1. Chem 106, Prof. J.T. Spencer 1 CHE 106: General Chemistry u CHAPTER ONE Copyright © James T. Spencer 1995 - 1999 All Rights Reserved

2. Chem 106, Prof. J.T. Spencer 2 u What is Chemistry –Study of the “Physical” Properties Matter (Form and Function) –Study of How Matter Changes (Reactivity) u Benefits of Chemistry –Pharmaceuticals –Enhanced food production (fertilizers, herbicides, etc...) –Plastics and Polymers u Why Study Chemistry –Core requirement (?) –Central Science u Employment –Many fields Matter Chapt. 1.1 BIO Physics Medicine GEO Engr CHEM Law also: environmental economics electronics agriculture politics etc... S.U. B.S.

3. Chem 106, Prof. J.T. Spencer 3 Chemistry; Common Chemicals acetic acid.........................................vinegar calcium hypochloride......................bleaching powder calcium sulfate.................................plaster of paris carbon tetrachloride.......................cleaning fluid ferric oxide.......................................iron rust graphite............................................pencil lead magnesium sulfate..........................Epsom salts naphthalene......................................mothballs silicon dioxide...................................sand sodium bicarbonate.........................baking soda sodium borate...................................borax sodium hydroxide............................lye sulfuric acid......................................battery acid sucrose...............................................cane sugar

4. Chem 106, Prof. J.T. Spencer 4 Chemistry; Chemical Production H 2 SO 4 N2N2N2N2 O2O2O2O2 C2H4C2H4C2H4C2H4 CaO NH 3 C3H6C3H6C3H6C3H6 NaOH H 3 PO 4 Cl 2 1995 Chemical and Engineering News

5. Chem 106, Prof. J.T. Spencer 5 Nanoscale Chemistry Use simpler molecular units are molecular-architectural elements

6. Chem 106, Prof. J.T. Spencer 6 Nanoscale Chemistry

7. Chem 106, Prof. J.T. Spencer 7 Nanosystems

8. Chem 106, Prof. J.T. Spencer 8 Nanomachines red blood cell Interstellar Space Travel - Significant concepts in this area include: launch vehicles, the space elevator, interplanetary transportation, the swarm concept, smart dust, extraterrestrial materials utilization, terraforming, suspended animation, space telescopes and virtual sample return. Human Therapeutics - Nanotechnology has caused scientists to re-examine the problems of the human body from the perspective of atomic- engineering. By assuming a nanotechnological point of view, the resolution of therapeutic ailments becomes simple. Nano-Robots and Nano-Computers with advanced Artificial Intelligence - Nanotechnology will operate under the control of nano-sized computers which will manage the process of Molecular Manufacturing. In order to achieve this, it will be necessary to devise advanced Artificial Intelligence that will be able to automate and regulate Molecular Manufacturing systems.

9. Chem 106, Prof. J.T. Spencer 9 Matter; A Review Chapt. 1.2 u Definition of Matter –anything that occupies space and has mass u States –gas (vapor); no fixed volume or shape, compressable –liquid; fixed volume no fixed shape, mostly incompressable –solid; fixed volume and shape, incompressable u Forms –Substances (pure or single); has a fixed composition and distinct properties. Most things encountered are mixtures of substances. u Properties –Physical Properties; can be measured without changing the substance, i.e., color, density, melting point, etc... –Chemical Properties; the way a substance changes (reacts), i.e., combustion

10. Chem 106, Prof. J.T. Spencer 10 Matter; Elements and Compounds Chapt. 1.2 u Substances –Elements - substances which cannot be decomposed into simpler substances (see periodic table) –Compounds- substances which can be separated into two or more elements u Elements –110 Known (periodic table to be revisited) –make up all matter and composed of “subatomic particles” –symbols used for abbreviations (from older or common names) u Compounds –Elements combined in a definite proportion by mass (law of definite proportion) – properties different than consititutent elements Water; example of mixtures, compound and elements?

11. Chem 106, Prof. J.T. Spencer 11 Matter; Elements and Periodic Table See Website: http://the-tech.mit.edu/Chemicool Periodic Table

12. Chem 106, Prof. J.T. Spencer 12 Matter; A Review Chapt. 1.2 u Mixtures; combinations of substances –Mixture –Mixture- combination of two or more substances in which each retains its own chemical identity (and properties). Vary widely by composition (infinite possibilities of combining ratios), can be separated using the different physical properties of the component substances. –Homogeneous –Homogeneous - appears the same throughout (solutions), liquid, gas and solid solutions are possible. –Heterogeneous –Heterogeneous - mixtures which do not have the same (uniform) appearance throughout.

