Why Mixtures Are Harder to Identify Using IR Spectroscopy

A sample produces a strong IR absorption near \(1715\ \mathrm{cm}^{-1}\) and a broad absorption around \(3350\ \mathrm{cm}^{-1}\). You might identify it as a hydroxyketone: one molecule containing both a carbonyl group, \(\ce{C=O}\), and an alcohol group, \(\ce{O-H}\).

But there’s another possibility. The sample could contain two separate compounds: a ketone and an alcohol. The ketone produces the carbonyl absorption, while the alcohol produces the O-H absorption.

Both explanations can produce the same functional-group evidence.

This is why mixtures can be harder to identify by IR than pure compounds. An IR spectrometer records the combined absorptions of all the IR-active substances in a sample. When their absorption bands overlap, merge, or hide weaker features, it becomes difficult to determine which compound produced which part of the spectrum.

The challenge isn’t simply recognising functional groups. It’s working out whether the observed pattern belongs to one compound or several.

01Why an IR spectrum becomes more complicated when substances are mixed

Imagine two musicians playing notes at almost exactly the same pitch. You hear a stronger combined sound, but you cannot necessarily distinguish the individual musicians.

Something similar happens when two substances absorb infrared radiation at similar wavenumbers.

Each substance contributes its own molecular vibrations to the spectrum. If two components absorb in different regions, their contributions may be relatively easy to distinguish. If they absorb at similar wavenumbers, their bands can overlap and appear as one feature.

The analogy has a limit: an IR spectrometer measures radiation absorption, not sound. However, the important idea is that overlapping signals can conceal the individual sources.

Here’s a useful prediction to make:

If a mixture contains two compounds that both have carbonyl groups, should its spectrum necessarily show two separate carbonyl peaks?

No. If their carbonyl absorptions occur at sufficiently similar wavenumbers, the bands may overlap. The resulting spectrum could show one stronger or broader absorption rather than two clearly separated peaks.

Notice that the mixture doesn’t produce a separate spectrum for each substance. It produces one spectrum containing the combined information.

A pure compound can also produce many absorption bands because its molecules have several different vibrational modes. The important difference is that all those bands originate from the same molecular species.

A mixture introduces additional molecular species, each contributing its own pattern.

02What the IR spectrometer actually measures

To understand the problem properly, it helps to picture the experiment.

Consider a simplified Fourier-transform infrared (FTIR) transmission arrangement.

The infrared radiation follows this path:

1. An infrared source produces radiation across a range of wavenumbers.
2. An interferometer processes the radiation before it reaches the sample.
3. The radiation passes through a sample cell containing the unknown material.
4. A detector measures the radiation that emerges.
5. A computer processes the signal and produces an IR spectrum.

A background measurement is also collected so that the instrument’s response can be accounted for.

What should happen if the sample contains two different compounds?

Suppose compound A absorbs strongly around \(1700\ \mathrm{cm}^{-1}\), while compound B absorbs strongly around \(1200\ \mathrm{cm}^{-1}\).

The detector receives less radiation in both regions. The resulting spectrum contains evidence of both compounds.

However, the detector doesn’t know which molecule absorbed the radiation. It only records the combined effect.

For a simple transmission measurement in which the components absorb independently, their absorbance contributions approximately add:

\[ A_{\mathrm{mix}}=A_A+A_B \]

Here:

• \(A_{\mathrm{mix}}\) is the total absorbance of the mixture.
• \(A_A\) is the absorbance contributed by compound A.
• \(A_B\) is the absorbance contributed by compound B.

This is an idealised model. It assumes the components do not interact significantly and that the measurement follows Beer-Lambert behaviour.

The model explains why overlapping bands can become stronger in a mixture without necessarily producing additional visible peaks.

Real mixtures can be more complicated. Hydrogen bonding, chemical interactions, and sample conditions can change the positions, widths, and relative intensities of absorption bands.

