CHEM 1211 Lecture 4 Outline of Last Lecture I Examples II The Unit Factor Method III Percentage A Equation IV Density V Specific Gravity VI Chapter Two Atoms Molecules Ions A Subatomic particles a Protons b Neutrons c Electrons Outline of Current Lecture I Nuclide Structure II Isotopes A Mass Spectrum B Natural Abundance III Atomic Weight A Example IV The Periodic Table A Main Group Metals B Transition Metals C Metalloids V D NonMetals Forms of Carbon Current Lecture I Nuclide Structure A X Z 14 Si 28 0855 12 6 II C A Mass Number Z of Protons 14 Atomic Number Si Symbol of Element 28 0855 Mass Number All Atomic masses are based on Carbon 12 12C 12amu Mass of Carbon 1 99265 x 10 23g 1 99265 x 10 26kg 1 99265 x 10 23g 1 66 x 10 24g 12amu Isotopes Same element different mass 35 17 x 37 17 x X Cl Same Element Different number of neutrons different masses A Mass Spectrum Isotopes Chemically of same reactivity Relative Abundance Cl 35 100 Cl 37 32 B Natural Abundance Cl 35 a lot more in sample 3 1 Ratio Cl 35 100 132 x 100 75 7 75 Cl 35 100 Cl 37 32 132 x 100 24 2 25 Cl 37 32 Formula weight deals with ionic compound III Atomic Weight Weighted average is calculated from the isotopic masses Ne 20 19 9924u Ne 21 20 9938u Ne 22 21 9914u Ne Natural Abundance AW 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 11 X 1991 y 8009 10 811u IV The Periodic Table A Main Group Metals B Transition Metals C Metalloids D NonMetals V A B C Forms of Carbon Diamond Covalently bonded Graphite Pencil lead layered structure in sheets Buckyballs found in sot from high temperature sparks C 70 C 50 A B C are all allotropes
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