13. Chem 106, Prof. J.T. Spencer 13 Salt and Sand Mixture Ink from Cabbage Juice solubility and filtration chromatography Water from Salt Water Iron and Gold Mixture distillationmagnetic properties melting point differences chem. reactivity (acids) Iodine from Copper Chloride solubility and filtration Matter; A Review Chapt. 1.2 u Separating Mixtures using Physical Properties –How would you separate;

14. Chem 106, Prof. J.T. Spencer 14 Matter; A Review Chapt. 1.2 u Separating Mixtures using Physical Properties –How would you separate; Filtration Sand from Salt Filter Everyday Examples Everyday Examples; Auto Oil Filter Auto Air Filter Aquarium Water Filter Spaghetti Strainer Window Screens Registrar Flow

15. Chem 106, Prof. J.T. Spencer 15 Matter; A Review Chapt. 1.2 u Separating Mixtures using Physical Properties –How would you separate; Distillation Water from Salt Water NaCl(aq) NaCl(s) + H 2 O(l)

16. Chem 106, Prof. J.T. Spencer 16 Matter; A Review Chapt. 1.2 u Separating Mixtures using Physical Properties –How would you separate; Chromatograpgy Dyes from M&M’s BeforeAfter Dyes

17. Chem 106, Prof. J.T. Spencer 17 Matter; A Review Chapt. 1.3 u Changes –Physical - Changes in appearance but not identity, i.e., evaporation, melting (all changes of state) –Chemical - transformation into a different substance Chemical ChangesPhysical Changes burning melting C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 O H 2 O(s)H 2 O(l) C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 O H 2 O(s)H 2 O(l) chemical reactionssublimation NaOH + HCl H 2 O + NaCl H 2 O(s)H 2 O(g) NaOH + HCl H 2 O + NaCl H 2 O(s)H 2 O(g) corrosiondissolution 4Fe + 3O 2 2 Fe 2 O 3 H 2 O(l ) + NaCl(s)NaCl(aq) 4Fe + 3O 2 2 Fe 2 O 3 H 2 O(l ) + NaCl(s)NaCl(aq)

18. Chem 106, Prof. J.T. Spencer 18 Matter Matter Uniform ? HeterogeneousMixture Homogeneous Can be separated by physical methods Pure Substance Homogeneous Mixture (solution) Decomposed ? Compound Element Yes No No No Yes Yes

19. Chem 106, Prof. J.T. Spencer 19 Observations and Experiments Patterns and Trends Form and test hypothesis Theory Scientific Method

20. Chem 106, Prof. J.T. Spencer 20 Observations to Theory Observations Theory

21. Chem 106, Prof. J.T. Spencer 21 Observations to Theory Observations Theory

22. Chem 106, Prof. J.T. Spencer 22 Observations to Theory Observations Theory

23. Chem 106, Prof. J.T. Spencer 23 Matter; Measurement A B Which is True? A = B A > B A < B

24. Chem 106, Prof. J.T. Spencer 24 Matter; Measurement A B Which is True? A = B A > B A < B

25. Chem 106, Prof. J.T. Spencer 25 Matter; Measurement A B Which is True? A = B A > B A < B

26. Chem 106, Prof. J.T. Spencer 26 Matter; Measurement Chapt. 1.4 u Systems –Metric - base 10 –SI- international scientific system –massKilogram –lengthMeter –timeSecond –electric currentAmpere –temperatureKelvin –lightCandela –AmountMole u Factor label method for conversions

27. Chem 106, Prof. J.T. Spencer 27 Matter; Measurement Chapt. 1.4 u Prefixes MegaM10 6 Kilok10 3 Decid10 -1 Centic10 -2 Millim10 -3 Micro  10 -6 Nanon10 -9

28. Chem 106, Prof. J.T. Spencer 28 Matter; Measurement Chapt. 1.4 Sample exercise: What fraction of a second is a picosecond, ps?

