Last updated: 17 August 2026
A chemical formula is more than a label attached to a substance. It tells us which elements are present and the proportions in which their atoms are combined. However, an experimental analysis usually gives masses or percentages rather than a formula directly.
The task is therefore to translate measurable quantities into an atomic ratio. The bridge between these two kinds of information is the mole. Once the amount of each element has been found in moles, the relative numbers of atoms can be compared.
This same reasoning allows us to determine an empirical formula from percentage composition, actual masses or combustion products. If the relative molecular mass is also known, the empirical formula can then be converted into the molecular formula.
Diagnostic Quiz
Try these five questions before reading the chapter. They check the mole, ratio and formula ideas that empirical and molecular formula calculations build upon.
- 5 quick prerequisite questions
- Focus on moles, ratios and chemical formulae
- Every answer option is explained
Empirical and molecular formulae
The empirical formula and molecular formula describe the composition of the same substance at different levels of detail.
| Feature | Empirical formula | Molecular formula |
|---|---|---|
| Meaning | Simplest whole-number ratio of atoms of each element | Actual number of atoms of each element in one molecule |
| What it shows | Relative proportions of the elements | The composition of an individual molecule |
| Can its subscripts have a common factor? | No | Yes |
| Glucose | CH₂O | C₆H₁₂O₆ |
| Benzene | CH | C₆H₆ |
| Water | H₂O | H₂O |
What does the empirical formula tell us?
The empirical formula gives a ratio, not necessarily the actual number of atoms in one molecule. For example, the empirical formula CH₂O means that carbon, hydrogen and oxygen atoms are present in the ratio:
Several different molecular formulae could have this same empirical formula. These include CH₂O, C₂H₄O₂ and C₆H₁₂O₆ because each set of subscripts simplifies to 1 : 2 : 1.
What does the molecular formula tell us?
The molecular formula gives the actual number of atoms of each element in one molecule. For example, C₆H₁₂O₆ tells us that one glucose molecule contains six carbon atoms, twelve hydrogen atoms and six oxygen atoms.
Dividing all three subscripts by 6 gives the empirical formula CH₂O.
The molecular formula of water is H₂O. Since the subscripts 2 and 1 have no common factor greater than 1, its empirical formula is also H₂O.
The relationship between empirical and molecular formulae
The molecular formula must be a whole-number multiple of the empirical formula. If the empirical formula is represented by EF:
Here, n is a positive whole number such as 1, 2, 3 or 4. It cannot be a fraction because a molecule cannot contain a fractional number of empirical formula units.
The value of n is found by comparing the relative molecular mass, Mᵣ, with the mass represented by the empirical formula.
Once n has been calculated, every subscript in the empirical formula is multiplied by this value.
Empirical formula: the simplest ratio after all subscripts have been reduced as far as possible.
Molecular formula: the actual formula obtained by multiplying every empirical-formula subscript by the same whole number.
Why masses must be converted into moles
A formula represents numbers of atoms. However, atoms of different elements have different masses. Equal masses of two elements therefore do not usually contain equal numbers of atoms.
For example, 12.0 g of carbon atoms and 16.0 g of oxygen atoms each represent 1.00 mol of atoms. The masses are different, but the numbers of atoms are the same.
Converting each mass into moles removes the effect of the different atomic masses. The resulting mole values can then be compared to determine the ratio of atoms.
The empirical formula is based on the ratio of numbers of atoms. Since the amount in moles is proportional to the number of atoms, the mass of each element must first be divided by its relative atomic mass.
A reliable method for calculating an empirical formula
The same central method can be used whether the question provides actual masses or percentage composition.
- Write down the mass or percentage of each element.
- Convert each value into moles by dividing by the element's Aᵣ.
- Divide every mole value by the smallest mole value.
- Interpret the resulting ratio.
- Multiply the entire ratio if necessary to obtain whole numbers.
- Use the whole-number ratio as the formula subscripts.
Why percentage composition can be treated as mass
Suppose a compound contains 40.0% carbon by mass. This means that every 100 g of the compound contains 40.0 g of carbon.
