1. Percent Composition, Empirical and Molecular Formulas

2. Calculating Percentage Composition Calculate the percentage composition of potassium chromate, K 2 CrO 4.

3. Calculating Percentage Composition Calculate the percentage composition of potassium chromate, K 2 CrO 4.

4. Calculating Percentage Composition Calculate the percentage composition of potassium chromate, K 2 CrO 4.

5. Calculating Percentage Composition Calculate the percentage composition of potassium chromate, K 2 CrO 4.

6. Formulas ❑ molecular formula = (empirical formula) n ❑ molecular formula = C 6 H 6 = (CH) 6 ❑ empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

7. Empirical Formula Determination 1.Given the percent of each element, base calculation on 100 grams of compound. Determine moles of each element in 100 grams of compound. 2.Divide each value of moles by the smallest of the values. (This gives one element the smallest number of moles necessary) 3.Multiply each number by an integer to obtain all whole numbers.

8. Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? 1.Treat % as mass, and convert grams to moles

9. Empirical Formula Determination 2. Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

10. Empirical Formula Determination 3. Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 3 52 Empirical formula: C3H5O2C3H5O2

11. Caffeine has the following percent composition: carbon 49.48%, hydrogen 5.19%, oxygen 16.48% and nitrogen 28.85%. Its molecular weight is 194.19 g/mol. What is the empirical formula? 11

12. Molecular Formulas

13. 13 A molecular formula is equal or a multiple of its empirical formula has a molar mass that is the product of the empirical formula mass multiplied by a small integer molar mass = a small integer empirical mass is obtained by multiplying the subscripts in the empirical formula by the same small integer Relating Molecular and Empirical Formulas

14. 14 Some Compounds with Empirical Formula CH 2 O

15. Calculating a Molecular Formula from an Empirical Formula 15

16. 16 Determine the molecular formula of a compound that has a molar mass of 78.11 g and an empirical formula of CH. STEP 1 Calculate the empirical formula mass. Empirical formula mass of CH = 13.02 g STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer. 78.11 g = 5.999 ~ 6 13.02 g Finding the Molecular Formula

17. 17 STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula. Multiply each subscript in C 1 H 1 by 6. Molecular formula = C 1x6 H 1x6 = C 6 H 6 Finding the Molecular Formula (continued)

18. 18 A compound has a molar mass of 176.1g and an empirical formula of C 3 H 4 O 3. What is its molecular formula? 1) C 3 H 4 O 3 2) C 6 H 8 O 6 3) C 9 H 12 O 9 Learning Check

19. 19 STEP 1 Calculate the empirical formula mass. C 3 H 4 O 3 = 88.06 g/EF STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer. 176.1 g (molar mass) = 2 88.06 g (empirical formula mass) STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula. molecular formula = 2 x empirical formula C 3x2 H 4x2 O 3x2 = C 6 H 8 O 6 (2) Solution

20. 20 A compound contains 24.27% C, 4.07% H, and 71.65% Cl. The molar mass is about 99 g. What are the empirical and molecular formulas? Molecular Formula

21. 21 STEP 1 Calculate the empirical formula mass. 24.27 g C x 1 mol C = 2.021 mol of C 12.01 g C 4.07 g H x 1 mol H = 4.04 mol of H 1.008 g H 71.65 g Cl x 1 mol Cl = 2.021 mol of Cl 35.45 g Cl Solution

22. 22 Solution (continued) 2.021 mol C =1 mol of C 2.021 4.04 mol H =2 mol of H 2.021 2.02 mol Cl=1 mol of Cl 2.021 Empirical formula = C 1 H 2 Cl 1 = CH 2 Cl Empirical formula mass (EM) CH 2 Cl = 49.48 g

23. 23 Solution (continued) STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer. Molar mass = 99 g = 2 Empirical formula mass 49.48 g STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula. 2 x (CH 2 Cl) C 1x2 H 2x2 Cl 1x2 = C 2 H 4 Cl 2

24. 24 A compound is 27.4% S, 12.0% N, and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula? Learning Check

25. 25 STEP 1 Calculate the empirical formula mass. In 100 g, there are 27.4 g S, 12.0 g N, and 60.6 g Cl. 27.4 g S x 1 mol S = 0.854 mol of S 32.07 g S 12.0 g N x 1 mol N = 0.857 mol of N 14.01 g N 60.6 g Cl x 1mol Cl = 1.71 mol of Cl 35.45 g Cl Solution

26. 26 STEP 2 Divide the molar mass by the empirical formula mass to obtain a small integer. 0.854 mol S = 1.00 mol of S 0.854 0.857 mol N = 1.00 mol of N 0.854 1.71 mol Cl = 2.00 mol of Cl 0.854 empirical formula = SNCl 2 empirical formula mass = 116.98 g Solution (continued)

27. STEP 3 Multiply the empirical formula by the small integer to obtain the molecular formula. Molar mass = 351 g = 3 Empirical formula mass 116.98 g Molecular formula = (SNCl 2 ) 3 = S 3 N 3 Cl 6 Solution (continued)

28. Concept Map 28