29. Chem 106, Prof. J.T. Spencer 29 Matter; Measurement Chapt. 1.4 Sample exercise: What fraction of a second is a picosecond, ps? 10 -12 second

30. Chem 106, Prof. J.T. Spencer 30 Matter; Measurement Chapt. 1.4 Common Units: 3Length and Mass 3Length - unit of distance measured in meters 3Mass - measures the amount of matter in an object in grams 3Temperature 3Kelvin 3Celsius°C = 5/9 (°F -32) K = °C + 273.15

31. Chem 106, Prof. J.T. Spencer 31 Matter; Measurement Chapt. 1.4 Sample exercise: Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5°C. What is the freezing point in a) K b) °F

32. Chem 106, Prof. J.T. Spencer 32 Matter; Measurement Chapt. 1.4 Sample exercise: Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5°C. What is the freezing point in a) K b) °F K = °C + 273.15 = -11.5 + 273.15

33. Chem 106, Prof. J.T. Spencer 33 Matter; Measurement Chapt. 1.4 Sample exercise: Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5°C. What is the freezing point in a) K b) °F K = °C + 273.15 = -11.5 + 273.15 = 261.65 K

34. Chem 106, Prof. J.T. Spencer 34 Matter; Measurement Chapt. 1.4 Sample exercise: Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5°C. What is the freezing point in a) K b) °F K = °C + 273.15 = -11.5 + 273.15 = 261.65 K = 261.7 K

35. Chem 106, Prof. J.T. Spencer 35 Matter; Measurement Chapt. 1.4 Sample exercise: Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5°C. What is the freezing point in a) K b) °F °C = 5/9 (°F -32)

36. Chem 106, Prof. J.T. Spencer 36 Matter; Measurement Chapt. 1.4 Sample exercise: Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5°C. What is the freezing point in a) K b) °F °C = 5/9 (°F - 32) -11.5 = 5/9(x - 32)

37. Chem 106, Prof. J.T. Spencer 37 Matter; Measurement Chapt. 1.4 Sample exercise: Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5°C. What is the freezing point in a) K b) °F °C = 5/9 (°F - 32) -11.5 = 5/9(x - 32) 9(-11.5) + 32 = x 5

38. Chem 106, Prof. J.T. Spencer 38 Matter; Measurement Chapt. 1.4 Sample exercise: Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5°C. What is the freezing point in a) K b) °F °C = 5/9 (°F - 32) -11.5 = 5/9(x - 32) 9(-11.5) + 32 = x 5 11.3°F = x

39. Chem 106, Prof. J.T. Spencer 39 Matter; Measurement Chapt. 1.4 Derived Units: 3Volume 3Length x length x length 3measured in cm 3, which is equal to mL

40. Chem 106, Prof. J.T. Spencer 40 Matter; Measurement Chapt. 1.4 Derived Units: 3Density 3amount of mass per unit volume 3measured in g/cm 3, or g/mL

41. Chem 106, Prof. J.T. Spencer 41 Matter; Measurement Chapt. 1.4 Sample exercise: A student needs 15.0 g of ethanol (ethyl alcohol) for an experiment. If the density of the alcohol is 0.789 g/mL, how many milliliters of alcohol are needed?

42. Chem 106, Prof. J.T. Spencer 42 Matter; Measurement Chapt. 1.4 Sample exercise: A student needs 15.0 g of ethanol (ethyl alcohol) for an experiment. If the density of the alcohol is 0.789 g/mL, how many milliliters of alcohol are needed? D = m/V so V = m/D

43. Chem 106, Prof. J.T. Spencer 43 Matter; Measurement Chapt. 1.4 Sample exercise: A student needs 15.0 g of ethanol (ethyl alcohol) for an experiment. If the density of the alcohol is 0.789 g/mL, how many milliliters of alcohol are needed? D = m/V so V = m/D = 15.0 g 0.789 g/mL

44. Chem 106, Prof. J.T. Spencer 44 Matter; Measurement Chapt. 1.4 Sample exercise: A student needs 15.0 g of ethanol (ethyl alcohol) for an experiment. If the density of the alcohol is 0.789 g/mL, how many milliliters of alcohol are needed? D = m/V so V = m/D = 15.0 g 0.789 g/mL = 19.0 mL

45. Chem 106, Prof. J.T. Spencer 45 Matter; Uncertainty in Measurement Chapt. 1.5 u Precision and Accuracy –Precision - how closely individual measurements agree –Accuracy- how closely the measurements agree with the true value u Significant Figures –All measurements are inaccurate intrinsically –measured quantities are reported such that the last figure is uncertain

46. Chem 106, Prof. J.T. Spencer 46 Matter; Uncertainty in Measurement Good Precision Poor Accuracy Good Precision Good Accuracy Poor Precision Poor Accuracy

