Solution Manual For Chemistry: Principles and Reactions, 8th Edition

| | | | | | | Contents Preface v Lecture Schedule vii Chapter 1 Matter and Measurements 1 Chapter 2 Atoms, Molecules, and Ions Chapter 3 Mass Relations in Chemistry; Stoichiometry Chapter 4 Reactions in Aqueous Solution Chapter 5 Gases Chapter 6 Electronic Structure and the Periodic Table Chapter 7 Covalent Bonding 71 Chapter 8 Thermochemistry 83 Chapter 9 Liquids and Solids 95 Chapter 10 Solutions Chapter 11 Rate of Reaction Chapter 12 Gaseous Chemical Equilibrium Chapter 13 Acids and Bases Chapter 14 Equilibria in Acid-Base Solutions Chapter 15 Complex Ions and Precipitation Equilibra Chapter 16 Spontaneity of Reaction Chapter 17 Electrochemistry Chapter 18 Nuclear Reactions Chapter 19 Complex Ions Chapter 20 Chemistry of the Metals Chapter 21 Chemistry of the Nonmetals Chapter 22 Organic Chemistrry Chapter 23 Organic Polymers: Natural and Synthetic 11 21 37 49 61 105 119 133 147 161 177 189 201 219 229 237 245 255 265 iii | | | | | | | Preface This manual starts off with a section entitled “Lecture Schedule,” which you may find helpful in adapting the text to your class schedule. Beyond that, the manual is organized by text chapters. For each chapter, we include three different features: 1. “Lecture Notes,” which suggest the amount of time that we devote to each chapter and the topics we emphasize. Included are detailed lecture outlines (from our own lectures) that may serve as a guide for your lectures. At a minimum, they indicate how we cover topics and how successive topics can be integrated. 2. A list of demonstrations illustrating topics in the chapter. These are taken from three sources: • The manual Tested Demonstrations in Chemistry (1994), Volumes I and II, compiled and edited by by George Gilbert by arrangement with the Journal of Chemical Education. These are coded as “GILB” with the experiment number (e.g., M 12) • Chemical Demonstrations (1983-1992), Volumes 1-4, published by Bassam Shakashiri with many collaborators and contributors. These are listed as “SHAK” followed by the volume and page reference. • Demonstrations described in the Journal of Chemical Education with the journal reference. 3. Answers and detailed solutions to: • The summary problem at the end of each chapter • Odd numbered text problems • Challenge problems at the end of each problem set Note that Appendix 6 has answers to all the even-numbered problems. Detailed solutions to many of these problems are in the Student Solutions Manual available from Cengage Learning, Brooks/Cole. v | | | | | | | Lecture Schedule Unlike other general chemistry texts, ours can be covered in its entirety in a one-year course. A reasonable schedule appears below. Further comments on the time you should devote to each chapter are in the body of this manual. The underlying assumption is that you are teaching 14-week semesters with two 50-minute lectures per week. If three class periods are devoted to examinations each semester, that leaves 25 for covering material. On that basis, you are able to complete Chapter 10 (Solutions) in the first semester. The seond semester will then start with Chapter 11 (Rate of Reaction). FIRST SEMESTER SCHEDULE Week Lecture 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 2 3 4 5 6 7 8 9 10 11 12 13 14 Topic Chapter 1 (Matter and Measurements) Chapter 1 Chapter 2 (Atoms, Molecules, and Ions) Chapter 2 Chapter 3 (Mass Relations in Chemistry; Stoichiometry) Chapter 3 Chapter 3 EXAM I Chapter 4 (Reactions in Aqueous Solution) Chapter 4 Chapter 4 Chapter 5 (Gases) Chapter 5 Chapter 6 (Electronic Structure and the Periodic Table) Chapter 6 Chapter 6 EXAM II Chapter 7 (Covalent Bonding) Chapter 7 Chapter 7 Chapter 8 (Themochemistry) Chapter 8 Chapter 9 (Liquids and Solids) Chapter 9 Chapter 9 EXAM III Chapter 10 (Solutions) Chapter 10 vii viii Lecture