We may therefore imagine a 100 g sample. Each percentage then becomes the same numerical value in grams.
| Percentage by mass | Mass in an assumed 100 g sample |
|---|---|
| 40.0% carbon | 40.0 g carbon |
| 6.7% hydrogen | 6.7 g hydrogen |
| 53.3% oxygen | 53.3 g oxygen |
The choice of a 100 g sample is simply a convenient mathematical device. It does not mean that the experiment was necessarily carried out using exactly 100 g of the compound.
Worked example: empirical formula from percentage composition
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Determine its empirical formula.
Step 1: assume a 100 g sample
The sample therefore contains:
- 40.0 g of carbon;
- 6.7 g of hydrogen; and
- 53.3 g of oxygen.
Step 2: convert each mass into moles
| Element | Mass / g | Aᵣ | Amount / mol |
|---|---|---|---|
| C | 40.0 | 12.0 | 40.0 ÷ 12.0 = 3.333 |
| H | 6.7 | 1.0 | 6.7 ÷ 1.0 = 6.700 |
| O | 53.3 | 16.0 | 53.3 ÷ 16.0 = 3.331 |
Step 3: divide by the smallest mole value
The smallest value is approximately 3.331 mol.
| Element | Mole value | Value divided by 3.331 |
|---|---|---|
| C | 3.333 | 1.001 ≈ 1 |
| H | 6.700 | 2.011 ≈ 2 |
| O | 3.331 | 1.000 ≈ 1 |
The simplest whole-number ratio is therefore:
Empirical formula: CH₂O
Experimental data and rounded percentage values may produce results such as 1.001, 2.011 and 1.000. These are sufficiently close to 1 : 2 : 1 to justify the empirical formula CH₂O.
Worked example: empirical formula from actual masses
A compound contains 2.24 g of iron combined with 0.96 g of oxygen. Determine its empirical formula.
| Element | Mass / g | Aᵣ | Amount / mol |
|---|---|---|---|
| Fe | 2.24 | 56.0 | 2.24 ÷ 56.0 = 0.0400 |
| O | 0.96 | 16.0 | 0.96 ÷ 16.0 = 0.0600 |
Dividing both values by the smaller value, 0.0400 mol, gives:
A ratio of 1 : 1.5 still contains a fractional value. Multiplying the entire ratio by 2 gives:
Empirical formula: Fe₂O₃
Incorrect: changing 1 : 1.5 directly into 1 : 2.
Correct: multiply both parts of the ratio by 2, giving 2 : 3. Rounding 1.5 to 2 would change the composition represented by the experimental data.
Handling ratios that are not whole numbers
After dividing by the smallest mole value, a ratio may contain values such as 1.5, 1.33 or 1.25. These values should not automatically be rounded to the nearest whole number because they may represent simple fractions.
| Approximate decimal in the ratio | Likely fraction | Multiply the entire ratio by |
|---|---|---|
| 0.50 or 1.50 | ½ | 2 |
| 0.33, 0.67, 1.33 or 1.67 | ⅓ or ⅔ | 3 |
| 0.25, 0.75, 1.25 or 1.75 | ¼ or ¾ | 4 |
| 0.20, 0.40, 0.60 or 0.80 | Fifths | 5 |
Example: a ratio containing thirds
Suppose the calculated ratio is:
The value 1.33 is close to 1⅓, or 4/3. Multiplying every value by 3 gives:
The empirical formula would therefore be C₃H₄O₃.
Multiplying only the decimal value would alter the relative proportions. A ratio can be scaled without changing its meaning only when every value is multiplied by the same number.
When is ordinary rounding acceptable?
Values very close to whole numbers can usually be interpreted as those whole numbers. For example, 0.998 : 2.01 : 1.00 may reasonably be interpreted as 1 : 2 : 1.
Values such as 1.20, 1.35 or 1.50 should not simply be forced to the nearest whole number. Instead, consider whether the decimals correspond to simple fractions or whether an earlier calculation has been rounded too severely.
When is oxygen found by difference?
It is important to distinguish between a hydrocarbon and an oxygen-containing compound.
