Identify a molecular-ion peak

A molecular ion is the intact molecule after one-electron removal, M⁺•.

GCE A-Level H3 Chemistry 9813-2027

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Learning objectives

Identify a molecular-ion peak

Find the intact formula mass

A molecular ion is the intact molecule after one-electron removal, M⁺•.

For C₄H₁₀O its nominal m/z is 74 when singly charged.

Locate a molecular-ion candidate near the high-mass end, then test it against the molecular formula, isotope pattern and every proposed fragment. The highest visible peak is not automatically M⁺•.

Distinguish parent from fragments

The molecular ion retains every atom; m/z 59 and 31 in a C₄H₁₀O spectrum are fragments.

The highest abundant peak need not be M⁺•.

The molecular ion contains the same atoms as the original molecule and has lost one electron. Its nominal m/z therefore gives the molecular relative mass when the ion is singly charged.

Use isotope satellites correctly

M+1 is an isotopic partner of M, not a second molecular formula.

For a chlorine compound, M+2 can be an isotope partner rather than the parent.

A molecular-ion peak may be weak or absent when the radical cation fragments readily. Conversely, an isotope peak such as M + 1 can lie above M, so the rightmost small peak need not be the monoisotopic molecular ion.

Once M is assigned, subtract fragment m/z values to propose neutral losses, then check whether those losses and fragment formulas preserve the starting connectivity.

Assign C₄H₁₀O

Peaks 31,59,74 include M⁺• at 74 because its nominal mass matches the formula.

Peak 31 may be CH₂OH⁺ but cannot be the intact parent.

For C₄H₁₀O, M⁺• at m/z 74 matches 48 + 10 + 16. The m/z 59 peak is 74 − 15, consistent with loss of CH₃•, and the base peak at m/z 31 is consistent with CH₂OH⁺ or CH₃O⁺.

Try this

A compound of formula C₄H₁₀O gives peaks at m/z 31 (relative abundance 100), 59 (40), 74 (8) and 75 (0.3). Identify the molecular ion and explain the role of each other peak.

Check your answer

M⁺• is the weak peak at m/z 74, because C₄H₁₀O has nominal mass 4 × 12 + 10 × 1 + 16 = 74 and the molecular ion keeps every atom. The tiny peak at 75 is its ¹³C partner: four carbons give about 4 × 1.1% ≈ 4% of the M peak. m/z 59 is M − 15, a fragment after loss of CH₃•. The base peak at m/z 31 is the most abundant ion, a small fragment such as CH₂OH⁺, not the intact molecule.

Reject a contaminant

A small peak at m/z 91 cannot be M⁺• for supplied C₅H₁₀O, Mr 86.

Check calibration/contamination instead of choosing the largest m/z blindly.

Use the nitrogen rule only if it belongs to the stated course context and the ion type is appropriate; otherwise rely on the supplied formula and isotope/fragment evidence rather than an unsupported shortcut.

Try this

A sample of a C₅H₁₀O compound gives peaks at m/z 43 (relative abundance 100), 71 (35), 86 (6), 87 (0.3) and 91 (2). A student assigns m/z 91 as the molecular ion because it is the highest peak. Evaluate the assignment.

Check your answer

C₅H₁₀O has nominal mass 60 + 10 + 16 = 86, so M⁺• is the weak peak at m/z 86, and 87 is its ¹³C partner at about 5 × 1.1% of it. No ion from this molecule can be heavier than M apart from small isotope partners, and 91 is five units above M, which no likely isotope combination of C, H and O explains. The m/z 91 peak must come from another substance, so check the sample and the instrument background rather than choosing the highest m/z.

Recognise weak parent

A weak m/z 86 peak in a C₅H₁₀O spectrum can still be M⁺• while m/z 43 is base peak.

Intensity reports abundance, not intactness.

When two structural isomers share the same molecular ion, explain why M identifies molecular mass but not connectivity. Distinguish them with their fragment ions and complementary evidence.

Try this

Pentan-2-one and pentan-3-one are both C₅H₁₀O. One spectrum has peaks at m/z 43 (relative abundance 100), 58 (12), 71 (10) and 86 (15); the other has peaks at m/z 29 (75), 57 (100) and 86 (20). Assign the molecular ion and the base peak in each spectrum, and explain why the molecular ion alone cannot tell the isomers apart.

Check your answer

In both spectra M⁺• is at m/z 86 (60 + 10 + 16), and it is weaker than the base peak, which is a fragment. Because both isomers have the same formula, the molecular ion gives the molecular mass but not the connectivity. The base peaks differ: m/z 43 (CH₃CO⁺) fits pentan-2-one, which has a methyl group on the carbonyl carbon, and m/z 57 (C₂H₅CO⁺) fits pentan-3-one, whose carbonyl carbon carries two ethyl groups.

Common mistake: highest-peak rule

An isotope satellite or contaminant can lie above M; rapid fragmentation can make M weak.

Formula, cluster spacing and plausible fragments decide the parent.

The base peak is simply the most intense peak and is assigned relative abundance 100. It may be a fragment; it does not mean ‘peak containing the whole molecule’.

Try this

Correct this statement, giving two different counterexamples: ‘The peak with the highest m/z is always the molecular ion.’

Check your answer

The highest-m/z peak can lie above M: an isotope partner such as M + 1 from ¹³C or M + 2 from ³⁷Cl, or a peak from a contaminant, such as m/z 91 in a C₅H₁₀O sample. It can also lie below M: when the molecular ion fragments completely, as for many alcohols and branched alkanes, the highest peak is a fragment and no M⁺• peak appears. Assign M as the highest-mass peak that fits the molecular formula and is not an isotope partner, then check that the fragments are sensible losses from it.

Cross-check the whole spectrum

For C₅H₁₀O, M⁺• 86 and losses to 71 (−15) and 43 support a coherent assignment.

Next: isotope patterns.

A defensible assignment states the candidate M⁺• formula, checks nominal mass and isotope companions, and accounts for important fragments as balanced losses.

Try this

A student assigns a C₅H₁₀O spectrum as follows: m/z 86 is M⁺•; m/z 71 is formed by loss of CH₃•; m/z 43 is CH₃CO⁺, formed by loss of C₃H₇•. Check each mass and decide whether the assignment is coherent.

Check your answer

C₅H₁₀O has nominal mass 60 + 10 + 16 = 86, so M⁺• at m/z 86 is consistent. 86 − 71 = 15 matches CH₃•. CH₃CO⁺ has m/z 12 + 3 + 12 + 16 = 43, and 86 − 43 = 43 matches C₃H₇•. Every fragment is lighter than M and each neutral loss is a sensible radical, so the assignment is coherent; a methyl ketone such as pentan-2-one fits it.