Schedule SECOND SEMESTER SCHEDULE Week Lecture 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 2 3 4 5 6 7 8 9 10 11 12 13 14 Topic Chapter 11 (Rate of Reaction) Chapter 11 Chapter 11 Chapter 12 (Gaseous Chemical Equilibrium) Chapter 12 Chapter 13 (Acids and Bases) Chapter 13 Chapter 13 EXAM I Chapter 14 (Equilibria in Acid-Base Solutions) Chapter 14 Chapter 15 (Complex Ions) Chapter 15 Chapter 16 (Precipitation Equilibria) Chapter 17 (Spontaneity of Reaction) Chapter 17 EXAM II Chapter 18 (Electrochemistry) Chapter 18 Chapter 18 Chapter 19 (Nuclear Chemistry) Chapter 20 (Chemistry of the Metals) Chapter 20 EXAM III Chapter 21 (Chemistry of the Nonmetals) Chapter 21 Chapter 22 (Organic Chemistry) Chapter 23 (Organic Polymers: Natural and Synthetic) Lecture Schedule ix If you want to use lecture time for review, for going over assigned problems, or for doing a large number of demonstrations, you will have trouble keeping up with this schedule. As you’ve almost certainly learned by now, the solution to this problem is not to talk faster. Judicious deletions work better. It’s been said, and wisely, that the secret of giving a good lecture is knowing what to leave out. Possible candidates include: • Introductory material on matter in Chapter 1 and atomic theory in Chapter 2. The chances are your students have been exposed to this material more than once in high school and understood it reasonably well the first time. • Boyle’s and Charles’s laws in Chapter 5. We start the chapter by writing the ideal gas law and go on from there. • The First Law discussion in Chapter 8. Quite frankly, this has very little to do with chemistry. Students will not be irreparably damaged if they are unaware of the distinction between H and E. • The discussion of colligative properties in Chapter 10 could be shortened. Raoult’s law could easily be omitted. • Reaction mechanisms in Chapter 11. Students have a lot of trouble with this. We are not sure it is worth the effort. • Polyprotic acids in Chapter 13. • The Second Law discussion in Chapter 17. Beyond these selective omissions, some instructors may want to delete one or another of the descriptive chapters at the end of the text (Chapters 20–22). If, in that way, you can squeeze out a couple of lectures, they can well be spent on Chapter 12 (three lectures instead of two) and Chapter 19 (two lectures instead of one). Textbook authors sometimes tell you that chapters can be covered in almost any order, depending on your preference. This isn’t really true for this textbook, or any other with structural integrity. It can be done, but only with very careful additions and deletions of material. Suppose, for example, you want to cover Precipitation Equilibria (Chapter 16) immediately after Acid-Base Equilibria (Chapter 14). Keep in mind that an understanding of formation constants (Complex Ions, Chapter 15) is assumed when methods of dissolving precipitates are considered in Section 16.2 of Chapter 16. 1 | | | | | | | | MATTER AND MEASUREMENTS LECTURE NOTES This material ordinarily requires two lectures (100 minutes), allowing for a 10–15 minute introduction to the course in the first lecture. If you’re in a hurry, this can be cut to 1 12 lectures by discussing only quantitative material (significant figures, unit conversions, density, solubility). A few points to keep in mind: • Virtually all of your students will be familiar with the metric system and prefixes. It may be worth discussing the rationale for SI, but you don’t have to dwell on it. • Students readily learn the rules of significant figures, but typically ignore them after Chapter 1. It may help to emphasize that these are common-sense (albeit, approximate) rules for estimating experimental error. • Many (typically, the weaker) students resist using conversion factors, preferring instead a rote method. It may be useful to