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TRY ME!

TRY ME!. The distance between the carbon atoms in diamond is 1.54 x 10 -8 cm. What is this distance in picometers (pm)? How many cm 3 is in 1 m 3 ?. 1 cm = 10 -2 m AND 1 pm = 10 -12 m . Or… 1 m = 10 2 cm = 10 12 pm. 10 6 cm 3. RVCC Fall 2009

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TRY ME!

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  1. TRY ME! • The distance between the carbon atoms in diamond is 1.54 x 10-8 cm. What is this distance in picometers (pm)? • How many cm3 is in 1 m3? 1 cm = 10-2 m AND 1 pm = 10-12 m Or… 1 m =102 cm = 1012 pm 106 cm3

  2. RVCC Fall 2009 CHEM 103-1&2 – General Chemistry I Chapter 2:Atoms and Elements Chemistry: The Molecular Science, 3rd Ed. by Moore, Stanitski, and Jurs

  3. Evolution of Thought Concerning the Structure of the Atom… “It often happens that the mind of a person who is learning a new science, has to pass through all the phases which the science itself has exhibited in its historical evolution.” -Stanislao Cannizzaro, 1860 Oliver Sacks, Uncle Tungsten: Memories of a Chemical Boyhood Alfred A. Knopf, Canada, 2001.

  4. The Law of Definite Proportions (Ch. 1.9) Proust, 1799 - • The law of constant composition states that a specific chemical compound obtained from any source always contains the same proportions by mass of its elements. • Example: HgS is the same no matter where you get it from

  5. The Law of Multiple Proportions (Ch.1.9) • The law of multiple proportions states that the masses of element Y that combine with a fixed mass of element X to form two or more different compounds, are in the ratios of small whole numbers. • Examples: NO, NO2, N2O, N2O5, etc.

  6. Atomic Theory Dalton –1803 • An element is composed of tiny indivisible particles called atoms. • All atoms of a given element have the same chemical properties and characteristic mass unique to that element. • Compounds are formed by the combination of two or more different kinds of atoms. • A chemical reaction involves joining, separating, or rearranging atoms. Hg Hg S S cinnabar

  7. Atomic Theory - Electrons Thomson –1897 Thompson saw the deviation of ‘cathode rays’ between charged plates, proving the rays consisted of charged particles. Based on the fields applied, he calculated the ratio of mass to charge for electrons. 1750 Ben Franklin (tape demo)

  8. Rutherford, 1911 - Atomic Theory – Protons and Neutrons The majority of the mass of an atom is concentrated in a very small nucleus (protons and neutrons) at the center. Chadwick, 1932 - fired -particles at Be atoms and neutral particles (neutrons) were ejected.

  9. The Mass and Charge of the Electron, Proton, and Neutron Particle Mass (g) (amu) Charge(C) atomic Electron(e) 9.11 x 10-28 0.00055 -1.60x10-19 -1 Proton(p) 1.67 x 10-24 1.0073+1.60x10-19+1 Neutron(n) 1.67 x 10-24 1.0087 0 0 - - + + - The Helium atom. 2 protons, 2 neutrons, 2 electrons

  10. Atomic Structure - shorthand • The atomic number(Z) is the number of protons • in the nucleus. • The mass number (A) is the number of protons and neutrons in the nucleus. A=p+n • 1H = Hydrogen atom has • 1 proton • 1 electron • mass of 1 • 0 neutrons 1 Z=p

  11. A=p+n 1H 1 Z=p 63 29 40 18 Cu Ar Atomic Structure 29 p+ = 29 e- (neutral atom: e- = p+) 63−29 = 34 n0 or… “Copper-63” 18 p+ = 18 e- (neutral atom: e- = p+) 40−18 = 22 n0 or… “Argon-40 “

  12. A=p+n 1H 1 Z=p Atomic Structure protons electrons neutrons 8 8 8 (16- 8) 47 47 61 (108-47) 24 24 28 35 35 46 35 35 44

  13. Isotopes Isotopesare atoms of the same element whose nuclei have the same atomic number (Z, # protons) but different mass numbers (A, #protons+#neutrons). Isotopes have the same number of protons but different numbers of neutrons (#n0=A-Z).

