Empirical and molecular formulas: what a peptide's formula tells you

An empirical formula is the simplest whole-number ratio of atoms in a compound; a molecular formula gives the actual number of each atom. Glucose's molecular formula is C₆H₁₂O₆ and its empirical formula is CH₂O. This guide shows how to find an empirical formula from percentages or masses, how to get from there to the molecular formula, and what a formula such as C₁₆H₃₀N₄O₄ on a peptide's product page does and doesn't describe.

Terms are defined in the glossary.

What is a molecular formula?

A molecular formula lists each element in a molecule with the number of its atoms written after it. The number 1 is left out.

• Water: H₂O, two hydrogen atoms and one oxygen.
• Acetic acid: C₂H₄O₂, two carbons, four hydrogens and two oxygens.
• Glucose: C₆H₁₂O₆, six carbons, twelve hydrogens and six oxygens.

A molecular formula tells you what is in a molecule and, with atomic weights, its molar mass. It doesn't tell you how the atoms are joined. Glucose and fructose are both C₆H₁₂O₆, but they are different molecules. Compounds that share a formula but differ in structure are isomers.

Hill order

To make formulas easy to look up, chemists write the elements in a fixed order, known as Hill order:

1. Carbon (C) first.
2. Hydrogen (H) second.
3. Every other element in alphabetical order of its symbol.

So a peptide with copper is written C₁₄H₂₂CuN₆O₄: carbon, hydrogen, then copper, nitrogen and oxygen in alphabetical order. Compounds with no carbon are written fully alphabetically. Databases such as PubChem use this order, which is why the same formula always looks the same.

What is an empirical formula?

An empirical formula gives only the simplest whole-number ratio of the atoms in a compound. To get it, divide all the numbers in the molecular formula by the largest number that goes into each of them.

Acetic acid and glucose share the empirical formula CH₂O but are very different compounds. An empirical formula on its own can't identify a substance.

The name comes from how it was first found: by experiment, or "empirically". Burning a sample and weighing the products gives the percentage of each element, and that gives the ratio of atoms, but not the actual size of the molecule.

How to work out an empirical formula

The method turns percentages by mass into a ratio of atoms. Here it is for a compound found to be 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass.

1. Assume 100 g of the compound. The percentages become grams: 40.0 g C, 6.7 g H, 53.3 g O.
2. Convert grams to moles by dividing by each element's atomic weight:
   • Carbon: 40.0 ÷ 12.011 = 3.33 mol
   • Hydrogen: 6.7 ÷ 1.008 = 6.65 mol
   • Oxygen: 53.3 ÷ 15.999 = 3.33 mol
3. Divide by the smallest number (3.33):
   • Carbon: 1.00
   • Hydrogen: 2.00
   • Oxygen: 1.00
4. Round to whole numbers. The ratio is 1 : 2 : 1, so the empirical formula is CH₂O.

If step 3 gives a number such as 1.5 or 1.33, multiply all the values by 2 or 3 to reach whole numbers before rounding. Small errors in the measured percentages mean the numbers rarely come out exactly whole, so round only values that are very close.

For the mole arithmetic in step 2, see moles, Avogadro's number and percentage yield.

How to find an empirical formula from mass

If you are given masses in grams rather than percentages, skip step 1 and start from the masses. Suppose a 4.60 g sample contains 2.40 g of carbon, 0.60 g of hydrogen and 1.60 g of oxygen.

1. Convert each mass to moles:
   • Carbon: 2.40 ÷ 12.011 = 0.200 mol
   • Hydrogen: 0.60 ÷ 1.008 = 0.595 mol
   • Oxygen: 1.60 ÷ 15.999 = 0.100 mol
2. Divide by the smallest number (0.100): carbon 2.00, hydrogen 5.95, oxygen 1.00.
3. Round to whole numbers: 2 : 6 : 1, so the empirical formula is C₂H₆O.

The masses don't need to add up to a round number, and you don't need the sample's total mass, only the mass of each element in it.