A hydrocarbon contains only carbon and hydrogen. If a substance is explicitly described as a hydrocarbon, there is no oxygen in the original compound and no oxygen calculation is required.
A different situation arises when the compound is stated to contain only carbon, hydrogen and oxygen. If the masses of carbon and hydrogen are known, the mass of oxygen can be found by subtracting their masses from the total mass of the sample.
If percentage composition is used instead:
The calculated mass or percentage of oxygen must then be converted into moles by dividing by 16.0.
The unaccounted mass may be assigned to oxygen only if the question states or clearly implies that the compound contains only carbon, hydrogen and oxygen. If another element could be present, the remaining mass cannot automatically be treated as oxygen.
Using combustion data to find an empirical formula
In combustion analysis, a measured sample is burned completely in excess oxygen. Carbon atoms from the sample form CO₂, while hydrogen atoms from the sample form H₂O.
The combustion products therefore allow us to work backwards and calculate the amounts of carbon atoms and hydrogen atoms that were present in the original sample.
Finding carbon from carbon dioxide
Each CO₂ molecule contains one carbon atom. The number of moles of carbon atoms is therefore equal to the number of moles of CO₂ produced.
For example, if 0.0200 mol of CO₂ is formed:
Finding hydrogen from water
Each H₂O molecule contains two hydrogen atoms. One mole of H₂O therefore contains two moles of hydrogen atoms.
For example, if 0.0200 mol of H₂O is formed:
CO₂ contains one C atom per molecule: n(C atoms) = n(CO₂).
H₂O contains two H atoms per molecule: n(H atoms) = 2n(H₂O).
What happens after carbon and hydrogen have been found?
The next step depends on the type of compound described in the question.
- If the compound is a hydrocarbon, it contains only carbon and hydrogen. Use the mole ratio of C atoms to H atoms directly.
- If the compound contains only carbon, hydrogen and oxygen, calculate the masses of carbon and hydrogen and find the oxygen mass by difference.
A step-by-step combustion method
- Convert the mass of CO₂ into moles of CO₂.
- Use n(C atoms) = n(CO₂).
- Convert the mass of H₂O into moles of H₂O.
- Use n(H atoms) = 2n(H₂O).
- If the compound is a hydrocarbon, simplify the C : H mole ratio.
- If the compound contains only C, H and O, calculate the masses of C and H.
- Find the mass of O by subtracting the C and H masses from the sample mass.
- Convert the mass of O into moles and find the simplest C : H : O ratio.
Combustion is carried out in O₂, so the oxygen atoms in the CO₂ and H₂O products may have come from the supplied oxygen gas, from the original compound, or from both. The amounts of CO₂ and H₂O therefore tell us directly about carbon and hydrogen, but not directly about the oxygen content of the original sample.
Worked example: a compound containing carbon, hydrogen and oxygen
A 0.600 g sample of a compound containing only carbon, hydrogen and oxygen is completely burned. It produces 0.880 g of CO₂ and 0.360 g of H₂O. Determine the empirical formula of the compound.
Step 1: find the amount of carbon atoms
Each mole of CO₂ contains one mole of carbon atoms:
The mass of carbon in the original sample is:
Step 2: find the amount of hydrogen atoms
Each mole of H₂O contains two moles of hydrogen atoms:
The mass of hydrogen in the original sample is:
Step 3: find the amount of oxygen atoms
The compound is stated to contain only carbon, hydrogen and oxygen. The mass not accounted for by carbon and hydrogen must therefore be the mass of oxygen in the original sample.
Step 4: obtain the simplest ratio
| Element | Amount of atoms / mol | Divide by 0.0200 |
|---|---|---|
| C | 0.0200 | 1 |
| H | 0.0400 | 2 |
| O | 0.0200 | 1 |
Empirical formula: CH₂O
The amounts of CO₂ and H₂O give the amounts of carbon atoms and hydrogen atoms in the original sample. Once their masses have been calculated, the remaining sample mass is oxygen because the question states that only C, H and O are present.
