1. 1 © 2008 Brooks/Cole Chapter 2:Atoms and Elements Chemistry: The Molecular Science Moore, Stanitski and Jurs

2. 2 © 2008 Brooks/Cole Atoms are composed of subatomic particles: electron (e - ), proton (p + ) and neutron (n 0 ). Key discoveries: Radioactivity Becquerel (1896) U ore emits rays that “fog” a photographic plate. Marie and Pierre Curie (1898) Isolated new elements (Po & Ra) that did the same. radioactivityMarie Curie called the phenomenon radioactivity. Atomic Structure and Subatomic Particles

3. 3 © 2008 Brooks/Cole Radioactivity Electrical behavior: “+” attracts “-” but like charges repel Atoms must contain smaller sub-units. Radioactive material Beam of , , and  Electrically charged plates β-particles (“–”) Gamma ray (γ) No charge, no deflection α-particle (“+” ) Heavier, deflected less than β – +

4. 4 © 2008 Brooks/Cole Electric and magnetic fields deflect the beam. Gives mass/charge of e - = −5.60 x 10 -9 g/C Coulomb (C) = SI unit of charge Electrons Thomson (1897) discovered the e - : “Cathode rays” Travel from cathode (-) to anode (+). Negative charge (e − ). Emitted by cathode metal atoms. fluorescent screen – high voltage + cathode ray

5. 5 © 2008 Brooks/Cole Electrons Millikan (1911) studied electrically-charged oil drops.

6. 6 © 2008 Brooks/Cole Charge on each droplet was: n (−1.60 x 10 -19 C) with n = 1, 2, 3,… n (e - charge) Modern value = −1.60217653 x 10 -19 C. = −1 “atomic units”. These experiments give: Modern value = 9.1093826 x 10 -28 g = (-1.60 x 10 -19 C)(-5.60 x 10 -9 g/C) = 8.96 x 10 -28 g m e = charge x mass charge Electrons

7. 7 © 2008 Brooks/Cole Protons Atoms gain a positive charge when e - are lost. Implies a positive fundamental particle. Hydrogen ions had the lowest mass. Hydrogen nuclei assumed to have “unit mass” protonsCalled protons. Modern science: m p = 1.67262129 x 10 -24 g m p ≈ 1800 x m e. Charge = -1 x (e - charge). = +1.602176462 x 10 -19 C = +1 atomic units

8. 8 © 2008 Brooks/Cole The Nuclear Atom How were p + and e - arranged? Thompson: Ball of uniform positive charge, with small negative dots (e - ) stuck in it. The “plum-pudding” model. 1910 1910 Rutherford fired α-particles at thin metal foils. Expected them to pass through with minor deflections.

9. 9 © 2008 Brooks/Cole But … But … some had large deflections. α particles Rutherford “It was about as credible as if you had fired a 15-inch shell at a piece of paper and it came back and hit you.” The Nuclear Atom

10. 10 © 2008 Brooks/Cole ≈10,000 times smaller diameter than the entire atom. e - occupy the remaining space. α particles nucleus Most of the mass and all “+” charge is concentrated in a small core, the nucleus. The Nucleus

11. 11 © 2008 Brooks/Cole Neutrons Atomic mass > mass of all p + and e - in an atom. Rutherford proposed a neutral particle. m n ≈ m p (0.1% larger). m n = 1.67492728 x 10 -24 g. Present in all atoms (except normal H). neutrons Chadwick (1932) fired  -particles at Be atoms. Neutral particles, neutrons, were ejected:

12. 12 © 2008 Brooks/Cole Nucleus Contains p + and n 0 Most of the atomic mass. Small (~10,000x smaller diameter than the atom). Positive (each p + has +1 charge). Small light particles surrounding the nucleus. Occupy most of the volume. Charge = -1. Atoms are neutral. Number of e − = Number of p + Electrons The Nuclear Atom

13. 13 © 2008 Brooks/Cole Scanning Tunneling Microscopy

14. 14 © 2008 Brooks/Cole Scanning Tunneling Microscopy Fe atoms arranged on Cu. “Atom” (Chinese characters)

15. 15 © 2008 Brooks/Cole Sizes of Atoms and Units Atoms are very small. 1 tsp of water contains 3x as many atoms as there are tsp of water in the Atlantic Ocean! Impractical to use pounds and inches... Need a universal unit system The metric system. The SI system (Systeme International) - derived from the metric system.

