What if the percentages do not add up to 100%?

The remainder is typically assumed to be oxygen if analyzing an organic compound via combustion analysis, but standard problems will specify all components or state this explicitly.

How do I handle a decimal ratio like 1.25?

Multiply all element ratios in the compound by 4 to convert the 1.25 into a whole number (5).

Is the empirical formula always the molecular formula?

No, the molecular formula is an integer multiple of the empirical formula. They are only the same if that integer multiple is 1.

How to find an empirical formula from percent composition

The empirical formula of a compound is the simplest whole-number ratio of its constituent atoms. To find it from percent composition, assume a 100 g sample, convert the resulting masses to moles, and divide by the smallest molar value to establish the ratio.

The setup

Assume a 100 g sample of the compound so that the given percent values translate directly into mass in grams. Obtain the atomic mass for each element from the periodic table.

The steps

Convert the percentage of each element to grams (e.g., 40.0% becomes 40.0 g). 2. Convert the mass of each element to moles using $n = m / M$ , where $M$ is the atomic mass. 3. Divide all calculated mole values by the smallest mole value in the set. 4. If the resulting ratios are not whole numbers, multiply all ratios by the smallest integer that converts them into whole numbers (e.g., multiply by 2 for a 0.5 ratio, or by 3 for a 0.33 ratio).

Checking the result

Verify that all subscripts in the final formula are integers. If a molar mass is provided in the problem, ensure that the empirical formula mass is a factor of the total molar mass.

Common errors

Rounding intermediate mole ratios too early or incorrectly is a common mistake (e.g., rounding 1.33 to 1 instead of multiplying the entire set by 3). Failing to verify that the initial percentages sum exactly to 100% can also lead to incorrect mass assumptions.

Worked example

Determine the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O by mass.

Assume a 100 g sample. Mass of C = 40.0 g, Mass of H = 6.71 g, Mass of O = 53.28 g. Convert to moles: Moles C = 40.0 g / 12.01 g/mol = 3.33 mol. Moles H = 6.71 g / 1.008 g/mol = 6.66 mol. Moles O = 53.28 g / 16.00 g/mol = 3.33 mol. Divide by the smallest value (3.33 mol): C ratio = 3.33 / 3.33 = 1. H ratio = 6.66 / 3.33 = 2. O ratio = 3.33 / 3.33 = 1. The ratios are whole numbers. The empirical formula is $CH_2O$ .

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References: OpenStax Chemistry 2e, Chapter 3.2: Determining Empirical and Molecular Formulas · Zumdahl Chemistry, 10th Edition, Chapter 3: Stoichiometry