Chemistry (9701) | Mass spectrometry

Mass Spectrometry: The Chemical Weighing Scale

Welcome to one of the most powerful tools in a chemist's arsenal: Mass Spectrometry! Don't worry if the name sounds intimidating—it's essentially a highly specialized chemical balance that helps us figure out exactly what substances are made of, and how much each component weighs.

In this chapter, we will learn how to 'read' the data generated by a mass spectrometer, allowing you to:

• Calculate the relative atomic mass of an element from its natural isotopes.
• Determine the size (molecular mass) of an organic molecule.
• Identify unique elements like chlorine and bromine based on their distinctive isotopic patterns.

Think of mass spectrometry as taking a chemical fingerprint. Every molecule produces a unique pattern of peaks that reveals its identity.

1. Fundamentals of the Mass Spectrum

Although the syllabus states that knowledge of the working of the mass spectrometer itself (ionisation, acceleration, deflection) is not required, you must understand the output it produces: the mass spectrum.

What is a Mass Spectrum?

A mass spectrum is a graph that plots the abundance (how much of it there is) of different ions against their mass-to-charge ratio (\(m/e\)).

Since the ions measured almost always carry a single positive charge (\(z = +1\)), the \(m/e\) ratio is effectively equal to the mass of the ion.

$$m/e \approx \text{mass}$$

Key Terms to Understand

• \((m/e)\) Value: The mass-to-charge ratio of the ion being detected.
• Relative Abundance: The height of the peak, indicating how common that specific ion fragment is compared to the most abundant ion (the base peak).
• Base Peak: The tallest peak in the spectrum. It is assigned a relative abundance of 100%. All other peaks are measured relative to this peak. (It represents the most stable ion fragment formed.)

Quick Review: The mass spectrum shows the masses of charged fragments (ions) produced when the sample breaks apart.

2. Determining Relative Atomic Mass (\(A_r\)) from Isotopes

Mass spectrometry is the definitive method for finding the average mass of an element, taking into account the natural proportions of its isotopes. This directly addresses syllabus point 22.2.2.

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons (and thus different masses).

Calculation Steps

To calculate the Relative Atomic Mass (\(A_r\)) of an element from its mass spectrum, you perform a weighted average calculation:

Step 1: Identify Mass (\(m/e\)) and Abundance.
Read the \(m/e\) values (which correspond to the isotopic masses) and their corresponding relative abundances (peak heights) from the spectrum.

Step 2: Calculate the Total Weighted Mass.
Multiply the mass of each isotope by its relative abundance, and then sum the results.

$$ \text{Total Weighted Mass} = \sum (\text{Isotope Mass} \times \text{Relative Abundance}) $$

Step 3: Calculate the Total Abundance.
Add up all the relative abundances (peak heights).

Step 4: Calculate \(A_r\).
Divide the Total Weighted Mass by the Total Abundance.

$$ A_r = \frac{\text{Total Weighted Mass}}{\text{Total Abundance}} $$

Analogy: This is just like calculating your average grade if one test is worth 70% and another is worth 30%. You must weight each mass according to its contribution (abundance).

Example: Neon (Ne)
A spectrum shows Ne-20 (Abundance 90.5%) and Ne-22 (Abundance 9.5%).

$$ A_r = \frac{(20 \times 90.5) + (22 \times 9.5)}{90.5 + 9.5} $$ $$ A_r = \frac{1810 + 209}{100} = 20.19 $$

The Relative Atomic Mass of Neon is 20.19.

Key Takeaway: Mass spectrometry provides the accurate masses and ratios needed to calculate the natural weighted average mass of an element.

3. Mass Spectrometry for Organic Molecules

When studying organic chemistry, mass spectrometry helps us determine the structure and formula of an unknown compound by finding its molecular mass and identifying key fragments.

3.1 The Molecular Ion Peak \([M]^+\) (Syllabus 22.2.3)

The molecular ion peak, labelled as \([M]^+\), is the key peak for determining the molecular mass.

It is formed when the intact molecule loses just one electron during the ionisation process:

$$\text{Molecule (M)} \rightarrow \text{M}^+ + e^-$$

• The \([M]^+\) peak corresponds to the Relative Molecular Mass (\(M_r\)) of the compound.
• Its \(m/e\) value gives the mass of the entire original molecule (the highest mass unit shown in the typical organic spectrum).

Warning: In complex or unstable molecules, the \([M]^+\) peak can sometimes be very small or even undetectable because the molecule fragments immediately upon ionisation.

3.2 Fragmentation (Syllabus 22.2.4)

After forming the molecular ion \([M]^+\), the ion often has excess energy and breaks up into smaller, more stable pieces. This process is called fragmentation. These smaller, positively charged pieces show up as peaks on the spectrum.

$$ \text{M}^+ \rightarrow \text{Fragment}_A^+ + \text{Fragment}_B $$

Analyzing these peaks helps us suggest the structure of the original molecule, as specific functional groups tend to break in predictable ways.