16. 16 © 2008 Brooks/Cole Metric Units Prefix Factor Example megaM10 6 1 megaton = 1 x 10 6 tons kilok10 3 1 kilometer (km) = 1 x 10 3 meter (m) decid10 -1 1 deciliter (dL) = 1 x 10 -1 liter (L) centic10 -2 1 centimeter (cm) = 1 x 10 -2 m millim10 -3 1 milligram (mg) = 1 x 10 -3 gram (g) microμ 10 -6 1 micrometer (μm) = 1 x 10 -6 m nanon10 -9 1 nanogram (ng) = 1 x 10 -9 g picop10 -12 1 picometer (pm) = 1 x 10 -12 m femtof10 -15 1 femtogram (fg) = 1 x 10 -15 g A decimal system. Prefixes multiply or divide a unit by multiples of ten.

17. 17 © 2008 Brooks/Cole 1 pm = 1 x 10 -12 m ;1 cm= 1 x 10 -2 m Metric Units How many copper atoms lie across the diameter of a penny? A penny has a diameter of 1.90 cm, and a copper atom has a diameter of 256 pm. x 1 x 10 -2 m 1 cm = 7.42 x 10 7 Cu atoms 1 pm 1 x 10 -12 m x 1.90 cm = 1.90 x 10 10 pm Number of atoms across the diameter: 1.90 x 10 10 pm x 1 Cu atom 256 pm

18. 18 © 2008 Brooks/Cole Length Length1 kilometer= 0.62137 mile 1 inch= 2.54 cm (exactly) 1 angstrom (Å)= 1 x 10 -10 m Volume Volume1 liter (L) = 1000 cm 3 = 1000 mL = 1.056710 quarts 1 gallon= 4 quarts = 8 pints Mass Mass1 amu= 1.66054 x 10 -24 g 1 pound= 453.59237 g = 16 ounces 1 ton (metric)= 1000 kg 1 ton (US)= 2000 pounds Some Common Unit Equalities

19. 19 © 2008 Brooks/Cole 5.0 lb Report the mass of a 5.0 lb bag of sugar in kilograms. 1 lb = 453. g = 2265 g x 453. g 1 lb = 2.3 x 10 3 g = 2.3 kg Some Common Unit Equalities

20. 20 © 2008 Brooks/Cole 165 mg dL A patient’s blood cholesterol level measured 165 mg/dL. Express this value in g/L 1 mg = 1 x 10 -3 g ; 1 dL = 1 x 10 -1 L x 1 x10 -3 g 1 mg = 1.65 g/L x 1 dL 1 x10 -1 L Some Common Unit Equalities

21. 21 © 2008 Brooks/Cole All measurements involve some uncertainty. one Reported numbers include one uncertain digit. Uncertainty and Significant Figures Consider a reported mass of 6.3492 g Last digit (“2”) is uncertain Close to 2, but may be 4, 1, 0 … significant figuresFive significant figures in this number.

22. 22 © 2008 Brooks/Cole Read numbers from left to right. starting Count all digits, starting with the 1 st non-zero digit. Allexcept All digits are significant except zeros used to position a decimal point (“placeholders”). 0.00024030 5 sig. figs. (2.4030 x 10 -4 ) placeholders significant Uncertainty and Significant Figures

23. 23 © 2008 Brooks/Cole Number Sig. figs.Comment on Zeros 2.123 4.5004 Not placeholders. Significant. 0.0025414 Placeholders (not significant). 0.001003 Only the last two are significant. Ambiguousor 500 1, 2, 3 ? Ambiguous. May be placeholders or may be significant. 500.3 Add a decimal point to show they are significant. 5.0 x 10 2 2 No ambiguity. Uncertainty and Significant Figures

24. 24 © 2008 Brooks/Cole dp = 4 dp = 3 Addition and subtraction Find the decimal places (dp) in each number. answer dp = smallest input dp. Add: 17.245 + 0.1001 17.3451 3 Rounds to: 17.345(dp = 3) Significant Figures in Calculations

25. 25 © 2008 Brooks/Cole dp = 2 dp = 4 Subtract 6.72 x 10 -1 from 5.00 x 10 1 Use equal powers of 10: 5.00 x 10 1 – 0.0672 x 10 1 4.9328 x 10 1 2 Rounds to:4.93 x 10 1 dp = 2 Significant Figures in Calculations

