2. John C. Kotz State University of New York, College at Oneonta John C. Kotz Paul M. Treichel John Townsend http://academic.cengage.com/kotz Chapter 2 Atoms, Molecules, and Ions

3. Important – Read Before Using Slides in Class Instructor: This PowerPoint presentation contains photos and figures from the text, as well as selected animations and videos. For animations and videos to run properly, we recommend that you run this PowerPoint presentation from the PowerLecture disc inserted in your computer. Also, for the mathematical symbols to display properly, you must install the supplied font called “Symb_chm,” supplied as a cross-platform TrueType font in the “Font_for_Lectures” folder in the "Media" folder on this disc. If you prefer to customize the presentation or run it without the PowerLecture disc inserted, the animations and videos will only run properly if you also copy the associated animation and video files for each chapter onto your computer. Follow these steps: 1.Go to the disc drive directory containing the PowerLecture disc, and then to the “Media” folder, and then to the “PowerPoint_Lectures” folder. 2.In the “PowerPoint_Lectures” folder, copy the entire chapter folder to your computer. Chapter folders are named “chapter1”, “chapter2”, etc. Each chapter folder contains the PowerPoint Lecture file as well as the animation and video files. For assistance with installing the fonts or copying the animations and video files, please visit our Technical Support at http://academic.cengage.com/support or call (800) 423-0563. Thank you. http://academic.cengage.com/support

4. 3 © 2009 Brooks/Cole - Cengage A TOMS, Molecules, & Ions

5. 4 © 2009 Brooks/Cole - Cengage ATOMIC COMPOSITION ProtonsProtons –+ electrical charge –mass = 1.672623 x 10 -24 g –relative mass = 1.007 atomic mass units (u) ElectronsElectrons – negative electrical charge –relative mass = 0.0005 u NeutronsNeutrons – no electrical charge –mass = 1.009 u

6. 5 © 2009 Brooks/Cole - Cengage ATOM COMPOSITION protons and neutrons in the nucleus.protons and neutrons in the nucleus. the number of electrons is equal to the number of protons.the number of electrons is equal to the number of protons. electrons in space around the nucleus.electrons in space around the nucleus. extremely small. One teaspoon of water has 3 times as many atoms as the Atlantic Ocean has teaspoons of water.extremely small. One teaspoon of water has 3 times as many atoms as the Atlantic Ocean has teaspoons of water. The atom is mostly empty space PLAY MOVIE

7. 6 © 2009 Brooks/Cole - Cengage Atomic Number, Z All atoms of the same element have the same number of protons in the nucleus, Z 13 Al 26.981 Atomic number, Z Atom symbol Atomic weight

8. 7 © 2009 Brooks/Cole - Cengage Atomic Weight This tells us the mass of one atom of an element relative to one atom of another element.This tells us the mass of one atom of an element relative to one atom of another element. OR — the mass of 1000 atoms of one relative to 1000 atoms of another.OR — the mass of 1000 atoms of one relative to 1000 atoms of another. For example, an O atom is approximately 16 times heavier than an H atom.For example, an O atom is approximately 16 times heavier than an H atom. Define one element as the standard against which all others are measuredDefine one element as the standard against which all others are measured Standard = carbonStandard = carbon

9. 8 © 2009 Brooks/Cole - Cengage Mass Number, A C atom with 6 protons and 6 neutrons is the mass standardC atom with 6 protons and 6 neutrons is the mass standard = 12 atomic mass units (u)= 12 atomic mass units (u) Mass Number (A) = # protons + # neutronsMass Number (A) = # protons + # neutrons A boron atom can have A = 5 p + 5 n = 10 uA boron atom can have A = 5 p + 5 n = 10 u

10. 9 © 2009 Brooks/Cole - Cengage Boron in Death Valley Death Valley has been a major source of borax and other boron-containing minerals.Death Valley has been a major source of borax and other boron-containing minerals. Borax was transported out of Death Valley in wagons pulled by teams of 20 mules.Borax was transported out of Death Valley in wagons pulled by teams of 20 mules.

