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1 Chapter 7 Chemical Quantities 7.4 Percent Composition and Empirical Formulas Basic Chemistry Copyright © 2011 Pearson Education, Inc. The label on a.

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Presentation on theme: "1 Chapter 7 Chemical Quantities 7.4 Percent Composition and Empirical Formulas Basic Chemistry Copyright © 2011 Pearson Education, Inc. The label on a."— Presentation transcript:

1 1 Chapter 7 Chemical Quantities 7.4 Percent Composition and Empirical Formulas Basic Chemistry Copyright © 2011 Pearson Education, Inc. The label on a bag of fertilizer states the percentages of N, P, and K.

2 2 Percent Composition Percent composition is the percent by mass of each element in a formula. Example: Calculate the percent composition of CO 2. CO 2 = 1C(12.01g) + 2O(16.00 g) = 44.01 g/mol 12.01 g C x 100 = 27.29% C 44.01 g CO 2 32.00 g O x 100 = 72.71% O 44.01 g CO 2 100.00 % Basic Chemistry Copyright © 2011 Pearson Education, Inc.

3 Calculating Percent Composition 3 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

4 4 What is the percent composition of lactic acid, C 3 H 6 O 3, a compound that appears in the blood after vigorous activity? Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

5 5 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 Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

6 6 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 Solution (continued) Basic Chemistry Copyright © 2011 Pearson Education, Inc.

7 7 Learning Check The chemical isoamyl acetate C 7 H 14 O 2 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 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

8 8 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 Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

9 9 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 = 84.07 g C x 100% = 64.58 %C 130.18 g Solution (continued) Basic Chemistry Copyright © 2011 Pearson Education, Inc.


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