The empirical formula shows the simplest whole number ratio of elements in a compound. It's determined from percent composition or mass data and represents the most reduced form of the molecular formula.
1. % → grams
Assume 100g sample
2. grams → moles
n = mass / molar mass
3. Divide by smallest
Find mole ratios
4. Make whole numbers
Multiply if needed
Assume 100.0 g sample, so % = grams
Example: 40.0% C = 40.0 g C in 100 g sample
Use molar mass: n = mass / molar mass
Example: 40.0 g C ÷ 12.01 g/mol = 3.33 mol C
Find the element with fewest moles, divide all by this number
This gives mole ratios (subscripts in formula)
If ratios aren't whole numbers, multiply all by same factor
• 0.5 → multiply by 2
• 0.33 or 0.67 → multiply by 3
• 0.25 or 0.75 → multiply by 4
40.0% C = 40.0 g C
6.7% H = 6.7 g H
53.3% O = 53.3 g O
nC = 40.0 g ÷ 12.01 g/mol = 3.33 mol C
nH = 6.7 g ÷ 1.008 g/mol = 6.65 mol H
nO = 53.3 g ÷ 16.00 g/mol = 3.33 mol O
C: 3.33 ÷ 3.33 = 1.00
H: 6.65 ÷ 3.33 = 2.00
O: 3.33 ÷ 3.33 = 1.00
C : H : O = 1 : 2 : 1
Empirical Formula: CH₂O
This is the formula for formaldehyde or sugars like glucose (C₆H₁₂O₆)
Step 1-2: 72.0g C → 6.00 mol, 12.0g H → 11.9 mol, 16.0g O → 1.00 mol
Step 3: Divide by 1.00 (smallest)
C: 6.00 ÷ 1.00 = 6.00
H: 11.9 ÷ 1.00 = 11.9 ≈ 12.0
O: 1.00 ÷ 1.00 = 1.00
Step 4: Already whole numbers!
Empirical Formula: C₆H₁₂O
| If ratio is... | Multiply all by... | Example |
|---|---|---|
| 1.0, 2.0, 3.0 | 1 (already whole) | 1:2:3 → C₁H₂O₃ |
| 1.0, 1.5, 2.0 | 2 | 2:3:4 → C₂H₃O₄ |
| 1.0, 1.33, 1.67 | 3 | 3:4:5 → C₃H₄O₅ |
| 1.0, 1.25, 1.75 | 4 | 4:5:7 → C₄H₅O₇ |
| 1.0, 1.2, 1.4 | 5 | 5:6:7 → C₅H₆O₇ |
Must convert grams to moles first! Subscripts represent mole ratios, not mass ratios.
Keep extra decimal places during calculations. Only round at the very end when determining whole numbers.
If you multiply one ratio by 2 to make it whole, you MUST multiply all ratios by 2.
Percents should add to 100%. If not, oxygen might be missing or there's an error in the data.
Empirical shows the simplest ratio (CH₂O). Molecular shows the actual formula (C₆H₁₂O₆). Molecular = empirical × integer.
Makes math easy! If it's 40% C, then in 100g there are 40g of C. You can use any mass, but 100g makes percent = grams.
Round to the nearest whole number if it's very close (within ±0.1). 1.99 becomes 2. This accounts for rounding errors in the data.
Look for common fractions: 0.5 = 1/2 (multiply by 2), 0.33 = 1/3 (multiply by 3), 0.25 = 1/4 (multiply by 4), etc.
Yes! For H₂O, CO₂, NH₃, the empirical and molecular formulas are identical because they're already in simplest ratio.