Last updated: 21 August 2026
The molecular formula C₂H₆O tells us that a molecule contains two carbon atoms, six hydrogen atoms and one oxygen atom. What it does not tell us is how those atoms are connected. C₂H₆O could describe ethanol, CH₃CH₂OH, or methoxymethane, CH₃OCH₃.
This chapter therefore links composition with structure. We will first learn what the different types of formula actually tell us. We will then work backwards from experimental masses, percentage compositions and combustion products to find empirical formulae and, when Mᵣ is known, molecular formulae.
Try these five questions before reading the chapter. Use the formulae and structures carefully — the same molecule can be represented in several different ways.
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What Information Does Each Type of Formula Give?
Different formulae answer different chemical questions. Before interpreting a formula, ask what information the representation is designed to communicate.
Empirical and molecular formulae describe composition. Structural, displayed, skeletal and stereochemical formulae add information about connectivity, bonds and spatial arrangement.
| Representation | What it tells you | Example or description |
|---|---|---|
| Empirical formula | Simplest whole-number ratio of atoms | CH₂O |
| Molecular formula | Actual number of atoms in one molecule | C₂H₄O₂ |
| Structural formula | How atoms or groups of atoms are connected | CH₃COOH |
| Displayed formula | Every atom and every covalent bond | All individual bonds are shown |
| Skeletal formula | Carbon framework using lines and vertices | Most C and C–H labels are omitted |
| Stereochemical formula | Three-dimensional arrangement of selected bonds | Normal, wedge and dashed-wedge bonds may be used |
Different representations may describe exactly the same molecule. The amount of information shown changes, not necessarily the identity of the compound.
Empirical Formula: The Simplest Ratio
The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound.
Hydrogen peroxide has molecular formula H₂O₂. The ratio of H atoms to O atoms is 2 : 2, which simplifies to 1 : 1.
Ethanoic acid has molecular formula C₂H₄O₂. Dividing all three subscripts by 2 gives:
Different compounds can share an empirical formula
Ethene, C₂H₄, and propene, C₃H₆, are different compounds, but both reduce to the same empirical formula, CH₂.
Ethene: C₂H₄ → CH₂.
Propene: C₃H₆ → CH₂.
When is the empirical formula the same as the molecular formula?
If all the subscripts have no common factor greater than 1, the molecular formula is already in its simplest ratio.
For ethanol, C₂H₆O, the subscripts 2, 6 and 1 cannot all be divided by the same whole number greater than 1. Its empirical formula is therefore also C₂H₆O.
Molecular Formula: The Actual Number of Atoms
The molecular formula gives the actual number of atoms of each element in one molecule.
For molecular substances, the molecular formula must be a whole-number multiple of the empirical formula.
The multiplier n is found by comparing Mᵣ with the relative formula mass of the empirical formula.
Worked example: empirical formula CH₂
A hydrocarbon has empirical formula CH₂ and Mᵣ = 56.0.
Relative formula mass of CH₂ = 12.0 + 2(1.0) = 14.0. Therefore n = 56.0 ÷ 14.0 = 4. Multiplying every subscript in CH₂ by 4 gives the molecular formula C₄H₈.
Multiply the entire empirical formula by n. If n = 3 and the empirical formula is CH₂O, the molecular formula is C₃H₆O₃.
Structural Formulae: Showing Connectivity
A molecular formula does not necessarily tell us which atoms are bonded together. A structural formula provides this extra information.
Both compounds have molecular formula C₂H₆O, but the oxygen atom is connected differently.
Molecular formula: tells us the types and numbers of atoms.
Structural formula: additionally tells us how those atoms or groups of atoms are connected.
Condensed structural formulae
A structural formula need not show every individual bond. Groups may be written together provided that their connectivity remains unambiguous.
The brackets indicate that the –OH group is attached to the middle carbon atom. Writing only C₃H₈O would lose this structural information.
Displayed Formulae: Showing Every Atom and Every Bond
A displayed formula shows every atom and every covalent bond.
Single, double and triple bonds must therefore be represented using the correct number of bond lines.
| Bond | Representation | Shared electron pairs |
|---|---|---|
| Single bond | One line | 1 |
| Double bond | Two lines | 2 |
| Triple bond | Three lines | 3 |
Bonds to hydrogen must also be shown
If a question asks for a displayed formula, bonds involving hydrogen cannot be hidden. For an alcohol, this includes the O–H bond.
CH₃CH₂OH is a structural formula, not a complete displayed formula. A displayed formula must show every C–H, C–C, C–O and O–H bond.
Skeletal Formulae: Reading the Carbon Framework
Large organic molecules become cumbersome if every carbon atom and every hydrogen bonded to carbon is written explicitly. Skeletal formulae provide a compact way of showing the carbon framework.
