3.5: Empirical Formulas

EMPIRICAL FORMULAS

Smallest whole number ratio of atoms in a compound

(Molecular formula) C6H12O6 ---> CH2O (Empirical Formula) "GLUCOSE"

Ratio = 1:2:1 with molar mass: 180.6 grams

(Molecular formula) C2H4O2 ---> CH2O (Empirical Formula) "VINEGAR"

Ratio = 1:2:1 with molar mass: 60.6 grams

From the two examples above, although they have the same empirical formulas, they are completely different compounds. That is why, the molar mass is always needed to identify the compound.

Exercise:

Given 7.7% of Hydrogen and 92.3% of Carbon, find the empirical formula.

Step 1

Assuming 100 grams of sample, change the percent to grams.

7.7 grams Hydrogen

92.3 grams Carbon

Step 2

Change to moles, and divide both by the smallest amount of moles between them:

For every 1 molecule of H, I have one molecule of C, therefore my empirical formula is: CH

PERCENT COMPOSITION OF COMPOUNDS

Identifies how much percent of each compound there is in a given formula.

Example:

What is the percent composition of each element in potassium permanganate (KMnO4)?

Step 1

Obtain the molar mass of the compound.

39.10 grams of K +

54.94 grans of Mn 

4 x 16 grams of O4

158.04 grams of KMnO4

Step 2

Obtain the percent of each compound by diving each element mass by the total mass and multiply it by 100%