03Read both spectrum axes before identifying anything

Before interpreting a mixture spectrum, check the horizontal and vertical axes.

Misreading either one can lead to an incorrect interpretation.

The horizontal axis: wavenumber

The horizontal axis normally represents wavenumber, measured in reciprocal centimetres:

\[ \mathrm{cm}^{-1} \]

A typical organic IR spectrum runs from approximately \(4000\ \mathrm{cm}^{-1}\) on the left to \(400\ \mathrm{cm}^{-1}\) on the right.

The values decrease as you move from left to right.

For example:

• \(3400\ \mathrm{cm}^{-1}\) appears towards the left.
• \(1700\ \mathrm{cm}^{-1}\) appears around the middle.
• \(1100\ \mathrm{cm}^{-1}\) appears further towards the right.

Always check the actual axis labels rather than assuming the direction.

The vertical axis: percentage transmittance or absorbance?

This distinction becomes particularly important when interpreting overlapping bands.

An IR spectrum can display either percentage transmittance or absorbance.

On a percentage-transmittance spectrum, a deeper downward trough means stronger absorption.

On an absorbance spectrum, a taller upward peak means stronger absorption.

These are two different ways of displaying the same underlying measurement.

The mathematical relationship is:

\[ T=\frac{I}{I_0} \]

where:

• \(T\) is the fraction of radiation transmitted.
• \(I_0\) is the intensity of the incoming radiation.
• \(I\) is the intensity of the transmitted radiation.

Absorbance is related to transmittance by:

\[ A=-\log_{10}(T) \]

Remember that \(T\) is expressed as a fraction in this equation, not as a percentage.

This relationship explains why absorbance contributions add approximately for independent absorbers, while their transmitted fractions multiply.

Band position, width, and intensity all matter

A common mistake is to focus only on the position of an absorption band.

In a mixture, the shape and intensity can be just as important.

Consider three features:

• Position: The wavenumber at which absorption occurs. This helps identify the type of bond involved.
• Width: The range of wavenumbers over which absorption occurs. A broad band can overlap with several other absorptions.
• Intensity: How strongly the sample absorbs radiation. This depends on concentration and how strongly the particular vibration absorbs.

Two compounds can have absorption bands in similar regions but produce different band shapes.

Likewise, a strong absorption does not automatically indicate that a particular compound is present in a large concentration. Different molecular vibrations have different absorption strengths.

04Three reasons mixtures are particularly difficult to identify

1. Different compounds can have overlapping absorption bands

This is probably the most important problem.

Consider a mixture containing a carboxylic acid and a hydrocarbon.

The carboxylic acid contains:

• A carbonyl bond, \(\ce{C=O}\).
• A hydroxyl bond, \(\ce{O-H}\).
• Carbon-hydrogen bonds, \(\ce{C-H}\).

The hydrocarbon also contains carbon-hydrogen bonds.

The carboxylic acid produces a very broad O-H absorption, commonly extending across approximately \(2500\)-\(3300\ \mathrm{cm}^{-1}\).

The hydrocarbon produces C-H absorptions that can occur within this same general region.

Consequently, the two contributions can overlap.

A student searching for a neat, isolated O-H peak might incorrectly conclude that the sample contains no carboxylic acid.

The shape of the entire absorption region is much more informative than simply looking for a separate peak.

Here are some useful approximate absorption ranges:

These are approximate guide ranges. Use the reference data supplied with your examination question where applicable.

Notice how many of these regions overlap.

In particular, a carboxylic acid’s broad O-H absorption can overlap with C-H absorptions. An alcohol’s C-O absorption can also overlap with absorptions from other oxygen-containing compounds.

This is why you should interpret a combination of features rather than assigning every absorption to a separate substance.

2. A strong component can hide a weaker component

Imagine a mixture containing:

• A large amount of an ester.
• A small amount of an alcohol.

The ester may produce a strong carbonyl absorption around \(1740\ \mathrm{cm}^{-1}\).