26. 26 © 2008 Brooks/Cole sig. fig. = 4 sig. fig. = 5 Multiplication and Division Answer sig. fig = smallest input sig. fig. 17.245 x 0.1001 1.7262245 4 Rounds to:1.726 sig. fig. = 4 Multiply 2.346, 12.1 and 500.99 Rounds to: 1.42 x 10 4 (3 sig. fig.) = 14,221.402734 Significant Figures in Calculations

27. 27 © 2008 Brooks/Cole Round 37.663147 to 3 significant figures. 1 st non-significant digit Examine the 1 st non-significant digit. If it: > 5, round up. < 5, round down. = 5, check the 2 nd non-significant digit.  round up if absent or odd; round down if even. last retained digit 1 st non- significant digit Rounds up to 37.7 2 nd non- significant digit Rules for Rounding

28. 28 © 2008 Brooks/Cole Round the following numbers to 3 sig. figs. 1 st non-sig.2 nd non-sig.Rounded Number digit digit Number 3 2.123 2.123- 2.12 72 51.372 51.372 51.372 51.4 5 131.5131.5- 132. 52 24.752 24.752 24.752 24.7 51 24.751 24.751 24.751 24.8 4 0.067440.06744- 0.0674 Rules for Rounding

29. 29 © 2008 Brooks/Cole Significant figures? 99.12444 – 6.321 27.5256 = 92.80344 27.5256 = 3.37153195571 = 3.3715 (5 sig. figs.) dp = 5dp = 3 Answer dp = 3. 92.803 is the significant result. (5 sig. figs). 6 sig. figs. Rules for Rounding

30. 30 © 2008 Brooks/Cole To avoid rounding errors Carry additional digits through a calculation. Use the correct number of places in the final answer.Note Exact conversion factors: (100 cm / 1 m) or (2H / 1 H 2 O) Have an infinite number of sig. figs. Rules for Rounding

31. 31 © 2008 Brooks/Cole Same element - same number of p + Atomic number Atomic number (Z) = number of p + 1 amu = 1.66054 x 10 -24 g ParticleMass Mass Charge (g) (amu) (atomic units) e − 9.1093826 x 10 -28 0.000548579 −1 p + 1.67262129 x 10 -24 1.00728 +1 n 0 1.67492728 x 10 -24 1.00866 0 Atomic Numbers & Mass Numbers Atomic mass unit (amu) = (mass of C atom) that contains 6 p + and 6 n 0. 1 12

32. 32 © 2008 Brooks/Cole The sum of the number of p + and n 0 in an atom is: A = mass number A = mass number A 12 or X e.g. C orX-A e.g. carbon-12 (Z is constant for a given element) AZAZ 12 6 For element X, write: X e.g. C Atomic Numbers & Mass Numbers A ≈ mass (in amu) of an atom

33. 33 © 2008 Brooks/Cole How many p +, n 0 and e - are in the following elements: Cu 63 29 Mg 25 29 p + = 29 e - (neutral atom: e - = p + ) 12 p + = 12 e - (periodic table; neutral) 63−29 = 34 n 0 25−12 = 13 n 0 AlAl 27 13 p + = 13 e - 27−13 = 14 n 0 Atomic Numbers & Mass Numbers

34. 34 © 2008 Brooks/Cole Mass Spectrometer

35. 35 © 2008 Brooks/Cole Isotopes same element Atoms of the same element with different A. equal numbers of p + different numbers of n 0 deuterium (D) tritium (T) Hydrogen isotopes:H1 p +, 0 n 0 1111 2121 H1 p +, 1 n 0 3131 H1 p +, 2 n 0 Isotopes and Atomic Weight

36. 36 © 2008 Brooks/Cole Most elements occur as a mixture of isotopes. Magnesium is a mixture of: 24 Mg 25 Mg 26 Mg number of p + 121212 number of n 0 121314 mass/ amu23.98524.98625.982 Isotopes and Atomic Weight

37. 37 © 2008 Brooks/Cole Isotopes and Atomic Weight percent abundance For most elements, the percent abundance of its isotopes are constant (everywhere on earth). The periodic table lists an average atomic weight. Example Boron occurs as a mixture of 2 isotopes, 10 B and 11 B. The abundance of 10 B is 19.91%. Calculate the atomic weight of boron.

38. 38 © 2008 Brooks/Cole Isotopes and Atomic Weights Atomic weight for B = 1.994 + 8.817 amu = 10.811 amu Atomic mass = Σ (fractional abundance)(isotope mass) (11.0093 amu) = 8.817 amu 11 B 80.09 100 (10.0129 amu) 10 B 19.91 100 = 1.994 amu % abundance of 11 B = 100% - 19.91% = 80.09% Boron occurs as a mixture of 2 isotopes, 10 B and 11 B. The abundance of 10 B is 19.91%. Calculate the atomic weight of boron.

