Chemistry Clarity · chemistryclarity.com

Infrared Spectroscopy and Mass Spectrometry

How can chemists identify a compound's functional groups and molar mass from its spectra?

IntermediateAnalytical ChemistryLast reviewed 6 October 2026
Download lesson (PDF)

What is it?

Spectroscopy and spectrometry let chemists identify a compound without a chemical test, using only a few milligrams.

  • Infrared (IR) spectroscopy shines infrared light through a sample. Covalent bonds vibrate (stretch and bend), and each type of bond absorbs IR at a characteristic wavenumber, measured in cm⁻¹. The absorptions show which functional groups are present.
  • Mass spectrometry (MS) turns molecules into positive ions, breaks some of them into fragments, and sorts the ions by their mass-to-charge ratio, m/zm/z. The heaviest ion usually gives the molar mass.

The wavenumber is the number of waves per centimetre, the reciprocal of the wavelength: ν~=1/λ\tilde{\nu} = 1/\lambda. A higher wavenumber means a higher frequency and more energy.

Key idea

IR answers “which functional groups?”; mass spectrometry answers “how heavy, and in what pieces?” Together with NMR, they let chemists work out the structure of almost any organic compound.

Why does it matter?

  • Identifying unknowns. Forensic laboratories, drug testing and quality control identify substances from their spectra.
  • Breathalysers measure ethanol in breath using its IR absorption.
  • Climate. Carbon dioxide, methane and water vapour absorb infrared radiation from the Earth, which is the greenhouse effect.
  • Space and medicine. Mass spectrometers analyse the atmospheres of planets, detect drugs in sport and identify proteins in disease.

How does it work?

1. Reading an IR spectrum

An IR spectrum plots transmittance (%) against wavenumber, with wavenumber decreasing from left (about 4000 cm⁻¹) to right (about 500 cm⁻¹). Absorptions appear as dips (“peaks” pointing down).

BondWhere foundWavenumber (cm⁻¹)Appearance
O–Halcohols3200–3550strong, broad
O–Hcarboxylic acids2500–3300very broad
N–Hamines, amides3300–3500medium
C–Halmost all organic compounds2850–3100medium–strong
C≡Nnitrilesabout 2250medium, sharp
C=Oaldehydes, ketones, acids, esters1680–1750strong, sharp
C=Calkenes1620–1680weak–medium
C–Oalcohols, esters, ethers1000–1300strong

The region below about 1500 cm⁻¹ contains many overlapping absorptions. It is called the fingerprint region: it is unique to each compound, so it can be matched against a database to confirm an identity.

2. Reading a mass spectrum

A mass spectrum is a bar chart of relative abundance against m/zm/z (for ions with a 1+ charge, m/zm/z is simply the mass).

  • The molecular ion, MX+\ce{M+}, is the peak at the highest m/zm/z (ignoring small isotope peaks). Its m/zm/z equals the molar mass.
  • The tallest peak is the base peak (set to 100 %). It comes from the most stable fragment.
  • Fragments reveal pieces of the molecule: m/zm/z 15 (CHX3X+\ce{CH3+}), 29 (CX2HX5X+\ce{C2H5+} or CHOX+\ce{CHO+}), 31 (CHX2OHX+\ce{CH2OH+}), 43 (CX3HX7X+\ce{C3H7+} or CHX3COX+\ce{CH3CO+}), 45 (COOHX+\ce{COOH+}).
  • Isotope patterns: a compound with one Cl atom shows peaks at M and M+2 in a 3 : 1 ratio (X35X2235Cl\ce{^{35}Cl} and X37X2237Cl\ce{^{37}Cl}); one Br atom gives M and M+2 in a 1 : 1 ratio.

Think of it like this

IR spectroscopy is like listening to an orchestra: each type of bond “plays” at its own pitch, so you can hear which instruments (functional groups) are present. Mass spectrometry is like weighing a vase and then the pieces it breaks into: the whole vase gives the total mass, and the pieces hint at how it was put together.

More precisely

A bond absorbs IR only if its vibration changes the dipole moment of the molecule, which is why NX2\ce{N2} and OX2\ce{O2} do not absorb but COX2\ce{CO2} and HX2O\ce{H2O} do. Stronger bonds and lighter atoms vibrate faster, so they absorb at higher wavenumbers: C≡C absorbs higher than C=C, which absorbs higher than C–C. In MS, the most common method (electron impact) knocks an electron out of the molecule; the small M+1 peak comes mainly from carbon-13 (1.1 % of carbon atoms), and high-resolution MS measures masses to four decimal places, enough to fix the molecular formula.

