Subatomic particles isotopes and mass spectrometry

Atomic Structure and Subatomic Particles

Every atom contains a dense central nucleus of protons and neutrons, surrounded by electrons that occupy energy levels (shells) at relatively large distances from the nucleus.

The properties of the three subatomic particles are summarised below.

The atomic number (Z), also called the proton number, is the number of protons in the nucleus and defines which element the atom is.

The mass number (A) is the total number of protons and neutrons in the nucleus.

The number of neutrons in any atom is therefore $(A-Z)$. In a neutral atom, the number of electrons equals the number of protons.

For ions, electrons are gained or lost while protons stay fixed. A cation has fewer electrons than protons, and an anion has more electrons than protons.

The standard nuclide notation places the mass number as a superscript and the atomic number as a subscript before the element symbol, for example ³⁵₁₇Cl.

For ³⁵₁₇Cl⁻: protons = 17, neutrons = 35 − 17 = 18, electrons = 17 + 1 = 18.

MisconceptionStudents often subtract the wrong way for ions. Cations have lost electrons, so they have fewer electrons than protons. Anions have gained electrons, so they have more electrons than protons. Protons never change.

Exam TipState the charge change before counting electrons.

Isotopes and Their Properties

Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons, giving them different mass numbers.

Because isotopes contain the same number of protons, they also contain the same number of electrons in a neutral atom. This means their electronic configurations are identical.

Identical electron arrangements result in identical chemistry. Isotopes of an element therefore react in the same way and form the same compounds.

Different neutron numbers give different masses. Isotopes therefore have different physical properties such as density and rate of diffusion, and they appear at different m/z positions in a mass spectrometer.

For example, ³⁵Cl contains 17 protons and 18 neutrons, while ³⁷Cl contains 17 protons and 20 neutrons. Both have the electron configuration 2,8,7 and react identically with sodium to form NaCl.

Mass Spectrometer Principles and Spectrum Interpretation

A mass spectrometer separates positive ions according to their mass-to-charge ratio (m/z) and produces a spectrum showing relative abundance against m/z.

The four stages of operation are:

1. Ionisation: the sample is vaporised and bombarded with high-energy electrons, knocking out one (or rarely two) electrons from each atom or molecule to form positive ions.
2. Acceleration: the positive ions are accelerated by an electric field to give them all the same kinetic energy.
3. Deflection: the ion beam passes through a magnetic field; ions are deflected according to their m/z value, with lighter ions and more highly charged ions deflected more.
4. Detection: ions of a chosen m/z reach the detector, generating a current proportional to the abundance of that ion.

The x-axis of a mass spectrum shows mass-to-charge ratio (m/z), and the y-axis shows relative abundance. Each peak corresponds to a distinct ion.

For singly charged ions (1+), m/z equals the mass of the ion. This is the most common case.

Some ions in a mass spectrometer carry a 2+ charge. These appear at half the expected m/z, so a ⁵⁶Fe²⁺ ion is detected at m/z = 28, not 56.

For an element, each peak represents an isotope, and the peak heights give the relative abundances. For a molecule, the peak at the highest m/z is the molecular ion peak (M⁺), and its m/z equals the relative molecular mass (Mr).

The relative atomic mass is calculated as a weighted average of isotope masses using their abundances.

Key Equations

Relative atomic mass:

$${A}_{r}=\frac{\sum (\text{isotope mass}\times \text{relative abundance})}{\sum \text{relative abundance}}$$

Variables: isotope mass = mass number of each isotope; relative abundance = % or ratio of each isotope.

SI unit: dimensionless (relative quantity).

Worked Example: Calculating Aᵣ from a Mass Spectrum

Scenario

A mass spectrum of element X shows two peaks: m/z = 63 with abundance 69.2%, and m/z = 65 with abundance 30.8%. Calculate the relative atomic mass.

Equation used

$${A}_{r}=\frac{\sum (\text{isotope mass}\times \text{relative abundance})}{\sum \text{relative abundance}}$$

Given

$$\text{isotope mass}=63, \text{abundance}=69.2\%$$

$$\text{isotope mass}=65, \text{abundance}=30.8\%$$

Working

$${A}_{r}=\frac{(63\times 69.2)+(65\times 30.8)}{100}$$

$${A}_{r}=\frac{4359.6+2002}{100}$$

$${A}_{r}=\frac{6361.6}{100}$$

Answer

$${A}_{r}=63.6$$

Interpretation

An Aᵣ of 63.6 matches copper, confirming element X is copper.

Examiner InsightWhen the question gives abundances as relative peak heights (e.g. 9:6:1) rather than percentages, divide by the sum of the heights, not by 100. Read the y-axis units carefully.

Exam TipCheck whether abundances are in % or as raw ratios before dividing.

Predicting Mass Spectra of Diatomic Molecules

For a diatomic molecule containing an element with two isotopes, the mass spectrum shows several molecular ion peaks because each atom in the molecule may independently be either isotope.

The relative peak heights are calculated by combining the abundances of each isotope pair using simple probability.

For Cl₂, with ³⁵Cl at 75% (= 3 parts) and ³⁷Cl at 25% (= 1 part), three molecular combinations are possible.

The mass spectrum of Cl₂ therefore shows three peaks at m/z = 70, 72 and 74 in the ratio 9 : 6 : 1.

The factor of 2 for the mixed combination arises because there are two equivalent ways to form ³⁵Cl–³⁷Cl: the ³⁵Cl atom may bond to a ³⁷Cl atom, or vice versa.

The same approach applies to any diatomic molecule of an element with two isotopes, including Br₂.

Worked Example: Predicting the Mass Spectrum of Br₂

Scenario

Bromine has two isotopes, ⁷⁹Br at 50.0% and ⁸¹Br at 50.0%. Predict the m/z values and relative peak heights of the molecular ion peaks for Br₂.

Step 1 — list the three possible isotope combinations:

• ⁷⁹Br–⁷⁹Br at m/z = 158
• ⁷⁹Br–⁸¹Br (either order) at m/z = 160
• ⁸¹Br–⁸¹Br at m/z = 162

Step 2 — calculate the probability of each combination using abundances of 1 : 1.

$$P(\text{⁷⁹Br–⁷⁹Br})=1\times 1=1$$

$$P(\text{⁷⁹Br–⁸¹Br})=2\times 1\times 1=2$$

$$P(\text{⁸¹Br–⁸¹Br})=1\times 1=1$$

Step 3 — express as a ratio.

Answer

Peaks at m/z = 158, 160, 162 in the ratio 1 : 2 : 1.

Interpretation

The characteristic 1 : 2 : 1 triplet pattern for Br₂ reflects the equal abundance of the two bromine isotopes, in contrast to the 9 : 6 : 1 pattern of Cl₂.