11. 10 © 2009 Brooks/Cole - Cengage Isotopes Atoms of the same element (same Z) but different mass number (A).Atoms of the same element (same Z) but different mass number (A). Boron-10 has 5 p and 5 n: 10 5 BBoron-10 has 5 p and 5 n: 10 5 B Boron-11 has 5 p and 6 n: 11 5 BBoron-11 has 5 p and 6 n: 11 5 B 10 B 11 B

12. 11 © 2009 Brooks/Cole - Cengage Hydrogen Isotopes Hydrogen has _____ isotopes 11H11H11H11H 21H21H21H21H 31H31H31H31H 1 proton and 2 neutrons, tritium radioactive 1 proton and 1 neutron, deuterium 1 proton and 0 neutrons, protium

13. 12 © 2009 Brooks/Cole - Cengage Isotope Composition IsotopeElectronsProtonsNeutrons Sulfur-32 Bromine- 79

14. 13 © 2009 Brooks/Cole - Cengage Isotopes & Their Uses Heart scans with radioactive technetium-99. 99 43 Tc Emits gamma rays

15. 14 © 2009 Brooks/Cole - Cengage Masses of Isotopes determined with a mass spectrometer See Active Figure 2.3

16. 15 © 2009 Brooks/Cole - Cengage Mass spectrum of C 6 H 5 Br

17. 16 © 2009 Brooks/Cole - Cengage Isotopes Because of the existence of isotopes, the mass of a collection of atoms has an average value.Because of the existence of isotopes, the mass of a collection of atoms has an average value. Average mass = ATOMIC WEIGHTAverage mass = ATOMIC WEIGHT Boron is 19.9% 10 B and 80.1% 11 B. That is, 11 B is 80.1 percent abundant on earth.Boron is 19.9% 10 B and 80.1% 11 B. That is, 11 B is 80.1 percent abundant on earth. For boron atomic weightFor boron atomic weight = 0.199 (10.0 u) + 0.801 (11.0 u) = 10.8 u = 0.199 (10.0 u) + 0.801 (11.0 u) = 10.8 u 10 B 11 B

18. 17 © 2009 Brooks/Cole - Cengage Isotopes & Atomic Weight Because of the existence of isotopes, the mass of a collection of atoms has an average value.Because of the existence of isotopes, the mass of a collection of atoms has an average value. 6 Li = 7.5% abundant and 7 Li = 92.5% 6 Li = 7.5% abundant and 7 Li = 92.5% –Atomic weight of Li = ______________ 28 Si = 92.23%, 29 Si = 4.67%, 30 Si = 3.10% 28 Si = 92.23%, 29 Si = 4.67%, 30 Si = 3.10% –Atomic weight of Si = ______________

19. 18 © 2009 Brooks/Cole - Cengage Periodic Table Dmitri Mendeleev (1834- 1907) developed the modern periodic table. Argued that element properties are periodic functions of their atomic weights.Dmitri Mendeleev (1834- 1907) developed the modern periodic table. Argued that element properties are periodic functions of their atomic weights. We now know that element properties are periodic functions of their ATOMIC NUMBERS.We now know that element properties are periodic functions of their ATOMIC NUMBERS.

20. 19 © 2009 Brooks/Cole - Cengage Periods in the Periodic Table PLAY MOVIE

21. 20 © 2009 Brooks/Cole - Cengage Groups/Families in the Periodic Table PLAY MOVIE

22. 21 © 2009 Brooks/Cole - Cengage Regions of the Periodic Table

23. 22 © 2009 Brooks/Cole - Cengage http://www.webelements.com/ Element Abundance Fe C Al O Si

24. 23 © 2009 Brooks/Cole - Cengage HydrogenHydrogen Shuttle main engines use H 2 and O 2 PLAY MOVIE

25. 24 © 2009 Brooks/Cole - Cengage Group 1A: Alkali Metals Li, Na, K, Rb, Cs Cutting sodium metal Reaction of potassium + H 2 O PLAY MOVIE

26. 25 © 2009 Brooks/Cole - Cengage Magnesium Magnesium oxide Group 2A: Alkaline Earth Metals Be, Mg, Ca, Sr, Ba, Ra PLAY MOVIE

27. 26 © 2009 Brooks/Cole - Cengage Group 3A: B, Al, Ga, In, Tl Al resists corrosion (here in nitric acid). Gallium is one of the few metals that can be liquid at room temp. CuAl

