What is the difference between empirical and molecular formula?

The empirical formula is the simplest whole-number ratio of atoms in a compound; the molecular formula is the actual number of each atom in one molecule. They can be the same (water is H2O both ways) or the molecular formula can be a whole-number multiple of the empirical one — glucose is CH2O empirically but C6H12O6 as a molecule, exactly 6 times the empirical formula. You need the compound's molar mass to get from the empirical formula to the molecular formula.

Why do I divide by the smallest number of moles?

Dividing every element's moles by the smallest value forces the least-abundant element to a ratio of exactly 1, which makes the other ratios easy to read as 'how many of this atom per one of that.' It's just a way of normalizing the ratio into its simplest starting form before you clean it up into whole numbers.

What do I do when a ratio comes out as 1.5 or 2.33?

Multiply every ratio by the smallest whole number that turns them all into integers. A 1.5 means multiply by 2 (giving 3); a 1.33 (which is 4/3) means multiply by 3 (giving 4); a 1.25 (5/4) means multiply by 4. The point is that atoms can only come in whole numbers, so a non-integer ratio just means your 'divide by the smallest' step landed on a fraction that needs scaling up.

How do I get the molecular formula once I have the empirical formula?

Add up the atomic masses in the empirical formula to get the empirical formula mass, then divide the compound's known molar mass by it. That gives a whole number (call it n), and the molecular formula is every empirical-formula subscript multiplied by n. For glucose: empirical CH2O has a mass of about 30, the molar mass is about 180, so n = 180/30 = 6, giving C6H12O6.

Can I enter masses in grams instead of percent?

Yes. Percent composition and grams are handled identically here, because assuming a 100 g sample turns each percent directly into grams. So whether a problem gives you '40.0% carbon' or '4.00 g of carbon,' you enter the number the same way and the mole ratio comes out the same.

Empirical Formula Calculator — and Molecular Formula, Free

Each element's amount is given as:

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Percent (%)

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Compound's molar mass (g/mol) — optional, for the molecular formula

Empirical formula

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How the empirical formula is found

The whole method turns "how much of each element" into "how many atoms of each, in the simplest whole-number ratio." Percent composition is treated as grams by imagining a 100 g sample, so 40% carbon just means 40 g of carbon. Convert each element's grams to moles (divide by atomic mass), then divide every mole value by the smallest one so the least-abundant element becomes 1. If any ratio lands on a fraction like 1.5 or 1.33, multiply them all by the smallest whole number that clears the fraction — atoms only come in whole numbers. Those integers are your subscripts. To go from there to the molecular formula, divide the real molar mass by the empirical formula's mass and multiply every subscript by that whole number.

Worked example: glucose (the default)

Composition: 40.0% C, 6.7% H, 53.3% O (molar mass 180.16 g/mol) Treat percents as grams (100 g sample): Step 1 — moles of each: C: 40.0 / 12.011 = 3.330 mol H: 6.7 / 1.008 = 6.647 mol O: 53.3 / 15.999 = 3.331 mol Step 2 — divide by the smallest (3.330): C: 3.330/3.330 = 1.00 H: 6.647/3.330 = 2.00 O: 3.331/3.330 = 1.00 Step 3 — already whole numbers → empirical formula = CH2O empirical formula mass = 12.011 + 2(1.008) + 15.999 = 30.03 g/mol Step 4 — molecular formula: n = 180.16 / 30.03 = 6 → multiply each subscript by 6 molecular formula = C6H12O6

Practice problems

1. A hydrocarbon is 82.7% C and 17.3% H, with a molar mass of 58.12 g/mol. Find the empirical and molecular formulas. (C = 12.011, H = 1.008)

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Moles: C = 82.7/12.011 = 6.885; H = 17.3/1.008 = 17.16 Divide by smallest (6.885): C = 1.00; H = 2.49 2.49 is close to 2.5 = 5/2, so multiply both by 2: C = 2, H = 5 → empirical formula = C2H5 empirical mass = 2(12.011) + 5(1.008) = 29.06 g/mol Molecular: n = 58.12/29.06 = 2 → C4H10 (butane)

2. Why can two completely different compounds share the same empirical formula?

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Because the empirical formula only fixes the ratio of atoms, not the actual count. Formaldehyde (CH2O), acetic acid (C2H4O2), and glucose (C6H12O6) all reduce to the same empirical formula CH2O — a 1:2:1 ratio of C:H:O — even though they are very different molecules. Only the molar mass tells you which multiple of the empirical formula you actually have, which is why the molecular formula needs it.

FAQ

Empirical vs. molecular formula?
Empirical is the simplest whole-number atom ratio; molecular is the actual atom count in one molecule. They match for water (H₂O) but differ for glucose (CH₂O empirical, C₆H₁₂O₆ molecular). You need the molar mass to get from one to the other.

Why divide by the smallest number of moles?
It forces the least-abundant element to a ratio of 1, making the other ratios easy to read as atoms-per-atom before you scale to whole numbers.

What if a ratio is 1.5 or 2.33?
Multiply every ratio by the smallest whole number that makes them all integers: 1.5 → ×2, 1.33 (4/3) → ×3, 1.25 (5/4) → ×4. Atoms only come in whole numbers.

How do I get the molecular formula?
Divide the real molar mass by the empirical formula mass to get a whole number n, then multiply every subscript by n.

Grams or percent?
Either — they're handled the same, because assuming a 100 g sample turns each percent directly into grams.