CHEM 1211 Lecture 4Outline of Last Lecture I. ExamplesII. The Unit Factor MethodIII. PercentageA. EquationIV. DensityV. Specific GravityVI. Chapter Two: Atoms Molecules & IonsA. Subatomic particlesa. Protonsb. Neutronsc. ElectronsOutline of Current Lecture I. Nuclide StructureII. IsotopesA. Mass SpectrumB. Natural % AbundanceIII. Atomic WeightA. ExampleIV. The Periodic TableA. Main Group MetalsB. Transition MetalsC. MetalloidsD. NonMetalsV. Forms of CarbonCurrent LectureI. Nuclide StructureA: Mass NumberZ: # of Protons14: Atomic NumberSi: Symbol of Element28.0855: Mass Number** All Atomic masses are based on Carbon-12 (12C) = 12amuMass of Carbon: 1.99265 x 10-23g 1.99265 x 10-26kg1.99265 x 10 -23 g = 1.66 x 10-24g 12amuII. Isotopes: Same element, different massX= ClSame Element… Different number of neutrons / different massesA. Mass Spectrum- Isotopes: Chemically of same reactivity- Relative Abundance:Cl-35: 100% Cl-35: a lot more in sampleCl-37: 32%B. Natural % AbundanceAZX14Si28.0855126C35173717xx3:1 Ratio: Cl-35: 100%Cl-37:32%**Formula weight deals with ionic compoundIII. Atomic Weight:- Weighted average is calculated from the isotopic massesAW= (%Ne-20)(19.9924u) + (%Ne-21)(20.9938u) + (%Ne-22)(21.9914u)A. Example: AW of B= 10.811u. Isotopic abundances B-10 and B-11 are 19.91% and 80.09%. What are the atomic masses?X: B-10 y: B-11X(.1991%) + y(.8009%) = 10.811uIV. The Periodic TableA. Main Group MetalsB. Transition MetalsC. MetalloidsD. NonMetalsCl-35: (100% / 132%) x 100% = 75.7% = 75%Cl-37: (32% / 132%) x 100% = 24.2% = 25% Ne-20: 19.9924uNe-21: 20.9938uNe-22: 21.9914u%Ne? = Natural % AbundanceV. Forms of CarbonA. Diamond- Covalently bondedB. Graphite(Pencil lead)- layered structure in sheetsC. Buckyballs- found in sot from high-temperature sparks (C-70, C-50)**A,B,C are all
View Full Document
Unlocking...