Carbon atoms occur at line ends and vertices
Do not count line segments and assume that this equals the number of carbon atoms. The lines represent bonds.
In the butane structure above, three C–C bond lines connect four carbon atoms.
Wrong approach: three line segments means three carbon atoms.
Correct approach: count the line ends and vertices representing carbon atoms.
Hydrogen atoms attached to carbon are usually hidden
Carbon normally forms four covalent bonds. Once the bonds already shown to a carbon have been counted, enough hydrogen atoms are assumed to complete four bonds.
A carbon at the end of one single bond therefore normally represents CH₃.
A carbon involved in a C=C double bond already uses two bond units in that double bond, so fewer hydrogen atoms are required.
Stereochemical Formulae: Adding Three-Dimensional Information
Molecules are three-dimensional even though chemical formulae are usually drawn on a flat page. A stereochemical formula can therefore show whether selected bonds lie in the plane of the page, project towards the viewer or project away.
| Bond representation | Meaning | Direction |
|---|---|---|
| Normal line | Bond lies approximately in the plane | In the plane of the page |
| Solid wedge | Bond projects out of the plane | Towards the viewer |
| Dashed wedge | Bond projects behind the plane | Away from the viewer |
Wedge and dashed-wedge notation communicates three-dimensional orientation. It does not indicate a different bond strength or bond order.
Finding an Empirical Formula from Masses
Formula subscripts represent ratios of numbers of atoms, but experimental data are often given as masses. Since different atoms have different masses, the masses cannot normally be compared directly.
For an element:
Worked example: magnesium oxide
A compound contains 2.40 g Mg and 1.60 g O. Use Aᵣ(Mg) = 24.0 and Aᵣ(O) = 16.0.
n(Mg) = 2.40 ÷ 24.0 = 0.100 mol, while n(O) = 1.60 ÷ 16.0 = 0.100 mol. Therefore Mg : O = 1 : 1 and the empirical formula is MgO.
Empirical formulae come from mole ratios, not mass ratios.
Finding an Empirical Formula from Percentage Composition
Percentage composition uses exactly the same reasoning. If a compound contains 40.0% carbon by mass, imagine that you have 100 g of the compound. You would then have 40.0 g of carbon.
A reliable four-step method
- Assume 100 g so that percentages become masses in grams.
- Convert every mass to moles by dividing by the relevant Aᵣ.
- Divide every mole value by the smallest to obtain the simplest ratio.
- Convert simple fractional ratios to whole numbers by multiplying the entire ratio where necessary.
Worked example: 85.7% carbon and 14.3% hydrogen
For a 100 g sample:
Convert each mass to moles:
Divide by 7.14:
Therefore the empirical formula is CH₂.
Dealing with Fractional Mole Ratios
After dividing by the smallest mole value, the result may not immediately contain exact integers. Values such as 1.50, 1.33 or 1.25 should not simply be rounded.
| Approximate value | Simple fraction | Multiply entire ratio by |
|---|---|---|
| 1.50 | 3/2 | 2 |
| 1.33 or 1.67 | 4/3 or 5/3 | 3 |
| 1.25 or 1.75 | 5/4 or 7/4 | 4 |
Example: 1 : 1.5 : 1
Suppose the calculated mole ratio is:
Since 1.5 = 3/2, multiply every value by 2:
The empirical formula is therefore C₂H₃O₂.
0.99 or 2.01: these may reasonably represent 1 or 2 because of experimental or rounding uncertainty.
1.50: this clearly represents a fractional ratio and should not be rounded directly.
Finding an Empirical Formula from Combustion Data
Combustion analysis is especially useful for compounds containing carbon and hydrogen.
During complete combustion, carbon from the compound forms CO₂ and hydrogen from the compound forms H₂O.
Do not forget the factor of 2. One mole of H₂O contains two moles of hydrogen atoms.
Worked example: a compound containing C, H and O
A 3.00 g sample containing only carbon, hydrogen and oxygen produces 4.40 g CO₂ and 1.80 g H₂O on complete combustion.
n(CO₂) = 4.40 ÷ 44.0 = 0.100 mol, so n(C) = 0.100 mol and mass of C = 1.20 g.
n(H₂O) = 1.80 ÷ 18.0 = 0.100 mol, so n(H) = 0.200 mol and mass of H = 0.200 g.
Since the compound contains only C, H and O: mass of O = 3.00 − 1.20 − 0.200 = 1.60 g. Therefore n(O) = 1.60 ÷ 16.0 = 0.100 mol.
The ratio C : H : O is 0.100 : 0.200 : 0.100 = 1 : 2 : 1. The empirical formula is CH₂O.