The alcohol produces an O-H absorption in the \(3230\)-\(3550\ \mathrm{cm}^{-1}\) region.

If the alcohol is present in a sufficiently small concentration, its absorption may be relatively weak. It could also overlap with another feature in the spectrum.

This creates a problem.

You might correctly identify the ester but overlook the alcohol.

The absence of an obvious second set of bands does not automatically establish that the sample contains only one compound.

The intensity of an absorption depends on more than the amount of a substance present. The strength of the particular molecular vibration also matters.

This means you cannot reliably estimate the number of components by counting the visible peaks.

3. Different combinations of functional groups can produce similar spectra

Return to the original example.

A strong C=O absorption and a broad alcohol O-H absorption could indicate:

• One compound containing both functional groups.
• A mixture of a ketone and an alcohol.

Both explanations are chemically reasonable.

The distinction becomes clearer when you examine other parts of the spectrum, particularly the fingerprint region, which lies approximately between \(1500\) and \(500\ \mathrm{cm}^{-1}\).

This region contains many complex absorptions that can help distinguish individual compounds.

However, mixtures can also produce overlapping features in the fingerprint region.

A reference spectrum for each suspected component, additional chemical evidence, or a separation technique may be needed to identify the substances confidently.

Also remember that hydrogen bonding between different components can alter O-H band positions and widths. A mixture’s spectrum does not always look like a perfect overlay of the spectra of its isolated components.

07Sampling can make identification even harder

So far, we’ve assumed that the sample entering the instrument represents the entire mixture.

That isn’t always true.

Consider a heterogeneous mixture containing two different powders.

One region might contain mostly compound A, while another contains mostly compound B.

If the instrument measures a small portion of the sample, the resulting spectrum may depend on exactly where that portion was taken.

This is particularly relevant when using attenuated total reflectance (ATR), where the sample must make suitable contact with the instrument’s crystal.

A measurement taken at one location may therefore give a different spectrum from a measurement taken elsewhere.

Compare the two situations:

A homogeneous mixture can still produce a complicated IR spectrum because it contains several substances.

A heterogeneous mixture can produce different spectra depending on where the measurement is taken.

These are separate issues: mixture classification describes how substances are distributed, while IR spectroscopy examines their molecular absorptions.

If you need to revisit the distinction, see Heterogeneous Mixtures: How to Identify Them in HSC Chemistry.

For a heterogeneous sample, collecting spectra from several representative locations can help establish whether the observed differences reflect genuine compositional variation or simply inconsistent sample contact.

08A reliable method for interpreting mixture spectra

When you are given an IR spectrum of an unknown mixture, use the following process.

1. Check the axes.

Confirm that the horizontal axis shows wavenumber and determine whether the vertical axis represents percentage transmittance or absorbance.

Remember that downward troughs indicate absorption on a %T spectrum, while upward peaks indicate absorption on an absorbance spectrum.

2. Identify the strongest diagnostic features.

Look for characteristic absorptions such as C=O, O-H, and C-O.

Consider both the position and the shape of each band.

3. Look for overlapping absorptions.

Check whether two functional groups could absorb within the same region.

Pay particular attention to broad O-H absorptions and the C-H region.

4. Consider multiple possible explanations.

Do not assume that all observed functional groups belong to one molecule.

Test whether a combination of separate compounds could explain the same pattern.

5. Examine the fingerprint region.

Look for additional features that could distinguish the proposed components.

A single absorption is rarely enough to establish the identity of an entire compound.

6. Check the strength of your conclusion.

Consider whether a weaker component could be hidden, whether the sample was representative, and whether interactions could have changed the absorption bands.

If two different compositions explain the available evidence, acknowledge that IR alone has not established a unique answer.

If the sample is known to contain one pure compound and you need to choose between candidate organic structures, see How to Choose Between Organic Structures Using Two IR Features.

09Questions and solutions

Question 1

A liquid sample is known to contain ethanol, \(\ce{CH3CH2OH}\), and ethyl ethanoate, \(\ce{CH3COOCH2CH3}\).

