1. Empirical Formulas
Empirical formulas: smallest whole number ratio of atoms present in a substance
Molecular formula: actual number of each type of atom present in a molecule
Many molecular compounds have different empirical and molecular formulas
N2O4 (molecular) NO2 (empirical)

2. Empirical Formulas
The empirical formula for a molecular compound can be found by dividing all subscripts in the molecular formula by the greatest common factor.
Examples:
C6H12O6
C4H10
Ca2SO4

3. Percent Composition
Empirical formulas are generally obtained by determining the percent composition of a compound:
Percent composition:
the percentage of the mass contributed by each element in a substance
% Element X = (# atoms of X)(AW) x 100%
FW of compound

4. % C = (# atoms of C)(AW) x 100%
Percent Compositon
Example: Calculate the % composition of C2H4O2 (i.e find %C, %H, and %O).
% C = (# atoms of C)(AW) x 100%
FW of compound
First, find the FW:
FW = 2(12.0 amu) + 4(1.0 amu) + 2(16.0 amu)
FW = 60.0 amu

5. Percent Composition Calculate the % composition for each element.
% C = 2(12.0 amu) x 100% = 40.0% C
60.0 amu
% H = 4(1.0 amu) x 100% = 6.7 % H
% O = 2(16.0 amu) x 100% = 53.3 % O

6. Percent Composition N2O4
Example: Calculate the % nitrogen in dinitrogen tetroxide.
N2O4
FW = 2(14.0 amu) + 4(16.0 amu) = 92.0 amu
% N = 2(14.0 amu) x 100% = 30.4%
92.0 amu

7. Calculating Empirical Formulas
Percent composition data is commonly used to determine the empirical formula of a compound:
Four steps:
percent to mass
mass to mole
divide by smallest
multiple ‘til whole

8. Calculating Empirical Formulas
Example: Calculate the empirical formula for a substance that contains 34.63% C, 3.875% H and 61.50% O.
Step 1: % to mass
If you assume g of substance, then
34.63 % C  g C
3.875 % H  g H
61.50 % O  g O

9. Calculating Empirical Formulas
Step 2: Mass to moles
# moles C = g C x 1 mole C = mol C
12.01 g C
# moles H = g H x 1 mole H = mol H
1.008 g H
# moles O = g O x 1 mole O = mol O
16.00 g O

10. Calculating Empirical Formulas
Step 3: Divide by smallest
(this gives the molar ratio of the elements)
C: moles C =
2.883 moles C
H: mol H =
2.883 mol C
O: mol O = 1.334

11. Calculating Empirical Formulas
Step 4: Multiply ‘til Whole
(this step is necessary only when step 3 does not give whole number molar ratios)
C: moles C =
2.883 moles C
H: mol H =
2.883 mol C
O: mol O = 1.334
x 3 = mol C
x 3 = mol H
x 3 = mol O

12. Calculating Empirical Formulas
Use these numbers to write the empirical formula:
3.000 mol C
4.000 mol H
C3H4O4
4.000 mol O

13. Calculating Empirical Formulas
Example: An iron compound contains % Fe and % O. Calculate its empirical formula.
Step 1: % to mass
% Fe  g Fe
% O  g O

14. Calculating Empirical Formulas
Step 2: Mass to Moles
mol Fe = g Fe x mol Fe = mol Fe
g Fe
mol O = g O x mol O = mol O
g O

15. Calculating Empirical Formulas
Step 3: Divide by smallest
O: mol O =
mol Fe
Fe: mol Fe =

16. Calculating Empirical Formulas
Step 4: Multiply ‘til whole
FeO1.5 doesn’t make sense.
mol O
mol Fe
x 2 = mol O
Fe2O3
x 2 = mol Fe

17. Using Empirical Formulas to Find Molecular Formulas
Empirical formula for a molecule is the smallest whole number ratio of atoms in the molecule.
The subscripts in the molecular formula must be some whole number multiple of the subscripts in the empirical formula:
CH2O C2H4O2, C3H6O3, C4H8O4
X 2
X 3
X 4

18. Using Empirical Formulas to Find Molecular Formulas
Steps:
Find the empirical formula
Calculate the formula weight for the empirical formula.
MW = the whole number
FW (empirical formula) ratio between MW
and FW
Multiply the subscripts in the empirical formula by the whole number ratio.

19. Using Empirical Formulas to Find Molecular Formulas
Example: A certain substances has an empirical formula of CH2O. If its molecular weight is amu, what is its molecular formula?
Step 1:
Empirical Formula = given = CH2O

20. Using Empirical Formulas to Find Molecular Formulas
Step 2:
FW = 1 (12.0 amu) + 2 (1.0 amu) + 1 (16.0 amu)
= amu
Step 3:
Whole number ratio = amu = 6
30.0 amu
Step 3:
Molecular formula: C(1x6)H(2x6)O(1x6) = C6H12O6