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Percent Composition, Empirical Formula and Molecular Formula 10.4 summary
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Percent Composition 1 pizza 2 slices pineapple 6 slices pepperoni 4 slices cheese What % of the pizza slices are pineapple? pepperoni? just cheese? (2/12) x 100 = 17% pineapple, (6/12) x 100 = 50% pepperoni, (4/12) x 100 = 33% cheese
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Percent Composition of a Compound from experimental data ex: a lab sample contains 24.02 g carbon and 64.00 g oxygen. compound mass = 24.02 g + 64.00 g = 88.02 g % C: (24.02 g / 88.02 g) x 100 = 27.29% C % O: (64.00 g / 88.02 g) x 100 = 72.71 % O good because 27.29% + 72.71% = 100%
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Percent Composition of a Compound from a chemical formula ex: find the percent composition of CO 2. MM of C = 12.01 g/mole MM of O = 16.01 g/mole molar mass of compound = 12.01 + 2(16.00) = 44.01 g/mole % C: (12.01 g / 44.01 g) x 100 = 27.29% % O: (2X16.00 g / 44.01 g) x 100 = 72.71% good because 27.29% + 72.71% = 100%
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Empirical Formula Empirical means determined experimentally. If an unknown sample is analyzed and the amount of each element is determined, an “elementary” formula can be created. smallest, whole-number ratio of moles of each element in the compound. hydrogen peroxide = HO
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Calculating Empirical Formula 1.If given %, assume 100g sample. 2.Convert mass of each element to moles. 3.Divide each of these numbers by the smallest number to force one of them to = 1. 4.If necessary, multiply by the smallest number possible to make each a whole number. 5.These whole numbers are the subscripts in the empirical formula called mole ratio. empirical formula = X A Y B Z C
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Molecular Formula Actual number of atoms of each element in a molecule or formula unit. hydrogen peroxide = H 2 O 2 experimental molar mass needs to be provided.
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Calculating Molecular Formula 1.Find molar mass of empirical formula. 2.Divide experimentally determined molar mass (given), by molar mass of empirical formula. 1.Has to equal a whole number “n” 2.Multiply each subscript by n. molecular formula = X nxA Y nxB Z nxC
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