Worked example: molecular formula of a hydrocarbon
A 0.580 g sample of a hydrocarbon produces 1.760 g of CO₂ and 0.900 g of H₂O on complete combustion. The hydrocarbon has Mᵣ = 58.0. Determine its empirical and molecular formulae.
Since the substance is a hydrocarbon, it contains only carbon and hydrogen. There is no oxygen in the original compound to calculate.
Step 1: calculate the amount of carbon atoms
Step 2: calculate the amount of hydrogen atoms
Step 3: find the empirical formula
Dividing both amounts by 0.0400 gives:
Multiplying the entire ratio by 2 gives:
Empirical formula: C₂H₅
Step 4: use Mᵣ to find the molecular formula
Multiplying both empirical-formula subscripts by 2 gives:
Molecular formula: C₄H₁₀
Calculating a molecular formula from Mᵣ
An empirical formula alone is not enough to determine a unique molecular formula. The relative molecular mass must also be known.
- Calculate the mass represented by the empirical formula.
- Divide the compound's Mᵣ by the empirical formula mass.
- Check that the result is a whole number.
- Multiply every subscript in the empirical formula by that number.
Worked example: glucose
A compound has empirical formula CH₂O and Mᵣ = 180. Determine its molecular formula.
Every subscript in CH₂O must be multiplied by 6:
Answer: C₆H₁₂O₆
Why must the multiplier be a whole number?
The empirical formula already represents the smallest possible whole-number ratio. A molecule must contain an integer number of complete empirical-formula units.
For example, a multiplier of 2 means that the molecule contains two empirical formula units. A calculated value such as 2.98 may reasonably correspond to 3 if the small difference arises from rounded or experimental data. A value such as 2.40 should not simply be forced to 2 or 3.
If the empirical formula is CH₂O and n = 3, the molecular formula is C₃H₆O₃. The carbon and oxygen subscripts must also be multiplied even though their empirical-formula subscripts of 1 are not visibly written.
Checking whether a calculated formula is reasonable
A formula calculation should be checked rather than accepted immediately. Several quick checks can reveal arithmetic or interpretation errors.
Check 1: are all empirical-formula subscripts whole numbers?
An empirical formula cannot contain decimal subscripts. A ratio such as 1 : 1.5 : 1 must be scaled to give whole numbers before it is written as a formula.
Check 2: is the ratio in its simplest form?
A proposed empirical formula such as C₂H₄O₂ is not in its simplest form because all subscripts can be divided by 2. Its empirical formula is CH₂O.
Check 3: does the molecular-formula mass equal the stated Mᵣ?
After finding the molecular formula, calculate its Mᵣ. It should agree with the value supplied in the question.
Check 4: has every element been included?
When oxygen is found by difference, it must still be converted into moles and included in the final ratio. Do not stop after calculating only carbon and hydrogen.
A small deviation from a simple whole-number or fractional ratio: may result from rounded experimental data and can often be interpreted sensibly.
A large unexplained deviation: usually indicates that a mole conversion, stoichiometric factor or arithmetic step should be checked.
Precision and rounding during formula calculations
Premature rounding can change the apparent ratio and produce an incorrect formula. Keep several significant figures in intermediate mole values, especially before dividing by the smallest value.
For example, suppose three calculated amounts are 0.03325 mol, 0.06661 mol and 0.03331 mol. Rounding these immediately to 0.03 mol, 0.07 mol and 0.03 mol produces a distorted ratio. Using the fuller values gives a ratio close to 1 : 2 : 1.
The final empirical or molecular formula itself does not contain decimal places or significant figures. Its subscripts are exact whole numbers inferred from the experimental data.
Experimental masses may be quoted to three or four significant figures, but a formula such as C₂H₆O contains exact integer subscripts. The measured data are being used to infer that whole-number composition.
What the formula does and does not tell us
A molecular formula gives the types and numbers of atoms in a molecule, but it does not show how those atoms are connected.