From empirical formula to molecular formula

To find the molecular formula you also need the molar mass, usually from mass spectrometry.

1. Work out the mass of one empirical unit. CH₂O is 12.011 + 2 × 1.008 + 15.999 = 30.03 g/mol.
2. Divide the measured molar mass by it. If the compound's molar mass is 180.16 g/mol, then 180.16 ÷ 30.03 = 6.0.
3. Multiply the empirical formula by that number. CH₂O × 6 = C₆H₁₂O₆, glucose.

If the molar mass had been 60.05 g/mol, the multiplier would be 2, giving C₂H₄O₂: acetic acid. The same percentages fit both, which is why a molar mass is needed.

For how molar mass is calculated, see molar mass, molecular weight and monoisotopic mass, or use the molecular weight calculator.

Can the empirical and molecular formula be the same?

Yes. When the numbers in a molecular formula have no common factor other than 1, the molecular formula is already the simplest ratio. Water (H₂O), carbon dioxide (CO₂), methane (CH₄) and ethanol (C₂H₆O) all have the same empirical and molecular formula. KPV is a counter-example: every number in C₁₆H₃₀N₄O₄ divides by 2, so its empirical formula, C₈H₁₅N₂O₂, is different.

Practice questions

Work out the empirical formula of each:

1. Benzene, C₆H₆.
2. Ethene, C₂H₄.
3. Dinitrogen tetroxide, N₂O₄.
4. Octane, C₈H₁₈.
5. A compound that is 85.6% carbon and 14.4% hydrogen by mass.
6. A compound in which 2.80 g of nitrogen is combined with 1.60 g of oxygen.

Answers: 1. CH. 2. CH₂. 3. NO₂. 4. C₄H₉. 5. CH₂ (7.13 mol C and 14.3 mol H, a 1 : 2 ratio). 6. N₂O (0.200 mol N and 0.100 mol O, a 2 : 1 ratio).

Other ways to write a formula

For how to read each type, including a peptide's structure diagram, see reading a structure diagram.

For peptides, the sequence does the job of a condensed formula. Lys-Pro-Val tells you the order of the building blocks, and the molecular formula follows from it. See amino acid codes and how to read a peptide sequence.

Reading the formula on a product page

Take KPV, listed with the sequence Lys-Pro-Val and the formula C₁₆H₃₀N₄O₄.

• The atom count. Each molecule has 16 carbons, 30 hydrogens, 4 nitrogens and 4 oxygens.
• The molar mass. 16 × 12.011 + 30 × 1.008 + 4 × 14.007 + 4 × 15.999 = 342.44 g/mol, which matches the 342.4 shown on the page. Adding up the residue masses of lysine, proline and valine plus one water gives the same total.
• The composition by mass. Carbon makes up 56.1% of the mass, hydrogen 8.8%, nitrogen 16.4% and oxygen 18.7%. These are the figures an elemental analysis of the pure, dry peptide would be expected to give.
• The form. The formula describes the free peptide, with a free acid at the C-terminus. An amide or acetyl group at either end would change it. See terminal modifications.

The formula of the powder is different

The formula on a product page is for one peptide molecule on its own. The powder in a vial is rarely just that.

• Counter-ions. Peptides are usually supplied as salts. One acetate ion with KPV would add C₂H₄O₂ to the formula and about 60 g/mol to the mass. Trifluoroacetate adds fluorine as well. See TFA counter-ions and salt exchange.
• Water. Freeze-dried (lyophilised) powders hold some water, which adds hydrogen and oxygen.

That is why an elemental analysis of the real powder doesn't match the figures above exactly, and why the share of the powder that is peptide is reported separately. See net peptide content vs purity.

Related guides

For how mass spectrometry confirms the mass a formula predicts, see mass spectrometry and identity confirmation. For why one formula gives a cluster of peaks in a mass spectrum, see what is an isotope?. For looking up a compound's formula and identifiers, see CAS numbers, PubChem and the literature.