How to Compare Molecular Formulas Before Identifying Isomers

Two compounds can have exactly the same molecular formula and still be different substances. For example, ethanol and dimethyl ether both have the molecular formula \(\ce{C2H6O}\), but their atoms are connected differently.

Before claiming that two compounds are isomers, you need to pass three checks: compare their exact molecular formulas, check that both structures obey carbon valence, and compare how their atoms are connected. If the connectivity is identical, you then need to investigate their three-dimensional arrangement.

The key distinction is that the molecular formula tells you which atoms are present, but not how those atoms are arranged.

01The four checks for identifying isomers

Use this decision process whenever you are given two molecular or structural formulas.

Passing the first check is not enough. Two compounds with the same molecular formula are not automatically isomers.

For example, consider \(\ce{CH2O}\) and \(\ce{C2H4O2}\). Both have the empirical formula \(\ce{CH2O}\), but their molecular formulas are different. They cannot be isomers.

Remember the difference:

• Molecular formula: the actual number of each type of atom.
• Empirical formula: the simplest whole-number ratio of those atoms.
• Structural formula: how the atoms are connected.

The molecular formula is your first filter. The structural formula provides the information needed to investigate isomerism.

02A model-kit investigation: how can one formula produce two structures?

Imagine being given a molecular model kit and asked to build every possible structure consistent with the formula \(\ce{C2H6O}\).

The question being tested is simple: does knowing the number of each atom uniquely determine the molecule?

Setting up the model

The apparatus consists of:

• Two carbon pieces, each with four bonding positions.
• Six hydrogen pieces, each with one bonding position.
• One oxygen piece, with two bonding positions.
• Connectors representing single covalent bonds.

The model represents the usual valencies of neutral carbon, hydrogen, and oxygen.

Start by placing the two carbon pieces next to each other. Connect them with a single bond. Attach the oxygen to the second carbon, then attach one hydrogen to the oxygen. Fill the remaining carbon bonding positions with the other five hydrogen atoms.

You have constructed ethanol:

\[ \ce{CH3-CH2-OH} \]

Now rearrange the same pieces. Put the oxygen between the two carbon atoms, then attach three hydrogen atoms to each carbon.

You have constructed dimethyl ether:

\[ \ce{CH3-O-CH3} \]

What should the model demonstrate?

If the molecular formula completely determined the structure, you would expect only one possible arrangement.

Instead, two valid arrangements use exactly the same atoms.

The observation is that both structures satisfy the same molecular formula, but their atoms have different neighbours.

In ethanol, the oxygen is bonded to a carbon and a hydrogen. In dimethyl ether, the oxygen is bonded to two carbon atoms.

This difference in connectivity makes them structural isomers.

A quick valence calculation

You can also check the total bonding capacity of the atoms.

For a neutral \(\ce{C2H6O}\) molecule, use the usual valencies:

• Each carbon contributes 4.
• Each hydrogen contributes 1.
• Oxygen contributes 2.

The total number of valence units is:

\[ (2\times4)+(6\times1)+(1\times2)=16 \]

Each bond uses one valence unit from each of its two atoms. Therefore, the total bond-order count is:

\[ \frac{16}{2}=8 \]

Both ethanol and dimethyl ether contain eight single bonds.

This is a useful consistency check. It confirms that both structures use the available bonding capacity, but it does not prove that they are the same molecule.

What the model cannot establish

A molecular model demonstrates that multiple structures are possible. It does not tell you which structure is present in an unknown chemical sample.

There is also a limitation when investigating spatial arrangements. A model kit with freely rotating connectors might allow you to rotate a double bond unrealistically.

A revised model needs a rigid double-bond connector. This preserves the relative positions of the attached groups and allows you to investigate geometric isomerism.

The distinction between connectivity and spatial arrangement becomes important when comparing alkenes.

03Check carbon valence before accepting a structure

Carbon valence is one of the most useful checks when comparing organic structures.

In the ordinary neutral organic molecules considered here, carbon forms four bonds in total. However, you must count bond order rather than simply count the number of lines touching the atom.

