Edexcel A-Level Chemistry: Analytical Chemistry — Complete Revision Guide (9CH0)

Analytical chemistry is the topic where Edexcel 9CH0 becomes most like the real working life of a chemist. Synthesise a compound; how do you know you got the right molecule? Spectroscopy and chromatography are the answers. Mass spectrometry gives you the molecular formula and information about how the molecule fragments. Infrared spectroscopy gives you the functional groups present. NMR spectroscopy maps the carbon skeleton and the hydrogen environments. Chromatography separates mixtures and quantifies what is there.

Analytical chemistry is also one of the most pattern-driven topics on the specification. Once you know that a peak around 1715 cm^-1 in IR is a C=O stretch, that two NMR signals with a 3:2 ratio of integrations probably means three CH3 protons and two CH2 protons, and that an n+1 multiplet pattern reveals the number of neighbouring equivalent protons, every spectrum problem becomes a structured deduction.

This guide walks through the analytical content in 9CH0 topic by topic. It covers TOF mass spectrometry and fragmentation patterns; IR spectroscopy and the key wavenumbers; 1H and 13C NMR (chemical shift, integration, splitting); chromatography (TLC, column, GC, HPLC and Rf values); and the strategy for combining several spectra to identify an unknown molecule. For each topic you will find the core ideas, common pitfalls, a worked example and a link into the LearningBro Analytical Chemistry course.

What the Edexcel 9CH0 Specification Covers

Edexcel A-Level Chemistry (9CH0) is examined through Paper 1 (Inorganic and Physical, 1h45, 90 marks), Paper 2 (Organic and Physical, 1h45, 90 marks) and Paper 3 (General and Practical, 2h30, 120 marks). Analytical chemistry appears across all three papers, but most heavily on Paper 2 (organic structure determination) and Paper 3 (practical-based identification questions).

Sub-topic	Spec area	Typical paper weight
TOF mass spectrometry (revision)	Topic 1	3-5 marks
Fragmentation patterns	Topic 19	4-6 marks
IR spectroscopy and key wavenumbers	Topic 6 / 19	4-6 marks
1H NMR (chemical shift, integration)	Topic 19	6-10 marks
1H NMR splitting (n+1 rule)	Topic 19	4-6 marks
13C NMR	Topic 19	3-5 marks
Chromatography (TLC, column, GC, HPLC)	Topic 19	3-5 marks
Rf values	Topic 19	2-4 marks
Combined-spectra structure problems	Topic 19	8-12 marks

These weights are estimates, modelled on the Edexcel 9CH0 paper format. What is reliable is that a combined NMR + IR + mass spectrum problem — typically eight to twelve marks — appears on essentially every Paper 2.

TOF Mass Spectrometry (Revision)

Time-of-flight (TOF) mass spectrometry is covered in detail in atomic structure. For analytical purposes, the key facts are: ions are accelerated to the same kinetic energy and then drift through a field-free region, with lighter ions arriving sooner. The mass-to-charge ratio (m/z) of each ion gives its mass directly, since most ions are singly charged.

For a molecular sample, the rightmost significant peak is the molecular ion peak (M+). Its m/z gives the relative molecular mass. Smaller peaks at lower m/z are fragment ions formed when the molecular ion breaks apart.

A common pitfall is to identify the most abundant peak as M+. The most abundant peak is the base peak, which is often a fragment, not M+. M+ is the rightmost peak (sometimes accompanied by a small M+1 peak from 13C isotopes).

See the mass spectrometry lesson for fragmentation worked examples.

Fragmentation Patterns

Fragment ions reveal structural features. Common fragment masses you should recognise immediately:

m/z	Fragment lost or formed
15	CH3+
17	OH
18	H2O
28	CO or C2H4
29	CHO+ (aldehydes) or C2H5+
31	OCH3 or CH2OH+
43	C3H7+ or CH3CO+ (acetyl)
45	COOH+ or OC2H5
57	C4H9+
77	C6H5+ (phenyl, aromatic)

Specific fragmentation patterns identify functional groups. Aldehydes show a strong 29 peak (CHO+) and often a peak at M-29 (loss of CHO). Ketones lose the smaller alkyl group preferentially (the more stable acylium ion is favoured). Esters show characteristic peaks at 45 (-COOH+ or -OC2H5) and at the alkyl mass.

