An empirical formula gives the simplest whole-number ratio of atoms in a compound. It is what an elemental analysis tells you directly: a combustion or gravimetric analysis yields the mass – or mass percentage – of each element, and from there you can work out how many atoms of each kind there are relative to one another. Glucose, for example, contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass, which corresponds to the empirical formula CH₂O.
The method is always the same. Assume 100 g of compound so that percentages become grams, convert each mass to moles by dividing by the atomic weight, divide every result by the smallest one and, if some ratios end in .5, .33 or .25, multiply them all by 2, 3 or 4 to reach whole numbers. The molecular formula is a whole-number multiple of the empirical one; to find the multiplier you need the molar mass, for instance from mass spectrometry or a gas density.
Type one element per line with its percentage or mass. If oxygen was determined by difference, write “rest” instead of a number and the calculator fills in 100% minus the others. The step table shows moles, ratios and the multiplier exactly as you would write them in a lab report.
How to use the Empirical Formula Calculator
- Composition given as
- Elements and values (one per line) — Symbol or name, then the value, e.g. “C 40.0” or “Fe 2.233”. Write “O rest” to take oxygen by difference (percent mode).
- Molar mass of the compound (optional, 0 = skip)
Everything is calculated in your browser, so results appear instantly and nothing you type is sent to a server.
How it works
Each value is divided by the element’s standard atomic weight to get moles, all moles are divided by the smallest, and the calculator searches multipliers 1–12 for the smallest one that brings every ratio within 0.1 of a whole number. If you enter a molar mass, it is divided by the empirical formula mass and rounded to the nearest whole number n.
Worked example
A compound contains 40.0% C, 6.7% H and 53.3% O and has a molar mass of 180.16 g/mol. In 100 g: C 40.0 ÷ 12.011 = 3.330 mol, H 6.7 ÷ 1.008 = 6.647 mol, O 53.3 ÷ 15.999 = 3.331 mol. Dividing by the smallest (3.330) gives 1 : 1.996 : 1.000, so the empirical formula is CH₂O (30.026 g/mol). n = 180.16 ÷ 30.026 = 6.00, so the molecular formula is C₆H₁₂O₆ (glucose).
Multipliers for common decimal ratios
| Ratio ends in | Multiply all ratios by | Example |
|---|---|---|
| .5 | 2 | 1 : 1.5 → 2 : 3 (Fe₂O₃) |
| .33 or .67 | 3 | 1 : 1.33 → 3 : 4 (Fe₃O₄) |
| .25 or .75 | 4 | 1 : 1.25 → 4 : 5 |
| .2, .4, .6 or .8 | 5 | 1 : 2.5 is ×2; 1 : 1.2 → 5 : 6 |
Tips
- Always start from a balanced equation – coefficients give the mole ratio.
- Convert everything to moles first, then use the ratio, then convert back.
- Every field has its own unit menu – g, mg, mol, mmol, L and mL all work.
- Everything is calculated in your browser.
Common mistakes to avoid
- Using mass ratios instead of mole ratios.
- Forgetting water of crystallisation in the molar mass of hydrates.
- Confusing molarity (per litre of solution) with molality (per kilogram of solvent).
Formula
moles of element = mass (or % in 100 g) ÷ atomic weight · ratio = moles ÷ smallest moles · multiply by 2, 3 … until all ratios are whole · n = molar mass ÷ empirical formula mass · molecular formula = (empirical formula)ₙWorked example
Inputs — Composition given as: Percent by mass (%); Elements and values (one per line): C 40.0
H 6.7
O 53.3; Molar mass of the compound (optional, 0 = skip): 180.16 g/mol.
Results — Empirical formula = CH₂O; Empirical formula mass = 30.026 g/mol; Molecular formula = C₆H₁₂O₆; Multiplier n (molar mass ÷ empirical mass) = 6; Molar mass of the molecular formula = 180.156 g/mol; Step-by-step calculation = Element Mass % Moles (in 100 g) ÷ smallest Rounded Subscript
──────────────────────────────────────────────────────────────────────
C 40 3.3303 1 1 1
H 6.7 6.6468 1.996 1.996 2
O 53.3 3.3315 1 1 1.
Key terms
- Empirical formula
- The simplest whole-number ratio of atoms of each element in a compound.
- Molecular formula
- The actual number of atoms of each element in one molecule; a whole-number multiple of the empirical formula.
- Empirical formula mass
- The sum of the atomic weights in the empirical formula.
- Elemental analysis
- Experimental determination of the mass percentage of each element in a sample.
Frequently asked questions
How do you find an empirical formula from percentages?
Assume 100 g, divide each percentage by the atomic weight, divide all results by the smallest and round (or multiply first if you see .5, .33 or .25). 40.0% C, 6.7% H, 53.3% O gives 3.33 : 6.65 : 3.33 = 1 : 2 : 1, so CH₂O.
How do I get the molecular formula from the empirical formula?
Divide the molar mass by the empirical formula mass and multiply all subscripts by the result. For glucose, 180.16 ÷ 30.03 = 6, so (CH₂O)₆ = C₆H₁₂O₆.
What if a ratio comes out as 1.5?
Multiply every ratio by 2. A ratio of 1 : 1.5 becomes 2 : 3 (as in Fe₂O₃). For ratios ending in .33 or .67 multiply by 3; for .25 or .75 multiply by 4.
Can the empirical and molecular formulas be the same?
Yes – whenever n = 1, as for water (H₂O), carbon dioxide (CO₂) or silver oxide (Ag₂O).
What if the percentages do not add up to 100%?
The missing mass is usually an element that was not measured directly, most often oxygen. Enter it as "rest".
How precise must the data be?
Analytical data are typically ±0.3%. With rounded data, ratios such as 1.96 or 2.04 are normal; a ratio such as 1.8 usually means you need a larger multiplier or better data.
What is the empirical formula of benzene?
CH. Benzene is C₆H₆ (n = 6) and acetylene is C₂H₂ (n = 2) – both share the empirical formula CH.