For example, consider three possible bonding arrangements around carbon:

• Four single bonds: \(1+1+1+1=4\).
• One double bond and two single bonds: \(2+1+1=4\).
• One triple bond and one single bond: \(3+1=4\).

All three satisfy carbon’s usual valence.

Other common valencies to remember are:

Consider a carbonyl group, which contains a carbon-oxygen double bond.

The carbonyl carbon already uses two valence units in its double bond to oxygen. It has two more units available for other bonds.

This is particularly useful when comparing aldehydes and ketones, as you will see in the second worked example.

Important: Check each atom individually. A structure is not valid simply because its total number of atoms matches the molecular formula.

These rules assume ordinary neutral covalent organic molecules. Charged species and less common bonding arrangements require additional considerations.

04Connectivity is not the same as spatial arrangement

Once both structures pass the molecular formula and valence checks, you need to distinguish two types of structural difference.

Structural isomers: different connectivity

Structural isomers, also called constitutional isomers, have the same molecular formula but different connections between their atoms.

For example:

• Butane: \(\ce{CH3-CH2-CH2-CH3}\)
• 2-methylpropane: \(\ce{CH3-CH(CH3)-CH3}\)

Both have the formula \(\ce{C4H10}\), but one has a continuous chain of four carbon atoms, while the other has a branched carbon skeleton.

The atoms are connected differently, so these are structural isomers.

Stereoisomers: same connectivity, different spatial arrangement

Now consider but-2-ene.

Its molecular formula is \(\ce{C4H8}\), and its condensed structural formula is:

\[ \ce{CH3-CH=CH-CH3} \]

There are two possible arrangements around the carbon-carbon double bond.

In cis-but-2-ene, the two methyl groups are on the same side of the double bond.

In trans-but-2-ene, the two methyl groups are on opposite sides.

Both compounds have:

• Four carbon atoms.
• Eight hydrogen atoms.
• Identical atom-to-atom connectivity.
• The same double-bond position.

The difference is their spatial arrangement.

Why can’t one simply rotate the molecule to turn cis-but-2-ene into trans-but-2-ene?

Because a carbon-carbon double bond restricts rotation. Rotating around it would require breaking the additional bonding interaction associated with the double bond.

Therefore, cis-but-2-ene and trans-but-2-ene are geometric stereoisomers.

A different drawing does not always mean a different molecule

There is one more distinction to make.

Imagine two drawings of butane. In one, the carbon chain is drawn horizontally. In the other, it is drawn at an angle.

They are still the same molecule.

Even if you rotate around a carbon-carbon single bond, the atoms remain connected in exactly the same way. The molecule simply adopts a different conformation.

Advanced chemistry discusses conformational isomerism, but for the usual HSC-style task of identifying distinct compounds, ordinary single-bond rotations are not counted as separate structural isomers.

A useful rule is:

• Rotating the drawing on the page does not change the molecule.
• Rotating around an ordinary single bond changes its conformation.
• Changing the fixed arrangement around a double bond can produce a different stereoisomer.

If you need more practice distinguishing molecular connectivity from the way a structure is drawn, see How to Check if Two Skeletal Formulas Show the Same Molecule.

For help counting hidden carbon and hydrogen atoms before comparing formulas, see How to Read Skeletal Formulas and Write Condensed Formulas.

05Worked examples

06Questions and solutions

Question 1

Two compounds have the following molecular formulas:

• Compound A: \(\ce{CH2O}\)
• Compound B: \(\ce{C2H4O2}\)

Both have the same empirical formula, \(\ce{CH2O}\).

Can these compounds be classified as isomers? Explain your reasoning.

Solution 1

No. These compounds are not isomers because their molecular formulas are different.

An empirical formula gives the simplest ratio of atoms, not the actual number of atoms in a molecule.

For compound A, the molecular formula is:

\[ \ce{CH2O} \]

For compound B, the molecular formula is:

\[ \ce{C2H4O2} \]

Dividing the subscripts in the second formula by two gives:

\[ \ce{C2H4O2}\rightarrow\ce{CH2O} \]

This shows that their empirical formulas match.

However, isomers must have identical molecular formulas, not merely identical empirical formulas.