Its percentage-transmittance IR spectrum contains:

• A strong downward absorption near \(1740\ \mathrm{cm}^{-1}\).
• A broad downward absorption near \(3350\ \mathrm{cm}^{-1}\).
• Several absorptions between \(1050\) and \(1300\ \mathrm{cm}^{-1}\).

Identify the most likely source of each feature. Explain why the absorptions between \(1050\) and \(1300\ \mathrm{cm}^{-1}\) are difficult to use to distinguish the two components.

Solution 1

The absorption near \(1740\ \mathrm{cm}^{-1}\) is most likely due to the ester carbonyl group, while the broad absorption near \(3350\ \mathrm{cm}^{-1}\) is consistent with the alcohol O-H group.

The absorptions between \(1050\) and \(1300\ \mathrm{cm}^{-1}\) can receive contributions from both compounds.

Step 1: Identify the carbonyl absorption.

Ethyl ethanoate contains a carbonyl group, \(\ce{C=O}\).

Its strong absorption near \(1740\ \mathrm{cm}^{-1}\) is consistent with an ester.

Ethanol does not contain a carbonyl group, so it cannot account for this feature.

Step 2: Identify the O-H absorption.

Ethanol contains an alcohol O-H bond.

Alcohol O-H absorptions are typically broad and occur around \(3230\)-\(3550\ \mathrm{cm}^{-1}\).

This makes ethanol the most likely source of the broad absorption near \(3350\ \mathrm{cm}^{-1}\).

Step 3: Interpret the C-O region.

Both ethanol and ethyl ethanoate contain C-O bonds.

Consequently, both can contribute absorptions between \(1050\) and \(1300\ \mathrm{cm}^{-1}\).

Their contributions may overlap, making it difficult to assign individual absorptions to a particular component.

Conclusion: The combined evidence supports the presence of an ester and an alcohol. The carbonyl and O-H features are particularly useful because they distinguish the two functional groups, whereas the shared C-O region is less diagnostic.

Question 2

An unknown mixture produces a percentage-transmittance IR spectrum with the following features:

• A broad downward absorption extending approximately from \(2500\) to \(3150\ \mathrm{cm}^{-1}\).
• A strong downward absorption at \(1712\ \mathrm{cm}^{-1}\).
• Several narrower absorptions between \(2850\) and \(3000\ \mathrm{cm}^{-1}\).

A student argues that the broad absorption must be caused entirely by C-H bonds. They therefore rule out the presence of a carboxylic acid because there is no separate O-H peak.

Evaluate this interpretation. What is the most defensible conclusion from the available evidence?

Solution 2

The student’s conclusion is not justified. A carboxylic acid O-H absorption can be extremely broad and overlap with C-H absorptions, so the absence of a separate, isolated O-H peak does not rule out a carboxylic acid.

Step 1: Examine the broad absorption.

Carboxylic acids produce a very broad O-H absorption, commonly extending across approximately \(2500\)-\(3300\ \mathrm{cm}^{-1}\).

This overlaps with the region where many C-H absorptions occur.

Therefore, the broad feature could contain contributions from both O-H and C-H bonds.

Step 2: Examine the carbonyl absorption.

The strong absorption at \(1712\ \mathrm{cm}^{-1}\) is consistent with a carbonyl group.

Carboxylic acids contain a carbonyl group, so this observation supports the possibility of an acid.

However, other carbonyl-containing compounds can also absorb in this region.

Step 3: Combine the evidence.

A strong carbonyl absorption together with a very broad feature extending through the O-H region is consistent with a carboxylic acid.

The narrower C-H absorptions may come from the acid itself or from another component in the mixture.

There is no need for the O-H absorption to appear as a separate, neatly defined peak.

Conclusion: A carboxylic acid is a plausible component of the mixture, but the available features do not prove its presence uniquely. The student’s mistake is treating overlapping absorptions as though each bond must produce an isolated peak.

