Empirical Formula Calculator

Calculate empirical formulas from masses or percentages with clear steps and smart tolerance. Convert to molecular formulas using molar mass. Includes hydrates combustion analysis batch uploads CSV export and printable steps. Mobile friendly accessible and ideal for students teachers chemists and labs worldwide. Supports oxygen by difference element picker copy outputs and FAQ tips.

Rounds when |k·ratio − nearest integer| ≤ ±0.03
Enter element masses. Units can be mixed; the calculator normalizes internally. Add rows as needed.
Element Mass Unit
Enter element percentages. The total should be ~100%. Use the checkbox to compute oxygen by difference.
Element Percent (%)
Sum: 0.00% (should be 100% ±0.5%)
Provide masses of CO2 and H2O formed during complete combustion. Optionally provide original sample mass to compute oxygen by difference.
Component Moles Mass (g) Notes
· n H₂O
Set n > 0 to append ·nH₂O in the final formula.
Results
Empirical formula
—
Multiplier used: ×1
Molecular formula (if molar mass given)
—
n = —
Step-by-step solution
Element Atomic wt (g/mol) Input mass (g) Moles Divide by smallest After ×k Final subscript
🔎
Notes. We search multipliers k=1…8 to reduce fractional parts under your tolerance. Typical hints: ×2 for .5, ×3 for .33≈.67, ×4 for .25≈.75, ×5 for .2 .4 .6 .8. You can adjust the tolerance above.
Quick tutorial
  1. Choose Mass, Percentage, or Combustion mode and add inputs.
  2. Set rounding tolerance and optional molar mass for molecular formula.
  3. Click Compute empirical formula to see steps and the final formula.
  4. Use the buttons to copy, export CSV, or print/save as PDF.
  5. For batches, use Batch CSV to process multiple samples at once.
Data sources & precision
Atomic weights provided in rounded form by default for teaching clarity. Toggle to high precision if your work requires finer values. Always confirm with your lab or IUPAC references.
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Empirical Formula: What this calculator does and how it works

This tool computes the empirical formula—the simplest whole‑number ratio of atoms in a compound—using three practical input routes: Mass mode (enter grams of each element), Percentage mode (enter mass percentages), and a Combustion analysis helper (enter masses of CO2 and H2O formed on burning and, if available, the original sample mass to infer oxygen by difference). You can also attach waters of crystallization to show hydrates (e.g., CuSO4·5H2O) and, if you know the compound’s molar mass, the calculator upgrades the empirical formula to a molecular formula by multiplying subscripts by the integer factor n = (molar mass)/(empirical formula mass).

Algorithm in plain language

  1. Normalize inputs. In percentage mode, the calculator assumes a 100 g sample so 40.0% becomes 40.0 g. In mass mode, any combination of g/mg/kg is converted to grams.
  2. Convert to moles. Each element’s mass is divided by its atomic weight to obtain moles. You can toggle rounded or higher‑precision atomic weights.
  3. Form a ratio. Divide all mole values by the smallest (non‑zero) mole value. This produces a set of relative subscripts (which may be fractional).
  4. Fix small fractions. The tool searches integer multipliers k = 1…8 and uses your tolerance to decide when k × ratio is close enough to an integer (e.g., ×2 for 0.50, ×3 for 0.33/0.67). The chosen k and the rounded integers become the empirical formula subscripts.
  5. Optional: Molecular formula. If you supply the compound’s molar mass, the tool computes n and multiplies each empirical subscript by n.

Typical fractional ratios and multipliers

Observed ratioCommon multiplierReason
0.50, 1.50, 2.50 …×2Halves become whole numbers
0.33, 0.67×3Thirds become whole numbers
0.25, 0.75×4Quarters become whole numbers
0.20, 0.40, 0.60, 0.80×5Fifths become whole numbers
0.125, 0.375×8Eighths become whole numbers

Mini worked example (percent composition → CH2O)

Suppose a compound is 40.00% C, 6.71% H, and 53.29% O.

ElementMass (assume 100 g)Atomic wt (g/mol)MolesDivide by smallestInteger
C40.00 g12.0113.3293.329/3.329 = 1.0001
H6.71 g1.0086.6546.654/3.329 = 1.9982
O53.29 g15.9993.3303.330/3.329 = 1.0001

Result: CH2O (glucose’s empirical formula). If the molar mass were 180.16 g/mol, n ≈ 6 and the molecular formula would be C6H12O6.

Combustion analysis tips

When an unknown organic sample is combusted completely, all carbon ends up in CO2 and all hydrogen in H2O. The calculator converts CO2 and H2O masses to moles of C and H. If you also provide the original sample mass (and optionally the combined mass of other heteroatoms), the remaining mass is attributed to oxygen, letting the tool infer O’s contribution and finish the empirical formula.

Accuracy, rounding, and best practices

Finally, remember that different substances can share the same empirical formula (e.g., CH is the empirical formula for benzene C6H6 and acetylene C2H2). Use the molar mass input to distinguish compounds and reveal their full molecular formulas.

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Important Note: All the Calculators listed in this site are for educational purpose only and we do not guarentee the accuracy of results. Please do consult with other sources as well.