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Chemistry · General chemistry I · Concept

Empirical and molecular formulas

To find an empirical formula, treat each element’s mass percent as grams in a 100 g sample, convert each mass to moles, and divide by the smallest amount to get the simplest whole-number ratio. The molar mass then tells you how many empirical units make one molecule.

Two kinds of formula

An empirical formula gives the simplest whole-number ratio of atoms; a molecular formula gives the actual number of atoms in one molecule. Glucose, C₆H₁₂O₆, has the empirical formula CH₂O, and so does formaldehyde, CH₂O, whose molecular and empirical formulas coincide.

Percent composition from a formula

The mass percent of an element is its mass in one mole of the compound divided by the molar mass, times 100%. The percentages of all the elements add up to 100%, apart from rounding.

%element=n×M⁢(element)M⁢(compound)×100⁢%

From percentages to an empirical formula

Assume a 100 g sample, so each percentage becomes grams. Convert each mass to moles with the atomic mass, divide every amount by the smallest, and round to whole numbers only when the ratios are already close to them.

When a ratio ends in .5 or .33

A ratio such as 1.5 or 2.33 is not a rounding error. Multiply every ratio by the same small integer (2 for halves, 3 for thirds) until all are whole numbers.

From empirical to molecular formula

Divide the measured molar mass by the empirical formula mass. The result is a whole number n, and the molecular formula is the empirical formula with every subscript multiplied by n.

n=M⁢(molecular)M⁢(empirical)

Common mistakes

  • Using the percentages directly as mole ratios, without dividing by atomic masses.
  • Rounding a ratio such as 1.5 to 2 instead of multiplying every ratio by 2.
  • Dividing by the largest amount instead of the smallest.
  • Forgetting that the molecular formula is a whole-number multiple of the empirical formula.

Key terms

Empirical formula
The simplest whole-number ratio of atoms in a compound, such as CH₂O for glucose, C₆H₁₂O₆. It doesn’t tell you the molecule’s actual size or structure.
Molecular formula
The actual number of each kind of atom in one molecule, such as C₆H₁₂O₆. Different structures can share one molecular formula, so it doesn’t show how the atoms connect.
Percent composition
Each element’s share of a compound’s mass, as a percentage: element mass ÷ compound mass × 100. The percentages of all the elements add up to 100%.
Combustion analysis
A method for finding an empirical formula by burning a known mass of a compound in excess oxygen and weighing the CO₂ and H₂O formed.
Molar mass
The mass of one mole of a substance, in g/mol. It converts between grams and moles: moles = mass ÷ molar mass.
Formula mass
The sum of the atomic masses of all the atoms in a chemical formula, in atomic mass units (amu). For a molecular substance it is the molecular mass.

Work through an example

A compound is 40.0% carbon, 6.71% hydrogen and 53.3% oxygen by mass, and its molar mass is 180.16 g/mol. Find its empirical and molecular formulas.

Find an empirical and a molecular formula →

Find percent composition from a formula →

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