Empirical formula versus molecular formula
An empirical formula is the simplest whole-number ratio of atoms in a compound. A molecular formula is the actual count of atoms in one molecule. They are often different: glucose has the molecular formula C₆H₁₂O₆ but the empirical formula CH₂O, because 6:12:6 reduces to 1:2:1.
The distinction exists because of what each measurement can see. Elemental analysis measures mass fractions, and mass fractions only ever tell you a ratio — burning glucose and burning formaldehyde produce carbon dioxide and water in exactly the same proportion. To get from the ratio to the real molecule you need one extra piece of information: the molar mass, from mass spectrometry, from vapour density, or from a colligative property such as freezing-point depression.
Ionic compounds have no molecular formula at all. Sodium chloride is an extended lattice, so NaCl is an empirical formula and there is nothing to scale it up to. The same is true of most minerals and of network solids such as silicon dioxide.
The four steps, and why each one is there
Step one: assume 100 g. If your data are percentages, treat each percentage as a mass in grams. This is not an approximation — the ratio you are after is unchanged by the size of the sample, so choosing the most convenient sample size is free.
Step two: convert mass to moles. Divide each mass by that element's atomic weight. This is the only step that does real chemical work, because a formula is a ratio of atoms and a balance measures mass. Heavy elements are badly under-represented in mass terms: in water, oxygen is 88.8% of the mass but only a third of the atoms.
Step three: divide by the smallest. Dividing every mole figure by the smallest of them makes the least abundant element the unit of comparison, so the smallest ratio is exactly 1 and the others read directly as multiples of it.
Step four: scale to whole numbers. Ratios like 1 : 1.5 are real but cannot be subscripts, so you multiply everything by the smallest integer that clears the fractions — 2 in that case, giving 2 : 3. Recognise the common decimals: .5 needs ×2, .33 and .67 need ×3, .25 and .75 need ×4, .2 and .4 need ×5. This calculator searches multipliers up to 12 and takes the first that brings every ratio within 0.08 of a whole number.
Finally, reduce by any common factor. If the rounding produced 2 : 4 : 2, the empirical formula is the reduced 1 : 2 : 1 — that is what makes it empirical.
Worked example: a compound that is 40.00% C, 6.71% H and 53.29% O
This is the classic combustion-analysis result. Work it through.
- Assume 100 g: 40.00 g C, 6.71 g H, 53.29 g O.
- Moles of carbon: 40.00 ÷ 12.011 = 3.3303 mol.
- Moles of hydrogen: 6.71 ÷ 1.008 = 6.6567 mol.
- Moles of oxygen: 53.29 ÷ 15.999 = 3.3308 mol.
- Divide by the smallest (3.3303): C = 1.000, H = 1.999, O = 1.000.
- Round: the ratio is 1 : 2 : 1, so the empirical formula is CH₂O, with an empirical formula mass of 12.011 + 2(1.008) + 15.999 = 30.026 g/mol.
- Bring in the molar mass. Mass spectrometry gives 180.16 g/mol. Divide: 180.16 ÷ 30.026 = 6.000, so the molecular formula is 6 × CH₂O = C₆H₁₂O₆.
Note how little the last step costs and how much it buys. Without it, the composition is equally consistent with formaldehyde (CH₂O, 30.03 g/mol), acetic acid (C₂H₄O₂, 60.05) and glucose (C₆H₁₂O₆, 180.16). Confirm the molar mass with the molar mass calculator and check your answer by running it back through the percent composition calculator: C₆H₁₂O₆ returns 40.00% C, 6.71% H and 53.28% O, reproducing the data you started from to the precision the input carried.
How to judge whether the answer is trustworthy
Look at how far the ratios sit from whole numbers before rounding. A value of 1.99 rounding to 2 is comfortable. A value of 1.90 is not — that is a 5% discrepancy, larger than any competent elemental analysis, and it usually means an element is missing from your data or one figure was transcribed wrongly. As a working rule, treat anything more than about 0.08 away from an integer, after the multiplier has been applied, as a signal to re-check rather than to round harder.