Visualise it

Top: schematic IR spectrum of ethanoic acid, transmittance against wavenumber from 4000 to 500 per centimetre, with a very broad O–H absorption from 2500 to 3300, a C–H dip near 2950, a strong sharp C=O dip near 1710 and a C–O dip near 1250, and the fingerprint region below 1500 marked. Bottom: schematic mass spectrum of propanone with bars at m/z 15, 43 (the base peak, 100 per cent) and 58 (the molecular ion).
Top: IR spectrum of ethanoic acid. Bottom: mass spectrum of propanone. (Schematic.)

Worked example

Worked example: Wavenumber and wavelength

Question: A C=O absorption appears at 1715 cm⁻¹. What is its wavelength in micrometres?

λ=1ν~=11715 cm−1=5.831×10−4 cm=5.831×10−4 cm×104 μm1 cm=5.83 μm\small\begin{aligned} &\lambda = \frac{1}{\tilde{\nu}} = \frac{1}{1715\ \text{cm}^{-1}} \\[4pt] &= 5.831 \times 10^{-4}\ \text{cm} \\[4pt] &= 5.831 \times 10^{-4}\ \cancel{\text{cm}} \\[4pt] &\quad \times \frac{10^{4}\ \mu\text{m}}{1\ \cancel{\text{cm}}} = 5.83\ \mu\text{m} \end{aligned}

Worked example: Alcohol, acid or ketone?

Question: Three compounds give these IR absorptions. Identify the functional group in each. (a) broad 3350 cm⁻¹, 1050 cm⁻¹, nothing near 1700 cm⁻¹ (b) very broad 2500–3300 cm⁻¹ and strong 1710 cm⁻¹ (c) strong 1715 cm⁻¹, no absorption above 3100 cm⁻¹

  1. (a) O–H (alcohol) and C–O, no C=O: an alcohol.
  2. (b) Very broad acid O–H plus C=O: a carboxylic acid.
  3. (c) C=O but no O–H: a ketone or aldehyde (or ester).

Worked example: Average atomic mass from a mass spectrum

Question: The mass spectrum of chlorine atoms shows X35X2235Cl\ce{^{35}Cl} (34.969, 75.76 %) and X37X2237Cl\ce{^{37}Cl} (36.966, 24.24 %). Calculate the relative atomic mass of chlorine.

Ar=34.969×0.7576+36.966×0.2424=35.45\small\begin{aligned} &A_\text{r} = 34.969 \\[4pt] &\quad \times 0.7576 + 36.966 \times 0.2424 \\[4pt] &= 35.45 \end{aligned}

The 3 : 1 ratio of the isotopes is why chlorine compounds show M and M+2 peaks in a 3 : 1 ratio.

Worked example: Putting IR and MS together

Question: An unknown has the formula CX3HX6O\ce{C3H6O}. Its IR spectrum shows a strong absorption at 1715 cm⁻¹ and none at 3200–3550 cm⁻¹. Its mass spectrum shows MX+\ce{M+} at m/zm/z 58 and the base peak at m/zm/z 43. Identify it.

  1. MX+\ce{M+} = 58 matches CX3HX6O\ce{C3H6O} (58.08 g/mol).
  2. IR: C=O present, no O–H: an aldehyde or a ketone.
  3. MS: a base peak at 43 is CHX3COX+\ce{CH3CO+}, formed when the molecule loses CHX3\ce{CH3} (58 − 15 = 43).
  4. The compound is propanone, CHX3COCHX3\ce{CH3COCH3}. (Its isomer propanal would give a strong peak at 29, CHOX+\ce{CHO+} or CX2HX5X+\ce{C2H5+}.)

Common mistake

Common mistake: Reading the wavenumber axis the wrong way

IR spectra are drawn with wavenumber decreasing from left to right. An O–H band near 3300 cm⁻¹ is on the left; C=O near 1700 cm⁻¹ is in the middle.

Common mistake: Taking the tallest peak as the molar mass

The tallest peak is the base peak, usually a fragment. The molar mass is given by the molecular ion, the peak at the highest m/zm/z (apart from small isotope peaks).

Common mistake: Trying to identify everything in the fingerprint region

Below 1500 cm⁻¹ the absorptions overlap. Use the fingerprint region to match a known spectrum, and the region above 1500 cm⁻¹ to identify functional groups.

Notation note

  • Wavenumber: ν~\tilde{\nu}, in cm⁻¹ (read “per centimetre” or “reciprocal centimetres”).
  • MX+\ce{M+}: molecular ion; M+2: an ion 2 mass units heavier, usually from an isotope.