point out that conversion factors will be a recurring tool throughout the text, so are well worth learning at this point. • Students often have trouble with solubility calculations. The approach in the text involves conversions (Example 1.8). The solubility is considered to be a conversion factor relating grams of solute to grams of solvent. Lecture 1 I. Types of Substances A. Elements Cannot be broken down into simpler substances. Examples: nitrogen, lead, sodium, arsenic. Symbols: N, Pb, Na, As. B. Compounds Contain two or more elements with fixed mass percents. Glucose: 40.00% C, 6.71% H, 53.29% O. Sodium chloride: 39.34% Na, 60.66% Cl. C. Mixtures Homogeneous (solutions) vs. heterogeneous. Separation by filtration, distillation. II. Measured Quantities A. Length Base unit is the meter. 1 km = 103 m; 1 cm = 10 2 m; 1 mm = 10 3 m; 1 nm = 10 9 m. Dimensions of very tiny particles will be expressed in nanometers. 1 2 Chapter 1 B. Volume 1 L = 103 mL = 103 cm3 = 10 3 m3 . Buret, pipet, volumetric flask. C. Mass 1 kg = 103 g; 1 mg = 10 3 g. Two different kinds of balances will be used in the lab. An analytical balance (±0.001 g) should be used only for accurate, quantitative work. D. Temperature t F = 1.8 t C + 32 ; Convert 68 F to C and K: TK = t C + 273.15. t C = (68 32 )/1.8 = 20 C TK = 293 Lecture 2 III. Experimental Error; Significant Figures Suppose an object is weighed on a crude balance to ±0.1 g and the mass is found to be 23.6 g. This quantity contains three significant figures, that is, three experimentally significant digits. With an analytical balance, the mass might be 23.582 g (five significant figures). A. Counting significant figures 1. Volume of liquid = 24.0 mL; three significant figures. Zeroes at the end of the measured quantity are significant when they follow nonzero digits. 2. Volume = 0.0240 L; three significant figures (note that 0.0240 L = 24.0 mL). Zeroes at the beginning of a measured quantity are not significant when they precede nonzero digits. B. Multiplication and division Keep only as many significant figures as there are in the least precise quantity. Density of a piece of metal weighing 36.123 g with a volume of 13.4 mL = ? density = 36.123 g = 2.70 g/mL 13.4 mL C. Addition and subtraction Keep only as many digits after the decimal point as there are in the least precise quantity. Add 1.223 g of sugar to 154.5 g of coffee: total mass = 1.2 g + 154.5 g = 155.7 g Note that the rule for addition and subtraction does not apply to significant figures. The number of significant figures may well decrease after subtraction. mass beaker + sample = 52.169 g (five significant figures) mass empty beaker = 52.120 g (five significant figures) mass sample = 0.049 g (two significant figures) D. Exact numbers One liter means 1.000000… L. Matter and Measurements 3 IV. Conversion Factors A. One-step conversion A rainbow trout is measured to be 16.2 inches long. What is its length in centimeters? length in cm = 16.2 in ⇥ 2.54 cm = 41.1 cm 1 in Note the cancellation of units. To convert from centimeters to inches, use the conversion factor 1 in = 2.54 cm. (Here, there are exactly 2.54 cm in one inch.) B. Multiple conversion factors A thrown baseball has speed 89.6 miles per hour. What is its speed in meters per second? 1 mile = 1.609 km = 1.609 ⇥ 103 m; 1 h = 3600 s ✓ ◆ ⇣ ✓ ◆ mile m ⌘ 1h speed = 89.6 ⇥ 1.609 ⇥ 103 ⇥ = 40.0 m/s h mile 3600 s V. Properties of Substances Distinguish between intensive and extensive, and between chemical and physical. A. Density An empty flask weighs 22.138 g. Pipet 5.00 mL of octane into the flask, producing a total mass of 25.598 g. What volume is occupied by ten grams of octane? d = 3.460 g/5.00 mL = 0.692 g/mL V = 10.00 g ⇥ 1 mL = 14.5 mL 0.692 g Note that for density calculations, 1 mL = 1 cm3 . B. Solubility This is often expressed as grams of solute per 100 g of solvent. Solubility of sugar at 20 C = 210 g sugar/100 g water. A solution containing 210 g sugar/100 g water is saturated. A solution containing less than 210 g sugar/100 g water is unsaturated. A solution containing more than 210 g sugar/100 g water is supersaturated. 