Empirical Formula Finder
Determine the empirical formula of a compound from elemental percent composition. Supports C, H, O, and N.
About this calculator
Empirical formula determination converts a compound's elemental mass-percent composition -- typically obtained from combustion analysis -- into the simplest whole-number ratio of atoms it contains. The calculator first divides each element's mass percent by its atomic weight (standard values: Carbon 12.011, Hydrogen 1.008, Oxygen 15.999, Nitrogen 14.007 g/mol) to convert mass into relative moles, treating the input as a 100 g sample so mass percent and grams are numerically identical. It then divides every element's mole count by the smallest non-zero mole count present, which normalizes the least-abundant element to exactly 1 atom per formula unit.
Real experimental data rarely lands on clean whole numbers because of measurement rounding, so the calculator checks whether each ratio is close enough to an integer to round directly, and if not, tries multiplying every ratio by 2 or 3 to see whether that clears the fraction -- the same trial-and-multiply technique used in a manual stoichiometry calculation, since a ratio like 1.5 needs doubling to 3 rather than rounding to 2 to preserve the true atom count. Empirical Formula Weight sums each element's rounded atom count times its atomic weight, giving the molar mass of one empirical formula unit; dividing a compound's actual molecular weight (found separately, for example by mass spectrometry) by this number tells you how many empirical units make up one real molecule, which is the multiplier needed to go from empirical to molecular formula. The default values (Carbon 40.00%, Hydrogen 6.71%, Oxygen 53.29%) are glucose's real combustion-analysis composition, which reduces to the empirical formula CH2O -- glucose itself, C6H12O6, is six times that empirical unit.
Inputs
Results
Empirical Formula
CH2O
How to Use This Calculator
- Enter the percentage composition for carbon, hydrogen, oxygen, and nitrogen.
- Percentages must sum to 100% (or close to it with rounding).
- Review the Empirical Formula string and Empirical Formula Weight (g/mol).
- Divide the molecular weight by empirical weight to find the molecular formula multiplier.
What each input means
- Carbon (%)
- Mass percent of carbon from combustion analysis or elemental analysis
- Hydrogen (%)
- Mass percent of hydrogen
- Oxygen (%)
- Mass percent of oxygen (often found by difference: 100% minus other elements)
- Nitrogen (%)
- Mass percent of nitrogen — set to 0 if the compound contains no nitrogen
How this is calculated
Formula
Moles = %element / atomic weight; Ratio = moles / smallest molesWorked example, using the default values
- Identify Input Parameters4 parametersCarbon (%) = 40, Hydrogen (%) = 6.71, Oxygen (%) = 53.29, Nitrogen (%) = 0 = 4 input(s) provided
- Calculate Empirical FormulaCH2O = CH2O
- Calculate Empirical Formula WeightEmpirical Formula Weight30.03 = 30.03
- Calculate C Mole RatioC Mole Ratio1 = 1
Engine last updated . Checked against 1 independently-derived test — how we verify calculators. Built by Paul Gunder, a software engineer, not a licensed financial, medical, or legal professional.
Frequently Asked Questions
Why does the calculator divide each element's mole count by the smallest mole count?
Dividing every element's moles by the smallest non-zero value normalizes the ratio so the least-abundant element becomes exactly 1 atom per formula unit, which is how a mole ratio derived from mass percent turns into the whole-number subscripts of a chemical formula. Every other element's ratio then shows how many atoms of that element accompany each atom of the least-abundant one.
What does it mean if a mole ratio comes out close to 1.5 instead of a whole number?
A ratio near 1.5 usually means the true whole-number ratio needs a common multiplier before it clears to integers -- multiplying 1.5 by 2 gives exactly 3, so the calculator tries multiplying every element's ratio by 2 and then 3 to see whether that produces integers close enough to round cleanly, rather than rounding 1.5 itself, which would silently lose an atom from the formula.
Is the empirical formula the same thing as the molecular formula?
Not necessarily. The empirical formula gives only the simplest whole-number ratio of atoms, while the molecular formula gives the actual atom count in one molecule, which can be a whole-number multiple of the empirical formula -- glucose's empirical formula CH2O and molecular formula C6H12O6 are related by a multiplier of 6. Empirical Formula Weight is what you divide the compound's real molecular weight by to find that multiplier.
Why do the atomic weights used here have decimal values instead of whole numbers?
Atomic weights reflect the natural isotopic mixture of each element rather than a single idealized atom -- carbon's standard atomic weight of 12.011 g/mol, for example, averages the small natural abundance of carbon-13 into the dominant carbon-12 mass. Using these standard values (12.011 for C, 1.008 for H, 15.999 for O, 14.007 for N) instead of rounded whole numbers keeps the mole calculations accurate for real combustion-analysis data.
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