QiLing Tools

🔎Empirical Formula Calculator - From Percent Composition or Combustion Analysis

An empirical formula gives the simplest whole-number ratio of atoms in a compound; the molecular formula is a whole-number multiple of it. The method has four steps: treat mass percentages as grams in a 100 g sample, divide by atomic weights to get moles, divide everything by the smallest number of moles, and multiply by a small integer until every ratio is a whole number. For glucose (40.00% C, 6.71% H, 53.29% O): n(C) = 40.00 ÷ 12.011 = 3.330, n(H) = 6.71 ÷ 1.008 = 6.657, n(O) = 53.29 ÷ 15.999 = 3.331, a 1 : 2.00 : 1 ratio, so the empirical formula is CH₂O (30.03 g/mol); with a molar mass of 180.16 g/mol, n = 6 and the molecular formula is C₆H₁₂O₆. Ratios like 1.33, 1.5 or 1.25 must not be rounded — multiply by 3, 2 or 4 instead: ascorbic acid (40.92% C, 4.58% H, 54.50% O) gives H 1.334, so ×3 gives C₃H₄O₃. This calculator accepts mass percentages (with an option to make the remainder oxygen), masses in grams, or combustion analysis: from the sample mass and the masses of CO₂, H₂O and optionally N₂ it works out C, H and N, with oxygen by difference. It searches multipliers 1 to 10 within a tolerance you choose, flags near-misses like x.5 or x.33, and shows the full working table; add a known molar mass to get the molecular formula, written in Hill order for carbon compounds.

Free · No sign-up · Runs in your browserLast updated:

Loading tool...

How to use Empirical & Molecular Formula

  1. 1Choose an input: mass percent, element masses or combustion analysis, then enter data or pick an example
  2. 2Read the moles, ratios and multiplier to get the empirical formula
  3. 3Enter a known molar mass for the molecular formula, then copy it or check it in the molar mass calculator

FAQ

What is the difference between an empirical and a molecular formula?
The empirical formula only shows the simplest whole-number ratio of atoms; the molecular formula shows the actual number of atoms in one molecule and is n times the empirical formula. Glucose has the empirical formula CH₂O and the molecular formula C₆H₁₂O₆ (n = 6). Elemental analysis alone gives only the empirical formula — you also need the molar mass (from mass spectrometry, gas density and so on) to find n = M ÷ empirical formula mass.
What if the mole ratios are not whole numbers?
Ratios within about ±0.1 of a whole number can be rounded, since lab data have some error. Ratios near 1.5, 1.33, 1.67 or 1.25 mean you must multiply everything by 2, 3 or 4 — for example 1 : 1.33 × 3 = 3 : 4. The calculator tries multipliers 1 to 10 and picks the first one that puts every ratio within the tolerance, and it tells you which element forced the multiplier.
How does combustion analysis give the element masses?
On complete combustion all carbon ends up in CO₂ and all hydrogen in H₂O, so m(C) = m(CO₂) × 12.011 ÷ 44.009 and m(H) = m(H₂O) × 2 × 1.008 ÷ 18.015; nitrogen is measured as N₂. Oxygen is the sample mass minus C, H and N. For a 0.2000 g sample giving 0.2998 g CO₂ and 0.1227 g H₂O, that is 0.0818 g C, 0.0137 g H and 0.1045 g O, so the empirical formula is CH₂O.
How do I go the other way, from a formula to percent composition?
Mass percent of an element = number of atoms × atomic weight ÷ molar mass × 100%. Enter the formula in the molar mass calculator on this site to get the percent composition of every element — it is also a quick way to check the formula found here.