The empirical formula is the bridge between a number on a combustion-analysis report and a chemical identity. You hand a sample to an elemental analyzer; it hands you back mass percentages of carbon, hydrogen, nitrogen, oxygen. Those percentages don’t tell you what the compound is — they tell you the ratio of atoms it contains. CH₂O might be formaldehyde, glucose, ribose, or a dozen other sugars; they all share the same simplest ratio. Getting from the percentages to that ratio is four steps of straightforward arithmetic, plus one step that reliably trips students up: knowing when to round and when to multiply.

What an empirical formula actually is

The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula is some integer multiple of it.

Compound	Empirical	Molecular
Formaldehyde	CH₂O	CH₂O
Acetic acid	CH₂O	C₂H₄O₂
Glucose	CH₂O	C₆H₁₂O₆
Hydrogen peroxide	HO	H₂O₂

Combustion analysis gives you the empirical formula directly. To get the molecular formula you also need the molar mass — usually from mass spectrometry.

The recipe

Assume 100 g. Each “X%” becomes “X g”. This is a free conversion that turns percentages into something you can put on a balance.
Convert each mass to moles. Divide by the atomic weight from the periodic table (use 12.011 for C, not 12.000).
Divide every mole count by the smallest. This rescales the ratio so the least-abundant element comes out as 1.
Inspect the ratios. If they’re within ~0.05 of whole numbers, round. If they sit near a clean fraction (1.5, 1.33, 1.25, 1.67), multiply every ratio by the appropriate integer (2, 3, 4, 3) to clear the denominator.
Write the formula with whole-number subscripts.

To go further to the molecular formula, divide the experimental molar mass by the empirical formula mass; that integer (call it n) tells you how many empirical units make up one molecule.

Worked example: clean integer ratio

A compound is 40.0% C, 6.7% H, 53.3% O.

100 g basis: 40.0 g C, 6.7 g H, 53.3 g O
Moles: C = 40.0/12.011 = 3.332; H = 6.7/1.008 = 6.647; O = 53.3/15.999 = 3.331
Divide by smallest (3.331): C = 1.000, H = 1.995, O = 1.000
Round: 1 : 2 : 1

Empirical formula: CH₂O. (Could be formaldehyde, glucose, lactic acid — molar mass tells you which.)

Worked example: when to multiply instead of rounding

A compound is 43.6% P, 56.4% O.

Moles: P = 43.6/30.974 = 1.408; O = 56.4/15.999 = 3.525
Divide by smallest: P = 1.000, O = 2.503

The 2.503 is suspiciously close to 2.5, not 3 or 2. Don’t round — multiply both ratios by 2:

P = 2, O = 5.006 ≈ 5

Empirical formula: P₂O₅ (phosphorus pentoxide). Rounding 2.5 down to 2 would have given you PO₂, which doesn’t exist.

Worked example: combustion analysis to molecular formula

You combust 0.500 g of a CHO compound and collect 0.733 g CO₂ and 0.300 g H₂O. Mass spec gives M = 46.07 g/mol.

Step 1 — extract C and H from the products:

mol CO₂ = 0.733/44.01 = 0.01666 → mol C = 0.01666 (one C per CO₂)
mol H₂O = 0.300/18.015 = 0.01665 → mol H = 0.03330 (two H per H₂O)

Step 2 — find O by difference. You can’t read it off a combustion product because the analyzer flooded the chamber with O₂.

mass C = 0.01666 × 12.011 = 0.2001 g
mass H = 0.03330 × 1.008 = 0.0336 g
mass O = 0.500 − 0.2001 − 0.0336 = 0.2663 g
mol O = 0.2663/15.999 = 0.01665

Step 3 — ratio: C : H : O = 0.01666 : 0.03330 : 0.01665. Divide by smallest: 1 : 2 : 1. Empirical formula CH₂O, mass 30.026.

Step 4 — molecular: n = 46.07 / 30.026 = 1.534. That’s not an integer, which means the empirical formula is wrong — the H/O ratio must actually be 3:1, not 2:1, and the rounding hid it. The true ratio is C : H : O = 1 : 3 : 1 with empirical mass 31.034 — still not matching. Check the doubled formula instead: C₂H₆O has mass 46.07, exactly matching the molar mass. Molecular formula: C₂H₆O (ethanol).

The lesson: when n comes out non-integer, your empirical formula needs revisiting before you scale.

Worked example: a ratio that’s actually integer

A compound has empirical formula NO₂ and M = 46.01 g/mol.

Empirical mass = 14.01 + 32.00 = 46.01
n = 46.01 / 46.01 = 1
Molecular formula: NO₂ (nitrogen dioxide). Molecular and empirical coincide.

Worked example: three elements, full chain

A compound is 52.2% C, 13.0% H, 34.8% O. M = 46.07 g/mol.

Moles: C = 4.346; H = 12.897; O = 2.175
Divide by smallest: C = 1.999, H = 5.930, O = 1.000 → 2 : 6 : 1
Empirical formula: C₂H₆O, empirical mass 46.07
n = 46.07 / 46.07 = 1
Molecular formula: C₂H₆O — ethanol again, this time arrived at without the false start.

Traps people fall into

Rounding fractional ratios. 1.50 rounds to neither 1 nor 2 — it means multiply everything by 2. Memorize the multipliers: 0.25 → ×4, 0.33 → ×3, 0.50 → ×2, 0.67 → ×3, 0.75 → ×4.
Skipping the 100 g assumption. Percentages aren’t moles. The 100 g basis converts percent to grams losslessly, which lets you divide by atomic weight to get moles.
Sloppy atomic weights. Use 12.011 for C, 1.008 for H, 15.999 for O. Rounding to integers introduces errors that look like genuine non-integer ratios.
Forgetting that combustion analysis hides the oxygen. The O₂ stream feeding the combustion chamber masks any oxygen in the sample. You always extract O by subtraction from total sample mass.
Percentages that don’t sum to 100%. The shortfall is usually an unmeasured element — most often oxygen in a CHN analyzer.

Use the Empirical Formula Calculator to compute empirical and molecular formulas from mass percentages with the steps shown.

How to Determine the Empirical Formula