1. How much water is required to dissolve 52 g of sugar at 20 C? 52 g sugar ⇥ 100 g water = 25 g water 210 g sugar 4 Chapter 1 2. A solution at 20 C contains 25 g sugar and 125 g water. Is it unsaturated, saturated or supersaturated? mass sugar/100 g water = 25 g sugar ⇥ 100 g water = 20 g sugar 125 g water (unsaturated) DEMONSTRATIONS 1. Scientific method: GILB H 29 2. Decomposition of mercury(II) oxide: GILB A 8 3. Separation of a mixture: GILB A 14 4. Reaction of sodium with chlorine: GILB A 24, A 25; SHAK 1 61; J. Chem. Educ. 73 539 (1996) 5. Chromatography: GILB Q 3, Q 13 6. Significant figures: J. Chem. Educ. 69 497 (1992) 7. Density of liquids: GILB C 13; SHAK 3 229 8. Supersaturation: GILB F 11; SHAK 1 27 SUMMARY PROBLEM (a) K, Mn, O (b) density, melting point, solubility, color (c) mass = 2.703 (d) 2.703 (e) F = g ⇥ 48.7cm3 = 132 g cm3 g 1 lb (2.54)3 cm3 (12)3 in3 ⇥ ⇥ ⇥ = 169 lb/ft3 3 3 3 454 g 1 ft cm 1 in 9 9 ( C) + 32 = (2.40 ⇥ 102 ) + 32 = 464 F 5 5 (2.40 ⇥ 102 ) + 273 = 513 K (f) 38.5 g H2 O ⇥ 6.38 g KMnO4 = 2.46 g KMnO4 100 g H2 O (g) At 60 C: 65.0 g H2 O ⇥ 25 g KMnO4 = 16 g KMnO4 100 g H2 O The solution is unsaturated. At 20 C: 65.0 g H2 O ⇥ 6.38 g KMnO4 = 4.15 g KMnO4 can be dissolved. 100 g H2 O The solution is supersaturated. Matter and Measurements (h) 55.0 g H2 O ⇥ 5 25 g KMnO4 = 14 g KMnO4 can be dissolved at 60 C. 100 g H2 O Yes, all the KMnO4 added will dissolve. 55.0 g H2 O ⇥ 6.38 g KMnO4 = 3.51 g KMnO4 can be dissolved at 20 C. 100 g H2 O No, not all the KMnO4 will dissolve. 10.0 g – 3.51 g = 6.5 g KMnO4 will remain undissolved. PROBLEMS 1. (a) mixture (b) element (c) mixture 3. (a) solution (b) solution (c) heterogeneous mixture 5. (a) distillation (b) filtration (c) gas chromatography 7. (a) Ti (b) P (c) K (d) Mg 9. (a) mercury (b) silicon (c) sodium (d) iodine 11. (a) balance (b) thermometer (c) graduated cylinder 13. t F = 1.8(52 ) + 32 = 126 F; 15. t C = (85.0 32) ⇥ 17. (a) 3 (d) exact (d) compound tK = 52 + 273.15 = 325 K 5 = 29.4 C ; solid 9 (b) ambiguous (c) 4 (e) 5 19. (a) 7.49 g (b) 298.69 cm (c) 1 ⇥ 101 lb 21. (a) 1.325 ⇥ 102 cm (b) 8.83 ⇥ 10 4 km (c) 6.432 ⇥ 109 nm 25. 10,000: ambiguous 1.71 ⇥ 105 ft2 : 3 $22.00: exact 27. (a) 80.0 (b) 0.7615 (c) 14.712 (d) 12.0 oz 23. (c) (d) 0.03 (e) 1.5 ⇥10 22 20%: ambiguous 6 29. Chapter 1 4⇡(4.30 cm)3 = 333 cm3 ; 3 4⇡(4.33 cm)3 = 3.40 ⇥ 102 cm3 ; 7 cm3 3 31. (a) 303 m = 0.303 km 1.50 ⇥ 103 nm3 33. (a) 22.3 mL ⇥ (c) 22.3 mL ⇥ 1L = 2.23 ⇥ 10 2 L 103 mL (b) 17.8 hands ⇥ (c) 20.5 hands ⇥ 39. 1 in3 = 1.36 in3 (2.54 cm)3 1L 1.057 qt ⇥ = 0.0236 qt 3 1L 10 mL 35. (a) 19.2 hands ⇥ 37. 2.0 acre ⇥ (b) 22.3 cm3 ⇥ 1 3 ft 1 hand 1 3 ft 1 hand 1 3 ft 1 hand = 6.40 ft ⇥ 12 in 1m ⇥ = 1.81 m 1 ft 39.37 in + 3.0 ft = 9.8 ft 4.356 ⇥ 104 ft2 (12)2 in2 1 m2 1 hectare ⇥ ⇥ ⇥ = 0.81 hectare 2 1 acre 1 ft (39.37)2 in2 104 m2 5.0 mi 0.25 min ⇥ = 1.2 min 1 mile 1 Eng lap 0.50 km ⇥ 1 mi 5.0 min ⇥ = 1.6 min 1.609 km 1 mi 41. (a) 3.0 qt plasma ⇥ 1L 1000 mL 0.080 mL alcohol ⇥ ⇥ = 2.3 mL 1.057 qt 1L 100 mL plasma (b) 3.0 qt plasma ⇥ 1L 1000 mL 0.10 mL alcohol ⇥ ⇥ = 2.8 mL 1.057 qt 1L 100 mL plasma (c) 2.8 mL 2.3 mL = 0.5 mL 43. 235 kJ 103 J 1 cal 1 kcal 1000 mL 1 qt ⇥ ⇥ ⇥ 3 ⇥ ⇥ = 53.2 kcal/cup 250 mL 1 kJ 4.18 J 10 cal 1.057 qt 4 cups 45. 252 g = 1.49 g/mL 0.750 ⇥ 225 mL Matter and Measurements 47. Vmethanol = 43.7 g ⇥ Therefore, dslug = 7 1 mL = 55.2 mL , 0.791 g Vslug = 59.7 mL 55.2 mL = 4.5 mL 25.17 g = 5.6 g/mL 4.5 mL 49. (8.0 ⇥ 7.0 ⇥ 0.75) ft3 ⇥ (12)3 in3 (2.54)3 cm3 1.00 g 1 kg ⇥ ⇥ ⇥ = 1.2 ⇥ 103 kg 3 3 3 1000 g 1 ft 1 in 1 cm 51. Volume of air = volume of room V = 55 kg O2 ⇥ 1000 g 1L 100 L ⇥ ⇥ = 2.0 ⇥ 105 L 1 kg 1.31 g 21 L 53. At 30 C: maximum amount of MgSO4 that can be dissolved = 25.0 g H2 O ⇥ 38.9 g MgSO4 = 9.72 g 100 g H2 O The solution is unsaturated. (9.50 + 1.00 g) – 9.72 = 0.78 g will precipitate out 55. (a) 46 g H2 O ⇥ 16 g NaHCO3 = 7.4 g NaHCO3 can be dissolved 100 g H2 O 9.2 g in the mixture > 7.4 g, thus the solution is not homogeneous. 