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Electrical Charge is fundamental to our understanding of the atom

Atoms and Elements. Electrical Charge is fundamental to our understanding of the atom. Electrostatics had been discovered by the time of Benjamin Franklin (late 1700’s) Radioactivity

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Electrical Charge is fundamental to our understanding of the atom

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  1. Atoms and Elements Electrical Charge is fundamental to our understanding of the atom

  2. Electrostatics had been discovered by the time of Benjamin Franklin (late 1700’s) Radioactivity 1896- Henri Becquerel: Discovered uranium ore emitted rays that exposed photographic plates protected by black paper 1898- Marie/Pierre Curie: Isolated polonium and radium which emitted the same kind of rays

  3. Suggested particles emitted through atomic disintegration Contradicted Dalton’s atomic theory which states that atoms are indivisible

  4. ,  and  radiation Radiation Movie 1 Radiation Movie 2

  5. Components of the Atom Electrons Primarily discovered by J.J. Thomson in 1897 by discovery of “negatively charged” cathode rays using Cathode Ray Tube (Crookes tube) Determined charge to mass ratio of the electron.

  6. Thomson’s method for determining the charge to mass ratio (q/m) for the electron Thomson developed the “plum pudding” model of the atom

  7. Robert Millikan (1909): Millikan’s Oil-drop Experiment Discovered the charge on the electron (-1.602x10-19C)

  8. Millikan’s Experiment Movie

  9. Eugen Goldstein (1886) – Discovered positively charged “canal” rays (positively charged nuclei of atoms, with hydrogen having the smallest mass).

  10. Protons and the Nucleus 1910- Ernest Rutherford: Gold foil experiment-Also experimented with alpha irradiated gaseous elements: Alpha deflection proportional to atomic mass-Bombardment of nitrogen gas produced particles consistent with that of hydrogen (deemed a fundamental particle)-Developed Nuclear Model of the Atom1919- Proton officially proclaimed

  11. Gold Foil Movie 1 Gold Foil Movie 2

  12. The Neutron 1932- James Chadwick: Discovered the neutron by bombarding beryllium with alpha particles

  13. Atomic Number and Atomic Mass

  14. Atomic Mass Unit (u) = 1/12 the mass of a C-12 isotope. About the mass of a proton or neutron (1u = 1.661x10-24g)

  15. Isotopes Atoms of the same element (i.e. the same number of protons), that differ in their number of neutrons. (Compare to isobars (atoms with the same mass number but different atomic number) and isotones (atoms with the same number of neutrons but different number of protons)

  16. The mass number of an isotope is not the sum of the masses of the individual particles due to the mass defect (E =mc2), which is the binding energy. Equations: Percent abundance = (number of atoms of a given isotope / total number of atoms of all isotopes of that element)x100% Atomic weight (weighted average)= (fractional abundance of isotope 1)(mass of isotope 1) + …

  17. Isotopic Abundance isDetermined by mass spectroscopy

  18. Mass spectroscopy of a molecule

  19. Sample Problems a) Argon has three isotopes with 18, 20 and 22 neutrons, respectively. What are the mass number and symbols of these three isotopes? b) Gallium has two isotopes:Ga-69 and Ga-71. How many protons and neutrons are in the nuclei of each of these isotopes? If the abundance of Ga-69 is 60.1%, what is the abundance of Ga-71?

  20. Answers: Ga-71 39.9%

  21. Sample Problems • There are three naturally occurring isotopes of neon. Their percent abundances and atomic masses are: neon-20, 90.51%, 19.99244u; neon-21, 0.27%, 20.99395u; neon-22, 9.22%, 21.99138u. Calculate the weighted average atomic mass of neon. • The two naturally occurring isotopes of copper are copper-63, mass 62.9298u, and copper-65, mass 64.9278u. What is the percent abundances of each of the two isotopes?

  22. Answers: (.9051)(19.99244u)+ (.0027)(20.99395u) + (.0922)(21.99138u) = 20.1794 = 20.18u (62.9298u)(X) + (64.9278u)(1-X) = 63.546u 62.9298X + 64.9278u – 64.9278X = 63.546u -1.998X = -1.3818u X = .6916 Cu-63 = 69.16% Cu-65 = 30.84%

  23. 1 mole  The quantity of things as there are atoms in exactly 12.0g of the C-12 isotope. This must be determined experimentally. NA (Avogadro’s number) = 6.022x1023 = 1 mole

  24. Mass number of an element 1) Taken in amu’s (u) = atomic mass (mass of 1 atom) 2) Taken in grams (g) = molar mass (mass of 1 mole of atoms) 24.30Mg If you have 24.30g of Mg you have 6.022x1023 atoms of Mg or 1mole Determining the molar mass of a compoundAdd the molar masses of the individual elements that the compound contains.

  25. Sample Problem a) What is the mass, in grams, of 1.5 mol of silicon? b) What amount (moles) of sulfur is represented by 454g? How many atoms? c) What is the average mass (in grams) of one sulfur atom?

  26. Answers: a) g Si = 1.5mol Si (28.09g/mol) = 42.135g = 42g b) Mol S = 454g (1mol/32.07g) =14.1565 = 14.2mol # S atoms = 14.1565mol (6.022x1023) = 8.53x1024atoms c) Mass of 1 sulfur atom = 32.07g/mol (1 mol / 6.022x1023atom) = 5.325x10-23g

  27. Sample problem The density of gold, Au, is 19.32g/cm3. What is the volume (in cm3) of a piece of gold that contains 2.6x1024 atoms? If the piece of metal is a square with a thickness of 0.10cm, what is the length (in cm) of one side of the piece?

  28. Answers: 2.6x1024atoms (1mol/6.022x1023atoms) (197.0g/1mol)(1cm3 / 19.32g) = 44.02 = 44cm3 X2(0.10cm) = 44cm3 X = 20.982 = 21cm

  29. The Periodic Table Dmitri Mendeleev: Russian Schoolteacher Father of the modern periodic table (1869) Table based on mass instead of atomic number (prior to understanding of atomic particles) Trends allowed for prediction of elements that had not yet been discovered.

  30. Diatomic elements

  31. Allotropes

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