Empirical vs. Molecular Formula: What's the Difference and How to Find Each

Glucose, formaldehyde and acetic acid are three very different substances. One is a sugar, one is a toxic gas, one gives vinegar its sharpness. Yet if you analyze each of them in a lab, you’ll find exactly the same thing: 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass.

That coincidence is the whole story of empirical versus molecular formulas.

The two definitions

• The molecular formula tells you the actual number of each type of atom in one molecule. Glucose is C₆H₁₂O₆.
• The empirical formula is the simplest whole-number ratio of those atoms. Divide 6 : 12 : 6 by 6 and you get 1 : 2 : 1, so glucose’s empirical formula is CH₂O.

Formaldehyde’s molecular formula is CH₂O. Acetic acid’s is C₂H₄O₂. All three share the empirical formula CH₂O, which is why their percent compositions are identical. Mass percentages only reveal ratios; they can’t tell you how big the molecule is.

Some compounds have the same empirical and molecular formula (water, H₂O; carbon dioxide, CO₂). And ionic compounds like NaCl or Fe₂O₃ don’t form discrete molecules at all, so for them the “formula” is by definition an empirical one — the simplest ratio of ions.

Finding an empirical formula: the four steps

Here’s the method, using a classic example: a compound that analysis shows is 69.94% iron and 30.06% oxygen.

Step 1 — Assume 100 g. Percentages turn directly into grams: 69.94 g Fe and 30.06 g O.

Step 2 — Convert each mass to moles by dividing by the atomic mass:

• Fe: 69.94 ÷ 55.845 = 1.2524 mol
• O: 30.06 ÷ 15.999 = 1.8789 mol

Step 3 — Divide by the smallest number of moles:

• Fe: 1.2524 ÷ 1.2524 = 1.000
• O: 1.8789 ÷ 1.2524 = 1.500

Step 4 — Make the ratio whole. 1 : 1.5 isn’t a whole-number ratio, and you can’t have half an atom. Multiply everything by 2: 2 : 3.

The empirical formula is Fe₂O₃, iron(III) oxide — rust.

The rounding rule that trips everyone up

Step 4 is where most mistakes happen. When a ratio comes out close to a whole number (1.98, 3.02), round it. When it comes out close to a common fraction, don’t round — multiply:

Ratio looks like	It’s really	Multiply everything by
x.50	½	2
x.33 or x.67	⅓ or ⅔	3
x.25 or x.75	¼ or ¾	4
x.20, x.40, x.60, x.80	fifths	5

Rounding 1.5 up to 2 would have given FeO₂, a compound that doesn’t exist. A good rule of thumb: only round when you’re within about 0.1 of a whole number.

For example, magnetite is 72.36% Fe and 27.64% O. That gives moles of 1.2957 and 1.7276, a ratio of 1 : 1.333. Multiply by 3 and you get Fe₃O₄.

From empirical to molecular formula

To get the molecular formula you need one more piece of information: the compound’s molar mass, usually measured separately (by mass spectrometry, for instance).

1. Calculate the empirical formula mass. For CH₂O: 12.011 + 2(1.008) + 15.999 = 30.03 g/mol.
2. Divide the actual molar mass by it. For glucose: 180.16 ÷ 30.03 = 6.00.
3. Multiply every subscript by that number: C₁×₆H₂×₆O₁×₆ = C₆H₁₂O₆.

The result of step 2 should always be very close to a whole number. If you get something like 4.5, recheck your empirical formula — it’s probably missing a factor of 2.

A combustion analysis example

In real labs, organic compounds are often analyzed by burning them and weighing the CO₂ and H₂O produced. Suppose a sample of a compound containing C, H and O gives, after the calculation from those combustion products, 2.61 g carbon, 0.658 g hydrogen and 1.74 g oxygen.

• C: 2.61 ÷ 12.011 = 0.2173 mol
• H: 0.658 ÷ 1.008 = 0.6528 mol
• O: 1.74 ÷ 15.999 = 0.1088 mol

Divide by 0.1088: C = 2.00, H = 6.00, O = 1.00. Empirical formula: C₂H₆O. With a measured molar mass of 46.07 g/mol (exactly one empirical unit), the molecular formula is also C₂H₆O — ethanol, or its isomer dimethyl ether. And that points to the last limitation: even the molecular formula doesn’t tell you how the atoms are connected. For that you need a structural formula.

Quick answers

Can the empirical and molecular formula be the same? Yes, whenever the subscripts in the molecular formula have no common factor, like H₂O, CO₂ or C₂H₆O.

Why do we assume a 100 g sample? Only for convenience — percentages become grams directly. Any sample size gives the same ratio.

What if my percentages don’t add to 100? If one element wasn’t measured (often oxygen), it’s usually the remainder. Otherwise the data has errors, and you may need a little more tolerance in step 4.

Let the calculator do the arithmetic

The empirical formula calculator takes percentages or grams, shows the moles-and-ratio table step by step, handles the 1.5 and 1.33 cases automatically, and gives the molecular formula if you add the molar mass. To go the other way — from a formula to its percentages — use the percent composition calculator.