47. Chem 106, Prof. J.T. Spencer 47 u Determining Significant Figures –all non zero digits are significant –zeros between nonzero digits are significant –zeros to the left of first nonzero digit are not significant –zeros at the end of a number and to the right of a decimal point are significant –when a number ends in a zero but with no decimal point, the zero may or may not be signigicant (use scientific notation) Matter; Uncertainty in Measurement Chapt. 1.5

48. Chem 106, Prof. J.T. Spencer 48 u Determining Significant Figures 3.573 has 4 significant figures 0.073 has 2 significant figures 3.070 has 4 significant figures 0.003 has 1 significant figures - multiplication and division; result can have no more than the figure with the fewest significant figures - addition and subtraction; result can have the same number of decimal places as the term with the least number of decimal places Matter; Uncertainty in Measurement Chapt. 1.5

49. Chem 106, Prof. J.T. Spencer 49 Matter; Uncertainty in Measurement Chapt. 1.5 Sample exercise: A balance has a precision of 0.001 g. A sample that weighs about 25 g is weighed on this balance. How many significant figures should be reported for this measurement?

50. Chem 106, Prof. J.T. Spencer 50 Matter; Uncertainty in Measurement Chapt. 1.5 Sample exercise: A balance has a precision of 0.001 g. A sample that weighs about 25 g is weighed on this balance. How many significant figures should be reported for this measurement? 25.XXX

51. Chem 106, Prof. J.T. Spencer 51 Matter; Uncertainty in Measurement Chapt. 1.5 Sample exercise: A balance has a precision of 0.001 g. A sample that weighs about 25 g is weighed on this balance. How many significant figures should be reported for this measurement? 25.XXX 5 sig figs

52. Chem 106, Prof. J.T. Spencer 52 Matter; Uncertainty in Measurement Chapt. 1.5 Sample exercise: How many significant figures are in each of the following measurements? A) 3.549 g B) 2.3 x 10 4 cm C) 0.00134 m 3

53. Chem 106, Prof. J.T. Spencer 53 Matter; Uncertainty in Measurement Chapt. 1.5 Sample exercise: How many significant figures are in each of the following measurements? A) 3.549 g4 sig figs B) 2.3 x 10 4 cm C) 0.00134 m 3

54. Chem 106, Prof. J.T. Spencer 54 Matter; Uncertainty in Measurement Chapt. 1.5 Sample exercise: How many significant figures are in each of the following measurements? A) 3.549 g4 sig figs B) 2.3 x 10 4 cm2 sig figs C) 0.00134 m 3

55. Chem 106, Prof. J.T. Spencer 55 Matter; Uncertainty in Measurement Chapt. 1.5 Sample exercise: How many significant figures are in each of the following measurements? A) 3.549 g4 sig figs B) 2.3 x 10 4 cm2 sig figs C) 0.00134 m 3 3 sig figs

56. Chem 106, Prof. J.T. Spencer 56 Matter; Uncertainty in Measurement Chapt. 1.5 Sample exercise: There are exactly 1609.344 m in a mile. How many meters are in a distance of 1.35 mi?

57. Chem 106, Prof. J.T. Spencer 57 Matter; Uncertainty in Measurement Chapt. 1.5 Sample exercise: There are exactly 1609.344 m in a mile. How many meters are in a distance of 1.35 mi? 1.35 mi = 1 mi x1609.344 m

58. Chem 106, Prof. J.T. Spencer 58 Matter; Uncertainty in Measurement Chapt. 1.5 Sample exercise: There are exactly 1609.344 m in a mile. How many meters are in a distance of 1.35 mi? 1.35 mi = 1 mi x1609.344 m x = 2172.6144 m

59. Chem 106, Prof. J.T. Spencer 59 Matter; Uncertainty in Measurement Chapt. 1.5 Sample exercise: There are exactly 1609.344 m in a mile. How many meters are in a distance of 1.35 mi? 1.35 mi = 1 mi x1609.344 m 1.35 has 3 sig figs x = 2172.6144 m 1609.344 has 7 sig figs 1 is infinitely significant

60. Chem 106, Prof. J.T. Spencer 60 Matter; Uncertainty in Measurement Chapt. 1.5 Sample exercise: There are exactly 1609.344 m in a mile. How many meters are in a distance of 1.35 mi? 1.35 mi = 1 mi x1609.344 m 1.35 has 3 sig figs x = 2172.6144 m 1609.344 has 7 sig figs x = 2170 m 1 is infinitely significant

61. Chem 106, Prof. J.T. Spencer 61 u Use Units throughout the calculation (helps “guide” calculation. u Should always yield the proper units u Uses conversion factors u Example; How fast is 50 mph in in/sec.? Dimensional Analysis 50 mi.1 hour5280 ft12 in. =in 1 hour3600 sec.1 mi.1 ftsec.

