Mass Spectrometry

Chemistry: AQA A Level, topic 1.1.3

Workings of a Mass Spectrometer

The mass spectrometer is split into four stages: ionisation, acceleration, ion drift and detection.

1) Ionisation

1) Ionisation

  • The first step is ionisation of the sample. The main techniques are:
  • Electrospray ionisation.
  • Electron impact ionisation.
  • Electrospray ionisation is a gentler technique and prevents fragmentation.
  • It’s typically used for polymers and biological materials like DNA.
  • The sample is dissolved in a solvent and a high voltage is applied.
  • The high voltage rips a proton off the solvent and attaches it to the sample molecules.
  • The sample molecules are now positively charged ions.
1) Ionisation cont.

1) Ionisation cont.

  • In electron impact ionisation, the sample is first vaporised and then hit with electrons from an electron gun.
  • The electrons knock off electrons from the molecule
  • The molecules are now positively charged ions.
  • This method often causes the sample to fragment.
2) Acceleration

2) Acceleration

  • Molecules are accelerated to all have the same kinetic energy.
  • From standard equations, the kinetic energy is equal to half of the mass multiplied by the velocity squared:
  • Kinetic energy = 12\frac{1}{2}mv2
  • All the molecules have the same kinetic energy, so the speed is dependent on the mass of the molecule.
  • Lighter particles move faster and are detected before heavier particles.
3) Ion drift

3) Ion drift

  • The time of flight is given by:
  • Time of flight = distancevelocity\frac{distance}{velocity}
  • This leads to an equation for the time travelled that depends on mass:
  • Time = distance ÷ 2KEm\sqrt\frac{{2KE}}{m}
  • Lighter ions take less time as the time is dependent on the square root of the mass.
4) Detection

4) Detection

  • The ions hit a negatively charged plate.
  • This causes a current and the size of this current gives a measure of the number of molecules hitting the plate.
  • This gives the abundance of the molecule.

Analysis to Identify Molecules

Once a sample has passed through the mass spectrometer, we can analyse the data to identify the molecule.

Spectrum produced

Spectrum produced

  • When the sample has passed through the mass spectrometer, a spectrum is produced by the spectrometer.
  • On this spectrum:
  • The x-axis is mass/charge ratio.
  • The y-axis is % abundance.
Main peak

Main peak

  • The spectrum produces lots of peaks, but the most important is the molecular ion peak.
  • This is the peak of the greatest mass/charge ratio.
  • This represents the mass/charge value of the molecule we are analyzing.
Isotopes

Isotopes

  • Smaller peaks will cluster around the molecular ion peak.
  • These are from the same molecules but with different isotopes in them.
  • The isotopic molecules have different masses and so different mass/charge ratio values.
Fragmentation

Fragmentation

  • Any smaller and significantly lighter peaks in the spectrum are because of fragmentation.
  • The molecule can fragment in the spectrometer.

Analysis to Calculate

Once we have the mass spectrum, we can calculate the relative atomic mass.

Relative atomic mass

Relative atomic mass

  • Relative atomic mass is the average weighted mass of an atom relative to carbon-12.
  • The key word to look at here is "average".
  • This is because its value is calculated taking into account all of its isotopes and their relative abundances.
Using the spectrum

Using the spectrum

  • We can use the spectrum to view all the different isotopes and their relative abundance.
  • Isotopic mass is along the x-axis.
  • Isotopic abundance is along the y-axis.
Calculation

Calculation

  • We can then calculate the relative atomic mass since we have all of the isotopic masses and their relative abundances.
  • It is easiest to show how to carry out the calculation with an example - see the next slide.
Example - boron

Example - boron

  • This is the mass spectrum of a sample of elemental boron.
  • Boron has two isotopes, 10B and 11B.
  • You can see from the spectrum that approximately 20% of the boron is 10B and 80% is 11B.
  • You can use this to work out the relative atomic mass:
  • 80% × 11 + 20% × 10 = 10.8

Test yourself on Mass Spectrometry

  • Why is electrospray ionisation better than electron impact ionisation?

  • Which of these applies to the detection plate?

  • What do clustered small peaks around the molecular ion peak represent?

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