Empirical & Molecular Formulas

Every chemical compound has a definite composition that can be expressed as a formula. The empirical formula gives the simplest whole-number ratio of atoms of each element present in a compound, while the molecular formula gives the actual number of atoms of each element in one molecule. For example, glucose has the molecular formula C₆H₁₂O₆, but its empirical formula is CH₂O — both formulas share the same 1:2:1 ratio of carbon, hydrogen, and oxygen atoms. Understanding the relationship between these two types of formulas is a cornerstone of quantitative chemistry. Chemists determine empirical formulas experimentally through combustion analysis or direct elemental analysis, converting mass percentages of each element into mole ratios. Once the empirical formula is known, the molecular formula can be derived if the molar mass of the compound is also known — typically determined by mass spectrometry or other physical methods. The molecular formula is always a whole-number multiple of the empirical formula, expressed as: Molecular Formula = (Empirical Formula)ₙ, where n is a positive integer. Mastering empirical and molecular formula calculations requires fluency with molar mass, the mole concept, and ratio simplification. These skills are essential for identifying unknown compounds in analytical chemistry, verifying the purity of synthesized substances, and understanding the connection between macroscopic measurements and molecular-level composition. This concept sits at the heart of stoichiometry and underpins virtually all quantitative work in chemistry.

Key takeaways

Convert mass percentages to grams
Convert grams to moles
Divide by the smallest number of moles
Handle non-integer ratios
Calculate the empirical formula molar mass
Determine the multiplier n and write the molecular formula

Convert mass percentages to grams

Assume a 100 g sample. This converts each percentage directly into grams. For example, if a compound is 40.00% C, 6.72% H, and 53.28% O, you have 40.00 g C, 6.72 g H, and 53.28 g O. This assumption simplifies arithmetic without affecting the ratio.

Convert grams to moles

Divide the mass of each element by its molar mass (g/mol). Using standard atomic masses: C = 12.011 g/mol, H = 1.008 g/mol, O = 15.999 g/mol. So: n(C) = 40.00/12.011 = 3.330 mol; n(H) = 6.72/1.008 = 6.667 mol; n(O) = 53.28/15.999 = 3.330 mol.

Divide by the smallest number of moles

Find the smallest mole value among all elements and divide each mole amount by it. Here the smallest is 3.330 mol. Ratios: C = 3.330/3.330 = 1.000; H = 6.667/3.330 = 2.001 ≈ 2; O = 3.330/3.330 = 1.000. The empirical formula is therefore CH₂O.

Handle non-integer ratios

Sometimes division yields ratios like 1.5 or 1.33. Multiply all ratios by the smallest integer that converts them all to whole numbers. If ratios are 1 : 1.5 : 1, multiply by 2 to get 2 : 3 : 2. If ratios end in ~0.33 or ~0.67, multiply by 3. Common multipliers are 2, 3, 4, and 6. Always round only when the value is within ±0.05 of a whole number.

Calculate the empirical formula molar mass

Sum the atomic masses of all atoms in the empirical formula. For CH₂O: M(EF) = 12.011 + 2(1.008) + 15.999 = 30.026 g/mol. This value is used to find the multiplier n.

Determine the multiplier n and write the molecular formula

Use the relationship n = M(molecular) / M(empirical formula). If the molecular molar mass is given as 180.16 g/mol, then n = 180.16 / 30.026 ≈ 6.00, so n = 6. The molecular formula is (CH₂O)₆ = C₆H₁₂O₆, which is glucose. Always verify by checking that the molecular formula molar mass matches the given value.

The Key Equation

n = \frac{M_{\text{molecular}}}{M_{\text{empirical}}}

n — Formula multiplier (dimensionless): A positive integer (1, 2, 3, …) indicating how many empirical formula units make up one molecule
M_{\text{molecular}} — Molecular molar mass (g/mol): The molar mass of the actual molecular compound, determined experimentally (e.g., by mass spectrometry)
M_{\text{empirical}} — Empirical formula molar mass (g/mol): The sum of atomic masses of all atoms in one empirical formula unit

Worked Example

Problem. A compound contains 85.63% carbon and 14.37% hydrogen by mass. Its molar mass is determined to be 56.11 g/mol. Determine both its empirical and molecular formulas.

\text{Empirical: } \ce{CH2} \quad \text{Molecular: } \ce{C4H8}

Answer. Empirical formula: CH₂; Molecular formula: C₄H₈ (butene or cyclobutane)

Figure

Diagram

Origins · 1 minJöns Jacob Berzelius and how this was discovered

Jöns Jacob Berzelius

1779–1848 · Swedish

Berzelius was the first chemist to determine accurate atomic masses for nearly all elements known in his time, and he developed the modern system of chemical symbols and formulas (e.g., H₂O, CO₂) that we still use today. His meticulous gravimetric analyses allowed chemists to establish empirical formulas for hundreds of compounds with unprecedented precision, laying the quantitative foundation for stoichiometry. Berzelius discovered four elements — cerium, selenium, thorium, and silicon — and coined the chemical terms 'protein,' 'isomer,' 'polymer,' and 'catalyst,' profoundly shaping modern chemical vocabulary.

The Mole Concept and Avogadro's Number — Converting grams to moles is the essential first step in every empirical formula calculation, making the mole concept a direct prerequisite.
Percent Composition by Mass — Empirical formula calculations begin with percent composition data; understanding how to calculate and interpret mass percentages is foundational.
Molar Mass Calculations — Molar mass of both individual elements and the empirical formula unit must be calculated accurately to find the molecular formula.
Combustion Analysis — The most common experimental technique used to generate the mass data needed to determine empirical formulas of organic compounds.
Stoichiometric Calculations and Mole Ratios — Mole ratio skills used in empirical formula determination directly transfer to balancing equations and calculating reactant/product amounts.