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Principles of Chemistry, Chapt. 2: Atomic Structure and The Elements I.The Structure of Atoms protons, neutrons, and electrons II.Atomic Structure and.

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Presentation on theme: "Principles of Chemistry, Chapt. 2: Atomic Structure and The Elements I.The Structure of Atoms protons, neutrons, and electrons II.Atomic Structure and."— Presentation transcript:

1 Principles of Chemistry, Chapt. 2: Atomic Structure and The Elements I.The Structure of Atoms protons, neutrons, and electrons II.Atomic Structure and Properties—the Elements atomic mass, atomic number, isotopes III.The Mole Concept: 6.02 x IV.The Periodic Table 1 Homework: Chapt. 2 Problems 26, 29, 37, 43, 75

2 ~ meters = 1 angstrom (Å) _ m Smeared out electron charge cloud Protons and neutrons 2 Most of the mass is here Most of the Chemistry is here Atomic Theory in a single Slide

3 3 STM Image: Oxygen atoms at the surface of Al 2 O 3 /Ni 3 Al(111) S. Addepalli, et al. Surf. Sci. 442 (1999) 3464 Electronic charge cloud surrounding the nucleus

4 4 What’s inside the nucleus: ParticleMass (amu) Charge Proton (p + )1.007 amu +1 Neutron (n 0 )1.009 amu 0 What’s outside the nucleus: Electron (e-) amu-1 Note: mass ratio of electron/proton (M p+ /M e- ) = 1836 For any atom: # of electrons = # of protons: Why?

5 Atomic Theory: Late 19 th Century Atomic theory—everything is made of atoms—generally accepted (thanks to Ludwig Boltzman, and others). Mendeleev/periodic table—accepted, but the basis for periodic behavior not understood What are atoms made of? How are they held together? 5

6 6 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 – + Alpha particle  2 n 0 + 2p + Beta particel  electron (e-) Gamma  photon

7 Electric and magnetic fields deflect the beam. Gives mass/charge of e - = −5.60 x g/C Coulomb (C) = SI unit of charge Thomson (1897) discovered the e - : 7 “Cathode rays” Travel from cathode (-) to anode (+). Negative charge (e − ). Emitted by cathode metal atoms. fluorescent screen – high voltage + cathode ray

8 + _ + _ Essence of the Thompson Experiement (and old fashioned TV’s) Electric field exerts Force + plate repels +charged particles - Plate repels – charged particles F = Eq = ma d = displacement = ½ at 2 = Eq/m (t = L/V x ) Therefore, the greater the displacement, the lower the mass of the particle d x y Phosphor screen 8

9 Millikan (1911) studied electrically-charged oil drops. 9

10 Charge on each droplet was: n (−1.60 x C) with n = 1, 2, 3,… n (e - charge) Modern value = − x C. = −1 “atomic units”. These experiments give: 10 Modern value = x g = (-1.60 x C)(-5.60 x g/C) = 8.96 x g m e = charge x mass charge

11 Atoms gain a positive charge when e - are lost. 11 Implies a positive fundamental particle. Hydrogen ions had the lowest mass. Hydrogen nuclei assumed to have “unit mass” protonsCalled protons. Modern science: m p = x g m p ≈ 1800 x m e. Charge = -1 x (e - charge). = x C = +1 atomic units

12  How were p + and e - arranged? 12 Thompson: Ball of uniform positive charge, with small negative dots (e - ) stuck in it. The “plum-pudding” model Rutherford (former Thompson graduate student) fired α-particles at thin metal foils. Expected them to pass through with minor deflections.

13 Rutherford Scattering Experiment 13

14 Different Models of the Atom: different scattering results α particles “Plum pudding model” + and – charges evenly distrubted low, uniform density of matter No back scattering Rutherford’s explanation of results: Small regions of very high density + charge in the dense regions - Charges in region around it From wikipedia 14

15 Some Large Deflections were osbserved 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.” 15

16 ≈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. 16

17 17 Nucleus diameter~ Å = meters Mass ~ Kg Charge cloud Diameter ~ 1 Å Mass ~ kg

18 18 Most Chemistry involves rearrangement of outermost electrons, not nuclei Example: H  1p +, 1 e- H + H  H 2 +

19 19

20 7 Å Epitaxial Al2O3(111) film on Ni 3 Al(111) (Kelber group): Grown in UHV Uniform No Charging S. Addepalli, et al. Surf. Sci. 442 (1999) 3464 STM Start with ordered films  growth studies Proceed to amorphous films on Si(100) Surface terminated by hexagonal array of O anions 20