The distinction is important because two compounds can have the same simplest atom ratio while containing different total numbers of atoms.

Question 2

Two proposed organic compounds have the following structural formulas:

• Compound A: ethanoic acid, \(\ce{CH3C(=O)OH}\)
• Compound B: methyl methanoate, \(\ce{HC(=O)OCH3}\)

Determine whether these compounds are isomers.

Your answer should:

• Verify their molecular formulas.
• Check the valence of each carbonyl carbon.
• Explain the difference in connectivity.

Solution 2

Yes. Ethanoic acid and methyl methanoate are functional-group isomers with the molecular formula \(\ce{C2H4O2}\).

Step 1: Check the molecular formula of ethanoic acid.

Its structure is:

\[ \ce{CH3-C(=O)-OH} \]

Count the atoms:

• Carbon: 2.
• Hydrogen: 4.
• Oxygen: 2.

Therefore:

\[ A=\ce{C2H4O2} \]

Step 2: Check the molecular formula of methyl methanoate.

Its structure is:

\[ \ce{H-C(=O)-O-CH3} \]

Count the atoms:

• Carbon: 2.
• Hydrogen: 4.
• Oxygen: 2.

Therefore:

\[ B=\ce{C2H4O2} \]

The molecular formulas are identical.

Step 3: Check carbon valence.

In ethanoic acid, the carbonyl carbon has one double bond to oxygen, one single bond to the methyl carbon, and one single bond to the hydroxyl oxygen.

Its valence is:

\[ 2+1+1=4 \]

In methyl methanoate, the carbonyl carbon has one double bond to oxygen, one single bond to hydrogen, and one single bond to the other oxygen.

Its valence is also:

\[ 2+1+1=4 \]

The methyl carbon in each compound has three hydrogen bonds and one additional single bond.

All carbon atoms satisfy the usual valence of four. Both oxygen atoms in each structure also satisfy their usual valence of two.

Step 4: Compare connectivity.

Ethanoic acid contains a carboxylic acid group, where the hydroxyl oxygen is bonded to hydrogen.

Methyl methanoate contains an ester group, where an oxygen atom connects the carbonyl-containing part of the molecule to a methyl group.

The functional groups are different because the atoms are connected differently.

Therefore, the compounds are functional-group isomers.

Question 3

A student proposes four possible structures for compounds with the molecular formula \(\ce{C5H12}\):

• A: Pentane, \(\ce{CH3CH2CH2CH2CH3}\).
• B: 2-methylbutane, \(\ce{CH3CH(CH3)CH2CH3}\).
• C: 2,2-dimethylpropane, \(\ce{C(CH3)4}\).
• D: A central carbon written as \(\ce{CH(CH3)4}\).

Determine which structures are valid compounds with the molecular formula \(\ce{C5H12}\).

For every valid structure, explain why it represents a different connectivity arrangement. Explain why structure D must be rejected.

Solution 3

Structures A, B, and C are three valid structural isomers of \(\ce{C5H12}\). Structure D is invalid because its central carbon would have five single bonds.

Step 1: Check structure A.

Pentane has the condensed formula:

\[ \ce{CH3CH2CH2CH2CH3} \]

It contains five carbon atoms.

Its hydrogen count is:

\[ 3+2+2+2+3=12 \]

Therefore, its molecular formula is \(\ce{C5H12}\).

Every carbon satisfies its usual valence of four.

Step 2: Check structure B.

2-methylbutane has the formula:

\[ \ce{CH3CH(CH3)CH2CH3} \]

It contains five carbon atoms.

Its hydrogen count is:

\[ 3+1+3+2+3=12 \]

Its molecular formula is also \(\ce{C5H12}\).

The central carbon is attached to three other carbon atoms and one hydrogen atom, giving it four bonds.

The structure is valid.

Step 3: Check structure C.

2,2-dimethylpropane has the formula:

\[ \ce{C(CH3)4} \]

It contains one central carbon and four surrounding methyl groups.

The hydrogen count is:

\[ 4\times3=12 \]

Therefore, its molecular formula is \(\ce{C5H12}\).

The central carbon is bonded to four other carbon atoms. It has no hydrogen atoms, but it still satisfies carbon valence.

