# Mass spectroscopy (MS)

Mass spectroscopy (MS) is a sophisticated spectroscopic technique, capable of measuring the molecular masses of chemical compounds up to an accuracy of 1 part in 105. Scientists use the mass spectroscopic technique frequently in a diverse range of fields. Whether it is food contamination detection, drug testing, and discovery, or carbon dating, MS is an essential technological requirement in the modern scientific world. In this article, we will discuss all that you would like to know about mass spectroscopy (MS).

## What is mass spectroscopy (MS)

‘ Mass spectroscopy (MS) is an analytical technique for identifying the chemical substances present in a sample by determining their molecular masses. Gaseous ions are created followed by their deflection and acceleration under simultaneously applied electric and magnetic fields. The deflection is based on the mass-to-charge ratio (m/e) of ions ‘.

A mass spectrum is plotted as a graph of relative abundance versus mass-to-charge ratio which is then used for molecular mass determination and compound identification. The instrument used for mass spectrometry is called a mass spectrometer or a mass spectrograph. The mass spectrometer uses an electric field while photographic means are used for detection in the mass spectrograph.

Mass spectroscopy is a general term that includes both types of instruments. But, in this article, our focus will largely stay on the working principle of mass spectrometry. Thus, by MS we will mean mass spectrometry throughout the article to go with the popular opinion.

## What are the 5 components of a mass spectrometer

• Vaporization chamber
• Ion source
• Electromagnet
• Mass analyzer
• Ion detection source

Now let’s see how each of the above components works in sync to carry out a mass spectrometric analysis.

## How a mass spectrometer functions

Step I:  Vaporization

The compound placed in the sample handling system is first vaporized at the high temperature of the vaporization chamber. Strict vacuum conditions are maintained inside the instrument so that only a small quantity of the vapor enters the spectrometer at a time.

Step II: Ion formation

As the vapors enter the ion source, electrons are fired at the gaseous molecules. This is called electrospray ionization. As the fast-speed electrons collide with the gaseous molecules, they knock out some of the electrons from the molecules themselves. The molecules are thus converted into gaseous ions.

Step III: Deflection under an applied electric field

About 5-10 kV electric field is then applied which facilitates the acceleration of gaseous molecular ions at a speed of 2 x 105 m/s. In the electrostatic analyzer, positively charged ions get deflected towards the negatively charged plate.

Step IV: Acceleration under an applied magnetic field

Followed by deflection under an electric field, the gaseous ions are further accelerated by the magnetic field so that the ions reach the receiver/detector.

The deflection of ions under an applied magnetic field depends on the factors shown in equation 1.

$r=&space;\sqrt{\frac{2mV}{eB^{2$              Equation 1

where r= radius of the circular path in the magnetic field, m=mass of ion, V= accelerating voltage, e= charge present on the ion, and B= magnetic field strength.

Equation 1 shows that the deflection of ions depends on m/e, so lighter ions (having lower mass) are deflected more strongly and vice versa. Thus, the gaseous ions are separated on the basis of their mass-to-charge ratios. The deflection of ions also depends on B. As a result, the magnetic field strength inside the mass spectrometer is changed gradually so that ions of differing m/e values reach the detector one by one.

Step V: Detection

The separated ions are passed through a narrow slit and collected on a metallic plate connected to the detector. As the ions hit the collector plate, a current flows through the detector. More the ions reaching the detector, the larger the current produced. A mass spectrum is plotted as a graph of ion current against the electromagnetic current which is equivalent to relative abundance against m/e respectively.

## How to interpret a mass spectrum

Each peak in the mass spectrum represents a specific chemical component with a unique m/e value. The intensity/ height of each peak is proportional to the relative abundance of that component in the sample. However, you should note that the mass spectrum of an element is quite different from that of an organic compound.

The peaks on the mass spectrum of an element show the relative abundance of different isotopes of that element. For instance, there are 3 naturally occurring isotopes of carbon i.e., the most abundant 12C isotope, the less abundant 13C isotope, and the least occurring 14C isotope. The peak of the most abundant element has the highest intensity on the mass spectrum. The mass spectrum of carbon can help us determine its relative atomic mass.

On the other hand, the peaks on the mass spectrum of an organic compound represent the different fragments produced via the molecular ion break-up inside the spectrometer.

### Fragmentation of molecular ions in mass spectroscopy

As we discussed in the previous section that when an electron is fired onto a gaseous molecule in the spectrometer, the molecule is converted into a gaseous ion. This ion is represented by M+ or M+. , the dot sign shows an electron that is left unpaired after one electron is knocked out of the molecule. As this molecular ion is extremely unstable, so it will break up into several different parts called molecular fragments.

One fragment can be a positively charged ion while the other a free radical. In this way, the molecular ion can split in different patterns to produce many different fragments and free radicals at the same time. The uncharged free radicals will get lost in the spectrometer while the charged fragments will be deflected and accelerated to reach the detector.

Accordingly, many different peaks will be produced on a mass spectrum, each peak corresponding to a specific charged fragment. By predicting the fragmentation pattern and combining all the fragments together we can determine the structure of an unknown chemical compound.

There are the following main types of peaks in a mass spectrum that helps in determining the structure of an organic compound.

• Molecular ion (M+) peak: The M+ peak is the peak corresponding to the main molecular ion formed.
• Base/parent peak: The base peak is the highest intensity peak in a mass spectrum. It is characteristic of the most common/stable ion produced during molecular fragmentation. It is given an arbitrary value of 100 and all the other peaks are measured relative to this peak. It may be the same or different from the M+ peak.
• M+1 peak: A small peak that is next to the main molecular ion (M+) peak is called the M+1 peak. This peak appears due to the presence of a 13C isotope of carbon in the chemical compound.