Worked example. A mass spectrum has M+ at 88 and a strong base peak at 43, with a smaller peak at 45. Possible structure? M+ 88 could be C4H8O2 (an ester or carboxylic acid). The 43 peak suggests CH3CO+ (acetyl) or C3H7+. The 45 peak suggests COOH+ or OC2H5. Together these point to ethyl ethanoate (CH3COOC2H5): the molecular ion (88) fragments to give CH3CO+ (43) by loss of OC2H5 (45). The combination is diagnostic of an ester.

A common pitfall is to attempt to identify a molecule from M+ alone — fragments are essential. Another is to forget that loss of 17 typically means -OH, while loss of 18 means -H2O. See the fragmentation lesson.

IR Spectroscopy

Infrared spectroscopy measures the absorption of IR radiation by chemical bonds. Each bond absorbs at a characteristic frequency (or wavenumber, in cm^-1) corresponding to its vibrational frequency. Stronger bonds absorb at higher wavenumbers; bonds to lighter atoms also absorb at higher wavenumbers.

The standard wavenumbers to memorise:

Bond	Wavenumber (cm^-1)	Notes
O-H (alcohol, free)	3500-3700	Sharp, weak
O-H (alcohol, H-bonded)	3200-3550	Broad, medium-strong
O-H (carboxylic acid)	2500-3300	Very broad, strong
N-H (amine)	3300-3500	Sometimes split (1°)
C-H (alkane)	2850-3100	Strong
C≡N (nitrile)	2200-2260	Sharp, medium
C=O (aldehyde, ketone, acid, ester)	1680-1750	Strong, sharp
C=C (alkene)	1620-1680	Variable
C-O (alcohol, ester, ether)	1000-1300	Strong, broad

The fingerprint region below 1500 cm^-1 contains many overlapping peaks unique to a specific molecule and is used for identification by matching against a reference. The functional group region above 1500 cm^-1 contains the diagnostic peaks listed above.

Worked example. An IR spectrum shows a broad strong absorption from 2500-3300 cm^-1 and a strong sharp absorption at 1715 cm^-1. Functional group? The broad O-H from 2500-3300 is diagnostic of a carboxylic acid; the C=O at 1715 is consistent. Together these identify -COOH.

A common pitfall is to ignore the difference between alcohol O-H (sharper, higher) and carboxylic acid O-H (much broader, extending to lower wavenumber). See the IR lesson for spectrum-reading practice.

1H NMR: Chemical Shift, Integration

Proton NMR maps the magnetic environments of hydrogen atoms in a molecule. Each chemically distinct hydrogen environment gives a peak at a specific chemical shift (δ, in ppm relative to TMS at 0 ppm).

Key chemical shifts to memorise (approximate; refer to a data book for exact ranges):

Environment	δ (ppm)
TMS (reference)	0.0
R-CH3 (alkyl)	0.7-1.6
R-CH2-R	1.2-2.0
R-CO-CH3 (methyl on ketone)	2.0-2.6
R-CH2-Cl, -Br	3.2-4.0
R-O-CH3 / R-O-CH2	3.3-4.5
R-CHO (aldehyde)	9.5-10.0
R-COOH	10-12
Aromatic H	6.5-8.0
O-H (alcohol, broad)	0.5-5.5 (variable)

The integration of each peak (the area under the peak, given as a number or trace) is proportional to the number of equivalent protons in that environment. So the integrations across all peaks add up to the total number of H atoms in the molecule.

Worked example. A compound C4H8O2 shows three 1H NMR signals: at δ 1.2 (3H), δ 2.0 (3H) and δ 4.1 (2H), all of which fit the integrations adding to 8H. The 1.2 (3H) is consistent with -CH3 on alkyl; 2.0 (3H) with CH3 on a carbonyl; 4.1 (2H) with -OCH2-. The molecule contains a CH3CH2- group connected to an oxygen, and a CH3CO- group. Structure: CH3COOCH2CH3 — ethyl ethanoate.

A common pitfall is to forget that chemically equivalent protons appear as a single peak. The three Hs of a CH3 group are equivalent; they show as one signal with integration 3. Another is to confuse integration ratios with the absolute numbers — they only give the ratio, which must be scaled to fit the molecular formula. See the 1H NMR lesson.

