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.

10. 10 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 Empirical Formulas Basic Chemistry Copyright © 2011 Pearson Education, Inc.

11. 11 Some Molecular and Empirical Formulas The molecular formula is the same or a multiple of the empirical. Basic Chemistry Copyright © 2011 Pearson Education, Inc.

12. 12 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 Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

13. 13 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 Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

14. 14 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) SN 4 3) S 4 N 4 Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

15. 15 A compound has an empirical formula SN. If there are four atoms of N in one molecule, what is the molecular formula? Explain. 3) S 4 N 4 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. Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

16. Calculating Empirical Formulas 16 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

17. 17 A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula. Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

18. 18 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 Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

19. 19 STEP 3 Use the lowest whole-number ratio of moles as subscripts. NiBr 2 Solution (continued) Basic Chemistry Copyright © 2011 Pearson Education, Inc.

20. 20 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 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

21. Percent Composition Using 100 g 21 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

22. 22 Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula. Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

23. 23 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 = 5.00 mol of C 12.01 g C 4.5 g H x 1 mol H = 4.5 mol of H 1.008 g H 35.5 g O x 1mol O = 2.22 mol of O 16.00 g O Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

24. 24 Solution (continued) STEP 2 Divide by the smallest number of moles. 5.00 mol C= 2.25 mol of C 2.22 4.5 mol H = 2.0 mol of H 2.22 2.22 mol O = 1.00 mol of O 2.22 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

25. 25 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 C 9 H 8 O 4 Basic Chemistry Copyright © 2011 Pearson Education, Inc.