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Chemical Composition Notes Conversions This part is the logic that will help you remember Moles are a chemist’s dozen: 3 dozen eggs x 12 = 36 eggs Moles x (conversion factor) = new unit Conversion factors are o Atomic mass or molar mass Units of grams per mole (g / mol) Used for converting between grams and moles o Avogadro’s Number Units of atoms per mole (atoms / mol) Used for converting between moles and atoms Here are the equations Moles x AM = grams moles x Av# = atoms Grams / AM = moles atoms / Av# = moles Don’t forget Av# = 6.022 x 1023 Percent Composition Designated as a fraction Part Whole Example: CH3O AM = 31.034 The part is each element times its quantity The whole is the AM for the compound C CH 3O 3H CH 3O O CH 3O Make sure you use the mass of each 12.011 (100) = 38.703% Carbon 31.034 3.0237 (100) = 9.7432% Hydrogen 31.034 15.999 (100) = 51.553% Oxygen 31.034 Determining Empirical Formulas Steps: 1. 3. 2. Convert grams to moles (gives ugly mole ratio) Divide by all numbers by the smallest (sets smallest to 1) If decimals are present, multiply all numbers to create integers 4. Apply integers to the formula Given either mass of percentages of constituent elements it works the same Example: An unknown sample was found to have a percent composition as follows: 47.0% K, 14.5% C, and 38.5% O. Find the empirical formula. Given %, assume a 100 g sample so: 47.0 g K 14.5 g C 38.5 g O This is a mass ratio, but all analysis is done by mole ratio so convert each into moles 47.0 14.5 38.5 = 1.20 mol K = 1.21 mol C = 2.41 mol O 39.098 12.011 15.999 These are the right units, but we need to reduce our numbers Divide all by the smallest reducing the least down to one while holding the ratio 1.20 1.21 2.41 = 1K = 1C = 2O 1.20 1.20 1.20 So the final compound is KCO2 Determining Molecular Formulas Steps: 1. Fine the empirical mass (molar mass of the EF) 2. Molar mass / Empirical mass = multiplier 3. Empirical formula (multiplier) = molecular formula Example: Given the empirical formula above, determine the molecular formula if the molar mass of the compound is 166.214 g/mol First find the empirical mass (atomic mass of the empirical formula) 39.098 + 12.011 + 2 (15.999) = 83.107 g/mol Now divide the atomic mass by the empirical mass 166.214 / 83.107 = 2 This is your multiplier (n) value KCO2 x 2 = K2C2O4 Vocabulary, definitions, abbreviations, and units. Abbrev Units Atomic mass: mass of one mole of an atom AM g/mol Molar mass: mass of one mole of a molecule MM g/mol Mole: 6.022 x 1023 atoms make up one mole mol Avogadro’s Number: 6.022 x 1023 Av# Mass: what asking for mass, calculate grams g Atom: basic building block of matter atom % composition: the % by mass of each element in a compound Empirical formula: reduced form of a chemical formula EF Empirical mass: molar mass of an empirical formula EM Molecular formula: actual chemical formula MF atoms / mol % (element) g/mol