How do mass spectrometers measure the mass of a compound?

Mass spectrometry is a method of identifying the molecular mass of compounds by measuring its mass to charge ratio. Methods vary depending on the molecules being analysed, their phase, the wanted sample insertion and purification methods, cost of the machine and the analysis, accuracy, speed and other variables. Molecules inserted in a mass spectrometer undergo three different processes: ionisation, mass analysis and detection, all of which can be done through various different methods. I will explain only one method of each process, and the calculations that the software in the MS has to do to identify the mass of each molecule.


A mass spectrometer is a general term, given to machines that identify molecules by finding out their mass to charge ratio. Because of this any machine used to identify compounds in any context are mass spectrometers, and so they have a wide variety of both routine and research uses. Uses include: identifying and analysing peptides and proteins for biotechnological research; drug discovery, combinatorial chemistry and drug metabolism in pharmaceutics; identification of precise oil composition in geology; water quality and food contamination identification in environmental regulation; neonatal screening and haemoglobin analysis in the clinical industry; and most importantly drug testing and identification in sports and for law enforcement. All these purposes are generally done to an accuracy of 0.01%, making mass spectrometers very useful machines for society.


Sample Insertion: Electron impact (EI) ionisation is the most common and cheapest method, usually used in gas chromatography mass spectrometers (GC-MS). The sample is usually inserted into a gas chromatography column, where the individual components are vaporised and then more finely separated and purified. Each purified component is then sequentially inserted into the mass spectrometer ionisation chamber. The whole system is held under a vacuum with vacuum pumps to prevent contamination or unnecessary collisions of the electron beam or the created ions.

Electron Beam: The electron beam is produced by heating a metal filament with a high current to cause thermionic emission of electrons, and by having a positively charged anode on the other side of the ionisation chamber, which when paired with the cathode that the electrons are emitted from, creates a potential difference that pushes electrons away from the cathode and towards the anode. The metal filament needs to have the highest surface area to volume ratio as possible so more electrons are surface electrons, and thus require less energy to remove, as well as a low work function so that the amount of energy needed to release electrons and cause a current to the cathode is low. This would allow relatively low currents to be used.

The emitted electrons, when they collide with the sample, can hit the molecules with enough energy to remove an electron from the sample, positively ionising it. For the compound to be ionised, the bond length of the most abundant bond has to be similar to the De Broglie wavelength of the electron impacting on the molecules:

De Broglie Wavelength (lambda)= h/rho     h=planck’s constant

rho= gamma x rest mass x velocity    rest mass= m0 and velocity= v

Therefore Lambda= h/(gamma x rest mass x velocity)

Since gamma here stands for the relativistic factor, gamma= square root of [1- (v^2/c^2)]

By squaring everything and then rearranging so that all v^2 terms are together, it can then be factorised and rearranged again to make v^2 the subject, and so the velocity can be found to be:

v= square root of (1/[(gamma x m0)^2/h^2]+1/c^2)

So the velocity v of an electron needed to efficiently ionise organic C-C bonds is:

Bond length =lambda= 1.54 x10^-10 m

Mass of an electron= me= 9.11x10^-31 kg

h= 6.63x10^-34 Js

c= 3x10^8 m/s

Substituting all this gives you a number of about 4.73x10^6 m/s

And the energy required to accelerate electrons to this speed is:

Kinetic energy= ½ mv^2 and when you substitute v in, KE=1.02x10^-17J=63.56eV

Due to the low percentage of the speed of light that the electron needs to travel at (only 1.6%) non-relativistic, Newtonian principles can be used.

This kinetic energy can be released when a tantalum filament is provided with:

Kinetic energy of e released= Energy provided – phi              phi= the work function=3.38eV for tantalum

So E=KE + phi=63.56+3.38=66.94eV need to be provided

This energy can be provided to the cathode with a potential difference and heating the filament. Higher energies are usually used (of about 70eV) to ensure that electrons can be energetically released, since 66.94eV is the minimum energy needed for an electron to be emitted with the necessary 63.56eV of kinetic energy from the filament. High energies can increase ion fragmentation (when bonds in a molecule are broken, they can form two separate and smaller fragmented molecules), which can make readings and identification of compounds much harder by creating peaks of lower m/z values than the parent molecules, so a value greater than, but close to, the optimum necessary energy to release a 63.56eV electron is ideal.

Exit: Once a particle is ionised it’s very simply expelled from the ionisation chamber and accelerated towards the mass analyser by the drawing-out plate; a positively charged plate at the beginning of the ionisation chamber, which is at or above 8kV, so as to repels the newly formed positive ions. The end plate is at 0V, creating a potential difference of 8kV between the ionisation chamber and the end of the drift tube through which the ions are accelerated.