62. Chem 106, Prof. J.T. Spencer 62 Dimensional Analysis Chapt. 1.6 Sample exercise: By using a conversion factor from the back inside cover, determine the length in kilometers of a 500.0 mi automobile race.

63. Chem 106, Prof. J.T. Spencer 63 Dimensional Analysis Chapt. 1.6 Sample exercise: By using a conversion factor from the back inside cover, determine the length in kilometers of a 500.0 mi automobile race. 500.0 mi

64. Chem 106, Prof. J.T. Spencer 64 Dimensional Analysis Chapt. 1.6 Sample exercise: By using a conversion factor from the back inside cover, determine the length in kilometers of a 500.0 mi automobile race. 500.0 mi 1 km 0.62137 mi

65. Chem 106, Prof. J.T. Spencer 65 Dimensional Analysis Chapt. 1.6 Sample exercise: By using a conversion factor from the back inside cover, determine the length in kilometers of a 500.0 mi automobile race. 500.0 mi 1 km = 804.674 km 0.62137 mi

66. Chem 106, Prof. J.T. Spencer 66 Dimensional Analysis Chapt. 1.6 Sample exercise: By using a conversion factor from the back inside cover, determine the length in kilometers of a 500.0 mi automobile race. 500.0 mi 1 km = 804.674 km 0.62137 mi * answer can only have 4 sig figs; 804.7 km

67. Chem 106, Prof. J.T. Spencer 67 Dimensional Analysis Chapt. 1.6 Sample exercise: The distance between carbon atoms in a diamond is 154 pm. Convert this distance to millimeters.

68. Chem 106, Prof. J.T. Spencer 68 Dimensional Analysis Chapt. 1.6 Sample exercise: The distance between carbon atoms in a diamond is 154 pm. Convert this distance to millimeters. 154 pm

69. Chem 106, Prof. J.T. Spencer 69 Dimensional Analysis Chapt. 1.6 Sample exercise: The distance between carbon atoms in a diamond is 154 pm. Convert this distance to millimeters. 154 pm 1 m 10 3 mm 10 12 pm 1 m

70. Chem 106, Prof. J.T. Spencer 70 Dimensional Analysis Chapt. 1.6 Sample exercise: The distance between carbon atoms in a diamond is 154 pm. Convert this distance to millimeters. 154 pm 1 m 10 3 mm = 1.54 x 10 -7 mm 10 12 pm 1 m

71. Chem 106, Prof. J.T. Spencer 71 Dimensional Analysis Chapt. 1.6 Sample exercise: A car travels 28 mi to the gallon of gasoline. How many kilometers per liter will it go?

72. Chem 106, Prof. J.T. Spencer 72 Dimensional Analysis Chapt. 1.6 Sample exercise: A car travels 28 mi to the gallon of gasoline. How many kilometers per liter will it go? 28 mi gal

73. Chem 106, Prof. J.T. Spencer 73 Dimensional Analysis Chapt. 1.6 Sample exercise: A car travels 28 mi to the gallon of gasoline. How many kilometers per liter will it go? 28 mi 1 km gal 0.62137 mi

74. Chem 106, Prof. J.T. Spencer 74 Dimensional Analysis Chapt. 1.6 Sample exercise: A car travels 28 mi to the gallon of gasoline. How many kilometers per liter will it go? 28 mi 1 km 1 gal gal 0.62137 mi 3.7854 L

75. Chem 106, Prof. J.T. Spencer 75 Dimensional Analysis Chapt. 1.6 Sample exercise: A car travels 28 mi to the gallon of gasoline. How many kilometers per liter will it go? 28 mi 1 km 1 gal = 11.9041 km gal 0.62137 mi 3.7854 L L

76. Chem 106, Prof. J.T. Spencer 76 Dimensional Analysis Chapt. 1.6 Sample exercise: A car travels 28 mi to the gallon of gasoline. How many kilometers per liter will it go? 28 mi 1 km 1 gal = 11.9041 km gal 0.62137 mi 3.7854 L L * 2 sig figs = 12 km L

77. Chem 106, Prof. J.T. Spencer 77 u Matter: Chemical and Physical Changes u Elements and Compounds u Units of Measurement u Uncertainty and Significant Figures u Precision and Accuracy u “Factor Label” Method (Dimensional Analysis) Chapter One; Review