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Chapter 2: Atoms (1) Dalton’s Atomic Theory * John Dalton’s proposal by 1808 to explain major Laws * Consists of several postulates: I: Atoms are the smallest, indivisible particles of matter still retaining a chemical identity. II: All atoms of an element are identical. (not correct!) III: Atoms of different elements are chemically different. IV: Atoms can combine in integer ratios to form compounds. V: Chemical reactions involve a rearrangement of atoms, not a change in the atoms themselves. (2) Atomic Theory Applications * Helped explain the following major Laws: I: Law of Conservation of Mass: * Discovered by Antoine Lavoisier in 1774. * The total mass of reactants always equals the total mass of products during a chemical reaction. II: Law of Constant Composition: * Discovered by Joseph Proust in 1799. * The mass ratio of elements in a compound are always the same, regardless of how the compound was obtained. III: Law of Multiple Proportions: * Derived from Dalton’s Atomic Theory * Certain elements can combine to form more than one compound, but always with integer atomic ratios. * Example of N with O: N2O, NO, NO2 or N2O5 * The Atomic Theory led to major experimental efforts to find the atom. (3) Atomic Discoveries Ernest Rutherford 1909 Gold foil experiment Ernest Rutherford 1919 +, 1.007 amu nucleus James Chadwick 1932 neutral, 1.008 amu subatomic particles p, n and e– J.J. Thomson 1897 – , 0.0006 amu Cathode ray tubes (4) Isotopes: * Same # protons and electrons; different # neutrons * Use isotopic notation (2 types) Mass Number, A: # nucleons (protons + neutrons) 235 U 92 or U–235 Atomic Number, Z: # protons in nucleus # neutrons = A – Z Example #1: How many of each subatomic particle does an isotope of Am–241 have (used in smoke alarms). # protons = 95 # electrons = 95 (from the Periodic Table) (same as #p for a neutral atom) # neutrons = 241 – 95 = 146 (calculated) Example #2: How many of each subatomic particle does the following isotope have (used in bone scans): 85Sr2+ 38 # protons = 38 # electrons = 36 (from the notation or Periodic Table) (lost 2 e– due to positive cation) # neutrons = 85 – 38 = 47 (calculated) (5) The Periodic Table * There are several ways to organize the Periodic Table: I: Representative Elements (Main-Group or Type A) Transition Metals (Type B) Inner Transition Metals (Lanthanides & Actinides) II: Groups (columns) and Periods (rows) III: Metals, Nonmetals and Metalloids IV: Families: Group IA: Alkali Metals Group IIA: Alkaline Earth Metals Group VIA: Chalcogens Group VIIA: Halogens Group VIIIA: Noble Gases (6) The Mole * Used for counting large # of particles * Avogadro’s Number = 6.022 x 1023 particles mol–1 * A conversion factor between # particles and moles (7) Atomic Mass * Weighted average over all isotopes * Conversion factor between mass and # atoms or mol * Units are g/mol or amu/atom Example: What mass would 3.20 x 1024 atoms of iron have: 3.20 x 1024 Fe atom 1 mol Fe 6.022 x 1023 Fe atoms 55.845 g Fe = 296 .75 g Fe 1 mol Fe 3 SF Avogadro’s # Atomic Mass 297 g Fe Determining Atomic Masses * Averaged over all isotopes and based on 2 factors: 1) Isotopic Masses (mi) 2) Fractional Abundance (fi) = % / 100. Avg atomic mass = Σ fimi Example: What is the average atomic mass of Silicon, provided: Si–28 = 27.97693 u and is 92.23 %; Si–29 = 28.97649 u and is 4.67 %; and Si–30 = 29.97376 u and is 3.10 %. What is the average atomic mass of Si? at. mass = 92.23 27.97693 u + 4.67 28.97649 u + 3.10 29.97376 u 100 100 100 at. mass = 25.80¦31 u + 1.35¦32 u + 0.929¦18 u = 28.08¦54 u -----> 28.09 u