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Generated on: July 20, 2026 at 09:33 PM

Empirical Formula Calculator

Inputs

Element 1 percentage40 %
Element 1 molar mass12.011 g/mol
Element 2 percentage6.7 %
Element 2 molar mass1.008 g/mol
Element 3 percentage53.3 %
Element 3 molar mass16 g/mol

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Chemistry

Empirical Formula Calculator

Find the empirical formula of a compound from the percent composition of its elements. Enter each element’s mass percent and molar mass to get the simplest whole-number mole ratio.

Inputs

%
0 – 100 %
g/mol
%
0 – 100 %
g/mol
%
0 – 100 %
g/mol

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The percentages do not add up to about 100%. Empirical-formula results assume the percentages cover the whole compound.

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What the empirical formula tells you

The empirical formula gives the simplest whole-number ratio of atoms in a compound. It answers "in what proportions are the elements combined?" rather than "how many atoms are in one molecule?" Glucose, hydrogen peroxide, and acetic acid all reduce to compact empirical formulas — CH₂O, HO, and CH₂O — even though their molecules are larger. This calculator takes the mass percent of each element and returns that simplest ratio.

From percent composition to a ratio

The method rests on one convenient assumption: imagine exactly 100 g of the compound. Then each element's mass percent is also its mass in grams. Convert each mass to moles by dividing by the element's molar mass, then divide every mole count by the smallest of them.

ni=wiMiri=ninmin⁡n_i = \frac{w_i}{M_i} \qquad r_i = \frac{n_i}{n_{\min}}
SymbolQuantityUnit
wᵢMass percent of element i%
MᵢMolar mass of element ig/mol
nᵢMoles per 100 gmol
rᵢMole ratio—

The ratios rᵢ, rounded to small whole numbers, become the subscripts of the formula.

Worked example

A compound is found to be 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Taking a 100 g sample:

nC=40.012.011=3.33nH=6.71.008=6.65nO=53.316.00=3.33n_\text{C} = \frac{40.0}{12.011} = 3.33 \qquad n_\text{H} = \frac{6.7}{1.008} = 6.65 \qquad n_\text{O} = \frac{53.3}{16.00} = 3.33

The smallest count is 3.33 mol, so dividing through:

C:3.333.33=1H:6.653.33=2O:3.333.33=1\text{C}: \frac{3.33}{3.33} = 1 \qquad \text{H}: \frac{6.65}{3.33} = 2 \qquad \text{O}: \frac{3.33}{3.33} = 1

The empirical formula is CH₂O.

Empirical to molecular formula

The molecular formula is always a whole-number multiple of the empirical formula. To find the multiplier, divide the compound's measured molar mass by the empirical-formula mass:

k=MmolecularMempiricalk = \frac{M_\text{molecular}}{M_\text{empirical}}

The empirical formula CH₂O has a mass of 12.011 + 2(1.008) + 16.00 = 30.03 g/mol. Glucose has a molar mass of 180.16 g/mol, so k = 180.16 ÷ 30.03 ≈ 6. Multiplying every subscript by 6 gives the molecular formula C₆H₁₂O₆.

When the ratios are not whole numbers

The integer-rounding step is the method's main limitation. If a computed ratio lands near 1.5, 1.33, or 1.25, the values are not rounding errors — they point to a larger common multiple. Multiply every ratio by the smallest factor that clears the fraction: ×2 for halves, ×3 for thirds, ×4 for quarters. A ratio of 2.5, for instance, signals that the true subscripts are 5 and 2 after doubling. Only round directly when a value sits within about 0.1 of a whole number, the gap you expect from percentages reported to one decimal place.

Common reductions

CompoundMolecular formulaEmpirical formula
GlucoseC₆H₁₂O₆CH₂O
Hydrogen peroxideH₂O₂HO
BenzeneC₆H₆CH
Acetic acidC₂H₄O₂CH₂O
WaterH₂OH₂O

Frequently Asked Questions (FAQ)

What is an empirical formula?

An empirical formula gives the simplest whole-number ratio of atoms in a compound. For glucose (C₆H₁₂O₆), the empirical formula is CH₂O because the 6:12:6 atom ratio simplifies to 1:2:1. It does not tell you the actual number of atoms in a molecule — that is the molecular formula — only their relative proportions.

How do you find the empirical formula from percent composition?

Assume a 100 g sample, so each element’s mass percent becomes its mass in grams. Divide each mass by the element’s molar mass to get moles, then divide every mole count by the smallest. Rounding those quotients to whole numbers gives the subscripts. For 40% C, 6.7% H, 53.3% O the moles are 3.33, 6.65, 3.33; dividing by 3.33 gives 1, 2, 1, so the formula is CH₂O.

How is the empirical formula related to the molecular formula?

The molecular formula is a whole-number multiple of the empirical formula. Divide the compound’s actual molar mass by the empirical-formula mass to get that multiplier. The empirical formula of glucose is CH₂O (mass 30.03 g/mol); glucose’s molar mass is 180.16 g/mol, and 180.16 ÷ 30.03 ≈ 6, so the molecular formula is C₆H₁₂O₆.

What if the mole ratios are not whole numbers?

Small fractions usually mean you should multiply every ratio by a common factor rather than round. A ratio near 1.5 becomes 3 when both values are doubled; near 1.33 (4/3), multiply by 3; near 1.25 (5/4), multiply by 4. Only round when a value is within roughly 0.1 of a whole number, since that gap reflects rounded percentages rather than a real fraction.

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