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Chapter 7 Chemical Quantities 7.4 Percent Composition and Empirical Formulas The label on a bag of fertilizer states the percentages of N, P, and K. 1 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Percent Composition Percent composition • is the percent by mass of each element in a formula. Example: Calculate the percent composition of CO2. CO2 = 1C(12.01g) + 2O(16.00 g) = 44.01 g/mol 12.01 g C 44.01 g CO2 2 x 100 = 27.29% C 32.00 g O x 100 44.01 g CO2 = 72.71% O 100.00 % Basic Chemistry Copyright © 2011 Pearson Education, Inc. Calculating Percent Composition 3 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Learning Check What is the percent composition of lactic acid, C3H6O3, a compound that appears in the blood after vigorous activity? 4 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution STEP 1 Determine the total mass of each element in the molar mass of a formula. 3C(12.01) = 36.03 g of C + 6H(1.008) = 6.048 g of H + 3O(16.00) = 48.00 g of O = 90.08 g/mol 5 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution (continued) STEP 2 Divide the total mass of each element by the molar mass and multiply by 100%. %C = 36.03 g C x 100 = 40.00% C 90.08 g %H = 6.048 g H x 100 = 6.714% H 90.08 g %O = 48.00 g O x 100 = 53.29% O 90.08 g 6 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Learning Check The chemical isoamyl acetate C7H14O2 gives the odor of pears. What is the percent of carbon in isoamyl acetate? 1) 7.102 %C 2) 35.51 %C 3) 64.58 %C 7 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution STEP 1 Determine the total mass of each element in the molar mass of a formula. 7C(12.01) = 84.07 g of C + 14H(1.008) = 14.11 g of H + 2O(16.00) = 32.00 g of O Molar mass = 130.18 g/mol 8 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution (continued) STEP 2 Divide the total mass of each element by the molar mass and multiply by 100%. %C = total g C x 100% molar mass %C 9 Basic Chemistry = 84.07 g C x 100% = 64.58 %C 130.18 g Copyright © 2011 Pearson Education, Inc. Empirical Formulas The empirical formula • is the simplest whole number ratio of the atoms • is calculated by dividing the subscripts in the actual (molecular) formula by a whole number to give the lowest ratio C5H10O5 5 = C1H2O1 = CH2O actual (molecular) empirical formula formula 10 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Some Molecular and Empirical Formulas • The molecular formula is the same or a multiple of the empirical. 11 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Learning Check A. What is the empirical formula for C4H8? 1) C2H4 2) CH2 3) CH B. What is the empirical formula for C8H14? 1) C4H7 2) C6H12 3) C8H14 C. Which is a possible molecular formula for CH2O? 1) C4H4O4 2) C2H4O2 3) C3H6O3 12 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution A. What is the empirical formula for C4H8? 2) CH2 C4H8 4 B. What is the empirical formula for C8H14? 1) C4H7 C8H14 2 C. Which is a possible molecular formula for CH2O? 2) C2H4O2 13 Basic Chemistry 3) C3H6O3 Copyright © 2011 Pearson Education, Inc. Learning Check A compound has an empirical formula SN. If there are four atoms of N in one molecule, what is the molecular formula? Explain. 1) SN 2) SN4 3) S4N4 14 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution A compound has an empirical formula SN. If there are four atoms of N in one molecule, what is the molecular formula? Explain. 3) S4N4 In this molecular formula four atoms of N and four atoms of S and N are related 1:1. Thus, it has an empirical formula of SN. 15 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Calculating Empirical Formulas 16 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Learning Check A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula. 17 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution STEP 1 Determine the moles of each element. 7.31 g Ni x 1 mol Ni = 0.125 mol of Ni 58.69 g Ni 20.0 g Br x 1 mol Br = 0.250 mol of Br 79.90 g Br STEP 2 Divide by the smallest number of moles. 0.125 mol Ni = 1 mol of Ni 0.125 0.250 mol Br = 2 mol of Br 0.125 18 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution (continued) STEP 3 Use the lowest whole-number ratio of moles as subscripts. NiBr2 19 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Converting Decimals to Whole Numbers When the number of moles for an element is a decimal • greater than 0.1, but less than 0.9 • multiply the moles by a small integer to obtain whole numbers 20 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Percent Composition Using 100 g 21 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Learning Check Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula. 22 Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution STEP 1 Calculate the moles of each element. 100.0 g aspirin contains 60.0% C or 60.0 g of C, 4.5% H or 4.5 g of H, and 35.5% O or 35.5 g of O. 60.0 g C x 1 mol C 12.01 g C = 5.00 mol of C 4.5 g H 1 mol H 1.008 g H = 4.5 mol of H 1mol O 16.00 g = 2.22 mol of O x 35.5 g O x 23 Basic Chemistry O Copyright © 2011 Pearson Education, Inc. Solution (continued) STEP 2 Divide by the smallest number of moles. 5.00 mol C 2.22 4.5 mol H 2.22 2.22 mol O 2.22 24 Basic Chemistry = 2.25 mol of C = 2.0 mol of H = 1.00 mol of O Copyright © 2011 Pearson Education, Inc. Solution (continued) STEP 3 Use the lowest whole-number ratio of moles as subscripts. To obtain whole numbers of moles, multiply by a factor, in this case x 4. C: 2.25 mol of C x 4 = 9 mol of C H: 2.0 mol of H x 4 = 8 mol of H O: 1.00 mol of O x 4 = 4 mol of O Using these whole numbers as subscripts, the simplest formula is C9H8O4 Basic Chemistry 25 Copyright © 2011 Pearson Education, Inc.