28. 27 © 2009 Brooks/Cole - Cengage Gems & Minerals Sapphire: Al 2 O 3 with Fe 3+ or Ti 3+ impurity gives blue whereas V 3+ gives violet.Sapphire: Al 2 O 3 with Fe 3+ or Ti 3+ impurity gives blue whereas V 3+ gives violet. Ruby: Al 2 O 3 with Cr 3+ impurityRuby: Al 2 O 3 with Cr 3+ impurity

29. 28 © 2009 Brooks/Cole - Cengage Group 4A: C, Si, Ge, Sn, Pb Quartz, SiO 2 Diamond

30. 29 © 2009 Brooks/Cole - Cengage Group 5A: N, P, As, Sb, Bi White and red phosphorus Ammonia, NH 3 PLAY MOVIE

31. 30 © 2009 Brooks/Cole - Cengage Phosphorus Phosphorus first isolated by Brandt from urine, 1669

32. 31 © 2009 Brooks/Cole - Cengage Group 6A: O, S, Se, Te, Po Sulfuric acid dripping from snot-tite in cave in Mexico Elemental S has a ring structure.

33. 32 © 2009 Brooks/Cole - Cengage Group 7A: Halogens F, Cl, Br, I, At PLAY MOVIE

34. 33 © 2009 Brooks/Cole - Cengage Group 8A: Noble Gases He, Ne, Ar, Kr, Xe, Rn

35. 34 © 2009 Brooks/Cole - Cengage Transition Elements Lanthanides and actinides Iron in air gives iron(III) oxide

36. 35 © 2009 Brooks/Cole - Cengage Colors of Transition Metal Compounds IronCobaltNickelCopperZinc

37. 36 © 2009 Brooks/Cole - Cengage Molecules, Ions & Their Compounds NaCl, salt Buckyball, C 60 Ethanol, C 2 H 6 O PLAY MOVIE

38. 37 © 2009 Brooks/Cole - Cengage Compounds & Molecules COMPOUNDS are a combination of 2 or more elements in definite ratios by mass.COMPOUNDS are a combination of 2 or more elements in definite ratios by mass. The character of each element is lost when forming a compound.The character of each element is lost when forming a compound. MOLECULES are the smallest unit of a compound that retains the characteristics of the compound.MOLECULES are the smallest unit of a compound that retains the characteristics of the compound.

39. 38 © 2009 Brooks/Cole - Cengage MOLECULAR FORMULAS Formula for glycine is C 2 H 5 NO 2Formula for glycine is C 2 H 5 NO 2 In one molecule there areIn one molecule there are –2 C atoms –5 H atoms –1 N atom –2 O atoms

40. 39 © 2009 Brooks/Cole - Cengage WRITING FORMULAS Can also write glycine formula as –H 2 NCH 2 COOH to show atom ordering structural formulaor in the form of a structural formula

41. 40 © 2009 Brooks/Cole - Cengage MOLECULAR MODELING Ball & stick Space-filling Structural formula of glycine

42. 41 © 2009 Brooks/Cole - Cengage Molecular & Ionic Compounds Iron-porphyrinNaCl Fe N

43. 42 © 2009 Brooks/Cole - Cengage ELEMENTS THAT EXIST AS MOLECULES Allotropes of C

44. 43 © 2009 Brooks/Cole - Cengage ELEMENTS THAT EXIST AS DIATOMIC MOLECULES

45. 44 © 2009 Brooks/Cole - Cengage ELEMENTS THAT EXIST AS POLYATOMIC MOLECULES White P 4 and polymeric red phosphorus S 8 sulfur molecules

46. 45 © 2009 Brooks/Cole - Cengage IONS AND IONIC COMPOUNDS

47. 46 © 2009 Brooks/Cole - Cengage IONS AND IONIC COMPOUNDS IONS are atoms or groups of atoms with a positive or negative charge.IONS are atoms or groups of atoms with a positive or negative charge. Taking away an electron from an atom gives a CATION with a positive chargeTaking away an electron from an atom gives a CATION with a positive charge Adding an electron to an atom gives an ANION with a negative charge.Adding an electron to an atom gives an ANION with a negative charge.

