AP Chemistry Empirical & Molecular Formula
From percent composition to empirical formula to molecular formula, fully worked.
Finding a compound's formula from lab data is a two-stage process: percent composition gets you the empirical formula first, and only then, if you also know the molar mass, can you scale up to the real molecular formula. Here's both stages, worked through one complete example.
Stage 1: Empirical Formula from Percent Composition
- Assume a 100 g sample. Each percentage becomes that many grams directly.
- Convert each element's mass to moles using its molar mass.
- Divide every mole value by the smallest one. This gives you the ratio between elements.
- Round to whole numbers (multiplying everything by a small whole number first if needed) to get the empirical formula's subscripts.
Stage 2: Molecular Formula from Empirical Formula + Molar Mass
- Calculate the empirical formula's mass by adding up the atomic masses in it.
- Divide the compound's actual molar mass by the empirical formula mass. This gives a whole-number multiplier, n.
- Multiply every subscript in the empirical formula by n to get the molecular formula.
Full Worked Example
A compound is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass, with a molar mass of 180 g/mol.
Stage 1. Assume 100 g: 40.0 g C, 6.7 g H, 53.3 g O.
- C: 40.0 g ÷ 12.01 g/mol = 3.33 mol
- H: 6.7 g ÷ 1.008 g/mol = 6.65 mol
- O: 53.3 g ÷ 16.00 g/mol = 3.33 mol
Divide each by the smallest (3.33 mol): C = 1.00, H = 2.00, O = 1.00. Empirical formula: CH₂O.
Stage 2. Empirical formula mass: 12.01 + 2(1.008) + 16.00 = 30.03 g/mol. Multiplier: n = 180 g/mol ÷ 30.03 g/mol ≈ 6. Multiply every subscript in CH₂O by 6: molecular formula C₆H₁₂O₆, glucose.
Common Mistakes
- Stopping at the empirical formula when the question asks for the molecular formula. They're only the same when the multiplier n = 1.
- Dividing by the largest mole value instead of the smallest. Always divide by the smallest value in the set, that's what turns the ratio into whole numbers starting at 1.
- Rounding an obviously-fractional ratio (like 1.5 or 1.33) to the nearest whole number instead of multiplying through. A clean 1.5 needs everything doubled, not rounded to 2.
- Using the wrong molar mass in Stage 2. The multiplier uses the compound's actual (given) molar mass divided by the empirical formula's mass, not the other way around.
Related Resources
- Unit 1 Review: Atomic Structure and Properties
- AP Chemistry Naming Compounds
- AP Chemistry Limiting Reactant & Percent Yield
- AP Chemistry Heating Curve
- AP Chemistry Significant Figures Rules
- AP Chemistry Study Guide
Frequently Asked Questions
What is the difference between empirical and molecular formula?
The empirical formula shows the lowest whole-number ratio of atoms in a compound. The molecular formula shows the actual number of atoms in one real molecule of that compound. They can be the same formula, or the molecular formula can be a whole-number multiple of the empirical formula, like C₆H₁₂O₆ (molecular) being 6 times CH₂O (empirical).
Why do you assume a 100 g sample when starting from percent composition?
Because percent composition is already a percentage by mass, assuming exactly 100 g of sample turns each percentage directly into that many grams, with no extra conversion step. It's a convenience assumption, not a claim about the actual sample size in any real experiment.
What do you do if the mole ratio comes out close to but not exactly a whole number, like 1.5?
Multiply every mole ratio in the formula by the same small whole number until they all become whole numbers. A ratio of 1.5 usually means multiplying everything by 2 (giving 3); a ratio of 1.33 usually means multiplying by 3 (giving 4). Round only to correct for ordinary measurement error, not to force an answer that's clearly still fractional.
Can the molecular formula ever equal the empirical formula?
Yes, whenever the whole-number multiplier works out to 1. Water (H₂O) and carbon dioxide (CO₂) are both already in their simplest whole-number ratio, so their empirical and molecular formulas are identical.
Do you need combustion analysis data for these calculations on the AP exam?
You need to be able to work with mass or percent composition data however it's given, including data derived from combustion analysis (grams of CO₂ and H₂O produced). The underlying method, converting to moles and finding the smallest whole-number ratio, is the same regardless of how the mass data was originally collected.
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