1. Empirical Formulas
Molecular Formulas

2. Since the formula of a compound indicates the number of atoms in the compound, CH4 = 4H atoms and 1C atom we must first convert the masses of the elements to number of atoms g  moles  atoms (ratio)

3. Step 1: g  moles Given: 38.67% Carbon by mass
38.67g Carbon per 100g of the compound
Given: 16.22% Hydrogen by mass
16.22g Hydrogen per 100g of the compound
Given: 45.11% Nitrogen by mass
45.11g Nitrogen per 100g of the compound

4. Step 1: g  moles Given: 38.67% Carbon by mass
38.67g Carbon per 100g of the compound
38.67g C 1 mol C = mol C
12.01g C

5. Step 1: g  moles Given: 16.22% Hydrogen by mass
16.22g Hydrogen per 100g of the compound
16.22g H 1 mol H = mol H
1.008 H

6. Step 1: g  moles Given: 45.11% Nitrogen by mass
45.11g Nitrogen per 100g of the compound
45.11g N 1 mol N = mol N
14.01g N

7. Step 2: Which number is the smallest?
38.67g C 1 mol C = mol C
12.01g C
16.22g H 1 mol H = mol H
1.008 H
45.11g N 1 mol N = mol N
14.01g N

8. Step 3: Whole Number Ratio
The smallest number from step two becomes the denominator when solving for atoms
C: = = 1
3.220
H: = = 5
N: = = 1
CH5N
Can find molar mass
C – 1 x 12.01
H – 5 x 1.008
N – 1 x 14.01
Compare to given and divide by = 1, 1 = multiplier

9. Molecular Formula
Molar Mass (given) = Molecular Formula Multiplier Empirical Formula Mass
Given: ClCH2 has a molar mass of g/mol and an empirical formula mass of g/mol
MM = g/mol = 2
EFM g/mol
We now multiple (ClCH2) by 2 =
(ClCH2) 2  Cl2C2H4

10. Step 3: Whole Number Ratio
Example from Empirical Formula Notes
Can find molar mass
C – 1 x 12.01
H – 5 x 1.008
N – 1 x 14.01
Compare to given (31.06 g/mol) and divide (found molar mass) = 1, 1 = multiplier
CH5N
Can find molar mass
C – 1 x 12.01
H – 5 x 1.008
N – 1 x 14.01
Compare to given and divide by = 1, 1 = multiplier