2. Empirical Formula:
Smallest ratio of atoms of all elements in a compound
Molecular Formula:
Actual numbers of atoms of each element in a compound
Determined experimentally
The empirical and molecular formulas may or my not be the same.

3. Empirical and Molecular:
H2O
CO2
Empirical
CH2O
Molecular
C6H12O6

4. Using % Comp to Find Empirical Formulas
Steps:
Convert mass of each element into moles. If given a % element, that percentage is equal to grams (assume 100g)
Divide moles of each element by the smallest number of moles available to generate mole ratios
If ratios are not whole numbers, multiply to make them whole.
Write the empirical formula

5. Using % Comp to Find Empirical Formulas
A sample of a compound containing sulfur and oxygen contains 59.95% O and 40.05% S. What is the empirical formula?
55.95g O x 1mol O = 3.747mol O/ = 3
15.99g O
40.05g S x 1mol S = 1.249mol S/1.249 = 1
32.07g S
Empirical formula is SO3

6. Using % Comp to Find Empirical Formulas
Determine the empirical formula for methyl acetate, which has the following analysis: 48.64% C, 8.16% H, 43.20% O.
43.20g O x 1mol O = mol O/ = 1x2 = 2
15.99g O
48.64g C x 1mol C = mol C/2.700 = 1.5 x 2 =3
12.01g C
8.16g H x 1mol H = 8.10 mol H/2.700 = 3 x 2 = 6
1.008g H
Empirical formula is C3H6O2

7. Molecular Formula Determine the molar mass of the empirical formula
Find the molecular molar mass in the problem and underline it.
Divide the molecular molar mass by the empirical molar mass
Multiply all the subscripts in empirical formula by the number obtained from step #3.

8. Practice
Acetylene (molar mass 26.04g/mol) and benzene (molar mass 78.12g/mol) both have an empirical formula of CH. The mass of the empirical formula is 13.02g/mol. What are the molecular formulas for acetylene and benzene?
Acetylene = g/mol = 2 C2H2
13.02 g/mol
Benzene = g/mol = 6 C6H6

9. %Comp, Empirical, and Molecular Formulas
Succinic acid is composed of 40.68%C, 5.08%H, and 54.24% O, and has a molar mass of g/mol. What are the empirical and molecular formulas?
54.24g O x 1mol O = mol O/ = 1x2 = 2
15.99g O
40.68g C x 1mol C = mol C/3.387 = 1 x 2 =2
12.01g C
5.08g H x 1mol H = 5.04 mol H/3.387 = 1.5 x 2 = 3
1.008g H
Empirical Formula = C2H3O2

10. Problem Continued…… Molar Mass of empirical formula:
2mol C x 12.01g/mol = 24.02g C
3mol H x 1.008g/mol = 3.024g H
2mol O x 15.99g/mol = 32.00g O
59.04g/mol succinic acid
To produce a molecular formula, multiply empirical by:
118.1 g/mol = 2 C4H6O4
59.04 g/mol