Edexcel A-level Chemistry (9CH0) · Topic 7: Modern Analytical Techniques I
Mini-Lesson
Modern Analytical Techniques I
This mini-lesson covers Edexcel Topic 7: mass spectrometry of organic molecules — the molecular ion (M) peak, the M+1 peak, fragmentation and chlorine and bromine isotope patterns — and infrared spectroscopy of functional groups.
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Every calculation is worked through for you first. Press Start when you're ready.
Mass spectrometry
The molecular ion peak
In a mass spectrometer the molecule is ionised (usually by electron impact) to give the molecular ion, M+•:
CH3CH2OH + e− → CH3CH2OH+• + 2e−a high-energy electron knocks one electron out, leaving a positive radical cation
The molecular ion peak (the M peak) is at the highest m/z value (ignoring the small M+1 peak) and gives you the Mr of the compound directly.
There is always a small M+1 peak, because 1.1% of all carbon atoms are the 13C isotope.
The height of M+1 relative to M tells you how many carbon atoms there are.
number of carbon atoms ≈ (height of M+1 ÷ height of M) × 100 ÷ 1.1each carbon adds about 1.1% to the M+1 peak
High-resolution mass spectrometry measures m/z to 4 or 5 decimal places, so it can tell apart species with the same whole-number mass — e.g. C3H8 (44.062) and CO2 (43.990).
Calculate
Your turn — counting carbons
1In the mass spectrum of a hydrocarbon, the M peak has a height of 100.0 and the M+1 peak has a height of 6.6. How many carbon atoms does the molecule contain?
C atoms
Hint: The M+1 peak is (6.6 ÷ 100.0) × 100 = 6.6% of the M peak. Each carbon contributes about 1.1%, so divide 6.6 by 1.1.
Calculate
Your turn — counting carbons again
2In another spectrum the M peak has a height of 40.0 and the M+1 peak has a height of 1.32. How many carbon atoms does this molecule contain?
C atoms
Hint: M+1 as a percentage of M = (1.32 ÷ 40.0) × 100 = 3.3%. Now divide 3.3 by 1.1.
Fragmentation
Reading the fragment peaks
The molecular ion is unstable and some of them break apart — fragmentation. Only the positively charged fragment is detected; the neutral radical that is lost is invisible.
15 = CH3+ · 29 = C2H5+ or CHO+ · 43 = C3H7+ or CH3CO+
57 = C4H9+ · 77 = C6H5+ (a phenyl group)
A gap of 15 between two peaks means a CH3 group was lost; a gap of 29 means C2H5 or CHO.
Method: take the M peak as Mr, then subtract each fragment’s m/z from it. The differences tell you which groups fell off.
Quick check
Identify the fragment
3Propanone, CH3COCH3, has M = 58 and a very strong peak at m/z 43. Which ion causes the peak at 43?
Calculate
Your turn — predicting a fragment
4Butane has a molecular ion peak at m/z 58. It fragments by losing a methyl radical, •CH3. At what m/z does the resulting fragment ion appear?
m/z
Hint: The methyl radical has a mass of 12 + 3 = 15. The charged fragment keeps the rest: 58 − 15.
Isotope patterns
Chlorine and bromine give away their presence
Two halogens have isotopes abundant enough to show up clearly in the molecular ion region — and the ratio of the peak heights is the giveaway.
Chlorine:35Cl and 37Cl are in a 3 : 1 ratio (75% : 25%). One Cl atom gives M and M+2 in a 3 : 1 ratio.
Bromine:79Br and 81Br are in a 1 : 1 ratio (about 50% : 50%). One Br atom gives M and M+2 of roughly equal height.
Two chlorine atoms give M : M+2 : M+4 in a 9 : 6 : 1 ratio (from 0.75² : 2×0.75×0.25 : 0.25²).
Worked example — chloroethane, C2H5Cl
With 35Cl: Mr = (2 × 12) + (5 × 1) + 35 = 64 → the M peak
With 37Cl: 24 + 5 + 37 = 66 → the M+2 peak, one third as tall
Quick check
Which halogen?
