1. Empirical and Molecular Formulas SCH 3U

2. Types of Formulas The formulas for compounds can be expressed as an empirical formula and as a molecular (true) formula. Empirical Formula Molecular/True Formula Name CHC2H2C2H2 acetylene CHC6H6C6H6 benzene CO 2 Carbon dioxide CH 2 OC 5 H 10 O 5 ribose

3. 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.

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 O2) C 2 H 4 O 2 3) C 3 H 6 O 3

5. “Percent to mass Mass to mole Divide by small Multiply ‘til whole” Finding the Empirical Formula

6. A compound is Cl 71.65%, C 24.27%, and H 4.07%. What are the empirical and molecular formulas? The molar mass is known to be 99.0 g/mol. 1. “Percent to Mass” - state mass percentages as grams in a 100.00 g sample of the compound. Cl 71.65 gC 24.27 g H 4.07 g

7. 2. “Mass to Moles” 71.65 g Cl x 1 mol Cl = 2.02 mol Cl 35.5 g Cl 24.27 g C x 1 mol C = 2.02 mol C 12.0 g C 4.07 g H x 1 mol H = 4.04 mol H 1.01 g H

8. 3.“Divide by Small” Find the smallest whole number ratio by dividing each mole value by the smallest mole values: Cl: 2.02 mol = 1 Cl 2.02 mol C: 2.02 mol = 1 C 2.02 mol H: 4.04 mol = 2 H 2.02 mol 4. Write the simplest or empirical formula CH 2 Cl (“Multiply ‘til whole”)

9. Finding the Molecular Formula Multiplier: molar mass = a whole number empirical mass To get Molecular Formula, first calculate Empirical Formula, then multiply all subscripts by the multiplier Note: If your multiplier = 1, the molecular formula = empirical formula eg. if multiplier = 2, the MF = 2 x EF

10. We calculated Empirical Formula = CH 2 Cl & were given the molar mass = 99.0 g/mol 5. Calculate EM (empirical mass) = 1(C) + 2(H) + 1(Cl) = 49.5 g/mol 6. Calculate Multiplier: Molar mass = 99.0 g/mol = multiplier = 2 Empirical mass 49.5 g/mol 7. Molecular formula = Empirical Formula x multiplier (CH 2 Cl) x 2 = C 2 H 4 Cl 2 Back to the problem….

11. Learning Check Aspirin is 60.0% C, 4.5 % H and 35.5% O. Calculate its simplest formula. In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O. “Percent to mass Mass to mole Divide by small Multiply ‘til whole”

12. “Percent to Mass & Mass to Mole” 60.0 g C x 1 mol C = 5.00 mol C 12.0 g C 4.5 g H x 1 mol H = 4.5 mol H 1.01 g H 35.5 g O x 1mol O= 2.22 mol O 16.0 g O

13. 5.00 mol C = 2.25 x 4 = 9 C 2.22 4.5 mol H = 2.00 x 4 = 8 H 2.22 2.22 mol O = 1.00 x 4 = 4 O 2.22 Therefore, the Molecular Formula (MF) = C 9 H 8 O 4 “Divide by Small, Multiply ‘til Whole”

14. Finding Subscripts for Decimals A fraction between 0.1 and 0.9 must not be rounded. Multiply all results by an integer to give whole numbers for subscripts. FractionDecimalMultiply by:To Get: ½0.521 1/30.33331 ¼0.2541 ¾0.7643

15. 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?

16. Solution S = 0.853 mol & divide by 0.853 = 1 S N = 0.857 mol & divide by 0.853 = 1 N Cl = 1.71 mol & divide by 0.853 = 2 Cl Empirical Formula (EF) = SNCl 2 Empirical Mass (EM) = 117.1 g/mol Given Molar Mass = 351 g/mol Multiplier = 351g/mol = 3,  so MF = S 3 N 3 Cl 6 117.1g/mol