How to Read and Compare IR Spectra of Petroleum Samples

A refinery analyst receives two unknown petroleum-derived liquids. Both produce strong infrared (IR) absorptions between 2960 and 2850 cm⁻¹. However, one sample also has distinct bands near 1600 and 1500 cm⁻¹.

Predict what this difference might tell you. Does the second sample contain a different type of hydrocarbon? Could you identify the exact molecules in either sample?

The extra bands provide evidence of aromatic ring structures. However, neither spectrum identifies a unique molecule. Petroleum samples contain mixtures of hydrocarbons, and many of their absorption bands overlap.

This is the central challenge when interpreting petroleum IR spectra: you need to recognise structural patterns without claiming more than the evidence supports.

Important: Every spectrum, peak intensity, and calibration dataset in this resource is synthetic teaching data. These examples illustrate how to interpret IR results; they are not measurements from actual petroleum samples.

01What does an IR spectrum actually measure?

Molecules are constantly moving. Their bonds stretch, bend, and vibrate, and different types of bonds vibrate at different frequencies.

When infrared radiation passes through a sample, certain frequencies are absorbed by the molecules. Absorption occurs when the radiation matches an allowed molecular vibration that involves a change in the molecule’s dipole moment.

For example:

• A carbon-hydrogen bond can undergo stretching vibrations.
• A carbon-carbon double bond can undergo stretching vibrations.
• A carbonyl group, C=O, has a characteristic stretching vibration.

The instrument records which infrared frequencies are absorbed. These absorptions appear as bands on an IR spectrum.

An important detail: a band indicates that a particular type of molecular vibration is occurring. It does not automatically identify the molecule responsible.

Understanding the axes

IR spectra usually plot wavenumber, measured in cm⁻¹, against either percentage transmittance or absorbance.

Wavenumber describes the spatial frequency of the radiation. A higher wavenumber corresponds to a higher frequency and greater energy.

Most conventional IR spectra display wavenumber decreasing from left to right, commonly from 4000 to 500 cm⁻¹.

There are two common ways to display the vertical axis:

• Percentage transmittance (%T): absorption bands point downwards because less radiation passes through the sample.
• Absorbance: absorption bands point upwards because greater absorption produces a larger absorbance value.

Don’t assume every spectrum uses the same vertical convention. Check the axis before interpreting the peaks.

The position of a band is often more useful for identifying a structural feature than its height. Its intensity and shape can provide additional evidence, but they need careful interpretation.

02Recognising the main structural features in petroleum samples

Petroleum consists mainly of hydrocarbons, although its exact composition depends on its source, refining, and subsequent treatment.

Common structural components include:

• Alkanes, containing only single carbon-carbon bonds.
• Cycloalkanes, containing saturated carbon rings.
• Alkenes, containing carbon-carbon double bonds.
• Aromatic hydrocarbons, containing aromatic rings.

Some petroleum-derived products also contain oxygen-, nitrogen-, or sulfur-containing compounds. Additives, contamination, and oxidation during storage can introduce further functional groups.

Each structural feature contributes characteristic absorption bands.

1. Aliphatic hydrocarbons: C-H stretching

Alkanes contain carbon-hydrogen bonds attached to saturated carbon atoms.

Their C-H stretching absorptions typically occur between 2960 and 2850 cm⁻¹.

A petroleum sample with strong absorptions across this region is likely to contain substantial aliphatic hydrocarbon structures.

However, this region does not distinguish one alkane from another. Straight-chain, branched-chain, and cyclic hydrocarbons can all contribute to the same general absorption region.

Additional evidence comes from bending vibrations:

• Around 1470-1450 cm⁻¹: commonly associated with CH₂ and CH₃ bending.
• Around 1380-1370 cm⁻¹: commonly associated with CH₃ bending.
• Around 720 cm⁻¹: can indicate rocking vibrations of CH₂ groups in longer, uninterrupted carbon chains.

The approximately 720 cm⁻¹ band needs particular care. Other vibrations, including aromatic C-H out-of-plane bending, can also occur in this lower-wavenumber region.

2. Aromatic hydrocarbons: characteristic ring absorptions

Aromatic hydrocarbons contain conjugated ring structures, such as a benzene ring.