Why is oxygen found by difference?
The oxygen atoms in CO₂ and H₂O do not all come from the original compound. Oxygen gas is supplied during the combustion reaction, so external O₂ also contributes oxygen atoms to the products.
When the original compound is known to contain only C, H and O, the original oxygen mass can instead be obtained by subtracting the masses of C and H from the sample mass.
Compound contains only C, H and O: oxygen may be found from the remaining mass.
Another element may also be present: the remaining mass cannot automatically be assigned entirely to oxygen.
From Empirical Formula to Molecular Formula
Experimental composition normally gives only the empirical formula. Mᵣ provides the extra information needed to determine the molecular formula.
Suppose a compound has empirical formula CH₂O and Mᵣ = 90.0.
Composition data normally establish the simplest ratio first. Mᵣ tells us how many empirical formula units make up the actual molecule.
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Using masses directly as empirical formula subscripts | Mass ratios are not generally atom ratios because elements have different atomic masses. | Convert every mass to moles before comparing the values. |
| Writing C₂H₄O₂ as the empirical formula | The subscripts still have a common factor of 2. | Reduce all subscripts to the simplest whole-number ratio, CH₂O. |
| Rounding 1.50 directly to 2 | 1.50 represents the simple fraction 3/2 and is not a small rounding error. | Multiply the entire mole ratio by 2. |
| Multiplying only one subscript when finding the molecular formula | The whole empirical formula must be multiplied by the same value of n. | Multiply every subscript by n. |
| Using n(H) = n(H₂O) | Each H₂O molecule contains two H atoms. | Use n(H) = 2n(H₂O). |
| Assuming the remaining sample mass is always oxygen | Another element could contribute to the remaining mass. | Use mass by difference only when the composition information justifies it. |
| Counting skeletal line segments as carbon atoms | The lines represent bonds, not atoms. | Count carbon atoms at appropriate line ends and vertices. |
| Calling CH₃CH₂OH a displayed formula | It groups atoms together and does not show every individual bond. | Distinguish condensed structural formulae from fully displayed formulae. |
What you should be able to do
Distinguish empirical and molecular formulae
I can explain the difference between the simplest atom ratio and the actual numbers of atoms in a molecule.
Interpret structural representations
I can distinguish molecular, structural, displayed and skeletal formulae and state what additional information each provides.
Read skeletal formulae
I can count carbon atoms at line ends and vertices and infer the hydrogen atoms omitted from carbon.
Determine empirical formulae
I can convert masses or percentage compositions to mole ratios and reduce them to the simplest whole numbers.
Use combustion data
I can determine carbon from CO₂, hydrogen from H₂O and oxygen by difference when the composition information permits it.
Determine molecular formulae
I can use the empirical formula and Mᵣ to determine the whole-number multiplier and hence the molecular formula.
These ten questions test whether you can apply the chapter to calculations, skeletal structures and unfamiliar formulae.
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Frequently asked questions
Questions Students Often Ask About Empirical, Molecular and Structural Formulae
The same few ideas cause most of the confusion in this topic, especially the distinction between ratios, actual molecular composition and structural information.
Can an empirical formula and molecular formula be identical?
Yes. If the molecular formula cannot be simplified further, it is already the empirical formula. C₂H₆O is one example because 2, 6 and 1 have no common factor greater than 1.
Can two different compounds have the same molecular formula?
Yes. A molecular formula specifies only the numbers of each type of atom. The atoms may be connected differently. Ethanol, CH₃CH₂OH, and methoxymethane, CH₃OCH₃, both have molecular formula C₂H₆O.
Why should I not round a mole ratio of 1.50 directly to 2?
1.50 is not very close to either 1 or 2. It represents the simple fraction 3/2, so the correct approach is to multiply the entire ratio by 2.
Why can oxygen sometimes be found by difference in combustion analysis?
If the compound is known to contain only C, H and O, the part of the original sample mass that is not carbon or hydrogen must be oxygen. This method cannot automatically be used if another element may also be present.
Why are hydrogen atoms attached to carbon omitted from skeletal formulae?
Skeletal formulae are designed to show carbon frameworks compactly. Carbon is assumed to form four bonds, so the number of hydrogen atoms attached to each carbon can normally be inferred from the bonds already shown.
Empirical, molecular and structural formulae are different layers of the same chemical language. Experimental data lead us from masses to moles and then to the simplest atom ratio; Mᵣ allows us to determine the actual molecular formula; and structural, displayed and skeletal formulae reveal progressively more about how those atoms are connected. Once you can move confidently between these representations, the structures used throughout A-Level Organic Chemistry become much easier to interpret.