  14. Isotopes = proton = neutron Hydrogen Deuterium Tritium Isotopes have very similar chemical properties.

  15. Isotopes Percent Abundance: 99.99% 0.01% 0.00%

  16. 12C 13C Isotopes – Average Mass Carbon 98.89% 12C (12.0000 amu) 1.11% 13C (13.0034 amu) (<0.01% 14C) • What is carbon’s average atomic mass? Atomic mass on the periodic table is a mass average – weighted by the percent abundance of each isotope of the element. 12.01 amu

  17. Isotopes – Average Mass Gallium (Ga) has an average mass of 69.8 amu. What are the common isotopes of gallium? a)50% 69Ga and 50% 71Ga  b)10% 69Ga and 90% 71Ga  c)40% 69Ga and 60% 71Ga  d)60% 69Ga and 40% 71Ga

  18. Practice • Only two isotopes of copper occur naturally, 63Cu (atomic mass = 62.9296 amu; abundance 69.17%%) and 65Cu (atomic mass = 64.9278 amu; abundance 30.83%). Calculate the atomic weight or average mass of copper. • The element Pb consist of four naturally occurring isotopes with atomic masses 203.97302, 205.97444, 206.97587, and 207.97663 amu. The relative abundances of these four isotopes are 1.4, 24.1, 22.1, and 52.4%. Calculate the average mass of lead.

  19. Practice Given: 107Ag 106.90509, X 109Ag, 108.90476, What are the natural abundances? (mass average = 107.8682) 107.8682 = (X)106.90509 + (1-X) 108.90476 107.8682 = 106.90509 X+ 108.90476-108.90476 X 108.90476X-106.90509X = 108.90476-107.8682 1.99967X = 1.03656 X= 0.518 51.8% 48.2% 1-X

  20. Making Measurements • Ruler Exercise

  21. Significant Figures 9.12 one uncertain digit all certain digits +

  22. Significant Figures 140 120 100 80 60 40 20 58.0 mL 3 sf 92 mL 2 sf

  23. Uncertainty in Measurements The last digit is uncertain, a guess. 3 sf 5sf

  24. Rules for Counting Significant Figures Lab • To determine the number of significant figures, read the number from left to right and count all digits, starting with the first digit that is not zero. Do not count zeros that only position the decimal. 3456 4 significant figures

  25. Rules for Counting Significant Figures Lab 0.0486 3 significant figures • Leading zeros in decimal ARE NOT counted as significant figures, they only “position” the decimal.

  26. Rules for Counting Significant Figures Lab 9.300m 4 significant figures • Trailing zeros in decimal number ARE significant. (Not for placement.) 9300 mm 2 sf • Otherwise (no decimal), they are for placement.

  27. Rules for Counting Significant Figures Lab All the figures given in scientific notation ARE significant. 0.04260 m 4.260 x 10-2 15260 mm 0.001010 s 1.526 x 104 1.010 x 10-3

  28. Practice Lab How many significant figures are in the following numbers? 0.04550 g 100 lb 101.05 mL 350.0 g 4 sf 1 sf 5 sf 4 sf

  29. Rules for Significant Figures in Mathematical Operations Lab • Multiplication and Division: The number of significant figures in the result equals the number in the measurement with the least. 6.380km  2.0hr = 12.76km/hr  13km/hr 4sf 2sf 2 sf 4.5m x 5.66m = 25.47m2  2.5 m2

  30. Rules for Significant Figures in Mathematical Operations Lab • Addition and Subtraction: The number of significant figures in the result depends on the least precise measurement. 6.8 cm 1dp + 11.93 cm 2dp = 18.73 cm  18.7 cm 1dp answer only precise to one tenth of a cm 3 significant figures

  31. Rules for Counting Significant Figures Lab • Exact numbers, or a count, have an infinite number of significant figures 1000 mm/1 m 60 seconds = 1 minute 12 drops/test tube These are counts, not measurements. 4.2m x 1000mm = 4200 mm 1m

  32. Lab Practice What is the correct answer to the following problems? • 1.23 g - 0.567 g = ? g • 0.34442 m x 4.68 m = ? m2 • (1.245 g + 6.34 g)  4.6 mL = ? 0.6630→0.66 g 1.61189 →1.61 m2 7.585/4.6=1.6 g/mL

  33. Rules for Rounding Lab • Examine the 1st non-significant digit. > 5 (or >50…), round up. Round 37.663147 to 3 significant figures. Rounds up to 37.7 1st non-significant digit Round 37.650 to 3 significant figures. Rounds up to 37.7

  34. Rounding Lab If the digit to be dropped is less than 5 just drop it. Round the following numbers to 3 sf. 1.2532 = 1.25 108400 = 1.08 x 105 or 108000 0.8703 = 0.870

  35. Accuracy and Precision Lab When we take a measurement, it contains a degree of uncertainty due to the limits of instruments and to us. Accuracy indicates how well a measurement agree with an accepted (true) value (known standard is 100 g and you find 92g – not very accurate!).  Precision indicates how well the values of several measurements agree with each other (reproducibility of measurements) - Nothing to do with the accepted value: if the instrument does not work properly a measurement can be very precise but inaccurate!