Question 3

A mixture contains two compounds, P and Q. Both contain carbonyl groups, so their C=O absorptions overlap near \(1715\ \mathrm{cm}^{-1}\).

Each compound also has a separate diagnostic absorption:

• P has a distinctive absorption at \(1120\ \mathrm{cm}^{-1}\).
• Q has a distinctive absorption at \(820\ \mathrm{cm}^{-1}\).

Two aliquots are measured using the same transmission cell and path length.

The following absorbance measurements are obtained:

Assume that both diagnostic absorptions follow Beer-Lambert behaviour and that the measurement conditions are identical.

Determine how the relative concentrations of P and Q have changed.

A student also notices that the shared carbonyl absorption becomes stronger in aliquot 2. They conclude that Q must have increased in concentration. Explain why this conclusion is incorrect.

Solution 3

The concentration of P has doubled, while the concentration of Q has remained unchanged. The stronger shared carbonyl absorption does not prove that Q has increased because P also contributes to that absorption.

Step 1: Compare the diagnostic absorption for P.

Under the Beer-Lambert relationship, absorbance is proportional to concentration when the measurement conditions and path length remain constant.

Therefore:

\[ \frac{c_{P,2}}{c_{P,1}} = \frac{A_{P,2}}{A_{P,1}} \]

Substituting:

\[ \frac{c_{P,2}}{c_{P,1}} = \frac{0.30}{0.15} =2 \]

The concentration of P has doubled.

Step 2: Compare the diagnostic absorption for Q.

\[ \frac{c_{Q,2}}{c_{Q,1}} = \frac{0.30}{0.30} =1 \]

The concentration of Q has not changed.

Step 3: Interpret the shared carbonyl absorption.

Both compounds contain carbonyl groups.

Their contributions overlap near \(1715\ \mathrm{cm}^{-1}\), so the observed absorption represents the combined contributions of P and Q.

Increasing the concentration of P can therefore increase the total carbonyl absorbance even when Q remains unchanged.

The stronger absorption cannot be assigned exclusively to Q.

Conclusion: P has doubled in concentration, and Q has remained constant. The separate diagnostic absorptions allow us to distinguish their contributions. The shared carbonyl absorption alone cannot identify which component caused the increase.

Question 4

A heterogeneous powder mixture contains two organic compounds, A and B.

A technician collects ATR-FTIR spectra at three different locations on the same sample. After controlling the background measurement and maintaining similar contact pressure, the following patterns are observed:

• Location 1: strong carbonyl absorption and relatively weak O-H absorption.
• Location 2: strong broad O-H absorption and relatively weak carbonyl absorption.
• Location 3: both absorptions are clearly visible.

Repeated measurements at each location reproduce these patterns.

Explain the most likely reason for the differences between the spectra.

Describe how the technician should investigate the overall composition of the powder mixture before assigning one spectrum as representative of the entire sample.

Solution 4

The differences are most likely caused by non-uniform distribution of compounds A and B throughout the powder mixture. Each measurement samples a different local composition, so the spectra contain different relative contributions from the two compounds.

Step 1: Interpret the differences between locations.

Location 1 has a strong carbonyl absorption.

This suggests that the sampled region contains a relatively large contribution from compound A or another carbonyl-containing component.

Location 2 has a strong broad O-H absorption.

This suggests that the sampled region contains a relatively large contribution from compound B or another O-H-containing component.

Location 3 shows both features, indicating that both components contribute to the measured spectrum.

Step 2: Consider why ATR measurements can vary.

ATR-FTIR measurements depend on the material making contact with the crystal.

If the powder is unevenly distributed, different measurement locations can contain different proportions of the components.

Consequently, the spectrum may change depending on which particles are sampled.

Step 3: Design a more representative measurement.