48. 47 © 2009 Brooks/Cole - Cengage Forming Cations & Anions A CATION forms when an atom loses one or more electrons. An ANION forms when an atom gains one or more electrons Mg f Mg 2+ + 2 e- F + e- f F - PLAY MOVIE

49. 48 © 2009 Brooks/Cole - Cengage See Active Figure 2.17

50. 49 © 2009 Brooks/Cole - Cengage PREDICTING ION CHARGES In general metals (Mg) lose electrons cationsmetals (Mg) lose electrons f cations nonmetals (F) gain electrons anionsnonmetals (F) gain electrons f anions See Figure 2.18

51. 50 © 2009 Brooks/Cole - Cengage Charges on Common Ions +3 -4-2-3 +1 +2 By losing or gaining e-, atom has same number of e-’s as nearest Group 8A atom.

52. 51 © 2009 Brooks/Cole - Cengage Predicting Charges on Monatomic Ions

53. 52 © 2009 Brooks/Cole - Cengage METALS M n e- + M n+ M f n e- + M n+ where n = periodic group Na + sodium ion Mg 2+ magnesium ion Al 3+ aluminum ion Transition metals M 2+ or M 3+ are common Transition metals f M 2+ or M 3+ are common Fe 2+ iron(II) ion Fe 3+ iron(III) ion

54. 53 © 2009 Brooks/Cole - Cengage NONMETALSNONMETALS NONMETAL + n e - f X n- where n = 8 - Group no. C 4-,carbide N 3-, nitride O 2-, oxide S 2-, sulfide F -, fluoride Cl -, chloride Group 7AGroup 6A Group 4A Group 5A Br -, bromide I -, iodide Name derived by adding -ide to stem

55. 54 © 2009 Brooks/Cole - Cengage Ion Formation Reaction of aluminum and bromine PLAY MOVIE

56. 55 © 2009 Brooks/Cole - Cengage POLYATOMIC IONS Groups of atoms with a charge. MEMORIZE the names and formulas in Table 2.4, page 74. Celestite, SrSO 4

57. 56 © 2009 Brooks/Cole - Cengage Note: many O containing anions have names ending in –ate (or -ite).

58. 57 © 2009 Brooks/Cole - Cengage Polyatomic Ions HNO 3 nitric acid NO 3 - nitrate ion

59. 58 © 2009 Brooks/Cole - Cengage Polyatomic Ions NH 4 + ammonium ion One of the few common polyatomic cations

60. 59 © 2009 Brooks/Cole - Cengage Polyatomic Ions CO 3 2- carbonate ion HCO 3 - bicarbonate ion hydrogen carbonate

61. 60 © 2009 Brooks/Cole - Cengage PO 4 3- phosphate ion CH 3 CO 2 - acetate ion Polyatomic Ions

62. 61 © 2009 Brooks/Cole - Cengage SO 4 2- sulfate ion SO 3 2- sulfite ion Polyatomic Ions

63. 62 © 2009 Brooks/Cole - Cengage NO 3 - nitrate ion NO 2 - nitrite ion Polyatomic Ions

64. 63 © 2009 Brooks/Cole - Cengage CATION + ANION f COMPOUND COMPOUND CATION + ANION f COMPOUND COMPOUND A neutral compd. requires equal number of + equal number of + and - charges. A neutral compd. requires equal number of + equal number of + and - charges. COMPOUNDS FORMED FROM IONS Na + + Cl - f NaCl

65. 64 © 2009 Brooks/Cole - Cengage IONIC COMPOUNDS NH 4 + Cl - ammonium chloride, NH 4 Cl

66. 65 © 2009 Brooks/Cole - Cengage Some Ionic Compounds Mg 2+ + NO 3 - f Mg(NO 3 ) 2 magnesium nitrate Fe 2+ + PO 4 3- f Fe 3 (PO 4 ) 2 iron(II) phosphate Be sure to practice naming compounds. calcium fluoride Ca 2+ + 2 F - f CaF 2

67. 66 © 2009 Brooks/Cole - Cengage Properties of Ionic Compounds Forming NaCl from Na and Cl 2 A metal atom can transfer an electron to a nonmetal.A metal atom can transfer an electron to a nonmetal. The resulting cation and anion are attracted to each other by electrostatic forces.The resulting cation and anion are attracted to each other by electrostatic forces. PLAY MOVIE

68. 67 © 2009 Brooks/Cole - Cengage Electrostatic Forces The oppositely charged ions in ionic compounds are attracted to one another by ELECTROSTATIC FORCES. These forces are governed by COULOMB’S LAW. PLAY MOVIE

69. 68 © 2009 Brooks/Cole - Cengage Electrostatic Forces COULOMB’S LAW As ion charge increases, the attractive force _______________. As the distance between ions increases, the attractive force ________________. This idea is important and will come up many times in future discussions!