For example, ethanol and methoxymethane both have the molecular formula C₂H₆O, but their atoms are arranged differently. They are different compounds with different structural formulae.
| Type of formula | Information provided | Example for ethanol |
|---|---|---|
| Empirical formula | Simplest whole-number ratio of atoms | C₂H₆O |
| Molecular formula | Actual number of each type of atom in one molecule | C₂H₆O |
| Structural formula | How the atoms are connected | CH₃CH₂OH |
In this example, the empirical and molecular formulae happen to be the same because the subscripts 2, 6 and 1 have no common factor greater than 1.
Common mistakes in empirical and molecular formula calculations
| Common mistake | Why it is wrong | Correct approach |
|---|---|---|
| Using the mass ratio directly as the atomic ratio | Atoms of different elements have different masses | Divide each mass by the element's Aᵣ first |
| Dividing masses by the smallest mass | The simplest atomic ratio must be based on moles, not masses | Convert every mass into moles before simplifying |
| Rounding 1.5 directly to 2 | 1.5 represents the simple fraction 3/2 | Multiply the entire ratio by 2 |
| Multiplying only the decimal value in a ratio | This changes the relative proportions of the elements | Multiply every value in the ratio by the same factor |
| Rounding intermediate mole values too early | Premature rounding can distort the final ratio | Retain several significant figures until the ratio is obtained |
| Using n(H atoms) = n(H₂O) | Each H₂O molecule contains two hydrogen atoms | Use n(H atoms) = 2n(H₂O) |
| Using n(C atoms) = 2n(CO₂) | Each CO₂ molecule contains only one carbon atom | Use n(C atoms) = n(CO₂) |
| Finding the original oxygen directly from the oxygen atoms in CO₂ and H₂O | Combustion is carried out in O₂, so product oxygen does not come only from the compound | Find oxygen by difference when the compound contains only C, H and O |
| Trying to calculate oxygen in a hydrocarbon | A hydrocarbon contains only carbon and hydrogen | Use only the C : H mole ratio |
| Assuming all unaccounted sample mass must be oxygen | Another element could be present unless the composition is specified | Use oxygen by difference only when the elements present are known |
| Treating the empirical formula as the molecular formula | Several molecular formulae can share the same empirical formula | Use the given Mᵣ to calculate the multiplier n |
| Dividing the empirical formula mass by Mᵣ | This gives the reciprocal of the required multiplier | Use n = Mᵣ ÷ empirical formula mass |
| Multiplying only one empirical-formula subscript by n | The entire empirical unit must be multiplied | Multiply every subscript, including unwritten 1s |
| Leaving an empirical formula such as C₂H₄O₂ unsimplified | An empirical formula must show the simplest whole-number ratio | Divide all subscripts by their highest common factor |
Empirical and molecular formula checklist
Organise the data
- List every element present.
- Record each mass or percentage.
- Assume 100 g when percentages are given.
- Find any missing oxygen content only when justified.
Find the empirical formula
- Divide every mass by the correct Aᵣ.
- Keep several significant figures.
- Divide all mole values by the smallest.
- Scale simple fractional ratios to whole numbers.
Use combustion data
- Use n(C atoms) = n(CO₂).
- Use n(H atoms) = 2n(H₂O).
- Remember that a hydrocarbon contains only C and H.
- Find O by difference only for a compound containing only C, H and O.
Find the molecular formula
- Calculate the empirical formula mass.
- Use n = Mᵣ ÷ empirical formula mass.
- Check that n is a whole number.
- Multiply every empirical-formula subscript by n.
Mastery Quiz
Now apply the chapter to percentage composition, combustion analysis, fractional ratios and multi-step molecular-formula calculations.
- 10 application-focused questions
- Includes calculation and combustion reasoning
- Every answer option is explained
Frequently asked questions
Questions students often ask about empirical and molecular formulae
These questions address the distinctions and calculation choices that commonly cause confusion when composition data are presented in unfamiliar forms.
Can two compounds have the same empirical formula?
Yes. An empirical formula gives only the simplest atomic ratio, so different compounds can share it. Ethene, C₂H₄, and propene, C₃H₆, both have the empirical formula CH₂.
Their molecular formulae differ because their molecules contain different actual numbers of atoms.
Can the empirical and molecular formulae be the same?
