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lect_8oct13f08

# lect_8oct13f08 - Outline 8 mass mol 13-OCT-2008 1 Review...

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1 Outline 8 13-OCT-2008 1. Review- moles, calculating formulas mass ↔ mol 2. Determining Compound Formulas (cont). empirical vs. molecular, combustion analysis Examples 3. Writing Chemical Equations

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2 One mole of any element (compound) weighs its atomic (formula) mass in grams . Ex: One molecule CO 2 44.0 amu/molecule One mole CO 2 44.0 g/mol Moles mass conversion necessary for understanding quantitative nature of chemical equations. STOICHIOMETRY.
3 Mole calculations (Cont.) # moles “A” = mass “A” / molar mass of “A” Ex: Suppose have 5.75 moles of magnesium (Mg, 24.3 g/mol). What is its mass? g/mol) (24.3 moles) 75 . 5 ( Mg mass × = Mg of grams 140 =

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4 Review Mole vs. Molecule
5 Ex: Empirical formula of compound is CH 2 O; its “true” molar mass is 60.0 g/mol. Molar weight of empirical formula molar mass is 30.0 g/mol. formula mass/empirical mass = [60.0 g/mol] / [30.0 g/ emp. mol] = 2.00 emp. mol/ mol Did last time. C 2 H 4 O 2

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6 Formula (Combustion Analysis ) (ex:s1) Repeat from last time! Given: 0.250 g sample of compound (147.0 g/mol) contains C, H and Cl and gives 0.451 g CO 2 and 0.0617 g H 2 O from combustion analysis. Find: Molecular Formula Soln: moles compounds → moles C&H→ mass C&H → mass Cl → mol Cl → formula
Formula (Combustion Analysis) (ex:s2) #mol CO 2 = 0.451g CO 2 *(1 mol CO 2 /44.01g CO 2 ) = 0.01025 mol CO 2 #mol H 2 O = 0.0617g H 2 O*(1 mol H 2 O/18.02 g H

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lect_8oct13f08 - Outline 8 mass mol 13-OCT-2008 1 Review...

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