1H NMR Splitting: The n+1 Rule

NMR signals are usually split into multiplets by neighbouring non-equivalent protons. The number of peaks in the multiplet equals n+1, where n is the number of equivalent neighbouring protons (on adjacent atoms).

n (neighbouring H)	Multiplicity	Pattern
0	Singlet	One peak
1	Doublet	Two peaks (1:1)
2	Triplet	Three peaks (1:2:1)
3	Quartet	Four peaks (1:3:3:1)
4	Quintet	Five peaks

The intensity ratios in the multiplet follow Pascal's triangle.

Worked example. Ethanol (CH3CH2OH) shows three signals: -CH3 (3H, triplet at δ 1.2 because it has 2 H neighbours on the CH2), -CH2- (2H, quartet at δ 3.7 because it has 3 H neighbours on the CH3), and -OH (1H, broad singlet because the OH proton exchanges fast and does not couple normally; D2O addition makes this peak disappear).

A useful diagnostic is the D2O shake: adding D2O exchanges any -OH or -NH protons with deuterium, causing those peaks to disappear from the spectrum. This identifies which signals come from exchangeable protons.

A common pitfall is to apply n+1 to non-adjacent protons. The rule applies only to protons on directly bonded neighbouring carbons. Another is to forget that aromatic protons can give complex multiplets when not equivalent. See the splitting lesson.

13C NMR

Carbon NMR shows one peak per chemically distinct carbon environment. Chemical shifts span a wider range than 1H NMR (0 to ~220 ppm). The peaks are usually shown without splitting (proton-decoupled spectra), so the spectrum is simply a series of singlets giving the count and types of carbon environments.

Carbon environment	δ (ppm, 13C)
R-CH3 / R-CH2-R	0-50
R-CH-N or R-CH-O	30-90
C=C (alkene)	100-150
Aromatic C	110-160
C≡N	110-125
C=O (ester / acid)	160-185
C=O (aldehyde / ketone)	190-220

Worked example. Propanone (CH3COCH3) shows two 13C signals: one near δ 30 (the two equivalent methyl carbons) and one near δ 205 (the carbonyl carbon). The molecule has three carbons but only two environments because the methyls are equivalent.

A common pitfall is to expect the number of 13C signals to equal the number of carbons; symmetric molecules have fewer signals than carbons. See the 13C NMR lesson.

Chromatography: TLC, Column, GC and HPLC

Chromatography separates mixtures based on differential affinity between a stationary phase and a mobile phase. The components partition between the two phases as the mobile phase moves through; faster-moving components have lower affinity for the stationary phase.

Thin-layer chromatography (TLC) uses a silica or alumina coating on a plate as the stationary phase and an organic solvent as the mobile phase. Components travel different distances; visualisation by UV lamp or staining reagent shows the spots.

Column chromatography uses the same principle on a larger scale, with the stationary phase packed into a column and the mobile phase eluted through. Components can be collected as separate fractions for further analysis or use.

Gas chromatography (GC) uses a long capillary column with a high-boiling stationary liquid coated on the inside; an inert gas (helium or nitrogen) flows as the mobile phase. Components volatilise and travel through at different rates depending on their interaction with the stationary liquid. The detector at the end produces a chromatogram with peaks at characteristic retention times. Often coupled to MS (GC-MS).

High-performance liquid chromatography (HPLC) uses a packed column under high pressure with a liquid mobile phase. Suitable for non-volatile or thermally unstable analytes that cannot be analysed by GC.

The retention factor (Rf) in TLC is Rf = distance travelled by spot / distance travelled by solvent front. Rf values are characteristic of a compound under a specific solvent system and can be used to identify components by comparison with reference Rf values.

Worked example. A spot travels 4.2 cm; the solvent front travels 6.0 cm. Rf = 4.2 / 6.0 = 0.70.

A common pitfall is to mis-measure distances on a TLC plate (always measure from the original baseline, not from the bottom of the plate). Another is to assume Rf values are universal — they depend on solvent system and stationary phase. See the chromatography lesson.