Yes. They are the same when the molecular-formula subscripts are already in their simplest whole-number ratio. Water has molecular formula H₂O and empirical formula H₂O.
Why is C₂H₄O₂ not the empirical formula of ethanoic acid?
The subscripts 2, 4 and 2 share a common factor of 2. Dividing all of them by 2 gives the simplest ratio 1 : 2 : 1, so the empirical formula is CH₂O.
Does an empirical formula always represent an actual molecule?
No. An empirical formula represents a composition ratio. For glucose, CH₂O gives the simplest ratio, but an individual glucose molecule is C₆H₁₂O₆, not CH₂O.
For an ionic compound, the formula represents the simplest ratio of ions in the lattice rather than a discrete molecule.
Why is there no molecular formula for an ionic compound such as sodium chloride?
Sodium chloride forms a giant ionic lattice rather than separate NaCl molecules. NaCl represents the simplest ratio of Na⁺ ions to Cl⁻ ions in the lattice.
The term relative formula mass is therefore used for such substances rather than relative molecular mass.
Can I round a ratio of 1.49 to 1?
No. A value close to 1.50 usually represents the simple fraction 3/2. The entire ratio should normally be multiplied by 2 rather than rounding 1.49 to 1.
Ordinary rounding is appropriate when a value is genuinely close to a whole number, such as 0.998 or 2.01.
What should I do if the calculated value of n is 2.99?
A value of 2.99 would normally be interpreted as 3 because the small difference may arise from rounded Mᵣ values or experimental data. You should still check the empirical formula mass and your arithmetic before making that interpretation.
What should I do if n is not close to a whole number?
Recheck the empirical formula, relative atomic masses, mole conversions and stated Mᵣ. A molecular formula must contain a whole number of empirical formula units, so a result such as 2.40 usually indicates an error or inconsistent data rather than a value that should simply be rounded.
How is oxygen in the original compound determined from combustion data?
Complete combustion is carried out in oxygen gas, so the oxygen atoms in the CO₂ and H₂O products do not necessarily come from the original compound. Some may have come from the supplied O₂.
The combustion products are therefore used directly to determine the amounts of carbon atoms and hydrogen atoms. What happens next depends on the type of compound described.
If the substance is a hydrocarbon, it contains only carbon and hydrogen, so there is no oxygen in the original compound.
If the compound is stated to contain only carbon, hydrogen and oxygen, its oxygen mass is found by difference:
How do CO₂ and H₂O tell us the amounts of carbon and hydrogen?
The subscripts in the product formulae tell us how many atoms of the relevant element are present in each molecule.
Each CO₂ molecule contains one carbon atom. Therefore, one mole of CO₂ contains one mole of carbon atoms:
Each H₂O molecule contains two hydrogen atoms. Therefore, one mole of H₂O contains two moles of hydrogen atoms:
For example, 0.0200 mol of CO₂ contains 0.0200 mol of carbon atoms, while 0.0200 mol of H₂O contains 0.0400 mol of hydrogen atoms.
Why do we use the amounts of atoms rather than the amounts of CO₂ and H₂O directly?
The empirical formula represents the ratio of atoms in the original compound, not the ratio of product molecules formed during combustion.
We must therefore convert the amounts of CO₂ and H₂O into the corresponding amounts of carbon atoms and hydrogen atoms before forming the C : H ratio.
Does the molecular formula show how the atoms are joined?
No. It gives only the actual numbers of each type of atom. Different structural arrangements can have the same molecular formula.
For example, ethanol and methoxymethane both have molecular formula C₂H₆O, but they have different structural formulae and different chemical properties.
Calculating a formula becomes systematic once every piece of composition data is translated into moles. The empirical formula comes from the simplest whole-number mole ratio, while the molecular formula is found by comparing the empirical formula mass with Mᵣ. In combustion analysis, CO₂ reveals the amount of carbon atoms and H₂O reveals the amount of hydrogen atoms. A hydrocarbon contains only these two elements, whereas oxygen is found by difference when a compound is stated to contain only carbon, hydrogen and oxygen. With these distinctions secure, even unfamiliar formula calculations can be approached as a sequence of clear, checkable steps.