1. 5-6 Determining the Emprical Formula of a Compound (Sections 8.6, 8.7)
And you

2. After the % by mass is known, the next step in determining the chemical formula is finding the empirical formula, which is the smallest whole number ratio of atoms.
Molecular formulas contain the exact ratio of atoms, for example, benzene is C6H6. In contrast the empirical formula for benzene is CH. Another example is glucose, whose molecular formula is C6H12O6. Its empirical formula is CH2O. Sometimes the empirical and molecular formulas are the same, as they are for water.

3. Determining the empirical formula involves a 4 step process:
Example:
Determine the empirical formula of a compound that is 24.27% C, 4.07% H, and 71.65% Cl by mass.
1. masses
2. moles
3. Divide by smallest # Moles
4. Multiply by integer to get Smallest Whole # Ratio
(if needed)

4. Step 1: Masses 24.27% C, 4.07% H, and 71.65% Cl by mass.
In all empirical formula problems, you can make the assumption that you have a 100 g sample. Then, any %’s turn into grams.
C = g H = 4.07 g Cl = g

5. Step 2: Moles C = 24.27 g H = 4.07 g Cl = 71.65 g
Since a formula represents mole ratios, we must change grams to moles.
C: g x 1 mole = mol C (It’s best to keep 4 sig figs.)
12.01 g
H: g x 1 mole = mol H
1.01 g
Cl: g x 1mole = mol Cl
35.45 g

6. Step 3: Divide by Smallest # of Moles 2. 021 mol C, 4. 030 mol H, 2
Step 3: Divide by Smallest # of Moles mol C, mol H, mol Cl
C: /2.021 = 1
H: /2.021 = 1.99 = 2
Cl: /2.021 = 1
Our example has an empirical formula of CH2Cl.

7. Step 4 (if needed): Multiply by integer to get Smallest Whole # Ratio
If the ratio of atoms was 1 to 1.5 to 1,
then multiply by Empirical = X2Y3Z2
If the ratio of atoms was 1 to 1.25 to 1,
then multiply by 4. Empirical = X4Y5Z5
If the ratio of atoms was 1 to 1.33 to 1,
then multiply by 3. Empirical = X3Y4Z3
If the ratio of atoms was 1 to 1.67 to 1, then multiply by 3. Empirical = X3Y5Z3

8. 5-7 Determining the Molecular Formula of a Compound (Section 8.8)
The final step in determining a compound’s formula is to take the empirical formula and translate it into the molecular formula (the “real” formula). Using the molar mass, which is determined by an experiment, the molecular formula is an integral factor of the empirical formula.

9. 2 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! C2H4Cl2 = Molecular formula
Using our example above, if the empirical formula is CH2Cl and the molar mass is g/mol, determine how much larger the molecular formula is compared to the empirical:
Molar Mass = = =
Emp. Formula Mass (1.01)
2 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 Since the “real” molecular formula is 2 x larger than the empirical formula, simply multiply the subscripts of the empirical by 2 to get
C2H4Cl2 = Molecular formula

10. Practice: A white powder is analyzed and found to contain 43
Practice: A white powder is analyzed and found to contain % phosphorous and % oxygen by mass. The compound has a molar mass of g/mole. What are the compound’s empirical and molecular formulas?
1. masses
2. moles
3. Divide by smallest # Moles
4. Multiply by integer to get Smallest Whole
# Ratio (if needed)