39. 39 © 2008 Brooks/Cole B 10.811 Boron 5 Atomic number (Z) Symbol Name Atomic weight Periodic table: Isotopes and Atomic Weight

40. 40 © 2008 Brooks/Cole A counting unit – a familiar counting unit is a “dozen”: 1 dozen eggs= 12 eggs 1 dozen donuts= 12 donuts 1 dozen apples= 12 apples 1 mole (mol) = Number of atoms in 12 g of 12 C Latin for “heap” or “pile” 1 mol = 6.02214199 x 10 23 “units” Avogadro’s numberAvogadro’s number Amounts of Substances: The Mole

41. 41 © 2008 Brooks/Cole A green pea has a ¼-inch diameter. 48 peas/foot. (48) 3 / ft 3 ≈ 1 x 10 5 peas/ft 3. V of 1 mol ≈ (6.0 x 10 23 peas)/(1x 10 5 peas/ft 3 ) ≈ 6.0 x 10 18 ft 3 height = V / area, 1 mol would cover the U.S. to: U.S. surface area = 3.0 x 10 6 mi 2 = 8.4 x 10 13 ft 2 6.0 x 10 18 ft 3 8.4 x 10 13 ft 2 =7.1 x 10 4 ft = 14 miles ! Amounts of Substances: The Mole

42. 42 © 2008 Brooks/Cole 1 mole of an atom = atomic weight in grams. 1 Xe atom has mass = 131.29 amu 1 mol of Xe atoms has mass = 131.29 g and There are 6.022 x 10 23 atoms in 1 mol of He and 1 mol of Xe – but they have different masses. 1 He atom has mass = 4.0026 amu 1 mol of He has mass = 4.0026 g … 1 dozen eggs is much heavier than 1 dozen peas! Amounts of Substances: The Mole

43. 43 © 2008 Brooks/Cole Example How many moles of copper are in a 320.0 g sample? Cu-atom mass = 63.546 g/mol (periodic table) Conversion factor: 1 mol Cu 63.546 g = 1 n Cu = 320.0 g x 1 mol Cu 63.546 g = 5.036 mol Cu n = number of moles Molar Mass and Problem Solving

44. 44 © 2008 Brooks/Cole Calculate the number of atoms in a 1.000 g sample of boron. Molar Mass and Problem Solving n B = (1.000 g) 1 mol B 10.81 g = 0.092507 mol B B atoms = (0.092507 mol B)(6.022  10 23 atoms/mol) = 5.571  10 22 B atoms

45. 45 © 2008 Brooks/Cole The Periodic Table Summarizes Atomic numbers. Atomic weights. Physical state (solid/liquid/gas). Type (metal/non-metal/metalloid). Periodicity Elements with similar properties are arranged in vertical groups.

46. 46 © 2008 Brooks/Cole In the USA, “A” denotes a main group element… …”B” indicates a transition element. International system uses 1 … 18. The Periodic Table

47. 47 © 2008 Brooks/Cole Main group metal Transition metal Metalloid Nonmetal The Periodic Table

48. 48 © 2008 Brooks/Cole A period is a horizontal row Period number The Periodic Table

49. 49 © 2008 Brooks/Cole A group is a vertical column Group 7A Halogens Group 8A Noble gases Group 2A Alkaline earth metals Group 1A Alkali metals (not H) The Periodic Table

50. 50 © 2008 Brooks/Cole Alkali metals (group 1A; 1) Alkaline earth metals (group 2A; 2) Grey … silvery white colored. Highly reactive. Never found as native metals. Form alkaline solutions. The Alkali Metals and Alkaline Earth Metals

51. 51 © 2008 Brooks/Cole Transition Elements (groups 1B – 8B) Also called transition metals. Middle of table, periods 4 – 7. Includes the lanthanides & actinides. Lanthanides and Actinides Listed separately at the bottom. Chemically very similar. Relatively rare on earth.  (old name: rare earth elements) Transition Elements, Lanthanides & Actinides

52. 52 © 2008 Brooks/Cole Groups 3A to 6A Most abundant elements in the Earth’s crust and atmosphere. Most important elements for living organisms. Halogens (group 7A; 17) Very reactive non metals. Form salts with metals. Colored elements. Noble gases (8A; 18) Very low reactivity. Colorless, odorless gases. Groups 3A to 8A