Combined Spectra: Strategy for an Unknown Molecule

The most demanding analytical question gives you several spectra (mass, IR, 1H NMR, sometimes 13C NMR) and asks you to identify the molecule. The systematic approach:

1. Mass spectrum → use M+ to find the molecular formula (helped by degrees of unsaturation: DoU = (2C + 2 + N - H - X) / 2). Use fragment masses to identify likely functional groups.
2. IR spectrum → look for diagnostic functional groups (broad O-H, C=O, C-O, N-H, C≡N).
3. 1H NMR → count environments from the integrations; identify each environment from chemical shift; use multiplicity to map neighbouring protons.
4. 13C NMR (if given) → count carbon environments.
5. Combine all evidence into a candidate structure and check it against every spectrum.

Worked example. M+ at 60. IR has a broad, strong O-H from 2500-3300 cm^-1 and a sharp C=O at 1710. 1H NMR has signals at δ 1.2 (3H, triplet), 2.4 (2H, quartet) and 11 (1H, broad singlet). What is the molecule? Mr = 60 fits C2H4O2 (ethanoic acid). IR confirms COOH. NMR: triplet (3H) at 1.2 and quartet (2H) at 2.4 are an ethyl group (CH3CH2-); the broad singlet at 11 is the COOH proton. Wait — for ethanoic acid we would expect a singlet at 2.0 for the CH3 next to C=O, not an ethyl group. Re-examining: triplet 3H + quartet 2H + 1H broad fits propanoic acid CH3CH2COOH (Mr 74) better. Let us re-examine: M+ at 60 fits CH3COOH (ethanoic acid) better, in which case the 1H NMR should have only two signals: a singlet at δ 2.0 (CH3) and a broad singlet at δ 11 (COOH). The triplet/quartet pattern of an ethyl group does not fit M+ 60 with a COOH. So the data is internally inconsistent in this contrived example — in real questions Edexcel makes the data consistent. Always check that every piece of evidence is compatible with your proposed structure.

A common pitfall is to commit to a structure too early. Always verify against every spectrum before locking in. See the combined-spectra lesson for full worked deductions.

Common Mark-Loss Patterns

• Identifying the most abundant peak as M+ in mass spectrometry (M+ is the rightmost peak).
• Confusing IR alcohol O-H (sharper, higher) with carboxylic acid O-H (much broader, lower).
• Forgetting that 1H NMR equivalent protons give one signal (not separate signals per H).
• Mis-applying the n+1 rule to non-adjacent protons.
• Missing a D2O shake confirming -OH/-NH peaks.
• Counting carbons in 13C NMR from the molecular formula instead of distinct environments.
• Mis-measuring TLC distances from the bottom of the plate instead of the baseline.
• Failing to check a proposed structure against all the supplied spectra.
• Forgetting state symbols in mass-spectrometry equations (the ions are gas-phase).
• Quoting Rf values to too many significant figures (typically 2 sf).

How to Revise This Topic

• Memorise the IR wavenumber table cold. Recite it from memory three times a week.
• Build a 1H NMR chemical shift flashcard set with environment on one side and shift range on the other. Drill until automatic.
• Practise the n+1 rule by drawing out coupling patterns for ten compounds a day for a week.
• Drill combined-spectra problems under timed conditions. The strategy is more important than spotting individual features.
• Practise calculating M+ and degrees of unsaturation from molecular formulae. This is the entry point to most spectrum problems.
• Use the LearningBro practice quizzes to test under timed conditions.

Linking to Other Topics

Analytical chemistry is heavily synoptic. The molecules you analyse here are the products of reactions from organic foundations and organic advanced. Mass spectrometry first appeared in atomic structure for isotope analysis. The functional groups identified by IR and NMR are the same groups whose pH behaviour appears in acids and buffers. Practical chromatography and identification questions appear directly on Paper 3 alongside paper strategy techniques. Time spent here pays off across both organic-heavy papers.

Final Word

Analytical chemistry rewards memorisation of a small library of diagnostic features (key IR wavenumbers, NMR chemical shifts, common fragments) combined with disciplined deduction. Drill each spectrum type independently, then practise the combined-spectra extended response. The full LearningBro Analytical Chemistry course walks through every sub-topic with worked examples and AI tutor feedback. Get this section fluent and the structure-determination question on Paper 2 becomes a routine exercise.