Their IR spectra can show:

• Aromatic C-H stretching around 3100-3000 cm⁻¹.
• Aromatic ring stretching bands, commonly near 1600 and 1500 cm⁻¹.
• Additional aromatic C-H bending absorptions in the lower-wavenumber region.

The pair of bands near 1600 and 1500 cm⁻¹ is particularly useful when assessing whether a petroleum sample contains aromatic structures.

However, a single band near 1600 cm⁻¹ is not enough to identify a particular aromatic compound.

Look for supporting evidence from the rest of the spectrum.

3. Alkenes: carbon-carbon double bonds

Alkenes contain at least one carbon-carbon double bond.

Their characteristic features can include:

• C=C stretching around 1680-1620 cm⁻¹.
• =C-H stretching around 3100-3000 cm⁻¹.
• Additional out-of-plane bending absorptions below approximately 1000 cm⁻¹.

These features can overlap with aromatic absorptions.

For example, both alkenes and aromatic compounds can produce absorptions in the region above 3000 cm⁻¹. Aromatic ring vibrations can also complicate interpretation around 1600 cm⁻¹.

You should therefore use several bands together rather than treating every absorption as unique.

4. Oxygen-containing functional groups

Petroleum-derived samples can contain oxygenated compounds, particularly when additives, contamination, or oxidation products are present.

Two useful features are:

Carbonyl absorptions are often particularly useful because they are relatively strong.

For example, an oxidised petroleum-derived sample may develop carbonyl-containing compounds that produce an absorption near 1715 cm⁻¹.

This suggests the presence of carbonyl-containing structures. It does not establish which particular carbonyl compound is present.

For a refresher on individual functional-group absorption ranges, use the tutorgum IR functional-group guide alongside this resource. The focus here is how to interpret those features when several compounds contribute to the same spectrum.

04Comparing two petroleum-derived samples

Recognising one spectrum is useful. Comparing two spectra is often more revealing.

Suppose a laboratory analyses two petroleum-derived samples under identical conditions. Both spectra are scaled so that their strongest C-H stretching envelope has the same relative intensity.

The following synthetic data are obtained.

What differences would you expect between these samples?

Step 1: Identify what the samples have in common.

Both samples have strong absorptions in the aliphatic C-H stretching region.

This indicates that both contain substantial saturated hydrocarbon structures.

Step 2: Compare the aromatic features.

Sample B has stronger relative absorptions near 1600 and 1500 cm⁻¹.

These bands provide stronger evidence of aromatic ring structures in Sample B.

We can therefore reasonably infer that Sample B has a greater relative aromatic contribution to its overall structural pattern.

Step 3: Compare the aliphatic features.

Sample B also has stronger relative CH₂ bending and a stronger band near 720 cm⁻¹.

Together with its C-H stretching pattern, these features are consistent with a greater contribution from longer methylene sequences.

Sample A has a stronger CH₃ bending feature relative to its CH₂ bending feature. This may indicate a greater relative contribution from methyl groups, which can be associated with branching or shorter carbon chains.

However, these are structural inferences, not direct measurements of chain length or branching.

Step 4: Decide what cannot be concluded.

We cannot calculate the percentage of aromatic compounds in either sample from these relative intensities alone.

We also cannot conclude that Sample B contains a greater total mass or concentration of hydrocarbons.

The spectra have been scaled to comparable C-H intensities, and different molecular vibrations do not necessarily absorb infrared radiation with equal strength.

Conclusion: Sample B shows stronger relative evidence of aromatic structures and longer methylene sequences. Sample A shows a stronger relative methyl-group contribution. These observations allow a qualitative comparison, but not an exact determination of molecular composition.

05Why petroleum mixtures make identification difficult

Here is where interpreting petroleum spectra becomes less straightforward.

A pure organic compound may produce a recognisable combination of absorption bands. With a suitable reference spectrum, these features can help identify the compound.

Petroleum samples are different because they contain many compounds at once.

Imagine a sample containing:

• Several straight-chain alkanes.
• Branched-chain alkanes.
• Cycloalkanes.
• Aromatic hydrocarbons.
• Small amounts of oxygenated compounds.