  36. Accuracy and Precision Lab accurate, precise not accurate, precise not accurate, not precise

  37. Precision and Accuracy Lab The widths of copper lines in printed circuit boards must be close to a specified value. Three manufacturing plants were asked to prepare circuit boards with 0.500 µ (5.00 x 10-7 m) wide copper lines. Each manufacturer’s quality control department reported the following line widths on five sample circuit boards: Manufacturer #1 Manufacturer #2 Manufacturer #3 0.512 µ 0.514 µ 0.500 µ 0.508 µ 0.513 µ 0.501 µ 0.516 µ 0.514 µ 0.502 µ 0.504 µ 0.514 µ 0.502 µ 0.513 µ 0.512 µ 0.501 µ x = 0.511 0.513 0.501 s = 0.004 0.001 0.001

  38. Measurements in Scientific Studies SI Base Units – Systeme International d’Unites

  39. Unit Conversion Examples • Change 5.0km to miles. 1km=0.62137mi • Express 12.0 pounds in kg. 2.2 lbs = 1.0 kg • Express 1 pint of blood in mL. 3.1 mi 5.45 kg 473 mL 2 pints = 1 quart 1 L = 1.057 quarts 1000mL = 1L

  40. Atoms and Molecules 1 2 4 7

  41. Atomic Masses (amu) Atomic masses for each element are listed on the periodic table. 1 H atom = 1.01 amu 1 C atom = 12.01 amu To calculate mass of onemoleculeof H2SO4, take the sum of its parts: 2 hydrogen atoms = 2 x 1.01 amu = 2.02 amu 4 oxygen atoms = 4 x 16.00 amu =64.00 amu 1 sulfur atom = 1 x 32.07 amu = 32.07 amu 98.09 amu

  42. Molar masses for each element are listed on the periodic table. Molar Mass (g) = the mass of 1 mole 1 mole H atoms = 1.01 g 1 mole C atoms = 12.01 g To calculate mass ofone moleof H2SO4 molecules, take the sum of its parts: 2 mols hydrogen = 2 x 1.01 g = 2.02 g 4 mols oxygen = 4 x 16.00 g =64.00 g 1 mol sulfur = 1 x 32.07 g = 32.07 g 98.09 g/mol

  43. The Mole Avrogadro’s Number, NA Atomic MassMolar Mass 1 H atom = 1.01 amu 1 C atom = 12.01 amu 1 molecule H2SO4 = 98.09 amu 1 mole H atoms = 1.01 g 1 mole C atoms = 12.01 g 1 mole H2SO4 molecules = 98.09 g NA nanoscopic macroscopic

  44. Amounts of Substances: The Mole Lorenzo Romano Amedeo Carlo Avogadro the “mole” – Avogadro’s Number (NA) A counting unit: • 1 dozen eggs = 12 eggs • 1 reem paper = 500 sheets • 1 pack soda = 6 cans • 1 mol = 6.02 x 1023 “units” 1 mole (mol) = Number of atoms in 12.0000 g of 12C

  45. Practice Atomic Mass(amu) or Molar Mass(g) • NaNO3 • Mg3(PO4)2 • C7H5NO3S 85.01 amu or g 262.87 amu or g 183.20 amu or g

  46. We have New Conversion Factors! • 1 mol = 6.02 x 1023 atoms or molecules and • 1mol = MM (g) substance

  47. The Mole to Atoms(or molecules, or particles…) • One milliliter of seawater contains about 2.5x 10-14 moles of gold. How many atoms of gold are in a milliliter of seawater? 1mole = 6.02x1023 atoms atoms mole NA 1.5 x 1010 atoms of Au

  48. The Mole to Grams 1 mole = g substance mole g MM • One milliliter of seawater contains about 2.5x 10-14 moles of gold. How many grams of gold are in a milliliter of seawater? 4.9 x 10-12 grams of Au

  49. The Mole to Grams Example How many moles of copper are in a 320.0 g sample? 1 mol of Cu atoms = 63.546 g (periodic table) 5.036 mol Cu

  50. Atoms to Mole to Grams : - NA mole -MM : atoms g xNA xMM How many atoms are in 1g of carbon? 5 x 1022 atoms of C

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