A chemical formula can be reported at two levels of detail. The empirical formula gives the simplest whole-number ratio of atoms; the molecular formula gives the actual number of each atom in one molecule. This concept moves between percentage composition, the empirical formula, and the molecular formula.
1 — Percentage composition by mass. The mass percent of an element in a compound is (mass of that element in one formula unit / molar mass) × 100. In water, oxygen is 16 / 18 × 100 = 88.9%. Read it as: of every 100 g of the compound, this many grams are that element. Given the formula you can compute any element's percent; given the percent you can work backwards.
2 — Empirical formula from composition. The recipe (works from either percentages or actual combining masses):
- Step 1 — take the mass (or the percent, treated as grams per 100 g) of each element.
- Step 2 — divide each by its atomic mass to get moles (relative number of atoms).
- Step 3 — divide every mole value by the smallest of them to get a ratio.
- Step 4 — if the ratio is not already whole numbers, multiply all of them up by a small integer to clear it. A value ending in .5 multiplies by 2; .33 or .67 by 3; .25 or .75 by 4. Rounding a genuine .5 down to 1 is the classic error — clear it, do not round it.
3 — Empirical-formula mass (EFM). Add the atomic masses in the empirical formula. For CH₂O the EFM is 12 + 2(1) + 16 = 30. The EFM is the stepping stone to the molecular formula.
4 — Molecular formula. The molecular formula is a whole-number multiple of the empirical formula: molecular formula = n × (empirical formula), where
n = molar mass / empirical-formula mass.
n should come out to a whole number (round only tiny rounding error). If glucose has empirical formula CH₂O (EFM 30) and molar mass 180, then n = 180 / 30 = 6, so the molecular formula is C₆H₁₂O₆. You can also get the molecular formula directly from percentages plus the molar mass: find the empirical formula, then scale by n.
5 — Combustion / elemental analysis. Burning a compound of C, H (and possibly O) in excess oxygen converts all the carbon to CO₂ and all the hydrogen to H₂O. Then:
- mass of C =
(12 / 44) × mass of CO₂ (each CO₂ carries one C, of mass 12 out of 44);
- mass of H =
(2 / 18) × mass of H₂O (each H₂O carries two H, mass 2 out of 18);
- mass of O (if the compound contains oxygen) =
mass of sample − mass of C − mass of H (by difference — never from the CO₂/H₂O, whose oxygen came from the air).
Convert those masses to moles and run the empirical-formula recipe; use the molar mass to get n. …