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UNCW CHM 101 - The Mole and Molar Calculations

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CHM 101 1st Edition Lecture 5 Overview of Last Lecture- Stoichiometry- Balancing equations- Formula weightOverview of Lecture 5 I. MolesII. Molar WeightThe Mole- Amount of matter that contains the same number of entities (atoms, molecules, ions)- Example: Atoms in exactly 12 g of pure 12Co 12.00g 12C = 1 mol 12C = 6.022 x 1023 12C atoms Avagadro’s Number (NA) = 6.022 x 1023 mol-1  Mol-1 – per moleo Examples:  1 mol Fe = 6.022 x 1023 Fe atoms 1 mol H2O = 6.22 x 1023 H2O molecules  1 mol NaCl = 6.022 x 1023 NaCl units - The definition of a mole also provides a mass relationshipo 1 amu = 1.66 x 10-24 g  Example: the mass of one 12C atom = 12.00 amu = 12.00(1.66 x 10-24 g)- One 12C atom = 1.99 x 10-23 g o 1 mol 12C atoms = 12.00 g 12C x 12 C atom 1.99 x 10-23 g 12C 1 mol 12C = 6.022 x 1023 12C atoms- Just as the atomic mass of an element is the mass of one atom in amu the atomic mass in grams is the mass of one mole of that type of atomo 1 Cu atom has a mass of 63.55 amu 1 mole of Cu has a mass of 63.55 go This rule also applies to molecules and molecular weight 1 O2 molecule has a mass of 32.00 amu 1 mole of O2 has a mass of 32.00 gExamples What is the mass of 2 moles of H2O?- Calculate the molecular weight o 2(1.01 amu) + 16.00 amu = 18.02 amu –OR- 18.02 g/mol2 mol H2O x 18.02 g H2O = 36.0 g H2O Mol H2OHow many moles are in 12.0 g of CO2?12.01 + 2(16.00) = 44.01 g/mol12.01 g CO2 x mol CO2 = 0.273 mol CO244.01 g CO2 How many g of C are there in 42.0 g of ethanol (C2H5OH)?Molecular weight = 46.08 g/mol42.0 g C2H5OH x mol C2H5OH x 2 mol C 46.08 g C2H5OH mol C2H5OH = 21.9 g CQuantitative Information from Balanced Equations- Using the stoichiometry factors provided by coefficients in balanced chemical equations we can predict the amount of reactant consumed and the amount of product formed in a reaction.o Example: Combustion of Propane (C3H8)o C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (g)o How many moles of O2 are consumed when 5 oles of C3H8 are combusted? 5 mol C3H8 x 5 mol O2 Mol C3H8 = 25 mol O2- Complete and balance equations- Calculate the molecular weight of required substances- Set up as a dimensional analysis problem with the appropriate stoichiometric ratio- Double check


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