Molecular vs. Empirical Formulas
Every compound has a chemical formula, but not every formula tells you the same story. There are two types you need to know cold for the MCAT, and confusing them is one of the most common mistakes on test day.
Empirical Formula - The Simplest Ratio
The empirical formula gives you the smallest whole-number ratio of atoms in a compound. It tells you the relative proportions, but not the actual count.
For example, glucose has the empirical formula CH₂O. This tells you that for every carbon atom, there are two hydrogen atoms and one oxygen atom. But it does not tell you how many of each atom are actually in one molecule of glucose.
Molecular Formula - The Actual Count
The molecular formula tells you the exact number of each type of atom in one molecule. Glucose’s molecular formula is C₆H₁₂O₆ - six carbons, twelve hydrogens, six oxygens. That is 6 times the empirical formula (6 x CH₂O).
The molecular formula is always a whole-number multiple of the empirical formula:
Molecular formula = n x (Empirical formula)
where n = molecular weight / empirical formula weight.
How They Connect
| Compound | Empirical Formula | Molecular Formula | n |
|---|---|---|---|
| Glucose | CH₂O | C₆H₁₂O₆ | 6 |
| Acetic acid | CH₂O | C₂H₄O₂ | 2 |
| Formaldehyde | CH₂O | CH₂O | 1 |
| Hydrogen peroxide | HO | H₂O₂ | 2 |
| Benzene | CH | C₆H₆ | 6 |
Notice that glucose, acetic acid, and formaldehyde all share the same empirical formula. The empirical formula alone cannot distinguish between them - you need the molecular weight to determine n and therefore the molecular formula.
Determining the Empirical Formula from Percent Composition
This is a classic MCAT problem type. Here is the step-by-step process:
- Assume 100 g of the compound (so percentages become grams directly)
- Convert grams to moles for each element (divide by atomic mass)
- Divide all mole values by the smallest mole value
- Round to the nearest whole number (if you get values like 1.5 or 2.33, multiply all values by 2 or 3 to clear the fraction)
Example: A compound is 40.0% C, 6.7% H, and 53.3% O by mass. Find its empirical formula.
- C: 40.0 g / 12.0 g/mol = 3.33 mol
- H: 6.7 g / 1.0 g/mol = 6.7 mol
- O: 53.3 g / 16.0 g/mol = 3.33 mol
Divide by smallest (3.33): C = 1, H = 2, O = 1
Empirical formula: CH₂O
Determining the Molecular Formula
Once you have the empirical formula and the molecular weight (from mass spectrometry or other data), calculate n:
n = molecular weight / empirical formula weight
For the example above, if the molecular weight is 180 g/mol:
- Empirical formula weight of CH₂O = 12 + 2(1) + 16 = 30 g/mol
- n = 180 / 30 = 6
- Molecular formula: C₆H₁₂O₆ (glucose)