70. 69 © 2009 Brooks/Cole - Cengage Electrostatic Forces COULOMB’S LAW See Active Figure 2.21

71. 70 © 2009 Brooks/Cole - Cengage Importance of Coulomb’s Law NaCl, Na + and Cl -, m.p. 804 o C MgO, Mg 2+ and O 2- m.p. 2800 o C

72. 71 © 2009 Brooks/Cole - Cengage Molecular Compounds Compounds without Ions CH 4 methane CO 2 Carbon dioxide BCl 3 boron trichloride PLAY MOVIE

73. 72 © 2009 Brooks/Cole - Cengage Naming Molecular Compounds CH 4 methane BCl 3 boron trichloride CO 2 Carbon dioxide All are formed from two or more nonmetals. Ionic compounds generally involve a metal and nonmetal (NaCl) Ionic compounds generally involve a metal and nonmetal (NaCl) PLAY MOVIE

74. 73 © 2009 Brooks/Cole - Cengage Counting Atoms Mg burns in air (O 2 ) to produce white magnesium oxide, MgO. How can we figure out how much oxide is produced from a given mass of Mg? PLAY MOVIE

75. 74 © 2009 Brooks/Cole - Cengage Counting Atoms Chemistry is a quantitative science—we need a “counting unit.” 1 mole is the amount of substance that contains as many particles (atoms, molecules) as there are in 12.0 g of 12 C. MOLE 518 g of Pb, 2.50 mol

76. 75 © 2009 Brooks/Cole - Cengage Particles in a Mole 6.0221415 x 10 23 Avogadro’s Number There is Avogadro’s number of particles in a mole of any substance. Amedeo Avogadro 1776-1856

77. 76 © 2009 Brooks/Cole - Cengage Molar Mass 1 mol of 12 C = 12.00 g of C = 6.022 x 10 23 atoms of C 12.00 g of 12 C is its MOLAR MASS Taking into account all of the isotopes of C, the molar mass of C is 12.011 g/mol

78. 77 © 2009 Brooks/Cole - Cengage One-mole Amounts

79. 78 © 2009 Brooks/Cole - Cengage PROBLEM: What amount of Mg is represented by 0.200 g? How many atoms? Mg has a molar mass of 24.3050 g/mol. = 4.95 x 10 21 atoms Mg How many atoms in this piece of Mg?

80. 79 © 2009 Brooks/Cole - Cengage MOLECULAR WEIGHT AND MOLAR MASS Molecular weight = sum of the atomic weights of all atoms in the molecule. Molar mass = molecular weight in grams

81. 80 © 2009 Brooks/Cole - Cengage What is the molar mass of ethanol, C 2 H 6 O? 1 mol contains 2 mol C (12.01 g C/1 mol) = 24.02 g C 6 mol H (1.01 g H/1 mol) = 6.06 g H 1 mol O (16.00 g O/1 mol) = 16.00 g O TOTAL = molar mass = 46.08 g/mol

82. 81 © 2009 Brooks/Cole - Cengage Formula = Molar mass = TylenolTylenol C 8 H 9 NO 2 151.2 g/mol

83. 82 © 2009 Brooks/Cole - Cengage Molar Mass Note that the mass of water is included in the molar mass of a compound.

