1. (4.6/4.7) Empirical and Molecular Formulas SCH 3U

2. An empirical formula represents the simplest whole number ratio of the atoms in a compound. The molecular formula is the true or actual ratio of the atoms in a compound.

3. Types of Formulas The formulas for compounds can be expressed as an empirical formula and as a molecular (true) formula. eg. empirical formula = CH 2 O

4. 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. What is a molecular formula for CH 2 O? 1) CH 2 O 2) C 2 H 4 O 2 3) C 3 H 6 O 3

5. Finding the Empirical Formula a) A compound is 71.65% Cl, 24.27% C, and 4.07% H. What is the empirical formula? 1. Assume a 100 g sample. Determine the mass, in grams, of each element present. Cl 71.65 gC 24.27 g H 4.07 g

6. 2. Calculate the number of moles of each element. n Cl = 71.65 g = 2.021 mol 35.45 g/mol n C = 24.27 g = 2.021 mol 12.01 g/mol n H = 4.07 g = 4.03 mol 1.01 g/mol

7. 3. Divide each by the smallest number of moles to obtain the simplest whole number ratio. **If whole numbers are not obtained, multiply subscripts by the smallest number that will give whole numbers** Cl: 2.021 mol = 1.000 Cl (1) 2.021 mol C: 2.021 mol = 1.000 C (1) 2.021 mol H: 4.04 mol = 2.00 H (2) 2.021 mol 4. Write the simplest or empirical formula. CH 2 Cl

8. Learning Check Aspirin is 60.0% C, 4.5 % H and 35.5% O. Calculate the empirical formula.

9. n C = 60.0 g = 5.00 mol 12.01 g/mol n H = 4.5 g = 4.5 mol 1.01 g/mol n O = 35.5 g = 2.22 mol 16.00 g/mol

10. 5.00 mol C = 2.25 mol 2.22 4.5 mol H = 2.0 mol 2.22 2.22 mol O = 1.00 mol 2.22 Therefore, the Empirical Formula (EF) = C 9 H 8 O 4 X 4 = 9 mol C X 4 = 8 mol H X 4 = 4 mol O

11. a) A compound is Cl 71.65%, C 24.27%, and H 4.07%. What is the empirical formula? (from yesterday, CH 2 Cl) b) The molar mass is known to be 99.0 g/mol. What is the molecular formula? 1. Calculate EFM (empirical formula mass) 1(12.01g/mol) + 2(1.01 g/mol) + 1(35.45g/mol) = 49.48 g/mol 2. Calculate Multiplier: Molar mass (M) = 99.0 g/mol = 2.00 EFM 49.48 g/mol 3. Multiply the empirical formula subscripts by the multiplier (CH 2 Cl) x 2 = C 2 H 4 Cl 2 Finding the molecular formula

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

13. n S = 27.4 g = 0.855 mol 32.06 g/mol n N = 12.0 g = 0.857 mol 14.01 g/mol n Cl = 60.6 g = 1.71 mol 35.45 g/mol Solution

14. 0.855 mol S = 1.00 mol 0.855 0.857 mol N = 1.00 mol 0.855 1.71 mol Cl = 2.00 mol 0.855 Therefore, the Empirical Formula (EF) = SNCl 2 Solution

15. empirical formula mass (EFM) = 32.06 g/mol + 14.01 g/mol + (2)(35.45 g/mol) = 116.97 g/mol Molar Mass = 351 g/mol Multiplier = 351 g/mol = 3.00 116.97 g/mol  so MF is S 3 N 3 Cl 6