1. Empirical and Molecular Formulas
Let’s use those mole conversions!

2. Empirical and Molecular Formulas
Empirical formula tells you the lowest ratio of atoms within a molecule
Molecular formula tells you the actual ratio of atoms within a molecule
The empirical formula CAN equal the molecular formula, but it doesn’t have to.

3. Empirical and Molecular Formulas
Example:
A compound contains 39.99% carbon, 6.73% hydrogen, and 53.28% oxygen. The molar mass is g/mol. What is the empirical and molecular formulas for this compound?

4. Empirical and Molecular Formulas
Step 1:
Convert grams to moles
(If the problem gives you percents, assume that you have 100g of sample and use the percents as grams)
C: 𝑔 1 × 1 𝑚𝑜𝑙 12.01𝑔 =3.33 𝑚𝑜𝑙
H: 6.73𝑔 1 × 1 𝑚𝑜𝑙 1.01𝑔 =6.66 𝑚𝑜𝑙
O: 𝑔 1 × 1 𝑚𝑜𝑙 16.00𝑔 =3.33 𝑚𝑜𝑙

5. Empirical and Molecular Formulas
Step 2:
Divide all of the moles from Step 1 by the lowest number of moles from Step 1.
C: 𝑔 1 × 1 𝑚𝑜𝑙 12.01𝑔 =3.33 𝑚𝑜𝑙
H: 6.73𝑔 1 × 1 𝑚𝑜𝑙 1.01𝑔 =6.66 𝑚𝑜𝑙
Lowest #
O: 𝑔 1 × 1 𝑚𝑜𝑙 16.00𝑔 =3.33 𝑚𝑜𝑙

6. Empirical and Molecular Formulas
Step 2:
Divide all of the moles from Step 1 by the lowest number of moles from Step 1.
C: 𝑔 1 × 1 𝑚𝑜𝑙 12.01𝑔 = 3.33 𝑚𝑜𝑙 3.33 𝑚𝑜𝑙 = 1
These numbers tell you how many of each atom you have in the molecule.
H: 6.73𝑔 1 × 1 𝑚𝑜𝑙 1.01𝑔 = 6.66 𝑚𝑜𝑙 3.33 𝑚𝑜𝑙 =2
O: 𝑔 1 × 1 𝑚𝑜𝑙 16.00𝑔 = 3.33 𝑚𝑜𝑙 3.33 𝑚𝑜𝑙 =1

7. Empirical and Molecular Formulas
Sometimes you may end up with non-whole numbers in Step 2. If this happens:
multiply all of the moles by 2 if any end in .5
multiply all of the moles by 3 if any end in .3

8. Empirical and Molecular Formulas
Step 3:
Write the formula using the numbers found in Step 2.
This is the empirical formula:
C: 𝑔 1 × 1 𝑚𝑜𝑙 12.01𝑔 = 3.33 𝑚𝑜𝑙 3.33 𝑚𝑜𝑙 = 1
CH2O
H: 6.73𝑔 1 × 1 𝑚𝑜𝑙 1.01𝑔 = 6.66 𝑚𝑜𝑙 3.33 𝑚𝑜𝑙 =2
O: 𝑔 1 × 1 𝑚𝑜𝑙 16.00𝑔 = 3.33 𝑚𝑜𝑙 3.33 𝑚𝑜𝑙 =1

9. Empirical and Molecular Formulas
If the problems asks for molecular formula, move on to Step 4:
Find the molar mass of the empirical formula and divide the GIVEN molar mass by this number.
Molar mass of empirical formula:
CH2O
12.01g + 2*1.01g g = g/mol
GIVEN molar mass = g/mol
𝑔/𝑚𝑜𝑙 30.03𝑔/𝑚𝑜𝑙 =6
multiplier

10. Empirical and Molecular Formulas
Step 5:
Multiply the number of atoms in the empirical formula by the multiplier.
(CH2O) * 6 =
C6H12O6
𝑔/𝑚𝑜𝑙 30.03𝑔/𝑚𝑜𝑙 =6
multiplier
This is the molecular formula.