84. 83 © 2009 Brooks/Cole - Cengage Empirical & Molecular Formulas A pure compound always consists of the same elements combined in the same proportions by weight. Therefore, we can express molecular composition as PERCENT BY WEIGHT Ethanol, C 2 H 6 O 52.13% C 13.15% H 34.72% O Ethanol, C 2 H 6 O 52.13% C 13.15% H 34.72% O

85. 84 © 2009 Brooks/Cole - Cengage Percent Composition Consider some of the family of nitrogen- oxygen compounds: NO 2, nitrogen dioxide and closely related, NO, nitrogen monoxide (or nitric oxide) Structure of NO 2 Chemistry of NO, nitrogen monoxide PLAY MOVIE

86. 85 © 2009 Brooks/Cole - Cengage Percent Composition Consider NO 2, Molar mass = ? What is the weight percent of N and of O? What are the weight percentages of N and O in NO?

87. 86 © 2009 Brooks/Cole - Cengage How to Determine a Formula? Mass spectrometer

88. 87 © 2009 Brooks/Cole - Cengage Mass Spectrum of Ethanol

89. 88 © 2009 Brooks/Cole - Cengage Determining Formulas In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula. PROBLEM : A compound of B and H is 81.10% B. What is its empirical formula?

90. 89 © 2009 Brooks/Cole - Cengage Because it contains only B and H, it must contain 18.90% H.Because it contains only B and H, it must contain 18.90% H. In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H.In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H. Calculate the number of moles of each constituent.Calculate the number of moles of each constituent. A compound of B and H is 81.10% B. What is its empirical formula?

91. 90 © 2009 Brooks/Cole - Cengage Calculate the number of moles of each element in 100.0 g of sample. A compound of B and H is 81.10% B. What is its empirical formula?

92. 91 © 2009 Brooks/Cole - Cengage Now, recognize that atoms combine in the ratio of small whole numbers. 1 atom B + 3 atoms H f 1 molecule BH 3 or 1 mol B atoms + 3 mol H atoms f 1 mol BH 3 molecules Find the ratio of moles of elements in the compound. A compound of B and H is 81.10% B. What is its empirical formula?

93. 92 © 2009 Brooks/Cole - Cengage But we need a whole number ratio. 2.5 mol H/1.0 mol B = 5 mol H to 2 mol B EMPIRICAL FORMULA = B 2 H 5 Take the ratio of moles of B and H. Always divide by the smaller number. A compound of B and H is 81.10% B. What is its empirical formula?

94. 93 © 2009 Brooks/Cole - Cengage A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5. What is its molecular formula ? Is the molecular formula B 2 H 5, B 4 H 10, B 6 H 15, B 8 H 20, etc.? B 2 H 6 is one example of this class of compounds. B2H6B2H6

95. 94 © 2009 Brooks/Cole - Cengage A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5. What is its molecular formula ? We need to do an EXPERIMENT to find the MOLAR MASS. Here experiment gives 53.3 g/mol Compare with the mass of B 2 H 5 = 26.66 g/unit = 26.66 g/unit Find the ratio of these masses. Molecular formula = B 4 H 10

96. 95 © 2009 Brooks/Cole - Cengage DETERMINE THE FORMULA OF A COMPOUND OF Sn AND I Sn(s) + some I 2 (s) f SnI x

97. 96 © 2009 Brooks/Cole - Cengage Data to Determine the formula of a Sn—I Compound Reaction of Sn and I 2 is done using excess Sn.Reaction of Sn and I 2 is done using excess Sn. Mass of Sn in the beginning = 1.056 gMass of Sn in the beginning = 1.056 g Mass of iodine (I 2 ) used = 1.947 gMass of iodine (I 2 ) used = 1.947 g Mass of Sn remaining = 0.601 gMass of Sn remaining = 0.601 g See p. 93See p. 93

98. 97 © 2009 Brooks/Cole - Cengage Find the mass of Sn that combined with 1.947 g I 2. Mass of Sn initially = 1.056 g Mass of Sn recovered = 0.601 g Mass of Sn used = 0.455 g Find moles of Sn used: Tin and Iodine Compound

99. 98 © 2009 Brooks/Cole - Cengage Tin and Iodine Compound Now find the number of moles of I 2 that combined with 3.83 x 10 -3 mol Sn. Mass of I 2 used was 1.947 g. How many mol of iodine atoms ? = 1.534 x 10 -2 mol I atoms

100. 99 © 2009 Brooks/Cole - Cengage Tin and Iodine Compound Now find the ratio of number of moles of moles of I and Sn that combined. Empirical formula is SnI 4