1. mole (symbolized mol) = 6.02 x 1023 particles
(602,000,000,000,000,000,000,000)
Avogadro’s Number (NA)
molar mass – mass (usually in grams) of one
mole of a substance

2. Moles Mole Conversions Molar mass NA = 6.02 x 1023 atoms or molecules
atoms or
molecules
grams
Moles
Molar
mass NA
= 6.02 x 1023

3. Molarity (M) = moles of solute Liter of solution
Concentration Units:
Percent by Mass:
Mass % = mass of solute x 100
Total mass of solution
Recall:
Total mass of solution = mass of solute + mass of solvent
Molarity (M) = moles of solute
Liter of solution
Volume x Molarity = moles of solute
L x mol / L = moles

4. Empirical Formula Molecular Formula
Simplest whole number ratio of atoms in a compound
Molecular Formula
Actual formula (which is a whole number multiple of the empirical formula)

5. Empirical Formula Examples: H2O2 HO C6H12O6 CH2O C6H6 CH
Molecular Formula
Examples:
H2O HO
C6H12O6 CH2O
C6H CH

6. Finding the Empirical Formula
Find the moles of each element
Determine the mole ratio (largest/smallest)
Example:
Calculate the empirical formula of a
compound containing g Al and
53.18 g Cl.

7. 13.43 g x 1 mol = mol Al
27.0g
53.18 g x 1 mol = 1.50 mol Cl
35.5 g
1.50 = 3
3 mol Cl
1 mol Al
AlCl3

8. Practice Problem Calculate the empirical formula of a
compound containing % C, 13.12% H,
and % O.
52.14 g C / g/mol = (2)
13.12 g H / 1.0 g/mol = (6)
34.73 g O / g/mol = (1)
C2H6O

9. Note: Sometimes the ratio is not a whole number
Note: Sometimes the ratio is not a whole number. You must then convert it to a whole number ratio:
1.5/1 = 3/2
2.5/1 = 5/2
1.33 = 1 1/3 = 4/3
1.75 = 1 3/4 = 7/4

10. Practice Problem Calculate the empirical formula of a
compound that contains 43.7 % P and
56.3% O by mass.
43.7 g P / g/mol = (1)
56.3 g O / g/mol = (2.5)
P2O5

11. Finding the Molecular Formula from the Empirical Formula
Given the molecular mass, divide by the
empirical mass
Example: Empirical Formula: CH2O
(Empirical mass = molar mass = 30 g/mol)
Molecular Mass: 180 g/mol
Ratio: 180/30 = 6
Molecular Formula: 6 x (CH2O) => C6H12O6

12. Practice Problem Calculate the empirical and molecular formulas
of a compound that contains 80.0 % C,
20.0% H, and has a molar mass (molecular
mass) of g.
80.0 g C / g/mol = mol (1)
20.0 g H / 1.0 g/mol = mol (3)
Empirical CH3
30 = 2 15
Molecular C2H6

13. Do Problems: 70, 76, 80, 82, 84
pgs