Remember this

Remember this

  • IR: bonds absorb at characteristic wavenumbers. Broad 3200–3550: alcohol O–H. Very broad 2500–3300: acid O–H. Strong 1680–1750: C=O.
  • Below 1500 cm⁻¹: fingerprint region, unique to each compound.
  • MS: molecular ion (highest m/z) = molar mass; base peak = tallest; fragments show pieces.
  • Cl: M and M+2 in 3 : 1; Br: 1 : 1.
  • Combine: MS gives the mass, IR the functional groups, NMR the carbon–hydrogen framework.

Test yourself

Check your understanding before moving on.

Flashcards

10 cards

  1. Question
    What does an IR spectrum show?
    Answer

    Which bonds (functional groups) are present: each bond absorbs IR at a characteristic wavenumber.

  2. Question
    What is a wavenumber?
    Answer

    The number of waves per centimetre, 1/λ, in cm⁻¹. Higher wavenumber = higher energy.

  3. Question
    Where does a C=O bond absorb in IR, and what does it look like?
    Answer

    1680–1750 cm⁻¹, strong and sharp.

  4. Question
    How do the O–H absorptions of alcohols and carboxylic acids differ?
    Answer

    Alcohol: broad, 3200–3550 cm⁻¹. Acid: very broad, 2500–3300 cm⁻¹.

  5. Question
    What is the fingerprint region?
    Answer

    Below about 1500 cm⁻¹: complex, unique to each compound, used to match a known spectrum.

  6. Question
    In a mass spectrum, what is the molecular ion?
    Answer

    M⁺, the peak at the highest m/z (ignoring small isotope peaks); its m/z equals the molar mass.

  7. Question
    What is the base peak?
    Answer

    The tallest peak (100 %), from the most stable fragment.

  8. Question
    Give the ions for fragments at m/z 15, 29 and 43.
    Answer

    15: CH₃⁺. 29: C₂H₅⁺ or CHO⁺. 43: C₃H₇⁺ or CH₃CO⁺.

  9. Question
    What M : M+2 ratio shows one Cl atom? One Br atom?
    Answer

    Cl: 3 : 1. Br: 1 : 1.

  10. Question
    Why do N₂ and O₂ not absorb IR?
    Answer

    Their vibration does not change the dipole moment of the molecule.

Quiz

7 questions

  1. Question 1EasyA strong, sharp IR absorption at 1715 cm⁻¹ indicates which bond?
    Show answer

    Answer: C=O

    C=O bonds absorb strongly at 1680–1750 cm⁻¹ (aldehydes, ketones, acids, esters).

  2. Question 2EasyAn IR spectrum shows a broad absorption at 3350 cm⁻¹ and none near 1700 cm⁻¹. The compound is most likely:
    Show answer

    Answer: an alcohol

    A broad band at 3200–3550 cm⁻¹ is an alcohol O–H. No C=O rules out ketones and acids.

  3. Question 3EasyIn a mass spectrum, the peak at the highest m/z (ignoring tiny isotope peaks) is the:
    Show answer

    Answer: molecular ion

    The molecular ion M⁺ is the whole molecule minus one electron; its m/z gives the molar mass. The base peak is the tallest peak.

  4. Question 4MediumA compound shows peaks at m/z 78 and 80 in a 3 : 1 ratio. What does it contain?
    Show answer

    Answer: One chlorine atom

    ³⁵Cl and ³⁷Cl occur in about a 3 : 1 ratio, giving M and M+2 peaks 2 units apart in that ratio. One Br would give 1 : 1.

  5. Question 5MediumPropanone (M = 58) shows its base peak at m/z 43. Which fragment is lost?
    Show answer

    Answer: CH₃ (15)

    58 − 43 = 15: loss of a methyl group leaves CH₃CO⁺ at m/z 43.

  6. Question 6HardWhat is the wavelength of IR radiation with a wavenumber of 3300 cm⁻¹?
    Show answer

    Answer: 3.03 μm

    λ = 1/3300 cm⁻¹ = 3.03 × 10⁻⁴ cm = 3.03 μm (1 cm = 10⁴ μm).

  7. Question 7MediumWhich pair of IR absorptions would best identify a carboxylic acid?
    Show answer

    Answer: Very broad 2500–3300 cm⁻¹ and strong 1700–1725 cm⁻¹

    A carboxylic acid has both an O–H (very broad, overlapping the C–H region) and a C=O. The first pair is an alcohol; 2250 is C≡N; 1650 + 3050 suggests an alkene.

Notes and downloads

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/ir-and-mass-spectrometry/

Spotted a mistake? Let us know and we'll fix it.