21 Atomic mass > mass of all p + and e - in an atom. Rutherford proposed a neutral particle. 21 m n ≈ m p (0.1% larger). m n = x g. Present in all atoms (except ‘normal’ H). neutrons Chadwick (1932) fired  -particles at Be atoms. Neutral particles, neutrons, were ejected:

22 ~ meters = 1 angstrom (Å) _ m Smeared out electron charge cloud Protons and neutrons 22

23 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 23

24 24 A neutron can decay into a proton and electron: n 0  p + + e - This can cause decay of a radioactive element, e.g., 14 6 C # of p + + n 0 Atomic No. (# e- = # p + Elemental symbol (carbon) Carbon with 6 protons and 8 neutrons is unstable (radioactive) Carbon with 6 protons and 6 neutrons is stable (non-radioactive 14 6 C 12 6 C radioctivestable

25 25 An atom of 14 C can undergo decay to N as a neutron turns into a proton + an emitted electron 14 6 C 7 N + e- 1 p +  1 n 0 + an electron (emitted)

26  Same element - same number of p + 26 Atomic number Atomic number (Z) = number of p + 1 amu = x g ParticleMass Mass Charge (g) (amu) (atomic units) e − x −1 p x n x Atomic mass unit (amu) = (mass of C atom) that contains 6 p + and 6 n

27  Isotopes same element  Atoms of the same element with different A. equal numbers of p + different numbers of n 0 27 deuterium (D) tritium (T) Hydrogen isotopes:H1 p +, 0 n H1 p +, 1 n H1 p +, 2 n 0

28 ISOTOPES: SAME Element, Different numbers of neutrons C 12 C 14 Carbon: atomic no. = 6  6 protons in the nucleus+ 6 electrons Atomic mass = 12 amu (12 gr/mole) Therefore, 6 protons + 6 neutrons Atomic mass = 14 amu Therefore, 6 protons+ 8 neutrons 28

29 Isotopes Display the same chemical reactivities (which depend mainly on the outer arrangement of the electrons) 12 C + O 2  CO 2 14 C +O 2  CO 2 Isotopes display different nuclear properties C 12 stable C 14 Radioactive: spontaneously emits electrons. Half-life ~ 5730 years 29

30 Isotopes and Moles (more on this later) and isotope abundance: 1 mole = 6.02 x of anything! 1 mole of atoms = 6.02 x atoms Molar Mass (in grams) = average atomic mass (in amu) 1 mole of H atoms = gr. Why not gr??  most atoms are 1 H, but some are 2 H (deuterium) 30

31 Average atomic mass of H = amu 100 atoms have a mass of amu # of 2 H atoms = n # of 1 H atoms = 100 –n (assume these are the only two isotopes that matter) Mass of 100 atoms = n x (100-n) x = amu 2n n = n = 0.8 So, out of every 100 atoms, have H atoms Out of every 1000 atoms, have 8 2 H atoms Natural abundance of “heavy hydrogen (deuterium) is then 0.8% 31

32  Most elements occur as a mixture of isotopes. 32 Magnesium is a mixture of: 24 Mg 25 Mg 26 Mg number of p number of n mass/ amu

33 percent abundance  For most elements, the percent abundance of its isotopes are constant (everywhere on earth).  The periodic table lists an average atomic weight. 33 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.

34 34 Atomic weight for B = amu = amu Atomic mass = Σ (fractional abundance)(isotope mass) ( amu) = amu 11 B ( amu) 10 B = amu % abundance of 11 B = 100% % = 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.

35 B Boron 5 Atomic number (Z) Symbol Name Atomic weight Periodic table: 35

36  A counting unit – a familiar counting unit is a “dozen”: 36 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 = x “units” Avogadro’s numberAvogadro’s number

37  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 peas)/(1x 10 5 peas/ft 3 )  ≈ 6.0 x ft 3 37 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 ft x ft x ft 2 =7.1 x 10 4 ft = 14 miles !

38  1 mole of an atom = atomic weight in grams Xe atom has mass = amu 1 mol of Xe atoms has mass = g and There are x atoms in 1 mol of He and 1 mol of Xe – but they have different masses. 1 He atom has mass = amu 1 mol of He has mass = g … 1 dozen eggs is much heavier than 1 dozen peas!