This structure is also valid.

Step 4: Reject structure D.

The proposed structure is:

\[ \ce{CH(CH3)4} \]

The central carbon is connected to four methyl groups and one hydrogen atom.

That gives it five single bonds:

\[ 1+1+1+1+1=5 \]

This violates the usual valence of carbon in a neutral organic molecule.

It also contains 13 hydrogen atoms rather than 12.

Therefore, structure D cannot represent a valid compound with the proposed formula.

Step 5: Compare the connectivity of the valid structures.

The three valid compounds have different carbon skeletons:

• A has a continuous chain of five carbon atoms.
• B has a four-carbon chain with one methyl branch.
• C has one central carbon attached to four methyl groups.

They have the same molecular formula but different connectivity.

Therefore, A, B, and C are three distinct structural isomers of \(\ce{C5H12}\).

The important lesson is that a structure must pass both the molecular formula check and the individual valence checks.

Question 4

A student is given four molecular drawings.

The first pair represents cis-but-2-ene and trans-but-2-ene. Both have the molecular formula \(\ce{C4H8}\).

The second pair represents two different conformations of butane, \(\ce{C4H10}\). One drawing shows the terminal methyl groups positioned opposite each other around the central carbon-carbon single bond. The other shows them separated by approximately 60 degrees.

The student claims that both pairs represent different compounds because the spatial arrangements are different.

Evaluate this claim. Explain why the two pairs should be classified differently.

Solution 4

The but-2-ene pair represents two distinct geometric stereoisomers. The butane pair represents two conformations of the same compound, rather than two different structural isomers.

Step 1: Compare the but-2-ene structures.

Both compounds have the molecular formula \(\ce{C4H8}\).

Their atoms have identical connectivity:

\[ \ce{CH3-CH=CH-CH3} \]

The difference is the relative position of the two methyl groups.

In cis-but-2-ene, they are on the same side of the double bond.

In trans-but-2-ene, they are on opposite sides.

Step 2: Explain why the difference matters.

The carbon-carbon double bond restricts rotation.

The two arrangements cannot interconvert simply by rotating around the double bond. Doing so would require breaking the additional bonding interaction associated with the double bond.

Therefore, cis-but-2-ene and trans-but-2-ene are distinct geometric stereoisomers.

Step 3: Compare the butane conformations.

Both drawings represent the same molecular formula:

\[ \ce{C4H10} \]

They also have identical connectivity.

The difference comes from rotation around a carbon-carbon single bond.

Unlike a double bond, a carbon-carbon single bond can rotate without breaking the bond.

This allows the molecule to move between different conformations.

Step 4: Evaluate the student’s claim.

The student’s mistake is assuming that every different spatial arrangement represents a different compound.

The distinction is whether the arrangement can change through ordinary rotation around a single bond.

The butane structures are different conformations of the same compound. The cis/trans structures of but-2-ene are distinct geometric stereoisomers because the double bond restricts rotation.

Question 5

A student is investigating possible isomers of \(\ce{C4H8}\) and prepares eight molecular drawings:

• A: But-1-ene, \(\ce{CH2=CHCH2CH3}\).
• B: Cis-but-2-ene, with both methyl groups on the same side of the double bond.
• C: Trans-but-2-ene, with the methyl groups on opposite sides of the double bond.
• D: 2-methylpropene, \(\ce{CH2=C(CH3)2}\).
• E: Cyclobutane, a four-carbon ring containing four \(\ce{CH2}\) groups.
• F: Methylcyclopropane, a three-carbon ring with a methyl group attached.
• G: Cyclobutene, a four-carbon ring containing one carbon-carbon double bond.
• H: The exact same drawing as B, but rotated through 180 degrees on the page.

The student claims that all eight drawings represent different isomers because they look different.

Determine:

1. Which candidates have the molecular formula \(\ce{C4H8}\)?
2. How many distinct compounds with this molecular formula are actually represented?
3. Which pair represents stereoisomers?
4. Which candidates have different connectivity?
5. Why must G and H be treated differently from the other candidates?

Assume all structures represent neutral molecules and apply the usual carbon valence rules.