Each component contributes its own molecular vibrations.

Unfortunately, different compounds can absorb infrared radiation at similar wavenumbers.

For example, several different alkanes contribute to the strong C-H stretching envelope between 2960 and 2850 cm⁻¹. Their individual absorptions overlap, producing a combined pattern rather than clearly separated peaks for every molecule.

The fingerprint region, approximately 1500-500 cm⁻¹, can become particularly complicated because many bending and skeletal vibrations occur there.

Why similar spectra do not prove identical compositions

Consider two petroleum mixtures.

Both contain aliphatic and aromatic hydrocarbons. Their spectra therefore show:

• Strong C-H stretching.
• Aromatic ring absorptions.
• Similar bending features.

Their overall spectra may look remarkably similar.

However, one mixture could contain a different combination of individual hydrocarbons from the other.

The same broad functional-group patterns do not require the same molecules.

This distinction is important:

• IR spectroscopy can provide evidence about the types of bonds and structural features present.
• IR spectroscopy does not necessarily identify every individual compound in a complex mixture.
• Similar spectra support similar broad structural characteristics, not necessarily identical chemical compositions.

For more detailed identification of individual components, another analytical technique may be needed.

For example, chromatography can separate mixture components before mass spectrometry analyses their molecular masses and fragmentation patterns. This provides a different type of evidence from IR spectroscopy.

See tutorgum’s Mass Spectrometry: Identifying and Measuring Water Pollutants for an explanation of how mass spectrometry contributes to chemical identification.

06Why the absence of a peak does not prove a compound is absent

Suppose an IR spectrum has no obvious absorption near 1715 cm⁻¹.

A student might conclude:

“The sample contains no carbonyl compounds.”

That conclusion is too strong.

A missing absorption tells us that the instrument has not resolved a detectable band in that region under the conditions used.

It does not establish that the corresponding structure is completely absent.

There are several possible explanations.

1. The concentration is too low

A compound may be present in such a small concentration that its absorption is too weak to distinguish from the baseline or instrumental noise.

This is particularly relevant when analysing petroleum mixtures containing trace additives or oxidation products.

2. Another absorption overlaps the band

A band may be hidden beneath a stronger absorption from another component.

The instrument records the combined absorption, so two overlapping features may appear as one broader band.

3. The vibration produces a weak absorption

Not every molecular vibration produces an equally strong IR signal.

The intensity depends partly on how much the molecular dipole moment changes during the vibration.

A weakly absorbing vibration may be difficult to identify even when the relevant structure is present.

4. Experimental conditions limit detection

Sample thickness, concentration, sample preparation, instrument resolution, and baseline quality can all affect whether a band is detectable.

The measured spectrum is evidence from a particular experiment, not an unlimited test for every possible compound.

How should you describe an absent band?

Use language that matches the evidence.

There is an important distinction between evidence against a structure and proof that the structure is absent.

For example, carbonyl absorptions are often strong. If a sample has no detectable absorption in the 1850-1650 cm⁻¹ region, substantial concentrations of carbonyl-containing compounds become less likely.

But trace carbonyl compounds may still be present.

07Qualitative interpretation versus quantitative analysis

So far, we have used IR spectra to identify broad structural features and compare samples.

This is called qualitative analysis.

Qualitative analysis answers questions such as:

• Does the sample contain evidence of aliphatic hydrocarbons?
• Are aromatic ring structures likely to be present?
• Is there evidence of carbonyl-containing compounds?
• Which sample shows stronger relative evidence of a particular structural feature?

Quantitative analysis asks a different question:

How much of a particular compound is present?

A stronger absorption does not automatically mean that a sample contains a greater concentration of a particular compound.

Absorption intensity can also depend on:

• Sample concentration.
• Path length through the sample.
• Sample thickness.
• Contact with an ATR crystal, when using ATR-FTIR.
• Sample preparation and baseline correction.
• The absorbing ability of the particular molecular vibration.

To obtain reliable concentrations, the instrument needs an appropriate calibration.

