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Week 3 Lecture Notes

Week 3 Lecture Notes - Atomic Radius Left to Right...

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October 16, 2007 Atomic Radius : Left to Right: Decreasing Up to Down: Increasing Z eff IP (Ionization Potential) EA (Electron Affinity) Z eff : effective nuclear charge (high Z eff ~ smaller radius) Ionization Potential : energy required to remove e - Electron Affinity : energy release upon adding e - Mole = 6.0221 x 10 23 atoms Hydrogen – smallest mass = 1g Normalize : divide everybody (moles of each element) by smallest number Ionic : Molar mass is the sum of the molar masses of all atoms in the formula unit. Molecular (covalent bonds): Molar mass is the sum of the molar masses of all atoms in the molecular formula. Empirical Formula : ratio of the elements in a compound (relevant for both ionic and molecular compounds) Molecular Formula : actual # of atoms of each element in one molecule of a compound (only relevant for molecular compounds) Calculate Mass Percentage from Experimental Data 3.16 g sample of eucalyptol: 2.46 g C, 0.373 g H, 0.329 g O (Part/Whole) x 100% 77.8% C, 11.8% H, 10.4 % O Calculate Mass Percentage from Molecular Formula formula unit (ionic): AgNO 3 Calculate mass of each element per mole of compound: Ag: 1 x 107.87 g = 107.87 g 63.5% N: 1 x 14.01 = 14.01 8.2% O: 3 x 16 = 48 28.3% M of compound = 169.88 g/mol (mass of element per mole / compound M ) x 100% Determining Empirical Formulas : 1) Assume 100 g of compound, so mass % = # of g. 2) Divide mass of each element by M of that element to get molar proportions ( n per 100 g of compound). 3) Normalize ~ divide each proportion by smallest value. 4) Multiply all proportions by smallest integer that will make all values whole #s. Left to Right: Increase Up to Down: Decrease *Exception for 6 noble gases for EA (EA = 0)

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EX: eucalyptol : n C = 77.8 g / 12.01 g = 6.5 /.65 10 n H = 11.8 g / 1.008 g = 11.7 /.65 18 n O = 10.4 g / 16 g = 0.65 /.65 1 (in this case step 4 not necessary) empirical formula: C 10 H 18 O Heterogeneous Mixtures
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