A question I asked myself was; why don’t the ions get attracted to the cathode, and instead get repelled out into the drift tube? The electron gun is created mainly through thermionic emission, so the voltage used at the anode doesn’t need to provide much of the energy used to remove the electrons; it’s mainly used to direct the electrons. Taking this into account, and how light electrons are, this voltage is extremely low compared to the voltage used to accelerate the much heavier ions into the drift tube. Whereas the 8kV accelerate the heavy ions to high speeds, the small potential difference of about 100V (7.93kV at the cathode and 8.03kV at the anode) would barely affect the heavy ions, and definitely nowhere near as much as the 8kV potential difference.

Mass Analyser:

Drift Tube: Once in the drift tube, one more charged plate focuses the ions into a tight beam. This focusing plate has an intermediate voltage of about 7.5kV which, as the ions are accelerated towards it by the potential difference, pushes the ions away from the walls of the plate through repulsion and thus focusing them into a tight beam, and then accelerating them away from it again, just like the drawing-out plate and the ion repeller do. At the end of the tube there’s a final plate at 0V, at which point all the particles should have gained a constant kinetic energy, since all the electric potential was converted to kinetic energy once the electric potential is 0V. This also ensures that the particles only vary in mass and velocity.

Deflection: One method of mass analysis is magnetic analysis. This works by deflecting the ions in an arc with the use of a magnetic field as per Fleming’s rule, which is kept constant and only changed to change the mass of ions getting through the deflection tube. The final velocity of the compounds after being accelerated through the potential difference will vary depending on their mass, since force=mass x acceleration which means that the same force will cause a different acceleration on molecules with different masses and therefore reach a different final velocity in the same amount of time. Because of this the method for calculating the object’s mass to charge ratio cannot include velocity as a variable, since this quantity will be unknown. When a charge travels through a magnetic field, the movement of the charged particle creates a current, since current is the rate of transfer of charge. This current, if perpendicular to the magnetic field, will create a force on the object experiencing both the current and the magnetic field, in a direction perpendicular to both the current and the magnetic field. This means that the ion would experience a force as it moves through the magnetic field, which would move the particle, and in turn move the direction of the current and therefore the direction of the force, which would then move the ion again. This would constantly occur as the particle moves through the magnetic field, creating a circular motion, as the force is always perpendicular to the current and therefore the ion, while the magnetic field remains constant, creating a sort of spinning effect so that the magnetic field is basically providing a centripetal force. This can be represented as follows:

Fb=Bqv                    Fc=(mv^2)/r                           q=ze

So Fb=Bzev            Fb=Lorentz Force,              B=magnetic field strength/magnetic flux density (in Tesla, T)

                                    v= velocity, m=mass, r=radius of circular path, q= charge,

z=charge in multiples of e (chemical charge), e= charge of an electron (but positive)

so if Fb=Fc






and since all electric potential was made to kinetic energy:


So ½ mv^2=qV=zeV



v can then be square rooted to get the subject v, which can then be substituted into the m/z equation to get:


Eduardo S. A Level Maths tutor, GCSE Maths tutor, GCSE Physics tutor,...

1 year ago

Answered by Eduardo, an A Level Chemistry tutor with MyTutor



£20 /hr

Laura L.

Degree: Chemistry (MCHEM) (Masters) - Warwick University

Subjects offered:Chemistry, Maths+ 1 more


“Hi there! I'm a third-year Chemistry student, I tutor maths from 11+ to A level as well as GCSE Science, A level Biology, and Chemistry”

£20 /hr

Rhiannon L.

Degree: Medicine (Masters) - Cardiff University

Subjects offered:Chemistry, Maths+ 3 more



MyTutor guarantee

£30 /hr

Roshni R.

Degree: Dentistry (Bachelors) - Birmingham University

Subjects offered:Chemistry, Spanish+ 4 more

-Personal Statements-
-Medical School Preparation-

“Tutoring approach Professional but friendly approach.  Work with my students to identify the areas the student is finding most difficult. Review the topic with my hand-written notes.  Explain any misunderstood areas. Work through examp...”

About the author

Eduardo S.

Currently unavailable: for new students

Degree: Bioengineering (Masters) - Sheffield University

Subjects offered:Chemistry, Spanish+ 3 more


“About Me: I am a first year, Bioengineering MEng student at the University of Sheffield. I was born in Venezuela and have lived in 6 different countries in which I’ve learnt fluent English and proficient French, having already learnt ...”

You may also like...

Other A Level Chemistry questions

What is the meaning of the term 'structural isomers'?

What are amino acids?

How do I predict the shape and bond angle of an molecule?

What's the difference between an aldehyde and a ketone?

View A Level Chemistry tutors