• M+2 peak: This peak arises due to the presence of halogen atoms such as a chlorine (Cl) or bromine (Br) atom in the tested compound.
• M+4 peak: If two halogen atoms are present in an organic molecule then an M+4 peak appears on the mass spectrum. The halogens present could be two chlorine atoms (any of 35Cl or 37Cl) or two bromine atoms (any of 79Br or 81Br) or both.

By measuring the relative heights of the M and the M+1 peaks, the number of carbon atoms present in an organic molecule can be determined. Similarly, M+2 and M+4 peaks can be used to identify halogen-containing chemical compounds. Let’s understand these concepts better by using the examples given below.

Example 1: Determining relative atomic mass (Ar) of an element from the %age abundance of its isotopes displayed in the mass spectrum.

$Ar=&space;\frac{\sum(Abundance&space;\times&space;m/e)}{100}$

$Ar&space;=&space;\frac{(20.6&space;\times&space;70)+&space;(27.4&space;\times&space;72)+&space;(7.7\times&space;73)+&space;(36.7\times&space;74)+&space;(7.6\times&space;76)}{100}$

$Ar&space;=&space;72.7$

The relative atomic mass of Germanium (Ge) is closest to  72Ge because this isotope is present most abundantly in the element, as determined by mass spectroscopy.

Example 2: Finding the number of carbon atoms present in a molecule from M+ and M+1 peaks.

The formula used for finding the number of carbon atoms n is :

$n=\frac{100&space;(Abundance(M+1)peak)}{1.1&space;(Abundance&space;(M)peak)}$

According to the mass spectrum given above, the abundance of the M peak at m/e 120 is 23 %  while that of the M+1 peak at m/e 121  is 2% so the number of carbon atoms present in the molecule is:

n= 100/1.1 (2/23) = 7.91 . 7.91 is closest to the whole number 8 so there are 8 carbon atoms present in this molecule.

Example 3:  Finding the molecular formula of an organic compound.

The mass spectrum of an unknown organic compound containing carbon (C), hydrogen (H), and oxygen (O) atoms is given below. Its molecular formula is also given as C8H8O2. We can use the given data and all the information we discussed earlier to determine the molecular structure of this compound and identify it.

Step I: Acknowledge the different peaks present in the spectrum.

The highest intensity peak is recorded at m/e =105 so this is the base peak. Other significant peaks are recorded at m/e 15, 51, 77, and 136.

Step II: Identify the heaviest fragment

The heaviest fragment is the one with a m/e value of 136. The peak at m/e = 105 represents the loss of 136-105 = 31 units. This means a fragment with a  mass of about 31 units is removed from the heaviest fragment to yield the most stable one.

Step III: Play around with numbers and predict a C.H, O combination with a total mass of 31 units.

Could this be [CH2O]? Let us check.

Atomic mass (C) + [ 2 x atomic mass (H) ] + atomic mass (O) = 12 + 2(1) + 16 = 30. So, It’s a  no.

What about  [CH3O]? Well, 12+3 +16 = 31 so it is a possibility.

Thus, if [CH3O] is removed from C8H8O2, this leaves behind a charged fragment i.e., [C7H5O]+ which gives a peak at m/e= 7(12) + 5(1) + 16 =105.

Step IV: Identify the other major peaks

The peak at m/e= 77 represents a loss of 136-77 = 59 units from the heaviest peak. These 59 units could represent an uncharged molecular fragment such as [C2H3O2] because 2(12) +3 (1) + 2(16)= 59.

Therefore, the peak at m/e = 77 itself is due to a charged fragment formed by removing C2H3O2  from C8H8O2 i.e., [C6H5]+ which represents an aryl ring.

Step V: Combine the different fragments to determine the actual structure of the molecule.

Combining CH3O with the aryl ring C6H5 gives us C6H5-OCH3. Calculating the total molecular mass = 7 (12) + 8(1) + 16 = 108. There is still space to accommodate 136-108 = 28 units of mass. H is a light atom so we will first try to make some combinations with the heavier C and O atoms.

2 C-atoms = 2(12) = 24.

2 O-atoms = 2(16) = 32.

1 C and 1 O-atom = 12+ 16 = 28.

This means there is a CO functional group present between C6H5 and OCH3. Thus, the complete structure of the organic compound is C6H5-CO-OCH3 which is methyl benzoate.

## Why is mass spectroscopy important

Mass spectroscopy or mass spectrometry (MS) is important because it helps a scientist in performing the following tasks.

• A Mass spectrometer can be coupled as a detector with chromatographic techniques such as HPLC and GC.
• MS can help determine the relative atomic mass of an element which is made up of different isotopes.
• An unknown organic compound can be identified by structure determination using its mass spectrum.
• A very meaningful application of mass spectroscopy is in geochemistry. Determining the isotopic ratios of an element especially radioactive isotopes such as the 14C isotope present in it can help in finding the age of organic matter. This is called radiocarbon dating.

We recommend our article what are the uses of spectroscopy to learn other interesting spectroscopic applications in specific fields.

Here is a video tutorial on mass spectrometry that you can watch to revise the concepts learned in our article.

Practice the interpretation of mass spectra here.

To learn about a more advanced version of mass spectroscopy, check out our article, Quadrupole Mass Spectroscopy.

References

1. Griffiths, J. (2008). “A Brief History of Mass Spectrometry.” Analytical Chemistry 80(15): 5678-5683.

2. Mellon, F. A. (2003). MASS SPECTROMETRY | Principles and Instrumentation. Encyclopedia of Food Sciences and Nutrition (Second Edition). B. Caballero. Oxford, Academic Press: 3739-3749.