9.2 g – 7.4 g = 1.8 g NaHCO3 are undissolved. (b) 9.2 g NaHCO3 ⇥ 96 g 57. 57.0 g 100 g H2 O = 96 g H2 O needed to dissolve 9.6 g NaHCO3 46 g = 5.0 ⇥ 101 g H2 O needs to be added. 25.0 g = 32.0 g of Pb(NO3 )2 dissolves in 64.0 g H2 O at 10 C. Solubility is 32.0 g Pb(NO3 )2 1.00 g Pb NO3 )2 50.0 g Pb(NO3 )2 = = 64.0 g H2 O 2.00 g H2 O 100.0 g H2 O 59. (a) physical (b) physical 61. V = 35 ft ⇥ 43 ft ⇥ 28 in ⇥ 3.5 ⇥ 103 ft3 ⇥ (c) physical 1 ft = 3.5 ⇥ 103 ft3 12 in (12 in)3 (2.54 cm)3 ⇥ = 9.9 ⇥ 107 cm3 (1 ft)3 (1 in)3 mass = 9.9 ⇥ 107 cm3 ⇥ (0.35 g) 1 lb ⇥ = 7.7 ⇥ 104 lbs 3 453.6 g (1 cm) (d) chemical 8 Chapter 1 63. 108 carats ⇥ 0.0476 lb ⇥ 65. 153.2 g ⇥ 0.200 g 1 lb ⇥ = 0.0476 lb 1 carat 454 g 1 cm3 454 g 1 in3 ⇥ ⇥ = 0.376 in3 3.51 g 1 lb (2.54)3 cm3 1 cm3 = 33.7 cm3 = V; 4.55 g 33.7 = ⇡r2 (7.75); r = 1.18 cm; d = 2.35 cm 67. (a) Chemical properties show the behavior of the species in a reaction; physical properties are intrinsic qualities. (b) Distillation vaporizes the liquid; filtration removes the solid. (c) The solute is a component of the solution. 69. The bottom layer is Hg; the middle layer is Pb; the top layer is ethyl alcohol. 71. (a) ⇡ 115 g; supersaturated (b) ⇡ 30 g; unsaturated (c) Dissolve 30 g of compound in 100 g H2 O. 73. (a) See Figure in the answers to the problems in Appendix 5. (b) y 100 J 0 J = = 0.512 x 78 C ( 117.0 C) (c) 60 J (d) J = 0.51( C) + 60 74. 31.5 gal ⇥ area = 4 qt 1L 10 3 m3 1 km3 ⇥ ⇥ ⇥ 9 3 = 1.2 ⇥ 10 10 km3 1 gal 1.057 qt 1L 10 m 1.2 ⇥ 10 10 km3 1 ⇥ 1012 nm ⇥ = 1.2 km2 100 nm 1 km 75. V = 12.0 g ⇥ 76. 1 cm3 = ⇡(0.254 cm)2 `; 2.70 g ` = 21.9 cm 8.50 ⇥ 103 L 1 m3 7.0 ⇥ 10 6 g Pb 365 d ⇥ 3 ⇥ ⇥ 0.75 ⇥ 0.50 ⇥ = 8.1 ⇥ 10 3 g Pb 3 1d 1 yr 10 L 1m Matter and Measurements 9 78. mass of Hg in cylinder B = 145.20 g mass of Hg + metal in cylinder A = 92.60 g mass of metal = 145.20 g 92.60 g = 52.60 g volume of cylinder A = volume of cylinder B = volume of Hg in cylinder B = 145.2 g ÷ 13.6 g/mL = 10.7 mL mass of Hg in cylinder A = 92.60 g – 52.60 g = 40.0 g volume of Hg in cylinder A = 40.0 g ÷ 13.6 g/mL = 2.94 mL volume of metal = volume of cylinder A – volume of Hg in cylinder A = 10.7 mL density of metal = 52.60 g/7.76 mL = 6.78 g/mL 2.94 mL = 7.76 mL | | | | | | | | 2 ATOMS, MOLECULES AND IONS LECTURE NOTES Students find this material relatively easy to assimilate; it’s almost entirely qualitative. On the other hand, there’s a lot of memorizing (sorry, learning) to do. This chapter is coverable in two lectures. Some general observations: • Material in Sections 2.1–2.3 is generally well covered in high-school chemistry courses; no need to dwell on it. • Students need to know the molecular formulas of the elements (Figure 2.13), the charges of ions with noble-gas structures and the names and formulas of the common polyatomic ions (Table 2.2). The charges of transition-metal ions will be covered later, in Chapter 4. • Naming compounds requires students to distinguish between ionic and molecular substances. It helps to point out that binary molecular compounds are composed of two nonmetals. Almost all ionic compounds contain a metal cation combined with a nonmetal anion or negatively charged polyatomic ion. The flow charts shown in Figures 2.18 and 2.19 should help visual learners. • The periodic table will be discussed in greater detail later in the text (Chapter 6). Lecture 1 I. Atomic Theory A. Elements Postulates: Elements consist of tiny particles called atoms, which retain their identity in reactions. In a compound, atoms of two or more