Empirical & Molecular Formulas

Burn a sample, weigh the pieces, and work backwards to the formula. Here's the recipe.

High schoolIntro Gen ChemUni Year 1
⏱️ About 18 min

A chemist isolates an unknown white powder. How do they figure out its formula? They measure how much of each element it contains by mass, then run that backwards through the mole. The result is the compound's simplest atom ratio — and, with one more clue, its true formula.

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The big idea: The empirical formula is the simplest whole-number ratio of atoms in a compound; the molecular formula is the actual count in one molecule, always a whole-number multiple of the empirical one. You get the empirical formula by converting each element's mass to moles and reducing the ratio.
🎯 By the end, you'll be able to
  • Distinguish empirical from molecular formulas
  • Convert percent composition to an empirical formula via moles
  • Scale an empirical formula up to the molecular formula using the molar mass
📎 Helpful to know first

Empirical vs molecular: the ratio and the real thing

The empirical formula is the simplest whole-number ratio of the atoms in a compound. The molecular formula is how many atoms are actually in one molecule.

Glucose is the classic example. Its molecular formula is C₆H₁₂O₆, but the ratio 6 : 12 : 6 reduces to 1 : 2 : 1, so its empirical formula is CH₂O. The molecular formula is always a whole-number multiple of the empirical one — here, six times.

🔑 The three-step recipe
To go from percent composition to an empirical formula: (1) assume a 100 g sample, so each percent becomes grams; (2) convert each element's grams to moles (÷ its molar mass); (3) divide every mole value by the smallest of them to get the whole-number ratio.
📝 Worked example: A compound is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its empirical formula.
  1. Assume 100 g, so we have 40.0 g C, 6.7 g H, 53.3 g O.
  2. Convert to moles: C = 40.0 ÷ 12.01 = 3.33 mol; H = 6.7 ÷ 1.008 = 6.65 mol; O = 53.3 ÷ 16.00 = 3.33 mol.
  3. Divide by the smallest (3.33): C = 1.00, H = 2.00, O = 1.00.
  4. The ratio is 1 : 2 : 1, so the empirical formula is CH₂O.
✓ CH₂O (the empirical formula of glucose, among others).

From empirical to molecular

The empirical formula gives the ratio, but not the size. To find the real molecular formula you need one more piece of data: the compound's molar mass. Divide it by the empirical formula's mass to get the whole-number multiplier n, then multiply every subscript by n.

\[ n = \frac{\text{molar mass}}{\text{empirical formula mass}} \]
n is always (very close to) a whole number. Molecular formula = n × empirical formula.
📝 Worked example: The empirical formula CH₂O has a mass of 30.03 g/mol. A compound with this empirical formula has a molar mass of 180.16 g/mol. Find the molecular formula.
  1. Empirical formula mass of CH₂O = 12.01 + 2(1.008) + 16.00 = 30.03 g/mol.
  2. n = molar mass ÷ empirical mass = 180.16 ÷ 30.03 = 6.00.
  3. Multiply every subscript by 6: C₆H₁₂O₆.
✓ C₆H₁₂O₆ — glucose.
⚠️ Watch the rounding
When you divide by the smallest mole value, you're aiming for whole numbers. A result like 1.5 is not a rounding error — it means you should multiply everything by 2 to clear the fraction (1.5 : 1 becomes 3 : 2). But 1.98 or 2.02 really is just 2. Keep enough decimal places to tell the difference.
✏️ Practice: A compound's empirical formula is CH₂ (empirical mass 14.03 g/mol) and its molar mass is 42.08 g/mol. What is the whole-number multiplier n?
Solution
  1. n = molar mass ÷ empirical formula mass.
  2. n = 42.08 ÷ 14.03 = 3.00.
  3. So n = 3, and the molecular formula is C₃H₆ (multiply CH₂ by 3).
✏️ Practice: Analysis of a nitrogen oxide gives 1.40 g of nitrogen and 3.20 g of oxygen. How many oxygen atoms are there for each nitrogen atom (the O : N ratio)?
O per N
Solution
  1. Convert to moles: N = 1.40 ÷ 14.01 = 0.0999 mol; O = 3.20 ÷ 16.00 = 0.200 mol.
  2. Divide by the smaller (0.0999): N = 1.00, O = 2.00.
  3. So the ratio is 1 N to 2 O — the empirical formula is NO₂.

Check your understanding

1. Hydrogen peroxide is H₂O₂. What is its empirical formula?
The ratio 2 : 2 reduces to 1 : 1, so the empirical formula is HO. H₂O₂ is the molecular formula (n = 2).
2. Why do we assume a 100 g sample when starting from percentages?
In 100 g, a value like 40.0% is exactly 40.0 g. It's a convenient assumption — the final ratio comes out the same for any sample size.
3. Two compounds both have the empirical formula CH₂O. What extra information tells them apart?
Formaldehyde (CH₂O), acetic acid (C₂H₄O₂) and glucose (C₆H₁₂O₆) all share the empirical formula CH₂O. The molar mass sets n and reveals which one you have.
🔎 Related lessons
Previous: The Mole & Molar Mass Next: Balancing Chemical Equations
✅ Key takeaways
  • Empirical formula = simplest whole-number atom ratio; molecular formula = the actual counts.
  • The molecular formula is always a whole-number multiple (n) of the empirical formula.
  • Percent → empirical: assume 100 g, convert each element to moles, divide by the smallest.
  • A non-whole ratio like 1.5 means multiply everything to clear the fraction — not round it off.
  • Empirical → molecular: n = molar mass ÷ empirical formula mass, then scale the subscripts.
➡️ Formulas in hand, we can now write reactions — but a raw reaction equation usually doesn't obey conservation of mass yet. Balancing fixes that, and it's the next step before any reaction calculation.
Want to test yourself on this? Try the Chemistry practice test →