Solution 5

Exactly six distinct compounds in the set have the molecular formula \(\ce{C4H8}\): A, B, C, D, E, and F.

Candidate G has the molecular formula \(\ce{C4H6}\), while H is simply another drawing of B. The student has therefore counted two candidates incorrectly.

Step 1: Check the open-chain structures.

Candidates A, B, and C are but-1-ene and the two geometric forms of but-2-ene.

Each contains four carbon atoms and eight hydrogen atoms.

Their molecular formula is:

\[ \ce{C4H8} \]

Candidate D, 2-methylpropene, also contains four carbon atoms and eight hydrogen atoms.

Its molecular formula is also:

\[ \ce{C4H8} \]

All four structures satisfy carbon valence.

For example, in 2-methylpropene, the central carbon has one double bond and two single bonds:

\[ 2+1+1=4 \]

It therefore satisfies carbon’s usual valence.

Step 2: Check the ring structures.

Candidate E is cyclobutane.

It contains four \(\ce{CH2}\) groups joined in a ring.

Its total hydrogen count is:

\[ 4\times2=8 \]

Therefore, its formula is \(\ce{C4H8}\).

Each carbon has two single bonds to neighbouring carbon atoms and two single bonds to hydrogen atoms.

Candidate F is methylcyclopropane.

It contains three carbon atoms in a ring and one carbon atom in a methyl substituent.

The hydrogen count is:

\[ 1+2+2+3=8 \]

The ring carbon attached to the methyl group has three carbon-carbon bonds and one carbon-hydrogen bond.

The other two ring carbons each have two carbon-carbon bonds and two carbon-hydrogen bonds.

All carbon atoms satisfy their usual valence.

Therefore, E and F are also valid structures with the molecular formula \(\ce{C4H8}\).

Step 3: Investigate candidate G.

Candidate G is cyclobutene.

It contains four carbon atoms, but it has both a ring and a carbon-carbon double bond.

The two carbon atoms in the double bond each have one hydrogen atom. The other two ring carbons each have two hydrogen atoms.

Therefore, its hydrogen count is:

\[ 1+1+2+2=6 \]

Its molecular formula is:

\[ \ce{C4H6} \]

Candidate G is a chemically valid structure, but it does not have the required molecular formula.

This is an important distinction. A structure can satisfy carbon valence and still fail the molecular formula test.

You can also understand the hydrogen count by starting with the open-chain saturated alkane \(\ce{C4H10}\).

• Forming one ring removes two hydrogen atoms.
• Introducing one double bond removes another two hydrogen atoms.

Therefore:

\[ \ce{C4H10}\rightarrow\ce{C4H6} \]

Cyclobutene cannot be an isomer of the \(\ce{C4H8}\) compounds in this question.

Step 4: Investigate candidate H.

Candidate H is the same drawing as B, rotated through 180 degrees on the page.

Rotating a drawing does not change the molecular structure.

The relative positions of the two methyl groups remain the same, so H still represents cis-but-2-ene.

It is not a new compound.

Step 5: Classify the six valid compounds.

The six distinct compounds with formula \(\ce{C4H8}\) are:

Candidates B and C have identical connectivity but different fixed spatial arrangements. They are stereoisomers.

Candidates A, D, E, and F have different connectivity arrangements. Each is a structural isomer of the others and of B and C.

Candidate H is identical to B, while G has a different molecular formula.

Therefore, there are six distinct compounds with the molecular formula \(\ce{C4H8}\) in the supplied set.

The key to solving this problem is applying the checks in the correct order. Counting different drawings before checking their formulas, connectivity, and spatial arrangements leads to incorrect classifications.

07What to check next

Once you have established that two compounds are isomers, the next question is how their structural differences affect their chemical and physical properties.

For example, ethanol and dimethyl ether have the same molecular formula but different functional groups. This affects their polarity, intermolecular forces, and physical properties.

Similarly, structural differences can affect how molecules interact and how they participate in chemical reactions.

The useful sequence is to establish the molecular formula, verify valence, identify the type of isomerism, and then investigate how that structural difference changes the compound’s behaviour.