1. What Could It Be?
Finding Empirical
and Molecular
Formulas

2. It’s kind of like a fraction reduced to its lowest terms.
Empirical Formulas
The empirical formula is the simplest whole number ratio of the atoms of each element in a compound.
Note: it is not necessarily the true formula of the compound.
Example: The molecular formula for glucose is C6H12O6, but its empirical formula is CH2O
It’s kind of like a fraction reduced to its lowest terms.

3. Molecular Formulas
The molecular formula gives the actual numbers of each element, and thereby represents the true formula of the compound.

4. Calculating the Empirical Formula
Example 1: A compound is found to contain the following…
2.199 g copper
0.277 g oxygen
Calculate its empirical formula.

5. Calculating the Empirical Formula
Step 1: Convert the masses to moles.
Copper:
Oxygen:

6. Calculating the Empirical Formula
Step 2: Divide all the moles by the smallest value This gives the “mole ratio”

7. Calculating the Empirical Formula
Step 3: Round off these numbers, they become the subscripts for the elements.
Cu2O

8. Example 2: A material is found to be composed of 38. 7% carbon, 51
Example 2: A material is found to be composed of 38.7% carbon, 51.6% oxygen, and 9.7% hydrogen. By other means, it is known that the molecular weight is 62.0 g/mol Calculate the empirical and molecular formulas.
If you assume a sample weight of 100 grams, then the percents are really grams.

9. Now, divide all the moles by the smallest one, 3.22
Example 2: A material is found to be composed of 38.7% carbon, 51.6% oxygen, and 9.7% hydrogen. By other means, it is known that the molecular weight is 62.0 g/mol Calculate the empirical and molecular formulas.
Carbon:
Now, divide all the moles by the smallest one, 3.22
Oxygen:
Hydrogen:

10. Example 2: A material is found to be composed of 38. 7% carbon, 51
Example 2: A material is found to be composed of 38.7% carbon, 51.6% oxygen, and 9.7% hydrogen. By other means, it is known that the molecular weight is 62.0 g/mol Calculate the empirical and molecular formulas.
So, the empirical formula must be CH3O
The molecular weight of the empirical formula is….
C x = g/mol
H x 3 = g/mol
O x 1 = g/mol
31.04 g CH3O/mol CH3O

11. Remember, the empirical formula is not necessarily the molecular formula!
MW of the empirical formula = 31.04
MW of the molecular formula = 62.0

12. 2 x ( )
CH3O
= C2H6O2
Empirical
Formula
31.04 g/mol
Molecular
Formula
62.0 g/mol

13. Remember, the molecular formula represents the actual formula.

14. What if the mole ratios don’t come out even?

15. Example 3: A compound is analyzed and found to contain 2
Example 3: A compound is analyzed and found to contain 2.42 g aluminum and 2.15 g oxygen. Calculate its empirical formula.
Aluminum:
X 2 = 2
Oxygen:
X 2 = 3
Al2O3