Mass spectrometry and relative atomic massAQA A-Level Chemistry: Revision notes

A mass spectrometer gives accurate information about the relative isotopic mass and the relative abundance of each isotope in a sample. It can be used to identify elements and to determine relative molecular mass. The version in the specification is the time of flight (TOF) instrument. Ions are separated by their mass-to-charge ratio, m/z. For the simple spectra met here all ions have a 1+ charge, so m/z equals the relative mass of the ion.

The sample is vaporised and turned into positive ions. Two methods are used.

Electron impact (for elements and many small molecules): the gaseous sample is bombarded by high-energy electrons from an electron gun. Each particle loses one electron to form a 1+ ion.

X(g) + e⁻ → X⁺(g) + 2e⁻

Electrospray ionisation: the sample is dissolved in a volatile solvent and forced through a fine needle at high voltage. Each particle gains a proton, forming XH⁺(g), and the solvent evaporates.

X(g) + H⁺ → XH⁺(g)

Acceleration. An electric field accelerates the positive ions so that all ions have the same kinetic energy.

Ion drift. The ions then drift through a field-free flight tube. Since KE = ½mv², at the same kinetic energy a lighter ion has a greater speed. Time of flight is the distance divided by speed, so lighter ions reach the detector first.

Detection. Ions arriving at the detector gain electrons, producing a current proportional to the abundance of the ions.

Data analysis. A computer converts times of flight into m/z values and produces a mass spectrum of relative abundance against m/z.

Each peak in the spectrum of an element is a different isotope. The position (m/z) gives the isotopic mass, and the height gives the relative abundance. For example, the spectrum of chlorine has peaks at m/z 35 and 37 with abundances in the ratio about 3 : 1.

The pattern of peaks is characteristic of an element, so a spectrum can be used to identify an element by comparison with data. The molecular ion peak (the peak with the greatest m/z) gives the relative molecular mass of a molecule. With electrospray ionisation the peak is MH⁺, so Mr = m/z − 1.

Relative atomic mass (Ar) is the weighted mean mass of an atom of an element compared with 1/12 of the mass of an atom of ¹²C.

Ar = Σ(isotopic mass × abundance) ÷ Σ(abundance)

Worked example. Chlorine: ³⁵Cl 75.8%, ³⁷Cl 24.2%.

Ar = (35 × 75.8 + 37 × 24.2) ÷ 100 = 3548.4 ÷ 100 = 35.5

The result is not a whole number because it is an average over isotopes. Give the answer to the same number of significant figures as the data requires, usually 3 sf.