• Example: In ethanol (\(M_r = 46\)), we might see an \([M]^+\) peak at \(m/e = 46\).
• If the ion loses a methyl group (\(\text{CH}_3\), mass 15), a fragment peak will appear at \(m/e = 46 - 15 = 31\). This strong peak at 31 is characteristic of alcohols (\(\text{CH}_2\text{OH}^+\)).
• Losing just a hydrogen atom (\(m/e = 1\)) would produce a peak at \(m/e = 45\) (\([M-1]^+\) peak).

Key Takeaway: Fragmentation peaks help identify the sub-units that make up the molecule.

3.3 The \([M+1]^+\) Peak: Counting Carbon Atoms (Syllabus 22.2.5)

When looking closely at the molecular ion region, you will often see a tiny peak right after the \([M]^+\) peak, labelled \([M+1]^+\).

This peak exists because carbon has a minor heavy isotope, \(^{13}\text{C}\), which has a natural abundance of about 1.1%.

• The \([M+1]^+\) peak occurs when one atom in the molecule is \(^{13}\text{C}\) instead of the usual \(^{12}\text{C}\).
• The height of this peak is proportional to the number of carbon atoms, \(n\), in the molecule.

We can use the relative abundances of the \([M]^+\) and \([M+1]^+\) peaks to calculate the exact number of carbon atoms (n) using the following formula:

$$ n = \frac{100 \times \text{abundance of } [M+1]^+ \text{ ion}}{1.1 \times \text{abundance of } [M]^+ \text{ ion}} $$

Important Tip: The value 1.1% is based on the natural abundance of the \(^{13}\text{C}\) isotope. Remember that you must use the relative abundances (often given as percentages or ratios relative to the base peak) for this calculation. Your answer for \(n\) should be rounded to the nearest whole number.

Quick Review: The presence of \([M+1]^+\) is due to \(^{13}\text{C}\). This calculation is a vital tool for verifying the molecular formula.

3.4 The \([M+2]^+\) Peak: Identifying Halogens (Syllabus 22.2.6)

Some elements have heavy isotopes that are naturally abundant enough to create large, distinctive peaks two mass units higher than the main molecular ion peak. This is the \([M+2]^+\) peak.

The presence of Chlorine (Cl) or Bromine (Br) in a molecule is immediately visible due to the unmistakable ratios between the \([M]^+\) and \([M+2]^+\) peaks.

Chlorine (Cl)

Chlorine exists as two major isotopes:

• \(^{35}\text{Cl}\) (mass 35): ~75% abundance
• \(^{37}\text{Cl}\) (mass 37): ~25% abundance

If a molecule contains one chlorine atom, the molecular ion will consist of two peaks, \([M]^+\) (using \(^{35}\text{Cl}\)) and \([M+2]^+\) (using \(^{37}\text{Cl}\)), in a ratio of 3:1.

Memory Aid for Cl: Cl-35 is three times more common than Cl-37. The peak pattern looks like three tall sticks followed by one short stick (3:1 ratio).

Bromine (Br)

Bromine also exists as two major isotopes:

• \(^{79}\text{Br}\) (mass 79): ~50% abundance
• \(^{81}\text{Br}\) (mass 81): ~50% abundance

If a molecule contains one bromine atom, the molecular ion will consist of two peaks, \([M]^+\) (using \(^{79}\text{Br}\)) and \([M+2]^+\) (using \(^{81}\text{Br}\)), in an approximately 1:1 ratio (equal height).

Memory Aid for Br: The bromine isotopes have roughly equal abundance, resulting in two peaks of nearly equal height.

Did you know? Oxygen also has a heavy isotope (\(^{18}\text{O}\)), but its natural abundance is so low (0.2%) that it rarely produces a noticeable \([M+2]^+\) peak unless the molecule is very large.

Common Mistake Alert: Students sometimes confuse the \([M+1]^+\) peak (used for calculating carbon atoms) with the \([M+2]^+\) peaks used for identifying halogens. Remember:

• \([M+1]^+\) is small, due to \(^{13}\text{C}\).
• \([M+2]^+\) is large and distinctive (3:1 or 1:1), due to halogens (Cl or Br).

Summary: Mass Spectrometry Key Takeaways

For Elemental Analysis (Isotopes):

• It measures the relative abundance of different isotopes of an element.
• These measurements are used to calculate the weighted average of the isotopic masses, giving the element's Relative Atomic Mass, \(A_r\).

For Organic Structure Determination:

• The highest mass peak (often) is the molecular ion peak \([M]^+\), which gives the Relative Molecular Mass (\(M_r\)).
• Smaller peaks are due to fragmentation, helping identify structural groups (like \(\text{CH}_3\) or \(\text{OH}\)).
• The \([M+1]^+\) peak, caused by \(^{13}\text{C}\), allows calculation of the number of carbon atoms, \(n\).
• The \([M+2]^+\) peak is used to identify heavy halogens: 3:1 ratio means Chlorine (Cl), and 1:1 ratio means Bromine (Br).