The technician should:

• Collect spectra from several representative locations.
• Repeat measurements to check consistency.
• Where appropriate, homogenise a representative portion of the powder without changing its chemical composition.
• Compare the resulting spectra to determine whether the combined sample contains both characteristic sets of absorptions.

Conclusion: The three spectra are consistent with a heterogeneous mixture containing different local proportions of A and B. A single measurement should not be treated as the definitive spectrum of the entire sample.

The important distinction is between identifying the compounds present and obtaining a representative measurement of the bulk mixture.

Question 5

A laboratory records an IR spectrum of an unknown gas sample.

Its spectrum is normalised and compared with a reference spectrum of pure propanone vapour. The two spectra have matching absorption positions and matching relative band shapes.

Two possible explanations are proposed:

Model A: The sample contains only propanone vapour.

Model B: The sample contains propanone vapour mixed with nitrogen, \(\ce{N2}\), and oxygen, \(\ce{O2}\).

Assume the following:

• The propanone concentration is lower in Model B.
• The spectra have been independently normalised to the same maximum absorbance.
• Nitrogen and oxygen do not produce detectable absorptions in the measured IR region.
• Neither gas reacts with propanone.
• Pressure effects and interactions that could alter the propanone band shapes are negligible.
• The instrument has sufficient resolution to distinguish all relevant IR-active absorptions.

Can the IR spectrum distinguish Model A from Model B?

Explain why the apparently perfect match with pure propanone is insufficient to establish that the sample contains only one substance. Identify one independent analytical measurement that could help resolve the uncertainty.

Solution 5

No. Under the stated assumptions, IR spectroscopy cannot distinguish the two models using the normalised spectra. Nitrogen and oxygen do not produce detectable absorptions in the measured region, so the spectrum contains only the absorption pattern contributed by propanone.

This exposes a more fundamental limitation of IR spectroscopy: it cannot identify substances that do not produce detectable IR-active vibrations.

Step 1: Consider why propanone produces an IR spectrum.

Molecular vibrations can absorb infrared radiation when they involve a change in the molecule’s dipole moment.

Propanone contains polar bonds, including its carbonyl bond.

Its molecular vibrations produce detectable absorptions, giving rise to its characteristic spectrum.

Step 2: Consider nitrogen and oxygen.

Nitrogen and oxygen each exist as homonuclear diatomic molecules:

\[ \ce{N2} \]

\[ \ce{O2} \]

Because each molecule contains two identical atoms, stretching its bond does not produce a changing dipole moment.

Their vibrations are therefore IR-inactive under the model used here.

They do not contribute additional characteristic absorption bands to the measured spectrum.

Step 3: Consider the effect of dilution.

In Model B, propanone is present at a lower concentration.

Under Beer-Lambert conditions, reducing its concentration reduces its absorbance.

However, the question states that both spectra are independently normalised.

Normalisation removes the overall difference in signal strength, allowing their relative band patterns to match.

The nitrogen and oxygen contribute no additional detectable bands to distinguish Model B.

Step 4: Decide whether another measurement is needed.

The IR spectrum alone cannot establish whether nitrogen and oxygen are present.

A separate gas-composition measurement, such as gas chromatography using a suitable detector or mass spectrometry, could provide additional evidence about the substances in the sample.

If the absolute absorbance had been measured under calibrated conditions with a known path length, the lower propanone concentration might also have been detected. However, that would not, by itself, establish the identities of the additional gases.

Conclusion: Models A and B cannot be distinguished using the normalised IR spectra under the stated assumptions. A matching spectrum establishes that the detectable IR-active component has the same absorption pattern as propanone. It does not prove that every substance in the sample is propanone.

This is the key limitation to remember: IR spectroscopy provides evidence about molecular vibrations, not a complete inventory of every substance present.

When interpreting an unknown sample, the next decision is therefore whether you need to identify functional groups, identify individual components, or determine the actual composition of a mixture. IR spectroscopy is useful for the first task and can help with the second, but overlapping absorptions, hidden components, and IR-inactive substances may require an additional analytical technique.