Part 2: Mass spectra and isotope calculations

Use relative peak heights as weights, solve missing-abundance problems and predict the isotope combinations in a diatomic molecular ion.

What the axes tell you

A mass spectrum displays mass-to-charge ratio, m/z, horizontally and relative abundance or intensity vertically. Ions are detected, not neutral particles. For a singly charged atomic ion, the numerical m/z approximately equals its isotopic mass; ²⁵Mg⁺ gives m/z 25. A ²⁴Mg²⁺ ion instead has m/z 12 because its charge is two.

If an axis is normalised so the highest peak is 100, the heights do not necessarily add to 100. Divide a weighted total by the sum of all relevant peak heights. If isotope abundances are given as percentages summing to 100%, divide by 100. State the identity and charge of a peak, rather than just naming the element.

Calculate a weighted mean

Constructed magnesium data: the mass-24, mass-25 and mass-26 isotope peaks have relative heights 8.0, 1.0 and 1.0. Multiply each isotope mass by its own abundance, add, then divide by the total abundance. Aᵣ = [24(8) + 25(1) + 26(1)] ÷ 10 = 243 ÷ 10 = 24.3.

An unweighted average, (24 + 25 + 26) ÷ 3 = 25, incorrectly treats all three isotopes as equally abundant. A useful check is that the correct mean lies between 24 and 26 and is closest to the dominant mass-24 isotope. Use exact isotopic masses when supplied rather than replacing them with integers.

Work backwards from the mean

Constructed two-isotope sample: masses 10.0 and 11.0, with Aᵣ = 10.8. Let x be the fraction of mass-10 atoms. The other fraction is 1 − x, so 10x + 11(1 − x) = 10.8. Expanding gives 11 − x = 10.8; therefore x = 0.20. The abundances are 20% and 80%, consistent with a mean nearer 11.

For a missing isotope mass, first obtain its abundance. Suppose masses 40 and 44 have abundances 80% and 5%, and the mean is 40.50. The missing abundance is 15%. Let its mass be y: 40.50(100) = 40(80) + y(15) + 44(5). Hence 15y = 630 and y = 42.0. Check by substituting all three contributions back into the mean.

Pearson 8CH0/01 June 2023 Q1(c) required visible working for an unknown isotope calculation. Label the unknown, the weighting and the total; a plausible isotope guessed from the Periodic Table does not demonstrate the calculation.

Count both ways to make the mixed isotope molecule

Assume chlorine atoms have isotope probabilities p(³⁵Cl) = 0.75 and p(³⁷Cl) = 0.25 and combine randomly. For a Cl₂⁺ molecular ion, ³⁵Cl–³⁵Cl has probability 0.75² = 0.5625, the mixed pair has probability 2(0.75)(0.25) = 0.3750, and ³⁷Cl–³⁷Cl has probability 0.25² = 0.0625. The factor two counts light–heavy and heavy–light arrangements.

Their m/z values are 70, 72 and 74; the peak ratio is 9:6:1. This is a prediction for the molecular-ion cluster, not a claim that a whole chlorine spectrum has only three peaks. Atomic fragment peaks at 35 and 37 can also appear in a ratio near 3:1.

For two equally abundant bromine isotopes, ⁷⁹Br and ⁸¹Br, the analogous molecular peaks are at 158, 160 and 162 in a 1:2:1 ratio. For an unfamiliar diatomic element with isotope fractions p and q, use p²:2pq:q² and check that the probabilities add to (p + q)² = 1.

Swipe horizontally to view the whole diagram.

Use the molecular ion to find molecular mass

The molecular ion M⁺ is formed by removing an electron without breaking the molecule apart. For a singly charged molecular ion, its m/z gives the molecular mass represented by that peak. A molecular ion assigned to C₃H₈⁺ at m/z 44 is consistent with 3(12) + 8(1) = 44.

The tallest peak is the base peak: it may be a stable fragment rather than M⁺. Do not assume that the tallest peak gives Mᵣ. Nor is the largest m/z always the main molecular-ion mass: heavier isotopes give M + 1 or M + 2 peaks and an impurity can have a larger mass. Use the peak identified as the molecular ion and the supplied context. Detailed fragmentation is developed in Topic 7.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources