common denominator (O in this case): Element Moles Fraction C 0.07032 2.00 H 0.2111 6.00 O 0.0352 1 So the empirical formula is C 2 H 6 O Chem 6A Michael J. Sailor, UC San Diego 21 Water: H!+ 2!- O 105 H!+ Chem 6A Michael J. Sailor, UC San Diego 22
Sodium Chloride (NaCl) Crystal Cl - Na + Unit Cell Chem 6A Michael J. Sailor, UC San Diego 23 Dissolution of NaCl in water ion-dipole interactions replace ion-ion interactions Solvent Solute = H 2 O = Na + = Cl - Chem 6A Michael J. Sailor, UC San Diego 24
Ethanol in water dipole-dipole interactions: Like Dissolves Like Hydrogen Bonds Chem 6A Michael J. Sailor, UC San Diego 25 Problem: Determine molar concentration What is the molarity of a solution made by adding enough water to 20.0 g of NaCl to make 50.0 ml of solution? MW of NaCl is 58.44 g/mol. Set up but do not solve. Chem 6A Michael J. Sailor, UC San Diego 26
Solution: Determine molar concentration Molarity is defined as moles of solute/liter of solution: 20.0 g NaCl mol NaCl 58.44 g NaCl 0.050 L solution (= 6.84 M) Chem 6A Michael J. Sailor, UC San Diego 27 Problem: Dilution calculations What volume of 0.146 M sucrose solution is needed to make 2.00 L of a 0.05 M sucrose solution? Set up but do not solve. Chem 6A Michael J. Sailor, UC San Diego 28
Solution: Dilution calculations 1. How many moles of sucrose are needed? 0.05 mol sucrose 2 Liters Liter = 0.1 mol sucrose 2. What volume of solution do I need? 0.1 mol sucrose Liter 0.146 mol sucrose = 0.685 L of solution Chem 6A Michael J. Sailor, UC San Diego 29 Problem: Dilution calculations Calculate the volume (in liters) of a 0.1 M KMnO 4(aq) stock solution that is needed to prepare 250 ml of 1.50 x 10-3 M KMnO 4(aq) (set up but do not solve). Chem 6A Michael J. Sailor, UC San Diego 30
Solution: Dilution calculations Volume = 1.5 x 10-3 mol KMnO 4 250 ml L L L 1000 ml 0.1 mol KMnO 4(aq) (= 3.75 x 10-3 L) Chem 6A Michael J. Sailor, UC San Diego 31 Problem: Titration calculations A 0.202 g sample of iron ore (mixture of Fe 2 O 3 and SiO 2 ) is dissolved in hydrochloric acid and 22.3 ml of 0.0118 M KMnO 4 are required to reach the stoichiometric point. What is the percent by mass of iron present in the sample? The balanced net ionic equation for the titration reaction is given below: 5Fe 2+ + MnO 4 - + 8H + 5Fe 3+ + Mn 2+ + 4H 2 O a) 63.6 % b) 7.28 % c) 36.4 % d) 100 % e) none of the above Chem 6A Michael J. Sailor, UC San Diego 32
Solution: Titration calculations Find out how many moles of Fe are there: 1.18 x 10-2 mol KMnO 4 22.3 ml L 5 mol Fe 2+ L 1000 ml - 1 mol MnO 4(aq) = 1.316 x 10-3 mol Fe 2+ = 1.316 x 10-3 mol Fe. How many grams of Fe were there? 1.316 x 10-3 mol Fe 55.847 g Fe Mol Fe = 0.07347 grams Fe. So the % by mass Fe is: 0.07347g/0.202g x 100% = 36.4% Chem 6A Michael J. Sailor, UC San Diego 33 Sections SECTION TIME LOCATION D01 M 2-2:50 pm WLH 2115 D02 M 3-3:50 pm WLH 2115 D03 M 4-4:50 pm WLH 2115 D04 W 2-2:50 pm WLH 2115 D05 W 3-3:50 pm WLH 2115 D06 W 4-4:50 pm WLH 2115 D07 F 2-2:50 pm WLH 2115 D08 F 3-3:50 pm WLH 2115 D09 F 4-4:50 pm WLH 2115 28 Chem 6A Michael J. Sailor, UC San Diego