The Beer-Lambert law

For a suitable solution measured using a transmission cell, absorbance can be described by the Beer-Lambert law:

\[ A = \varepsilon lc \]

Where:

• \(A\) is absorbance, which is dimensionless.
• \(\varepsilon\) is the molar absorptivity, measured in L mol⁻¹ cm⁻¹.
• \(l\) is the path length through the sample, measured in cm.
• \(c\) is the concentration, measured in mol L⁻¹.

For a particular absorbing compound under suitable conditions, increasing its concentration increases its absorbance.

However, petroleum mixtures contain many compounds with different absorption characteristics.

The total absorption in a particular region can therefore reflect contributions from several different substances.

This is why a calibration based on a known compound cannot automatically measure the total concentration of every compound contributing to an overlapping band.

Using a calibration curve

A calibration curve is produced by measuring samples with known concentrations of the target compound.

The measured absorbances are plotted against their known concentrations.

An unknown sample can then be measured and its absorbance compared with the calibration relationship.

08Questions and solutions

The following questions progress from straightforward spectrum interpretation to more demanding problems involving overlapping absorptions and quantitative analysis.

All additional datasets are synthetic teaching examples.

Question 1

A petroleum-derived sample produces the following synthetic IR data:

(a) Identify the dominant structural features.

(b) Explain why the spectrum does not prove that the sample contains only alkanes.

Solution 1

(a) The sample is most consistent with an aliphatic-rich hydrocarbon mixture containing saturated structures.

The strong absorptions between 2960 and 2850 cm⁻¹ indicate aliphatic C-H stretching.

The bands near 1465 and 1377 cm⁻¹ provide supporting evidence of CH₂ and CH₃ bending vibrations.

Together, these features indicate substantial saturated hydrocarbon character.

(b) The spectrum cannot establish that the sample contains only alkanes because other compounds may be present below the detection limit or have overlapping absorptions.

The absence of a resolved aromatic band near 1600 cm⁻¹ provides no strong evidence of aromatic ring structures, but does not prove their complete absence.

Similarly, no resolved carbonyl absorption near 1715 cm⁻¹ means that carbonyl-containing compounds have not been detected clearly.

The correct conclusion describes the dominant structural pattern rather than claiming complete knowledge of the sample’s composition.

Question 2

Two petroleum-derived samples, P and Q, are analysed using the same IR method.

Their spectra are normalised to comparable C-H stretching intensities.

The synthetic results are:

A student makes two claims:

1. Sample P contains more aromatic structures than Sample Q.
2. Sample Q cannot be petroleum-derived because it contains oxygenated functional groups.

Evaluate both claims using the IR evidence.

Solution 2

The first claim is supported qualitatively, but the second claim is incorrect.

Sample P has stronger relative absorptions near 1600 and 1500 cm⁻¹. These bands are consistent with aromatic ring vibrations.

Because both samples have comparable C-H stretching intensities, the relative difference provides evidence that Sample P has a greater aromatic contribution to its structural pattern.

However, this does not establish an exact percentage of aromatic compounds.

The second claim confuses petroleum-derived mixtures with pure hydrocarbons.

Although petroleum consists mainly of hydrocarbons, petroleum-derived products can also contain oxygenated compounds.

For example, oxygenated compounds may arise through:

• Oxidation during storage or exposure to air.
• Additives incorporated during processing.
• Contamination during handling.
• Other chemical changes during refining or use.

Sample Q’s strong absorption near 1715 cm⁻¹ is consistent with carbonyl-containing compounds. Its absorption in the C-O region provides additional evidence of possible oxygenated structures.

However, these bands do not identify the specific oxygen-containing compounds.

The correct interpretation is that both samples contain substantial aliphatic hydrocarbon structures, while P has stronger relative aromatic features and Q has stronger evidence of oxygenated functional groups.

Question 3

A petroleum-derived sample produces the following synthetic IR observations:

• Strong absorptions between 2960 and 2850 cm⁻¹.
• No resolved absorption between 1850 and 1650 cm⁻¹.
• A weak, broad absorption between 3500 and 3200 cm⁻¹.
• A weak absorption between 1250 and 1050 cm⁻¹.

A laboratory report concludes:

“The sample contains no oxygen-containing compounds because no carbonyl absorption is present.”

Evaluate this conclusion.

Your response should:

(a) Explain what can reasonably be inferred from the missing carbonyl absorption.

