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22.2 Mass spectrometry

Syllabus
9701–2028–2029
Topic
22.2
Level
AS

Learning objectives

22.2.1Analyse mass spectra• Analyse mass spectra in terms of m/e values and isotopic abundances (knowledge of the working of the mass spectrometer is not required)22.2.2The atomic mass of an element• Calculate the relative atomic mass of an element given the relative abundances of its isotopes, or its mass spectrum22.2.3The molecular mass of an organic molecule• Deduce the molecular mass of an organic molecule from the molecular ion peak in a mass spectrum22.2.4Suggest the identity of molecules formed• Suggest the identity of molecules formed by simple fragmentation in a given mass spectrum22.2.5The number of carbon atoms, n, in a compound• Deduce the number of carbon atoms, n, in a compound using the [M + 1]+ peak and the formula n = 100 × abundance of [M + 1]+ ion 1.1 × abundance of M+ ion22.2.6The presence of bromine and chlorine atoms• Deduce the presence of bromine and chlorine atoms in a compound using the [M + 2] + peak

Read mass-spectrum position and height as different evidence

Spectrum feature Meaning
horizontal m/e value ion mass divided by its positive charge
vertical relative abundance amount of that ion relative to the chosen reference peak
base peak most abundant detected ion, assigned 100
peaks separated by isotope mass differences ions with the same structure but different isotopes

For the commonly shown singly charged ions, m/e equals the ion's nominal mass. Decide whether a peak belongs to the molecular ion, an isotopic partner or a fragment by combining its position with the rest of the pattern; peak height alone cannot assign identity.

The base peak is not automatically M⁺, and one m/e value can fit more than one possible formula. Instrument operation is explicitly not required; interpretation of mass/charge and abundance is.

Relative atomic mass is an isotope-abundance weighted mean

Ar=∑(isotopic mass×relative abundance)∑relative abundanceA_r=\frac{\sum(\text{isotopic mass}\times\text{relative abundance})}{\sum\text{relative abundance}}

Multiply every isotope mass by its abundance, add those contributions, then divide by the total abundance. If abundances are percentages, the denominator is 100; arbitrary relative peak heights may have a different total.

Ar=35×75+37×25100=35.5A_r=\frac{35\times75+37\times25}{100}=35.5

The weighted mean must lie between the lightest and heaviest isotope masses and closer to the more abundant isotope. Do not take an unweighted mean unless the abundances are equal.

The molecular-ion peak gives the intact molecule's mass

The molecular ion M⁺ is the intact molecule after loss of one electron. For charge +1, its m/e value equals the molecule's relative molecular mass, so identify M⁺ before reading the molecular mass.

Look for a plausible high-mass parent peak and its M+1 or M+2 isotopic partners, then check that lower-mass peaks can be fragments of that parent. The tallest peak is the base peak and need not be M⁺.

M+ at m/e=46⇒Mr=46(z=+1)M^+\text{ at }m/e=46\quad\Rightarrow\quad M_r=46\quad(z=+1)

An M+1 or M+2 isotope peak is not a second molecular mass, and a fragment such as m/e 31 cannot replace the identified parent. Some molecular-ion peaks are weak, so the full pattern must support the assignment.

Assign a simple fragment by mass, atoms, charge and complementary loss

  1. Calculate candidate formulas whose nominal mass equals the fragment m/e (for +1 ions). 2. Keep only candidates that can be cut from the given parent structure. 3. Subtract the fragment atoms/mass from M to identify a plausible neutral or radical loss. 4. Check that one fragment carries the positive charge and all parent atoms are accounted for.
Ethanol evidence Assignment Accounting
M⁺ at 46 C₂H₆O⁺ intact parent ion
peak at 31 CH₂OH⁺ 46 − 31 = 15, consistent with loss of CH₃•
peak at 29 C₂H₅⁺ 46 − 29 = 17, consistent with loss of •OH

A matching arithmetic mass is necessary but not sufficient: m/e 29 could represent different formulas in another molecule. Use the supplied parent connectivity and a chemically possible complementary loss rather than memorising one universal peak list.

Use the M+1/M abundance ratio to count carbon atoms

The [M+1]⁺ peak is mainly produced when one ¹²C in the molecular ion is replaced by naturally occurring ¹³C. Because ¹³C is about 1.1% of carbon, the [M+1]⁺ abundance grows approximately in proportion to the number of carbon atoms.

n=100×abundance of [M+1]+1.1×abundance of M+n=\frac{100\times\text{abundance of }[M+1]^+}{1.1\times\text{abundance of }M^+}

n=100×5.51.1×50=10n=\frac{100\times5.5}{1.1\times50}=10

Use peak heights or areas on the same relative-abundance scale; do not use their m/e positions. The result should be close to a chemically plausible whole-number carbon count.

Use the molecular-ion M⁺ abundance, not the base peak unless it happens to be M⁺. If the answer is far from an integer, recheck the M/M+1 assignment before rounding.

M and M+2 abundance ratios distinguish chlorine from bromine

One halogen atom Main isotope pair Approximate M : M+2 abundance Inference
chlorine ³⁵Cl : ³⁷Cl 3 : 1 chlorine present
bromine ⁷⁹Br : ⁸¹Br 1 : 1 bromine present

Find two molecular-ion cluster peaks separated by two m/e units and compare their heights. The +2 separation shows the isotope mass difference; the ratio identifies whether the element's heavier isotope is about one-third or about equal in abundance to the lighter isotope.

More than one Cl or Br atom produces a larger cluster (M, M+2, M+4, …), so a simple 3:1 or 1:1 pair is the one-atom pattern. Inspect the whole cluster before deciding.

An M+2 peak by itself does not identify the halogen. Use its abundance relative to M, and do not confuse this +2 isotope spacing with the +1 peak used for carbon counting.