Empirical & Molecular Formula Calculator

How this is worked out

Given a percentage composition by element (from combustion analysis, for example), treat each percentage as the mass in a 100 g sample, divide by each element's atomic mass to get moles, then divide every mole value by the smallest one — the resulting ratios, rounded to whole numbers (scaling up by a small common factor if needed, e.g. all multiplied by 2 to clear halves), give the empirical formula, the simplest whole-number ratio of atoms. If you also know the compound's actual molar mass, dividing it by the empirical formula's own mass tells you how many empirical units make up one real molecule, giving the true molecular formula (e.g. empirical CH₂O with molar mass 180.16 g/mol is a ×6 multiple, giving the molecular formula C₆H₁₂O₆ — glucose).

Frequently asked questions

Why might my percentages not add up to exactly 100%?

Real combustion-analysis data often has a little rounding or measurement error, and sometimes one element (commonly oxygen) is found "by difference" rather than measured directly — this tool treats whatever percentages you enter proportionally, so they don't need to sum to exactly 100.

Why is the empirical formula sometimes the same as the molecular formula?

Whenever the real molecule happens to already be in its simplest ratio — water, H₂O, is its own empirical formula, for instance, because 2:1 can't be simplified any further.

Related tools: Molar Mass Calculator, Stoichiometry Calculator, Chemical Equation Balancer.

Source / Last verified: standard general chemistry method (percentage composition → empirical formula → molecular formula), using the same atomic-weight table as this site's Molar Mass Calculator — read/checked 1 October 2026. Reproduces the classic textbook glucose example (40.0% C, 6.7% H, 53.3% O by mass, molar mass 180.16 g/mol) exactly: empirical formula CH₂O, molecular formula C₆H₁₂O₆.