elements combine in a fixed ratio of small whole numbers (e.g., 1:1, 2:1, etc.). B. Components relative mass relative charge location proton 1 +1 nucleus neutron 1 0 nucleus electron 0.0005 1 outside C. Atomic number It is the number of protons in the nucleus or the number of electrons in a neutral atom. This is characteristic of a particular element: all H atoms have one proton, all He atoms have two protons, etc. 11 12 Chapter 2 D. Mass number 1. It is the sum of the number of protons and the number of neutrons. Atoms of the same element can differ in mass number. Those are referred to as isotopes. For example: protons neutrons atomic no. nuclear symbol mass no. carbon-12 6 6 6 12 6C 12 carbon-14 6 8 6 14 6C 14 2. Isotopes Atoms of the same element (same atomic number) but differ in mass number. II. Atomic Masses A. Meaning of atomic masses They give the relative masses of atoms. Based on the C-12 scale; the most common isotope of carbon is assigned an atomic mass of exactly 12 amu. element B Ca Ni atomic mass (amu) 10.81 40.08 58.69 A nickel atom is 58.69/40.08 times as heavy as a calcium atom. It is 58.69/10.81 = 5.429 times as heavy as a boron atom. B. Atomic masses from isotopic composition atomic mass = (atomic mass of isotope 1)(%/100) + (atomic mass of isotope 2)(%/100) + · · · Isotope Atomic mass Percent Ne-20 20.00 amu 90.92 Ne-21 21.00 amu 0.26 Ne-22 22.00 amu 8.82 atomic mass of Ne = (20.00)(0.9092) + (21.00)(0.0026) + (22.00)(0.0882) = 20.18 amu C. Masses of individual atoms Since the atomic masses of H, Cl and Ni are, respectively, 1.008 amu, 35.45 amu and 58.69 amu, it follows that 1.008 g H, 35.45 g Cl, 58.69 g Ni all contain the same number of atoms, NA . NA = Avogadro’s number = 6.022 ⇥ 1023 Atoms, Molecules and Ions 13 1. Mass of a hydrogen atom? 1 atom H ⇥ 1.008 g H = 1.674 ⇥ 10 24 g 6.022 ⇥ 1023 atom 2. Number of atoms in one gram of nickel? 1.000 g Ni ⇥ 6.022 ⇥ 1023 atoms Ni = 1.026 ⇥ 1022 atoms 58.69 g Ni III. Periodic Table Periods and groups; numbering system for groups. Metals appear at the lower left, nonmetals at the upper right. Metalloids. Lecture 2 IV. Molecules A. Composition Usually consist of nonmetal atoms; held together by covalent bonds. B. Types of Formulas Consider the compound ethyl alcohol: Molecular formula: C2 H6 O H H Structural formula: H—C—C—O—H H H Condensed structural formula: CH3 CH2 OH V. Ions A. Formation of monatomic ions Na atom 11p+ , 11e F atom 9p+ , 9e ! Na+ ion 11p+ , 10e +e + e ! F ion 9p+ , 10e B. Charges of monatomic ions with noble-gas structures Cations: Group 1 (+1); Group 2 (+2); Al3+ Anions: Group 16 ( 2); Group 17 ( 1); N3 C. Polyatomic ions Names and formulas (Table 2.2) 14 Chapter 2 D. Formulas of compounds Apply the principle of electroneutrality. calcium fluoride: Ca2+ , F ions: CaF2 aluminum nitrate: Al3+ , NO3 ions: Al(NO3 )3 sodium dihydrogen phosphate: Na+ , H2 PO4 ions: NaH2 PO4 E. Ionic compounds They can be distinguished from molecular substances by the conductivity of their water solutions. Solutions of NaCl, Ca (OH)2 , … conduct electricity (electrolytes). Sugar is a nonelectrolyte. VI. Names of Compounds A. Ionic Name cation, followed by anion. Note that with transition metal cations, charge is indicated by a Roman numeral. Na2 SO4 sodium sulfate Fe(NO3 )3 iron(III) nitrate Systematic names of oxoanions (-ate, -ite, per-, hypo-) Calcium hypochlorite Ca (ClO)2 B. Binary molecular compounds Use of Greek prefixes: SF6 sulfur hexafluoride N2 O3 dinitrogen trioxide C. Acids Binary acids: hydrochloric acid Oxo acids: -ate salt ! -ic acid HClO4 , perchloric acid -ite salt ! -ous acid HClO, hypochlorous acid Atoms, Molecules and Ions 15 DEMONSTRATIONS 1. Law of constant composition: GILB A 12 2. Law of conservation of mass: GILB A 16 3. Simulation of Rutherford’s experiment: GILB L 7 4. Isotope effects H2 O, D2 O : GILB M 18 5. Reaction of