The big idea: Two different formulas describe a compound:
- The empirical formula is the simplest whole-number ratio of atoms. - The molecular formula is the actual number of each atom in one molecule.
The molecular formula is always a whole-number multiple of the empirical formula.
Same ratio, different size: Glucose has the molecular formula C6H12O6. Divide every subscript by 6 and you get the simplest ratio CH2O — that is its empirical formula.
So glucose's molecular formula = its empirical formula × 6.
Which is which?: - Empirical = the reduced ratio (like simplifying a fraction). - Molecular = the real count.
For an ionic compound the formula given (e.g. NaCl, CaCl2) is already an empirical formula — there are no separate molecules.
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Whether you are given masses or percentages by mass, the method is the same. Turn each one into an amount in moles with the given equation, then find the simplest ratio.
- amount of each element (mol)
- mass of that element (g)
- molar mass / relative atomic mass A_{r} (g mol⁻¹)
The four-step method: 1. Write the mass (or % — treat % as grams in a 100 g sample) of each element. 2. Divide each by its Ar to get moles (n = m/M). 3. Divide every answer by the smallest of those mole values. 4. Round to whole numbers — if you get a neat 0.5 or 0.33, multiply the whole ratio up to clear it.
Worked example — empirical formula from % composition
A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Determine its empirical formula. (Ar: C = 12.01, H = 1.01, O = 16.00.)
Solution
- Treat each % as grams in a 100 g sample, then convert to moles — formula first (n = m/M):
- Repeat for hydrogen:
- Repeat for oxygen:
- Divide every value by the smallest (3.33):
Final answer
Empirical formula = CH2O.
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In a combustion problem you are told the masses of CO2 and H2O produced. Every carbon atom ends up in a CO2, and every two hydrogen atoms end up in one H2O — so work back to the moles of C and H, then find the ratio.
Combustion shortcut: - n(C) = n(CO2) — one C per CO2. - n(H) = 2 × n(H2O) — two H per H2O.
If the compound also contains oxygen, find the mass of C and H, subtract from the sample mass to get the mass of O, then convert.
Worked example — empirical formula from combustion
Burning 4.60 g of a compound (containing only C, H and O) gives 8.80 g of CO2 and 5.40 g of H2O. Determine its empirical formula. (M: CO2 = 44.01, H2O = 18.02; Ar: C = 12.01, H = 1.01, O = 16.00.)
Solution
- Moles of C = moles of CO2 (one C per CO2):
- Moles of H = 2 × moles of H2O (two H per H2O):
- Find the mass of C and H, then the oxygen mass by difference:
- Convert oxygen to moles:
- Divide all by the smallest (0.100):
Final answer
Empirical formula = C2H6O.
To get the molecular formula, you also need the relative molecular mass Mr. Find how many times the empirical unit fits into Mr, then scale up.
- whole-number multiple (molecular = empirical × x)
- relative molecular mass of the compound
- M_{r} of one empirical-formula unit
Worked example — molecular formula from Mr
A compound has the empirical formula CH2 and a relative molecular mass of 56. Determine its molecular formula. (Empirical formula mass of CH2 = 14.03.)
Solution
- Formula first — find the whole-number multiple:
- Substitute the values:
- Multiply every subscript in the empirical formula by 4:
Final answer
Molecular formula = C4H8.
How this is tested: This skill is a Paper 1A and Paper 2 staple.
- Paper 1A (MCQ): 'which empirical formula matches this % composition?' or 'which formula is consistent with this Mr?'. - Paper 2: a part-marks determine — usually empirical formula from % or combustion, then the molecular formula once you are given Mr.
The classic trap: stopping at the empirical formula when the question asks for the molecular one — or forgetting to find oxygen by difference.
Marks you can always grab: (1) Always convert to moles before comparing — never compare masses directly. (2) Divide by the smallest mole value. (3) Read the verb: empirical (ratio) or molecular (uses Mr).
IB-style question — vitamin C analysis (a)
(a) Vitamin C contains 40.9% carbon, 4.58% hydrogen and 54.5% oxygen by mass. Determine its empirical formula. (Ar: C = 12.01, H = 1.01, O = 16.00.) [3]
Solution
- Treat each % as grams per 100 g and convert to moles (n = m/M):
- Hydrogen and oxygen:
- Divide each by the smallest (3.41):
- A 1.33 means a ×3 multiple — multiply all three by 3 to clear it:
Final answer
Empirical formula = C3H4O3.
IB-style question — vitamin C analysis (b)
(b) The relative molecular mass of vitamin C is 176. Determine its molecular formula. (Empirical formula mass of C3H4O3 = 88.06.) [1]
Solution
- Formula first — find the multiple:
- Multiply every subscript by 2:
Final answer
Molecular formula = C6H8O6.