Μass spectrometry is based on the formation of a beam of ionic fragments by bombardment of test mole — Physical Chemistry — Kinetics Chemistry Question

Μass spectrometry is based on the formation of a beam of ionic fragments by bombardment of test molecules usually with energetic electrons. The generated fragments are then separated by application of electrostatic or magnetic fields or by a combination of both. This separation depends on the mass-to-charge ratio (m/z) of each ionic fragment. In most cases fragments are singly charged (z = 1), therefore the separation depends on the mass of each ion.

The capability of a mass spectrometer to differentiate between masses is usually expressed in terms of its resolution, which is defined as R = m / Δm, where Δm is the mass difference between two adjacent peaks that are just resolved and m is the nominal mass of the first peak. For example, in order to discriminate the ionic species C₂H⁴⁺ and CH₂N⁺, which have the same nominal mass (m = 28), but different exact masses (28.0313 and 28.0187, respectively), an instrument with a resolution of at least R = 28 / (28.0313 – 28.0187) ≈ 2200 is required. Less expensive low resolution mass spectrometers (R ≈ 300 – 1000) can readily differentiate simple ions (of relatively low relative mass) of different nominal masses.

Isotope peaks in Mass Spectrometry
Even with the low resolution mass spectrometers, the same ionic fragment can generate multiple adjacent peaks of different nominal mass attributable to ions having the same chemical formula but different isotopic compositions. For example, the ion CH³⁺ consists of fragments of nominal mass ranging from 15 (fragment ¹²C ¹H³⁺) up to 19 (fragment ¹³C ²H³⁺).

The relative intensity of isotope peaks depends on the natural isotopic composition of each element. For C the per cent natural isotopic abundance is 98.90 % of ¹²C and 1.10 % of ¹³C, and for H 99.985 % of ¹H and 0.015 % of ²H. Therefore, the more intense peak (M=15) is attributed to the more abundant ¹²C ¹H³⁺, the next in intensity but much smaller than peak M (M⁺1=16) is attributed to both ¹³C ¹H³⁺ and ¹²C 1H₂ ²H⁺, whereas peak M⁺4, attributed to ¹³C ²H³⁺, has practically zero intensity due to the extremely low probability of occurrence.

Below is shown how the relative intensities of mass peaks for the ionic fragment CH₂Cl⁺ can be exactly (without approximations) calculated, taking into consideration the isotopic abundance for C, H and Cl (75.77 % of ³⁵Cl and 24.23 % of ³⁷Cl).

Fragment M = 49
¹²C 1H₂ ³⁵Cl : 0.989 × (0.99985)^2 × 0.7577 = 0.7491

Fragments M⁺1 = 50
¹³C 1H₂ ³⁵Cl : 0.011 × (0.99985)^2 × 0.7577 = 0.00833
¹²C ²H ¹H ³⁵Cl : 0.989 × 0.00015 × 0.99985 × 0.7577 = 0.00011
¹²C ¹H ²H ³⁵Cl: 0.989 × 0.99985 × 0.00015 × 0.7577 = 0.00011
Sum = 0.00855

Fragments M⁺2 = 51
¹³C ²H ¹H ³⁵Cl : 0.011 × 0.00015 × 0.99985 × 0.7577 = 1.25 × 10^-6
¹³C ¹H ²H ³⁵Cl : 0.011 × 0.99985 × 0.00015 × 0.7577 = 1.25 × 10^-6
¹²C 1H₂ ³⁷Cl : 0.989 × (0.99985)^2 × 0.2423 = 0.240
Sum = 0.240

Fragments M⁺3 = 52
¹³C 2H₂ ³⁵Cl : 0.011 × (0.00015)^2 × 0.7577 = 1.9 × 10^-10
¹³C 1H₂ ³⁷Cl : 0.011 × (0.99985)^2 × 0.2423 = 2.66 × 10^-3
¹²C ¹H ²H ³⁷Cl : 0.989 × 0.99985 × 0.00015 × 0.2423 = 3.59 × 10^-5
¹²C ²H ¹H ³⁷Cl : 0.989 × 0.00015 × 0.99985 × 0.2423 = 3.59 × 10^-5
Sum = 2.7 × 10^-3

Fragments M⁺4 = 53
¹³C ²H ¹H ³⁷Cl : 0.011 × 0.00015 × 0.99985 × 0.2423 = 4.0 × 10^-7
¹³C ¹H ²H ³⁷Cl : 0.011 × 0.99985 × 0.00015 × 0.2423 = 4.0 × 10^-7
¹²C 2H₂ ³⁷Cl : 0.989 × (0.00015)^2 × 0.2423 = 5.4 × 10^-9
Sum = 8.1 × 10^-7

Fragment M⁺5 = 54
¹³C 2H₂ ³⁷Cl : 0.011 × (0.00015)^2 × 0.2423 = 6 × 10^-11

The intensity of each peak (from M to M⁺5) is proportional to the relative population of each fragment and the calculation of the probability is based on the summation of the probabilities of occurrence of all combinations resulting into the same nominal mass. The most intense peak is called base peak and the relative intensities of the other peaks are commonly reported as % of base peak.

Obviously, for the example above (ionic fragment CH₂Cl⁺), the fragment M = 49 constitutes the base peak (relative intensity 100 %). The relative intensities of the other fragments can be easily calculated, and we have:
Relative intensity for M = 49: 100 %
Relative intensity for M⁺1 = 50: (0.00855 / 0.7491) × 100 = 1.14 %
Relative intensity for M⁺2 = 51: (0.240 / 0.7491) × 100 = 31.98 %
Relative intensity for M⁺3 = 52: (0.0027 / 0.7491) × 100 = 0.36 %
Relative intensity for M⁺4 = 53: (8.1 × 10^-7 / 0.7491) × 100 = 1 × 10^-4 %
Relative intensity for M⁺5 = 54: (6 × 10^-11 / 0.7491) × 100 = 8 × 10^-9 %

[A Java applet demonstrating the isotopic peaks encountered in mass spectrometry principles can be found at the Internet site http://www.chem.uoa.gr/applets/appletMS/appl_MS₂.html.]