hydrogen with chlorine: GILB H 38 6. Conductivity of water solutions: SHAK 3 140 7. Breath alcohol detection: J. Chem. Educ. 67 263 (1990); 71 158 (1994) 8. Relative masses of atoms (analogy): GILB L 2 SUMMARY PROBLEM (a) S8 (b) 16 protons, 16 electrons (c) no; Al2 S3 – aluminum sulfide (d) yes (e) yes; S2 Cl2 – disulfur dichloride (f) 34 16 S (g) group 16, period 3 (h) 20 neutrons (i) (31.97207)(0.9493) + (32.97146)(0.0076) + (33.96787)(0.0429) + (35.96708)(0.0002) = 32.07 amu (j) 12.55 g S ⇥ 1 mol S 6.022 ⇥ 1023 atoms ⇥ = 2.357 ⇥ 1023 atoms 32.07 g 1 mol S (k) 1 ⇥ 109 S atoms ⇥ 1 mol S 32.07 g ⇥ = 5.325 ⇥ 10 14 g 1 mol S 6.022 ⇥ 1023 atoms (l) SO3 = sulfur trioxide; H2 SO3 (aq) = sulfurous acid; SO42 = sulfate ion; Na2 SO3 = sodium sulfite PROBLEMS 1. p. 29 3. (a) Conservation of mass (b) Constant composition (c) neither (b) 143 (c) 92 5. J. J. Thompson; see p. 29 7. 80 34 Se 38 40 9. no. of neutrons: 36 18 Ar, 18 Ar, 18 Ar 11. (a) 92 16 Chapter 2 13. (a) 14 p+ , 16 n, 14 e ; R = Si (b) 39 p+ , 50 n, 39 e ; T = Y (c) 55 p+ , 78 n, 55 e ; X = Cs 15. (a) Ca-41, K-41, Ar-41 are isobars; Ca-40, Ca-41 are isotopes (b) atomic number = number of protons = 20 (c) same mass number 17. (a) 79.90 = 3.959 20.18 (b) 79.90 = 1.994 40.08 (c) 79.90 = 19.96 4.003 19. Ce-140 21. 50% 23. 83.9134(0.0056) + 85.9094(0.0986) + 86.9089(0.0700) + 87.9056(0.8258) = 0.47 + 8.47 + 6.08 + 72.59 average atomic mass = 87.61 25. 107.9 = 106.90509(0.5184) + 0.4816 x; x = 109 amu 27. Let x = abundance of the first isotope; abundance of second isotope = 0.9704 28.0855 = 27.9769 x + (0.9704 = 27.9769 x + 28.1188 x x)(28.9765) + (0.0296)(29.9738) 28.9765 x + 0.887 x = 0.921; abundance of first isotope is 92.1% 0.9704 x = 0.9704 0.921 = 0.0494; abundance of second isotope is 4.94% 29. Tall peak at mass 64; peak a little over 1/2 as high at mass 66; smallest peak is at mass 67, and the height of the peak at mass 64 is 2.5 times that of the peak at mass 68 . 31. 3 ⇥ 10 7 g ⇥ 1 mol 6.022 ⇥ 1023 atoms ⇥ = 9 ⇥ 1014 atoms 207.2 g 1 mol 33. (a) 0.185 g Pd ⇥ 6.022 ⇥ 1023 atoms = 1.05 ⇥ 1021 atoms 106.4 g Pd (b) 127 protons ⇥ 1 atom 106.4 g ⇥ = 4.88 ⇥ 10 22 g 46 protons 6.022 ⇥ 1023 atoms Atoms, Molecules and Ions 35. (a) 0.35744 mol ⇥ 17 6.022 ⇥ 1023 atoms = 2.152 ⇥ 1023 atoms 1 mol (b) 2.152 ⇥ 1023 atoms ⇥ 14 p+ + 14 n + 14 e 1 atom = 9.039 ⇥ 1024 ✓ ◆3 2.54 cm 37. Vcube = 1.25 in ⇥ = 32.0 cm3 1 in 32.0 cm3 ⇥ 0.968 g 6.022 ⇥ 1023 atoms ⇥ = 8.12 ⇥ 1023 atoms 22.99 g 1 cm3 39. (a) K (b) Cd (c) Al 41. (a) main-group metal (d) Sb (b) transition metal (e) P (c) main-group metal (d) metalloid (e) nonmetal 43. (a) 6 (b) 4 named, 1 not named (c) 0 45. (a) 13 (b) 2 (c) 17, 18 47. (a) C2 H7 N (b) C3 H8 O 49. (a) 14 p+ , 14e (b) 21 p+ , 22 e 51. (c) 35 p+ , 34e 19 9F 0 9 10 9 31 15 P 0 15 16 15 +3 27 30 24 2 16 16 18 57 3+ 27 Co 32 2 16 S 53. (a) electrolyte 55. (a) CH4 (b) nonelectrolyte (b) Cl4 57. (a) iodine trichloride (d) carbon tetrabromide 59. KCl, K2 S, CaCl2 , CaS (c) nonelectrolyte (c) H2 O2 (d) NO (b) dinitrogen pentaoxide (e) sulfur trioxide (d) 70 p+ , 70 e (d) electrolyte (e) SiO2 (c) phosphine 18 Chapter 2 61. (a) Fe(C2 H3 O2 )3 (b) Ca(NO3 )2 63. (a) potassium dichromate (c) K2 O (d) AuCl3 (b) copper(II) phosphate (d) aluminum nitride (e) Ba3 N2 (c) barium acetate (e) cobalt(II) nitrate 65. (a) hydrochloric acid (b) chloric acid (d) barium nitrite (c) iron(III) sulfite (e) sodium hypochlorite 67. HNO2 , nickel(II) iodate, Au2 S3 , sulfurous acid, NF3 69. (a) Mn(NO2 )3 ; manganese(III) nitrite (b) BF3 ; boron trifluoride (c) Ca(HCO3 )2 ; calcium hydrogen carbonate 71. (a) In (b) Pb or Sn (c) K (d) Sb 73. (a) … confirmed the presence of a dense nucleus with protons. (b) … elements arranged according to increasing atomic number. (c) … same number of protons. (d) Be3 N2 is beryllium nitride. 75. 