Empirical and Molecular Formula

Empirical/molecular formula questions are always the same move: convert the data into moles of atoms, simplify to a ratio, then (if needed) scale to match Mᵣ. This lesson covers both common data types: composition by mass and combustion data.

Build on the mole and Avogadro constant, using your course’s Stoichiometry topic navigation to review them when needed. Keep unit conversions and mole ratios together in your working.

Definitions (Must Know)

A. Empirical formula

The empirical formula is the simplest whole-number ratio of atoms in a compound.

B. Molecular formula

The molecular formula shows the actual number of each type of atom in one molecule.

C. Combustion data (combustion analysis)

Combustion data is the mass (or moles) of CO₂ and H₂O formed when a compound is completely combusted, used to deduce the compound’s formula.

Key Ideas (What Earns Marks)

• Convert everything to moles first.
• For percentage composition, assume 100 g first.
• Divide by the smallest mole value.
• If you get near-halves (e.g. 1.5), multiply to clear fractions (e.g. ×2).
• Molecular formula: multiplier = Mᵣ/Mᵣ(empirical).
• For combustion data:
  • moles of C = moles of CO₂
  • moles of H = 2 × moles of H₂O
  • if O is present, find O by mass difference (then convert to moles)

Detailed Explanations

A. Empirical formula from composition (workflow)

Because the subscripts in a formula represent the ratio of atoms, you must convert mass data into moles (proportional to number of atoms).

1. If given percentages, assume 100 g (so % becomes grams).
2. Convert each element’s mass to moles: n = m/Aᵣ.
3. Divide every mole value by the smallest to get the simplest ratio.
4. Multiply the ratio to clear fractions, then write the empirical formula.

Mini example:

• C = 24.0 g, H = 4.0 g → moles C = 2.00, H = 4.00 → ratio 1 : 2 → empirical formula CH₂.

B. Molecular formula from empirical formula (workflow)

Because the molecular formula is an integer multiple of the empirical formula:

1. Calculate Mᵣ(empirical).
2. Find the multiplier k = Mᵣ(molecule)/Mᵣ(empirical).
3. Multiply every subscript in the empirical formula by k.

Mini example:

• empirical CH₂O has Mᵣ = 30; if molecular mass is 180 then k = 180/30 = 6 → molecular formula C₆H₁₂O₆.

C. Empirical formula from combustion data (workflow)

For a compound containing C and H (and possibly O), combustion gives you CO₂ and H₂O, which lets you deduce moles of C and H directly.

1. Find n(CO₂) and n(H₂O) from their masses.
2. Convert to moles of atoms:
   • moles of C atoms = n(CO₂)
   • moles of H atoms = 2n(H₂O)
3. If oxygen is present in the compound, find it by mass difference:
   • mass of C in the sample = n(C) × 12.0
   • mass of H in the sample = n(H) × 1.00
   • mass of O in the sample = mass of sample − mass of C − mass of H
   • moles of O atoms = m(O)/16.0
4. Divide all mole values by the smallest to get the empirical ratio.

Worked Examples

Common Mistakes

• Rounding too early (keep 3 s.f. until the final ratio step).
• Forgetting to multiply ratios like 1 : 1.5 : 1 to whole numbers (→ 2 : 3 : 2).
• Using Mᵣ without checking that the multiplier is an integer.
• In combustion: using the mass of CO₂ as the mass of carbon (you must convert to moles first).
• In combustion: forgetting the factor of 2 for hydrogen atoms from H₂O.

Use the topic check in Practise and check below to practise and check your understanding.

Exam Tips

• If ratios look like 1.33 or 0.67, suspect thirds (multiply by 3).
• Show the ratio line explicitly; it is often a mark.

Mind Stretchers