An empirical formula tells you the ratio of atoms. A molecular formula tells you the count. CH₂O is the empirical formula of formaldehyde, acetic acid, glyceraldehyde, ribose, and glucose — same atom ratio, five different molecules. The information that disambiguates them is the molar mass, which you get from mass spectrometry, freezing-point depression, or osmometry. Once you have both numbers, the rest is one division: how many empirical units fit into the actual molecule? Round n to a whole number, multiply every subscript by it, and you’re done.

The relationship

molecular formula = n × empirical formula n = (experimental molar mass) / (empirical formula mass)

n must come out very close to an integer. If it doesn’t, your empirical formula needs another look.

Empirical	Empirical mass	Molar mass	n	Molecular	Compound
CH₂O	30.03	30.03	1	CH₂O	formaldehyde
CH₂O	30.03	60.05	2	C₂H₄O₂	acetic acid
CH₂O	30.03	180.16	6	C₆H₁₂O₆	glucose
HO	17.01	34.01	2	H₂O₂	hydrogen peroxide

The recipe

Get the empirical formula (from mass percentages, combustion analysis, or directly given).
Compute the empirical formula mass — sum atomic weights of every atom in the empirical formula.
Divide the experimental molar mass by the empirical formula mass. Round to the nearest integer (n).
Multiply every subscript in the empirical formula by n.
Sanity-check by adding up the molecular formula mass — it should match the input.

Worked example: percent composition + molar mass

A compound is 40.0% C, 6.7% H, 53.3% O. Molar mass = 180.2 g/mol.

Empirical step (100 g basis):

C: 40.0/12.01 = 3.33 mol
H: 6.7/1.008 = 6.65 mol
O: 53.3/16.00 = 3.33 mol
Divide by smallest: 1 : 2 : 1 → CH₂O

Molecular step:

Empirical mass = 12.01 + 2(1.008) + 16.00 = 30.03 g/mol
n = 180.2 / 30.03 = 6.00
Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ (glucose)

Worked example: combustion analysis end-to-end

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

C and H from the combustion products:

mol CO₂ = 0.733/44.01 = 0.01666 → mol C = 0.01666
mol H₂O = 0.300/18.015 = 0.01665 → mol H = 0.03330

O by difference (the analyzer’s O₂ stream masks the sample’s oxygen):

mass C = 0.01666 × 12.01 = 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 = 0.2663/16.00 = 0.01665

Mole ratio: C : H : O = 0.01666 : 0.03330 : 0.01665, which divides to 1.00 : 2.00 : 1.00 → empirical CH₂O, mass 30.03.

n = 46.07 / 30.03 = 1.53. That’s not an integer — alarm bells. The CHO ratio cannot really be 1:2:1 if the compound has molar mass 46.07. Try the doubled formula: C₂H₆O has mass 46.07. Bingo.

The hidden problem: combustion-analysis numbers got rounded coming out, and the rounding compressed C : H : O = 1 : 3 : 1 down to look like 1 : 2 : 1. The molar mass forced you back to look. Molecular formula: C₂H₆O (ethanol).

Always validate n. A non-integer is information — it means the empirical formula has been over-simplified.

Worked example: n = 1

Empirical formula NO₂, molar mass 46.01 g/mol.

Empirical mass = 14.01 + 32.00 = 46.01
n = 1
Molecular formula: NO₂ (nitrogen dioxide)

When n = 1 the empirical and molecular formulas coincide. Common for small inorganics.

Why n must be a whole number

A molecule contains a whole-number count of each atom. If your division gives 1.53, you can’t have 1.53 carbons in a molecule. Either the empirical formula is wrong (rounding artifact), the molar mass measurement is off, or both. Diagnose before scaling.

Acceptable tolerance: ±0.05 from an integer is normal experimental noise. Anything worse and you should re-examine the empirical step.

Traps people fall into

Rounding non-integer n. Treating 1.53 as 2 turns ethanol into a phantom compound. When n is wrong, the empirical formula is wrong.
Forgetting that combustion hides oxygen. The O₂ feed swamps any sample O. You always back out O by mass-difference from the original sample.
Using rough atomic weights. Use 12.011 for C and 1.008 for H, not 12.000 and 1.000. The 0.01–0.02 errors per atom compound up and can shift n by enough to look non-integer.
Skipping the sanity check. After multiplying subscripts by n, add up the molecular formula mass and confirm it equals the input molar mass. Catches transcription errors.

Practice

Try these against the Empirical Formula Calculator:

Empirical CH, molar mass 78.11 → molecular formula?
Combust 1.000 g of a CHO compound: 2.279 g CO₂, 0.467 g H₂O, M = 154.12 → molecular formula?
85.7% C, 14.3% H, M = 56.10 → molecular formula?
Empirical P₂O₅, M = 283.89 → molecular formula?
26.7% C, 2.2% H, 71.1% O, M = 90.03 → molecular formula?

How to Determine a Molecular Formula