1. Chemical Quantities Percent Composition and Empirical Formulas 1 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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 = 1 C(12.01g) + 2 O(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 % 2

3. Learning Check What is the percent composition of lactic acid, C 3 H 6 O 3, a compound that appears in the blood after vigorous activity? 3

4. Solution STEP 1 3C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol 36.03 g C + 6.048 g H + 48.00 g O STEP 2 %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 4

5. Learning Check The chemical isoamyl acetate C 7 H 14 O 2 gives the odor of pears. What is the percent carbon in isoamyl acetate? 1) 7.102 %C 2) 35.51 %C 3) 64.58 %C 5 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

6. Solution 3) 64.58 %C Molar mass C 7 H 14 O 2 = 7C(12.01) + 14H(1.008) + 2O(16.00) = 130.18 g/mol Total C = 7C(12.01) = g % C = total g C x 100 total g % C= 84.07 g C x 100 = 64.58 % C 130.18 g 6

7. 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. C 5 H 10 O 5  5 = C 1 H 2 O 1 = CH 2 O actual (molecular) empirical formula formula 7

8. Some Molecular and Empirical Formulas The molecular formula is the same or a multiple of the empirical. 8 Table 6.3 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

9. Learning Check A. What is the empirical formula for C 4 H 8 ? 1) C 2 H 4 2) CH 2 3) CH B. What is the empirical formula for C 8 H 14 ? 1) C 4 H 7 2) C 6 H 12 3) C 8 H 14 C. Which is a possible molecular formula for CH 2 O? 1) C 4 H 4 O 4 2) C 2 H 4 O 2 3) C 3 H 6 O 3 9

10. Solution A. What is the empirical formula for C 4 H 8 ? 2) CH 2 C 4 H 8  4 B. What is the empirical formula for C 8 H 14 ? 1) C 4 H 7 C 8 H 14  2 C. Which is a possible molecular formula for CH 2 O? 2) C 2 H 4 O 2 3) C 3 H 6 O 3 10

11. Learning Check A compound has an empirical formula SN. If there are 4 atoms of N in one molecule, what is the molecular formula? Explain. 1) SN 2) SN 4 3) S 4 N 4 11

12. Solution A compound has an empirical formula SN. If there are 4 atoms of N in one molecule, what is the molecular formula? Explain. 3) S 4 N 4 In this molecular formula 4 atoms of N and 4 atoms of S and N are related 1:1. Thus, it has an empirical formula of SN. 12

13. Learning Check A compound contains 7.31 amu Ni and 20.0 amu Br. Calculate its empirical (simplest) formula. 13

14. Solution Convert 7.31 g Ni and 20.0 amu Br to atoms. 7.31 amu Ni x 1 atom Ni = 0.125 atoms Ni 58.69 amu Ni 20.0 amu Br x 1 atoms Br = 0.250 atoms Br 79.90 amu Br Divide by smallest: 0.125 atoms Ni = 1 Ni0.250 atoms Br = 2 Br 0.125 0.125 Write ratio as subscripts: NiBr 2 14

15. Converting Decimals to Whole Numbers When the number of atoms for an element is a decimal, all the atoms are multiplied by a small integer to obtain whole number. 15 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Table 6.4

16. Learning Check Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula. 16

17. Solution STEP 1. Calculate the atoms of each element in 100 amu. 100 amu aspirin contains 60.0% C or 60.0 amu C, 4.5% H or 4.5 amu H, and 35.5% O or 35.5 amu O. 60.0 amu C x 1 atoms C = 5.00 atoms C 12.01 amu C 4.5 amu H x 1 atoms H = 4.5 atoms H 1.008 amu H 35.5 amu O x 1atoms O = 2.22 mol O 16.00 amu O 17

18. Solution (continued) STEP 2. Divide by the smallest number of atoms. 5.00 atoms C= 2.25 atoms C (decimal) 2.22 4.5 atoms H = 2.0 atoms H 2.22 2.22 atoms O = 1.00 atoms O 2.22 18

19. Solution (continued) 3. Use the lowest whole number ratio as subscripts When the atoms are not whole numbers, multiply by a factor to give whole numbers, in this case x 4. C: 2.25 atoms C x 4 = 9 atoms C H: 2.0 atoms Hx 4 = 8 atoms H O: 1.00 atoms O x 4 = 4 atoms O Using these whole numbers as subscripts the simplest formula is C 9 H 8 O 4 19