(b) Discuss the significance of the two weaker absorption regions.

(c) Suggest one appropriate follow-up investigation.

Solution 3

The conclusion is not justified because the absence of a carbonyl absorption does not establish the absence of all oxygen-containing compounds.

(a) The missing carbonyl absorption suggests that substantial concentrations of detectable carbonyl-containing compounds are unlikely under the measurement conditions.

Carbonyl groups typically produce relatively strong absorptions between 1850 and 1650 cm⁻¹.

The absence of a resolved band therefore provides evidence against a significant detectable carbonyl contribution.

However, trace carbonyl compounds may still be present below the detection limit.

(b) The broad absorption between 3500 and 3200 cm⁻¹ could indicate O-H stretching.

This may be associated with alcohols or water contamination.

The weak absorption between 1250 and 1050 cm⁻¹ could be associated with C-O stretching.

Together, these features suggest possible oxygen-containing structures, although neither region uniquely identifies a functional group in a complex petroleum mixture.

Water contamination, overlapping absorptions, and other sample components must also be considered.

(c) A suitable follow-up investigation would be to dry the sample under appropriate conditions and repeat the IR analysis.

If the broad O-H absorption decreases substantially, water contamination may have contributed to the original signal.

If it remains, further investigation of possible oxygen-containing compounds would be appropriate.

A more selective analytical method could then be used to investigate their identities.

The important distinction is that the absence of one functional group’s characteristic absorption does not establish the absence of every other functional group.

Question 4

A laboratory develops a calibration curve to determine the concentration of a particular carbonyl-containing marker in a petroleum-like matrix.

The synthetic calibration data are:

An unknown sample produces an absorbance of 0.320.

(a) Calculate the gradient of the calibration relationship.

(b) Determine the concentration of the marker in the unknown sample.

(c) Explain why this concentration cannot automatically be interpreted as the total concentration of carbonyl-containing compounds in the petroleum sample.

Solution 4

(a) The gradient is 0.060 L mg⁻¹.

Using the first and last calibration points:

\[ m = \frac{0.380-0.020}{6-0} \]

\[ m = \frac{0.360}{6} \]

\[ m = 0.060\ \mathrm{L\,mg^{-1}} \]

The calibration equation is:

\[ A = 0.060c+0.020 \]

(b) The unknown sample contains 5.0 mg/L of the calibrated marker.

Substitute the measured absorbance:

\[ 0.320 = 0.060c+0.020 \]

Rearrange:

\[ c = \frac{0.320-0.020}{0.060} \]

\[ \boxed{c=5.0\ \mathrm{mg/L}} \]

The result indicates a marker concentration of 5.0 mg/L under the calibration conditions.

(c) The result does not automatically represent total carbonyl concentration because other carbonyl-containing compounds may contribute to the same absorption region.

The calibration curve was developed using a particular marker.

Other compounds may have different molar absorptivities and therefore produce different absorbances at the same concentration.

If their absorptions overlap, the measured band can represent the combined contributions of several compounds.

Consequently, the calibration gives a reliable marker concentration only when the method adequately distinguishes that marker from interfering components.

It cannot automatically quantify every carbonyl-containing compound in the mixture.

Question 5

A laboratory analyses two petroleum-derived mixtures, U and V.

Both contain two possible carbonyl-containing compounds, X and Y. Their absorptions overlap at approximately 1715 cm⁻¹, and neither compound has another resolved diagnostic absorption in the recorded spectrum.

Under identical experimental conditions, the laboratory establishes the following synthetic relationship:

\[ A_{1715}=0.050c_X+0.025c_Y+0.010 \]

Where:

• \(A_{1715}\) is the measured absorbance near 1715 cm⁻¹.
• \(c_X\) is the concentration of compound X in mg/L.
• \(c_Y\) is the concentration of compound Y in mg/L.
• 0.050 and 0.025 are the respective calibration response coefficients, in L/mg.
• 0.010 is the background absorbance.

The measured absorbances are:

A student uses a calibration curve prepared with pure compound X.

The student calculates:

• Sample U: 6 mg/L.
• Sample V: 4 mg/L.

