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Dalton’s Atomic Theory

Dalton’s Atomic Theory. Elements - made up of atoms Same elements, same atoms. Different elements, different atoms. Chemical reactions involve bonding of atoms. The Atom. Made up of: Protons – (+) charged Electrons – (-) charged neutrons. Periodic Table. Alkaline Metals – Grps. I & II

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Dalton’s Atomic Theory

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  1. Dalton’s Atomic Theory • Elements - made up of atoms • Same elements, same atoms. • Different elements, different atoms. • Chemical reactions involve bonding of atoms

  2. The Atom • Made up of: • Protons – (+) charged • Electrons – (-) charged • neutrons

  3. Periodic Table • Alkaline Metals – Grps. I & II • Transition Metals • Non-metals • Halogens – Group VII • Noble Gases –Group VIII - little chemical activity

  4. Periodic Table • Atomic Mass - # at bottom • how much element weighs • Atomic Number - # on top • gives # protons = # electrons

  5. Periodic Table • Atomic Mass • number below the element • not whole numbers because the masses are averages of the masses of the different isotopes of the elements

  6. Ions • Are charged species • Result when elements gain electrons or lose electrons

  7. 2 Types of Ions • Anions – (-) charged • Example: F- • Cations – (+) charged • Example: Na+

  8. Highly Important! • Gain of electrons makes element (-) = anion • Loss of electrons makes element (+) = cation

  9. Isotopes • Are atoms of a given element that differ in the number of neutrons and consequently in atomic mass.

  10. Example Isotopes % Abundance 12C 98.89 % 13C 1.11 % 14C 11C

  11. For example, the mass of C = 12.01 a.m.u is the average of the masses of 12C, 13C and 14C.

  12. Determination of Aver. Mass • Ave. Mass = [(% Abund./100) (atomic mass)] + [(% Abund./100) (atomic mass)]

  13. Take Note: • If there are more than 2 isotopes, then formula has to be re-adjusted

  14. Sample Problem 1 • Assume that element Uus is synthesized and that it has the following stable isotopes: • 284Uus (283.4 a.m.u.) 34.6 % • 285Uus (284.7 a.m.u.) 21.2 % • 288Uus (287.8 a.m.u.) 44.20 %

  15. Solution • Ave. Mass of Uus = • [284Uus] (283.4 a.m.u.)(0.346) • [285Uus] +(284.7 a.m.u.)(0.212) • [288Uus] +(287.8 a.m.u.)(0.4420) • = 97.92 + 60.36 + 127.21 • = 285.49 a.m.u (FINAL ANS.)

  16. Oxidation Numbers • Is the charge of the ions (elements in their ion form) • Is a form of electron accounting • Compounds have total charge of zero (positive charge equals negative charge)

  17. Oxidation States • Are the partial charges of the ions. Some ions have more than one oxidation states.

  18. Oxidation States • - generally depend upon the how the element follows the octet rule • Octet Rule – rule allowing elements to follow the noble gas configuration

  19. Nomenclature • - naming of compounds

  20. Periodic Table • Rows (Left to Right) - periods • Columns (top to bottom) - groups

  21. Determination of Aver. Mass • Ave. Mass = [(% Abund./100) (atomic mass)] + [(% Abund./100) (atomic mass)]

  22. Sample Problem 1 • Assume that element Uus is synthesized and that it has the following stable isotopes: • 284Uus (283.4 a.m.u.) 34.6 % • 285Uus (284.7 a.m.u.) 21.2 % • 288Uus (287.8 a.m.u.) 44.20 %

  23. Solution • Ave. Mass of Uus = • [284Uus] (283.4 a.m.u.)(0.346) • [285Uus] +(284.7 a.m.u.)(0.212) • [288Uus] +(287.8 a.m.u.)(0.4420) • = 97.92 + 60.36 + 127.21 • = 285.49 a.m.u (FINAL ANS.)

  24. STOICHIOMETRY • For example, the mass of C = 12.01 a.m.u is the average of the masses of 12C, 13C and 14C.

  25. Formula Weight & Molecular Weight • The FORMULA WEIGHT of a compound equals the SUM of the atomic masses of the atoms in a formula. • If the formula is a molecular formula, the formula weight is also called the MOLECULAR WEIGHT.

  26. The MOLE • Amount of substance that contains an Avogadro’s number (6.02 x 10 23)of formula units.

  27. The MOLE • The mass of 1 mole of atoms, molecules or ions = the formula weight of that element or compound in grams. Ex. Mass of 1 mole of water is 18 grams so molar mass of water is 18 grams/mole.

  28. Formula for Mole Mole = mass of element formula weight of element

  29. Sample Mole Calculations 1 mole of C = 12.011 grams • 12.011 gm/mol • 0.5 mole of C = 6.055 grams • 12.011 gm/mol

  30. Avogadro’s Number • Way of counting atoms • Avogadro’s number = 6.02 x 1023

  31. Point to Remember One mole of anything is 6.02 x 1023 units of that substance.

  32. And…….. • 1 mole of C has the same number of atoms as one mole of any element

  33. Formula Weight & Molecular Weight • The FORMULA WEIGHT of a compound equals the SUM of the atomic masses of the atoms in a formula. • If the formula is a molecular formula, the formula weight is also called the MOLECULAR WEIGHT.

  34. Summary • Avogadro’s Number gives the number of particles or atoms in a given number of moles • 1 mole of anything = 6.02 x 10 23 atoms or particles

  35. Sample Problem 2 • Compute the number of atoms and moles of atoms in a 10.0 gram sample of aluminum.

  36. Solution • PART I: • Formula for Mole: • Mole = mass of element atomic mass of element

  37. Solution (cont.) • Part II: To determine # of atoms • # atoms = moles x Avogadro’s number

  38. Problem # 2 • A diamond contains 5.0 x 1021 atoms of carbon. How many moles of carbon and how many grams of carbon are in this diamond?

  39. Molar Mass • Often referred to as molecular mass • Unit = gm/mole • Definition: • mass in grams of 1 mole of the compound

  40. Example Problem • Determine the Molar Mass of C6H12O6

  41. Solution • Mass of 6 mole C = 6 x 12.01 = 72.06 g • Mass of 12 mole H = 12 x 1.008 = 12.096 g • Mass of 6 mole O = 6 x 16 = 96 g • Mass of 1 mole C6H12O6 = 180.156 g

  42. Problem #3 • What is the molar mass of (NH4)3(PO4)?

  43. Molar Mass • Often referred to as molecular mass • Unit = gm/mole • Definition: • mass in grams of 1 mole of the compound

  44. Sample Problem • Given 75.99 grams of (NH4)3(PO4), determine the ff: • 1. Molar mass of the compound • 2. # of moles of the compound • 3. # of molecules of the compound • 4. # of moles of N • 5. # of moles of H • 6. # of moles of O • 7. # of atoms of N • 8. # of atoms of H • 9. # of atoms of O

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