Check that percentages sum to about 100. Combustion analysis measures carbon, hydrogen and nitrogen directly and reports oxygen by difference, so a set that sums to 94% usually means an unmeasured element — a halogen, a metal, or sulfur — that has been left out entirely. Adding the missing 6% to oxygen would be wrong; you need the actual assay.
Sanity-check the multiplier against the molar mass. If the measured mass divided by the empirical formula mass comes out as 3.4 rather than close to 3 or 4, the two measurements disagree and at least one is wrong. A ratio that lands within about 0.1 of an integer is the standard for accepting a molecular formula.
Finally, remember that an empirical formula is not a structure. C₂H₆O is both ethanol and dimethyl ether; the two have the same composition, the same molar mass, and boiling points 100 °C apart. Composition constrains structure but never determines it.
Empirical formulas of familiar compounds
| Compound | Molecular formula | Empirical formula | Multiple | Empirical mass (g/mol) |
|---|---|---|---|---|
| Formaldehyde | CH₂O | CH₂O | 1 | 30.026 |
| Acetic acid | C₂H₄O₂ | CH₂O | 2 | 30.026 |
| Glucose | C₆H₁₂O₆ | CH₂O | 6 | 30.026 |
| Acetylene | C₂H₂ | CH | 2 | 13.019 |
| Benzene | C₆H₆ | CH | 6 | 13.019 |
| Ethane | C₂H₆ | CH₃ | 2 | 15.035 |
| Hydrogen peroxide | H₂O₂ | HO | 2 | 17.007 |
| Water | H₂O | H₂O | 1 | 18.015 |
| Sodium chloride | none (ionic lattice) | NaCl | — | 58.440 |
| Hematite | none (ionic lattice) | Fe₂O₃ | — | 159.687 |
The first three rows share a composition exactly, which is why a molar mass is required to choose between them.
Pitfalls that produce a wrong formula
- Dividing percentages by each other instead of converting to moles. Mass ratios are not atom ratios. This is the single most common error and it always inflates the heavy element.
- Rounding too early. Carry at least four significant figures through the mole step; rounding 3.3303 to 3.33 before dividing shifts the hydrogen ratio enough to matter.
- Rounding 1.5 down to 1. A half means multiply everything by two. Rounding it away produces a formula that no compound has.
- Forgetting an element determined by difference. If your percentages sum to well under 100, something was never measured. The gap does not automatically belong to oxygen.
- Using the empirical formula mass where the molar mass is needed. In a yield calculation the molecular formula mass is the one that counts.
- Assuming an empirical formula identifies the compound. It fixes composition, not structure and not molecular size.
Where the input data comes from
For organic compounds the numbers almost always come from combustion analysis. A few milligrams are burned in excess oxygen; all the carbon becomes carbon dioxide and all the hydrogen becomes water, and the masses of those two products are measured. Carbon mass is 12.011/44.009 of the carbon dioxide, and hydrogen mass is 2(1.008)/18.015 of the water. Nitrogen is measured in the same instrument by reducing nitrogen oxides to N₂. Oxygen is reported by difference, which is why oxygen carries the accumulated error of every other element.
For inorganic samples the route is usually a dissolution followed by an instrumental technique — ICP-OES or ICP-MS for metals, ion chromatography for anions — or a classical gravimetric precipitation. For hydrates, the water content is often found simply by heating to constant mass, and the mass lost divided by the molar mass of water gives the number of waters directly.
Once you have a formula, the rest of the quantitative chain follows: use the molar mass calculator to get g/mol, the moles-to-grams calculator to weigh a sample, and the theoretical yield calculator once the compound is a product rather than an unknown.
A historical note worth keeping: this procedure is essentially Justus von Liebig's, from the 1830s. His combustion train, with a potash bulb to trap carbon dioxide, made carbon and hydrogen determinations routine and turned organic chemistry into a quantitative science. The instrument is now automated and the sample is a thousand times smaller, but the arithmetic on this page has not changed.