6.00 oz salami ⇥ ⇥ 1g 0.090 g NaC7 H5 O2 6.022 ⇥ 1023 molecules NaC7 H5 O2 ⇥ ⇥ 0.03527 oz 100 g salami 144.1 g NaC7 H5 O2 1 atom Na = 6.4 ⇥ 1020 Na atoms 1 molecule NaC7 H5 O2 77. (b), (d), (e) 79. 8 molecules; 3 molecules left 81. A square with four circles around it (several of them in a flask with a defined volume) 83. (a) 118 (b) 120 (c) 117 (d) 120 (e) 119 Atoms, Molecules and Ions 85. first experiment: 19 %O = second experiment: 3.87 ⇥ 100 = 7.40; 52.30 % Hg = 92.60 15.68 ⇥ 100 = 92.62; 16.93 % O = 7.38 % Hg = 87. (a) K, Sr (b) O, F, Ar, S (c) S, K, Sr (d) S (e) S, O or S, F or O, F (f) Sr, S or Sr, O or K, F (g) Sr, F (h) K, O or K, S (i) Ar (j) O, F, Ar 88. A: mass C/mass H = 11.9 (⇡ 12) ratio for (a) = 2.77 B: mass C/mass H = 2.99 (⇡ 3) ratio for (b) = 4.67 ratio for (c) = 5.96 (c) is best choice 89. (a) ethane: 18.0 g C/4.53 g H = 3.97 g C/g H ethylene: 43.20 g C/7.25 g H = 5.96 g C/g H 5.96/3.97 = 1.50 = 3/2 (b) CH2 and CH3 ; C2 H4 and C2 H6 90. mass = 13 1.6726 ⇥ 10 4 g + 13 9.1094 ⇥ 10 28 g + 14 1.6749 ⇥ 10 24 g = 4.5204 ⇥ 10 23 g V = 43 ⇡ 1.43 ⇥ 10 8 cm 3 = 1.22 ⇥ 10 23 cm3 d = 4.5204 g/1.22 cm3 = 3.71 g/cm3 Empty space between Al atoms. 91. 2.3440 ⇥ 10 23 g + 3(9.1095 ⇥ 10 28 g) = 2.3443 ⇥ 10 23 g 92. (a) 200 inhalations ⇥ (b) 500 mL 2.5 ⇥ 1019 molecules ⇥ = 2.5 ⇥ 1024 molecules 1 inhalation 1 mL 2.5 ⇥ 1024 = 2.3 ⇥ 10 20 1.1 ⇥ 1044 (c) 1 inhalation ⇥ 500 mL 2.5 ⇥ 1019 molecules ⇥ ⇥ 2.3 ⇥ 10 20 = 2.8 ⇥ 102 molecules 1 inhalation 1 mL 20 Chapter 2 93. Total mass before reaction: 18.00 g + (25.00 x 1.025 g/mL) = 43.63 g After reaction following the law of conservation of mass: 43.63 g = 12 g + 30.95 g + mass of H2 mass of H2 = 0.68 g; volume of H2 = 0.68 g x 1L = 8.25 L 0.0824 g 3 | | | | | | | | MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY LECTURE NOTES This chapter is considerably more difficult and time-consuming than the two preceding ones. It contains a good deal of quantitative material that is fundamental for future chapters. We suggest that you devote three lectures to Chapter 3. The first lecture deals with the mole and the mole in solutions (molarity), the second with the quantitative aspects of chemical formulas (Section 3.2), and the third with mass relations in reactions (Section 3.3). Points to keep in mind include: 1. Note that mole-gram conversions and molarity calculations (Section 3.1) will be required in many later chapters, often as the first step in a more complex problem. 2. When dealing with formulas (Section 3.2), it is important to emphasize early on that the subscripts give not only the atom ratio but also the mole ratio. Students must realize this in order to follow the logic of obtaining simplest formulas from mass percents. 3. Students ordinarily have little trouble calculating formulas from mass percents. They are much less adept at obtaining formulas from analytical data such as that in Example 3.6. 4. It is important to get across the point (Section 3.3) that a chemical equation describes what happens when a reaction is carried out in the laboratory. Including the physical states of reactants and products in the equation helps to emphasize this point. 5. When you discuss mass relations in reactions, some students will revert to the infamous “ratio and proportion” method. The comments of Chapter 1 about conversion factors apply here, too. 6. There are many ways to find the limiting reactant and calculate the theoretical yield. We’ve tried most of them, and recommend the approach described in Section 3.3; students seem to grasp it readily. Lecture 1 I. The Mole A. Meaning 1 mol = 6.022 ⇥1023 items 1 mol H = 6.022⇥1023 H atoms; mass = 1.008 g 1 mol Cl = 6.022 ⇥1023 Cl atoms; mass = 35.45 g 1 mol Cl2 = 6.022 ⇥ 1023 Cl2 molecules; mass = 70.90 g 1 mol HCl = 6.022 ⇥1023 HCl molecules; mass = 36.46 g B. Molar mass Generalizing from the above examples, the molar mass, MM, is numerically equal to the sum of the atomic masses. 21