1. The Mole & Chemical Formulas A chemical formula represents the ratio of atoms that always exists for that compound Example: Water – H 2 O Always 2 H atoms to 1 O atom

2. The Mole & Chemical Formulas This also means if you had a mole of water… You would have 2 moles of Hydrogen atoms And 1 mole of Oxygen atoms

3. Empirical Formula This is the simplest ratio of atoms in a chemical formula. Example: Glucose, a sugar, has the molecular formula C 6 H 12 O 6 The empirical formula is CH 2 O

4. Empirical Formula Example: A chemical has 80.0% C atoms and 20.0% H atoms. What is the empirical formula of the compound? Use the %’s as masses to find moles…

5. Empirical Formula 80.0 g C 1 mol C 12.0 g C = 6.67 mol C 20.0 g H 1 mol H 1.0 g H = 20.0 mol H

6. Empirical Formula Use the moles to determine the ratio: Carbon: 6.67 mol = 1 Hydrogen: 20.0 mol 6.67 mol = 3 CH3H3 This is the Empirical Formula!

7. Molecular Formula This gives the actual number of atoms for each element in a compound. Example: If the molar mass of the “CH 3 ” compound from the previous problem is 30.0 g/mol, what is the molecular formula?

8. Molecular Formula Empirical formula mass for CH 3 is 15.0 g/mol Then find the ratio of molar to empirical mass. 30.0 g/mol (molecular) 15.0 g/mol (empirical) = 2.00

9. Molecular Formula Multiply the empirical formula by your new ratio: C (1x2) H (3x2) = C 2 H 6 (molecular formula)

10. Molecular Formula Example: A chemical is 48.4% C atoms, 8.12% H atoms and the rest is Oxygen. If the molecular mass is 222 g/mol, find the molecular formula. First find empirical formula…

11. Molecular Formula 48.4 g C 1 mol C 12.0 g C = 4.03 mol C 8.12 g H 1 mol H 1.0 g H = 8.12 mol H 43.5 g O 1 mol O 16.0 g O = 2.72 mol O

12. Molecular Formula Carbon: 4.03 mol 2.72 mol = 1.48 Hydrogen: 8.12 mol 2.72 mol = 2.99 x 2 = 3 C3C3 Oxygen: 2.72 mol = 1.00 x 2 = 6 x 2 = 2 H6H6 O2O2

13. Molecular Formula 222 g/mol (molecular) 74.0 g/mol (empirical) = 3.00 C 3x3 H 6x3 O 2x3 C 9 H 18 O 6 Molecular Formula!