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Mass Spectroscopy and Isotopes Chemistry Tutorial

Key Concepts

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Mass Spectrometer : General Principles

Step 1 sample is vaporised before entering mass spectrometer.
M(s) → M(g)
or M(l) → M(g)
Step 2 Sample enters the low pressure ionisation chamber where an electron beam ionises the sample.
M(g) → M+(g) + e-
Step 3 Ions are accelerated to high speeds by an electric field.
Step 4 Ions pass through a perpendicular magnetic field which bends the path of the ions dependent on their mass-to-charge ratio.
Heavier ions are deflected less than lighter ions.
Step 5 Ions with a particular mass reach the collector. Different masses can be collected by changing the magnetic or electric field.
Step 6 The detector identifies the mass of each ion from its path. Data is recorded as a mass spectrum.

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Example: Mass Spectrometer to Separate Isotopes

A sample of neon gas made up of 3 isotopes, 20Ne, 21Ne and 22Ne, is injected into the mass spectrometer.

The sample passes through the electron beam which ionises the atoms of each isotope:
20Ne → 20Ne+ + e-
(or   20Ne + e-20Ne+ + 2e-)
21Ne → 21Ne+ + e-
(or   21Ne + e-21Ne+ + 2e-)
22Ne → 22Ne+ + e-
(or   22Ne + e-22Ne+ + 2e-)

The ions are accelerated and pass through slits in negatively charged metal plates (the second plate is more highly charged than the first).

The ions enter the magnetic field where the path of the ions is deflected. Ions with the greatest mass (heavy isotopes) are deflected the least.
The path of 20Ne+ is deflected more than 21Ne+ which is deflected more than 22Ne+

The voltage of the slits in the negatively charged plates is slowly raised causing the neon ion beams to sweep across the detector one by one.

The ion detector is connected to a recorder which records the mass spectrum of the sample of neon gas.

The mass spectrum shows the three isotopes of neon as three separate peaks.

The height of each peak is related to the relative abundance of each isotope of neon.
20Ne+ has the highest peak so 20Ne is the most abundant isotope.
21Ne+ has the lowest peak so 21Ne is the least abundant isotope.

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Example: Using a Mass Spectrum to Determine an Approximate Relative Atomic Mass of an Element

The mass spectrum could be presented in two ways, either giving the actual signal intensity for each isotope, or, giving you the relative peak heights or percentage abundances.

Signal Intensity Mass Spectrum of Neon % Abundance Mass Spectrum of Neon

mass number

Relative atomic mass (r.a.m.) = [(%abundance isotope a × mass isotope a) + (%abundance isotope b × mass isotope b) +(%abundance isotope c × mass isotope c)] ÷ 100

Mass number (A) can be used to approximate the mass of an isotope.
(see Relative Atomic Mass tutorial for details)

For the isotopes of neon:
the mass of 20Ne ≈ A(neon-20) ≈ 20
the mass of 21Ne ≈ A(neon-21) ≈ 21
the mass of 22Ne ≈ A(neon-22) ≈ 22

So the calculation for relative atomic mass (or atomic weight), now looks like this:

Relative atomic mass (Neon) = [(%abundance neon-20 × A(neon-20)) + (%abundance neon-21 × A(neon-21)) +(%abundance neon-22 × A(neon-22))] ÷ 100

The % abundance of each isotope can be read directly off the second mass spectrum given above.

% abundance 20Ne = 90.92 %
% abundance 21Ne = 0.26 %
% abundance 22Ne = 8.82 %

Relative atomic mass (r.a.m.), or atomic weight, can then be calculated:

r.a.m. (Ne) = [(90.92 × 20) + (0.26 × 21) + (8.82 × 22)] ÷ 100
= [1818.4 + 5.46 + 194.04] ÷ 100
= 20.18

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Example: Using a Mass Spectrum to Determine the Relative Atomic Mass of an Element

If the data you are provided with give you the actual mass2 of each isotope (isotopic mass) as well as (or instead of) the mass number of each isotope, then you should use the actual isotopic masses in your calculation of relative atomic mass (r.a.m.).

Signal Intensity Mass Spectrum of Neon Tabulated Data
Mass Number Isotopic Mass % Abundance
20 19.992 90.92
21 20.994 0.26
22 21.991 8.82

Relative atomic mass (r.a.m.) = [(%abundance neon-20 × mass neon-20) + (%abundance neon-21 × mass neon-21) +(%abundance neon-22 × mass neon-22)] ÷ 100

r.a.m. (Ne) = [(90.92 × 19.992) + (0.26 × 20.994) + (8.82 × 21.991)] ÷ 100
= [1817.6 + 5.4584 + 193.96] ÷ 100
= 20.17

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1. Mass spectroscopy can also be used to determine the structure of molecules (Mass spectroscopy for structural determination)

2. The mass of an isotope is not a whole number because the measured mass of a proton and of a neutron and of an electron is not a whole number. (Find out more at Mass Defect)