They conclude that Sample U must contain a greater total concentration of carbonyl-containing compounds than Sample V.

Is this conclusion necessarily correct?

Demonstrate your reasoning by constructing a possible composition for each sample that reproduces the measured absorbances but reverses the student’s conclusion. Then explain what the laboratory can and cannot determine from the available IR measurements.

Solution 5

No. The student’s conclusion is not necessarily correct. Sample V could contain a greater total carbonyl concentration even though its measured absorbance is lower.

The problem is that compounds X and Y have different absorption response coefficients.

One mg/L of X produces a greater increase in absorbance than one mg/L of Y.

Therefore, the total absorbance does not depend only on the total concentration of carbonyl-containing compounds.

It also depends on the relative amounts of X and Y.

Step 1: Examine the student’s calculation.

The student uses the calibration relationship for pure X:

\[ A_{1715}=0.050c_X+0.010 \]

For Sample U:

\[ c_X=\frac{0.310-0.010}{0.050} \]

\[ c_X=6\ \mathrm{mg/L} \]

For Sample V:

\[ c_X=\frac{0.210-0.010}{0.050} \]

\[ c_X=4\ \mathrm{mg/L} \]

These calculations are mathematically correct if all the measured carbonyl absorption is attributed to X.

However, the samples may also contain Y.

The calculated values are therefore X-equivalent concentrations, not necessarily total carbonyl concentrations.

Step 2: Construct a possible composition for Sample U.

Suppose Sample U contains:

\[ c_X=6\ \mathrm{mg/L} \]

\[ c_Y=0\ \mathrm{mg/L} \]

Substitute into the full absorption relationship:

\[ A_{1715}=0.050(6)+0.025(0)+0.010 \]

\[ A_{1715}=0.310 \]

This exactly reproduces the measured absorbance of Sample U.

Its total carbonyl concentration is:

\[ c_{\mathrm{total}}=6\ \mathrm{mg/L} \]

Step 3: Construct a possible composition for Sample V.

Now suppose Sample V contains:

\[ c_X=0\ \mathrm{mg/L} \]

\[ c_Y=8\ \mathrm{mg/L} \]

Substitute:

\[ A_{1715}=0.050(0)+0.025(8)+0.010 \]

\[ A_{1715}=0.210 \]

This also exactly reproduces the measured absorbance.

However, its total carbonyl concentration is:

\[ c_{\mathrm{total}}=8\ \mathrm{mg/L} \]

Sample V therefore contains more total carbonyl-containing material than Sample U in this valid example.

Step 4: Explain why the available measurements cannot determine the actual compositions.

The laboratory has one measured absorbance for each sample, but two unknown concentrations in each mixture.

For Sample U, the available equation is:

\[ 0.300=0.050c_X+0.025c_Y \]

For Sample V:

\[ 0.200=0.050c_X+0.025c_Y \]

Each equation has two unknowns.

There are multiple possible combinations of X and Y that produce the same absorbance.

The single overlapping band therefore cannot uniquely determine both concentrations.

Step 5: Identify what additional evidence is required.

The laboratory needs another independent measurement that can distinguish X from Y.

Possible approaches include:

• Finding another IR absorption that responds differently to X and Y.
• Separating the compounds using chromatography before further analysis.
• Using a validated analytical method with suitable standards for both compounds.

Final conclusion: The student’s calculations give X-equivalent concentrations, not total carbonyl concentrations. Because X and Y absorb differently and their bands overlap, the measured absorbance cannot establish which sample contains more total carbonyl material.

This is an important limitation of IR spectroscopy in complex mixtures: a stronger band does not necessarily mean a greater total concentration of the functional groups contributing to it.

09Where to go next

When interpreting a petroleum IR spectrum, separate two questions:

1. What structural features are present? Use characteristic absorption regions and supporting bands to make qualitative inferences.
2. How much of a particular compound is present? Use an appropriate calibration and account for overlapping contributions from other compounds.

If the aim is to identify individual molecules in a complex petroleum mixture, IR spectroscopy may need to be combined with a separation technique and mass spectrometry.

The most reliable interpretation is not the one that assigns the most compounds to a spectrum. It is the one that makes the strongest conclusions the available evidence actually supports.