The empirical formula of a compound shows the simplest whole-number ratio of atoms of each element. The molecular formula shows the actual numbers. Combustion analysis and percentage composition calculations let chemists determine both from experimental data — a skill examined in every GCSE chemistry paper.
What is the difference between empirical and molecular formula?
The empirical formula gives the simplest integer ratio of atoms of each element in a compound. It is the most reduced form of the formula.
The molecular formula gives the actual number of atoms of each element in one molecule. It may be the same as the empirical formula, or it may be a simple multiple of it.
| Compound | Molecular formula | Empirical formula | Relationship |
|---|---|---|---|
| Glucose | C₆H₁₂O₆ | CH₂O | Molecular = 6 × empirical |
| Ethanoic acid | C₂H₄O₂ | CH₂O | Molecular = 2 × empirical |
| Water | H₂O | H₂O | Same (already simplest ratio) |
| Ethene | C₂H₄ | CH₂ | Molecular = 2 × empirical |
| Benzene | C₆H₆ | CH | Molecular = 6 × empirical |
Note that glucose and ethanoic acid share the same empirical formula (CH₂O) even though they are completely different compounds. The molecular formula is needed to distinguish them.
How do you find the empirical formula from percentage composition?
When given the percentage by mass of each element in a compound, follow these steps:
Step 1: Assume a 100 g sample. Convert each percentage to a mass in grams (e.g. 40% carbon → 40 g carbon).
Step 2: Convert each mass to moles using: moles = mass ÷ relative atomic mass (Ar).
Step 3: Divide every mole value by the smallest mole value to get a ratio.
Step 4: If the ratios are not whole numbers, multiply through by a small integer (2 or 3) until they are.
Step 5: Write the empirical formula using these whole-number ratios as subscripts.
Worked example 1: empirical formula from percentage composition
Question: A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find the empirical formula. (Ar: C = 12, H = 1, O = 16)
| Element | Mass in 100 g | Moles (mass ÷ Ar) | Ratio (÷ smallest) |
|---|---|---|---|
| C | 40.0 g | 40.0 ÷ 12 = 3.33 | 3.33 ÷ 3.33 = 1 |
| H | 6.7 g | 6.7 ÷ 1 = 6.7 | 6.7 ÷ 3.33 = 2.01 ≈ 2 |
| O | 53.3 g | 53.3 ÷ 16 = 3.33 | 3.33 ÷ 3.33 = 1 |
The ratio is C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O.
How do you find the molecular formula from the empirical formula?
To go from empirical to molecular formula, you need the relative molecular mass (Mr) of the compound — this would be given in the exam.
Step 1: Calculate the relative formula mass of the empirical formula.
Step 2: Divide the Mr of the compound by the Mr of the empirical formula.
Step 3: Multiply every subscript in the empirical formula by this number.
Worked example 2: molecular formula from empirical formula
Question: The empirical formula of a compound is CH₂O. The relative molecular mass is 180. Find the molecular formula. (Ar: C = 12, H = 1, O = 16)
- Mr of CH₂O = 12 + (2 × 1) + 16 = 30
- Multiplication factor = 180 ÷ 30 = 6
- Multiply each subscript by 6: C₁×₆H₂×₆O₁×₆ = C₆H₁₂O₆
The molecular formula is C₆H₁₂O₆ (glucose).
What if the ratio is not a whole number?
Sometimes dividing by the smallest gives ratios such as 1 : 1.5 or 1 : 1.33. This happens when the empirical formula has subscripts of 2 or 3.
| Ratio obtained | Multiply by | Result |
|---|---|---|
| 1 : 1.5 : … | ×2 | 2 : 3 : … |
| 1 : 1.33 : … | ×3 | 3 : 4 : … |
| 1 : 2.5 : … | ×2 | 2 : 5 : … |
Example: Ratios of C : H = 1 : 1.5. Multiply by 2 → C : H = 2 : 3. Empirical formula = C₂H₃ (found in polyacrylonitrile, for example).
Always check your final empirical formula: can the subscripts be divided further? For instance, C₂H₄O₂ can be simplified to CH₂O. The empirical formula must be the simplest whole-number ratio.
Worked example 3: empirical formula from masses
Question: 2.4 g of carbon combines with 0.8 g of hydrogen to form a hydrocarbon. Find the empirical formula. (Ar: C = 12, H = 1)
| Element | Mass | Moles | Ratio (÷ smallest) |
|---|---|---|---|
| C | 2.4 g | 2.4 ÷ 12 = 0.2 | 0.2 ÷ 0.2 = 1 |
| H | 0.8 g | 0.8 ÷ 1 = 0.8 | 0.8 ÷ 0.2 = 4 |
Empirical formula = CH₄ (methane — same as molecular formula for this gas).
Frequently asked questions
Why do we use 100 g as the assumed sample mass?
Assuming a 100 g sample is a mathematical convenience, not a physical requirement. Because percentage means "per hundred", a 100 g sample means the percentage figure and the mass in grams are numerically identical — 40% carbon becomes 40 g carbon, with no unit conversion needed. This simplifies the calculation. You could use any mass and divide by the correct fraction, but 100 g avoids one arithmetic step.
What does a combustion analysis experiment involve?
In combustion analysis, a known mass of a compound is burned completely in excess oxygen. The products (carbon dioxide and water) are collected and weighed. Because all the carbon in CO₂ came from the compound, and all the hydrogen in H₂O came from the compound, you can work backwards: moles of CO₂ = moles of C; moles of H₂O × 2 = moles of H. If the compound contains oxygen, you find the remaining mass by subtracting the masses of C and H from the original sample mass. This gives the masses needed for an empirical formula calculation.
Can two different compounds have the same empirical formula?
Yes, and this is one of the key limitations of empirical formula alone. Glucose (C₆H₁₂O₆), fructose (C₆H₁₂O₆), galactose (C₆H₁₂O₆), and many other carbohydrates share the empirical formula CH₂O. To distinguish between them, the molecular formula and structural formula are needed. This is why mass spectrometry, which gives the molecular ion peak (and therefore the Mr), is so valuable for compound identification.
How do I know if I have the simplest ratio?
After you have your whole-number ratio, check whether all the subscripts share a common factor. If C₂H₄O₂ is your result, all subscripts are divisible by 2, so the simplest ratio is CH₂O. If C₃H₆O₃ is your result, all are divisible by 3, giving CH₂O again. A correct empirical formula has subscripts with no common factor greater than 1. Common mistake: writing H₂O₂ when the empirical formula of hydrogen peroxide is HO.
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