5A compound gives molecular ion peaks at m/z 108 and m/z 110 of roughly equal height. What does this tell you?
Calculate
Your turn — the M+2 peak
6Chloroethane is C2H5Cl. Using Ar: C 12, H 1 and the isotope 37Cl, calculate the m/z of the M+2 peak.
m/z
Hint: Add up the masses using ³⁷Cl: (2 × 12) + (5 × 1) + 37.
Infrared
How infrared spectroscopy works
Covalent bonds vibrate (stretching and bending). A bond absorbs infrared radiation when the frequency matches its natural vibration frequency — and the frequency depends on the masses of the atoms and the strength of the bond. So every type of bond absorbs in a characteristic range of wavenumbers (cm−1).
Below 1500 cm−1 is the fingerprint region: too complex to interpret bond by bond, but unique to each compound, so a computer match against a database identifies the substance exactly.
Reading a spectrum
The two questions to ask
A carboxylic acid: a very broad O–H trough and a sharp, strong C=O spike.
When you are handed an IR spectrum, ask just two questions:
Is there a strong sharp trough near 1700? If yes there is a C=O — aldehyde, ketone, carboxylic acid, ester or amide.
Is there a broad trough above 2500? If yes there is an O–H. Broad and starting around 3300 = alcohol. Very broad, spread from 2500 to 3300 = carboxylic acid.
Both together (very broad O–H and C=O) = carboxylic acid. C=O with no O–H = aldehyde or ketone. O–H with no C=O = alcohol.
Quick check
Name the functional group
7An IR spectrum shows a very broad absorption from 2500 to 3300 cm−1and a strong sharp absorption at 1710 cm−1. Which functional group is present?
Quick check
Telling two compounds apart
8How would you use IR to tell ethanol (CH3CH2OH) apart from ethanal (CH3CHO)? Both have peaks near 2900 cm−1.
Calculate
Your turn — combine the evidence
9A compound has a molecular ion peak at m/z 60. Its IR spectrum shows a very broad absorption from 2500–3300 cm−1 and a strong peak at 1710 cm−1. It is a straight-chain carboxylic acid. How many carbon atoms does the molecule contain in total?
C atoms
Hint: The IR says carboxylic acid: CₙH₂ₙ₊₁COOH. Its Mₕ = 14n + 46. Set 14n + 46 = 60 to get n = 1 — then remember to add the carbon in the COOH group itself.
In the real world
Why infrared absorption matters beyond the lab
The same physics explains two things you will be asked about in synoptic questions:
The greenhouse effect. CO2, CH4 and H2O absorb infrared radiation emitted from the Earth’s surface, because their bonds vibrate at infrared frequencies. The energy is re-emitted in all directions, warming the atmosphere. N2 and O2 do not do this, because a symmetrical diatomic molecule’s vibration causes no change in dipole moment.
Breathalysers and exhaust-gas analysers. An infrared beam is passed through the sample; the absorbance at a wavenumber characteristic of a particular bond (for example the C–H bonds of ethanol near 2950 cm−1) is proportional to the concentration of that substance.
The rule: a bond only absorbs IR if the vibration changes the molecule’s dipole moment. That is exactly why the main components of air are not greenhouse gases.
Quick check
Greenhouse gases
10Why do CO2 and CH4 absorb infrared radiation, while N2 and O2 essentially do not?
Sort it
Which bond made that absorption?
Tap an absorption, then tap the bond that caused it.
🟫 O–H
🟩 C=O
🟪 C–H
Quick check
Why high resolution?
11A low-resolution mass spectrum gives a molecular ion at m/z 44. A high-resolution spectrum measures it as 44.0624. Why is that useful?
Match it
Match the clue to the deduction
Tap an item on the left, then its partner on the right.
Spectral clue
What it tells you
Recap
The big ideas to know
M peak: the highest m/z (ignoring M+1) — it gives Mr directly
M+1 peak: caused by 13C (1.1% abundant); number of C atoms = (M+1 ÷ M) × 100 ÷ 1.1
Fragmentation: only the cation is detected; the difference between peaks tells you what was lost (15 = CH3, 29 = C2H5 or CHO)