39 Example How many moles of copper are in a g sample? Cu-atom mass = g/mol (periodic table) Conversion factor: 1 mol Cu g = 1 n Cu = g x 1 mol Cu g = mol Cu n = number of moles 39

40  Calculate the number of atoms in a g sample of boron. 40 n B = (1.000 g) 1 mol B g = mol B B atoms = ( mol B)(6.022  atoms/mol) =  B atoms

41 Dimensional Analysis and Problem Solving Special Homework Problem: Due Tues. Recitation Density = mass/volume Problem: Assume that a hydrogen atom has a spherical diameter of 1 angstrom Assume that the nucleus (1 proton) has a diameter of angstrom 1.Calculate the densities of the nucleus, and of the electron charge cloud in kg/m 3 2.Calculate the ratio of the two densities: R = d nucleus /d electron cloud Mass of proton = 1.67 x kg Mass of electron = 9 x kg 41

42  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. 42

43 In the USA, “A” denotes a main group element… …”B” indicates a transition element. International system uses 1 … 18. The Periodic Table 43

44 Main group metal Transition metal Metalloid Nonmetal The Periodic Table 44

45 A period is a horizontal row Period number 45

46 A group is a vertical column Group 7A Halogens Group 8A Noble gases Group 2A Alkaline earth metals Group 1A Alkali metals (not H) 46

47  Alkali metals (group 1A; 1)  Alkaline earth metals (group 2A; 2) 47 Grey … silvery white colored. Highly reactive. Never found as native metals. Form alkaline solutions.

48  Transition Elements (groups 1B – 8B) Also called transition metals. Middle of table, periods 4 – 7. Includes the lanthanides & actinides. 48 Lanthanides and Actinides Listed separately at the bottom. Chemically very similar. Relatively rare on earth.  (old name: rare earth elements)

49  Groups 3A to 6A Most abundant elements in the Earth’s crust and atmosphere. Most important elements for living organisms. 49 Halogens (group 7A; 17) Very reactive non metals. Form salts with metals. Colored elements. Noble gases (8A; 18) Very low reactivity. Colorless, odorless gases.

50  Atoms are very small. 1 tsp of water contains 3x as many atoms as there are tsp of water in the Atlantic Ocean! 50 Impractical to use pounds and inches... Need a universal unit system The metric system. The SI system (Systeme International) - derived from the metric system.

51 A decimal system. Prefixes multiply or divide a unit by multiples of ten. 51 Prefix Factor Example megaM megaton = 1 x 10 6 tons kilok kilometer (km) = 1 x 10 3 meter (m) decid deciliter (dL) = 1 x liter (L) centic centimeter (cm) = 1 x m millim milligram (mg) = 1 x gram (g) microμ micrometer (μm) = 1 x m nanon nanogram (ng) = 1 x g picop picometer (pm) = 1 x m femtof femtogram (fg) = 1 x g

52 1 pm = 1 x m ;1 cm= 1 x m 52 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 m 1 cm = 7.42 x 10 7 Cu atoms 1 pm 1 x m x 1.90 cm = 1.90 x pm Number of atoms across the diameter: 1.90 x pm x 1 Cu atom 256 pm

53 Length Length1 kilometer= mile 1 inch= 2.54 cm (exactly) 1 angstrom (Å)= 1 x m Volume Volume1 liter (L) = 1000 cm 3 = 1000 mL = quarts 1 gallon= 4 quarts = 8 pints Mass Mass1 amu= x g 1 pound= g = 16 ounces 1 ton (metric)= 1000 kg 1 ton (US)= 2000 pounds 53

54 Example: What is the volume of a 2 gallon container in Liters? 1 gallon x 4 quarts/gallon = 4 quarts 4 quarts x 1Liter/1.057 quarts = Liters (L) 54

55 165 mg dL A patient’s blood cholesterol level measured 165 mg/dL. Express this value in g/L 1 mg = 1 x g ; 1 dL = 1 x L x 1 x10 -3 g 1 mg = 1.65 g/L x 1 dL 1 x10 -1 L 55

56  All measurements involve some uncertainty. one  Reported numbers include one uncertain digit. 56 Consider a reported mass of g Last digit (“2”) is uncertain Close to 2, but may be 4, 1, 0 … significant figuresFive significant figures in this number.

57  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”).   5 sig. figs.  ( x ) 57 placeholders significant

58 58

59 Round to 3 significant figures. 1 st non-significant digit  Examine the 1 st non-significant digit. If it: 59 > 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 nd non- significant digit


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