IB DP Chemistry -S1.2.3 Mass Spectrometry - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S1.2.3 Mass Spectrometry – Study Notes – New Syllabus
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Structure 1.2.3 – Mass Spectra and Relative Atomic Mass
Structure 1.2.3 – Mass Spectra and Relative Atomic Mass
Mass Spectrum
- A mass spectrum is a graph produced by a mass spectrometer that shows the relative abundance of isotopes of an element against their mass-to-charge ratio (m/z).
- For atoms with a +1 charge (which is common in simple ionization), the mass-to-charge ratio is effectively equal to the mass number of the isotope.
Purpose of Mass Spectrometry:
- Determine the number and abundance of isotopes of an element.
- Calculate the relative atomic mass \( A_r \) of an element from isotopic data.
Understanding a Mass Spectrum:
A typical mass spectrum for an element with multiple isotopes will show:
- Peaks: Each peak corresponds to an isotope.
- Height or intensity of peaks: Proportional to the relative abundance of the isotope.
- m/z value: The position of each peak on the x-axis corresponds to the mass number (for ions with +1 charge).
Monoatomic Elements
Diatomic Elements
Example Interpretation:
If an element has two peaks at:
- \( m/z = 20 \), abundance = 75%
- \( m/z = 22 \), abundance = 25%
This tells us that the element has two isotopes: mass numbers 20 and 22, with $75\%$ and $25\%$ relative abundance respectively.
Calculating Relative Atomic Mass from Mass Spectrum:
Use the formula:
\( A_r = \frac{(a_1 \times m_1) + (a_2 \times m_2) + \ldots}{100} \)
- \( a_1, a_2, \ldots \): % abundances of the isotopes
- \( m_1, m_2, \ldots \): mass numbers of the isotopes
Example
A sample of magnesium contains 3 isotopes with the following data:
- Magnesium-24: 78.6%
- Magnesium-25: 10.1%
- Magnesium-26: 11.3%
Calculate the relative atomic mass of magnesium.
▶️Answer/Explanation
Using the formula:
\( A_r = \frac{(24 \times 78.6) + (25 \times 10.1) + (26 \times 11.3)}{100} \)
\( A_r = \frac{1886.4 + 252.5 + 293.8}{100} = \frac{2432.7}{100} = 24.33 \)
Relative atomic mass of magnesium is 24.33
Operational Steps (Not Assessed but Helpful for Understanding):
Ionization – Making the Atoms Charged
- To analyze an atom in the mass spectrometer, it first needs to be converted into a positively charged ion.
- This is achieved by firing a high-energy beam of electrons at gaseous atoms. One of the atom’s electrons is removed, resulting in a positive ion.
- Purpose: Only charged particles can be manipulated and detected by the instrument.
Example: \( \text{X (g)} + \text{e}^- \rightarrow \text{X}^+ (g) + 2\text{e}^- \)
Acceleration – Giving the Ions Kinetic Energy
- Once ionized, the positive ions are accelerated through an electric field.
- This ensures all ions have similar kinetic energy as they enter the next section.
- Purpose: Equal speeds are important to accurately compare how ions behave based on their masses during deflection.
Deflection – Separating Ions by Mass
- As the ions move through a magnetic field, they are forced to change direction.
- Lighter ions are deflected more easily, while heavier ions resist bending and are deflected less.
- Analogy: Imagine ions being like paper and coins in a wind tunnel – paper (lighter ions) flies off more easily, while coins (heavier ions) resist the airflow.
- The strength of the magnetic field is gradually increased during the process so that ions with different mass-to-charge ratios are sequentially deflected into the detector.
Detection – Measuring and Recording the Ions
- Ions that are perfectly deflected reach the detector.
- On impact, each ion picks up an electron, generating a small electrical current.
- The greater the number of ions hitting the detector, the stronger the signal produced.
- These signals are sent to a computer and used to create a mass spectrum graph showing the relative abundance of each ion.
Continuous Sweep: The magnetic field is not static—it is gradually increased. As this happens, ions of different masses are detected at different times, allowing a complete spectrum to be generated from lightest to heaviest ions.
Final Output – The Mass Spectrum:
- The spectrum is a plot of relative abundance (y-axis) versus mass-to-charge ratio \( m/z \) (x-axis).
- Each peak corresponds to an isotope, and the height of the peak reflects its relative abundance in the sample.
MCQ Example
Which statement about mass spectrometry is correct?
- The height of each peak represents the mass of the isotope.
- The x-axis shows the charge of each ion.
- The mass spectrum shows both chemical and physical properties.
- The relative atomic mass is the weighted average based on isotopic abundances.
▶️Answer/Explanation
Correct Answer: D
Relative atomic mass is calculated as a weighted average of the isotope masses using their relative abundances.
Example
A sample of an unknown element has three isotopes with the following data from a mass spectrum:
- Isotope A: mass = 10, abundance = 20%
- Isotope B: mass = 11, abundance = 50%
- Isotope C: mass = 12, abundance = 30%
Calculate the relative atomic mass of the element and identify the most abundant isotope.
▶️Answer/Explanation
\( A_r = \frac{(10 \times 20) + (11 \times 50) + (12 \times 30)}{100} \)
\( A_r = \frac{200 + 550 + 360}{100} = \frac{1110}{100} = 11.1 \)
Most abundant isotope is B with mass 11
Example
Using mass spectra to determine the formula of an unknown compound containing bromine.
▶️Answer/Explanation
Analyze the Hard Mass Spectrum
- Peaks are observed at mass/charge (m/z) values of 1, 12, 79, and 81.
- This suggests the presence of:
- Hydrogen (\( ^1\text{H} \)) → mass = 1 amu
- Carbon (\( ^{12}\text{C} \)) → mass = 12 amu
- Bromine isotopes → \( ^{79}\text{Br} \) and \( ^{81}\text{Br} \)
- The observed intensity ratios (approximately 60% for H, 20% for C, and 10% + 10% for Br isotopes) suggest a 3:1:1 atomic ratio of H:C:Br.
Therefore, the empirical formula is \( \text{CH}_3\text{Br} \).
Analyze the Soft Mass Spectrum
- Two major peaks are observed at m/z = 94 and m/z = 96.
- This indicates two molecular species:
- \( \text{CH}_3^{79}\text{Br} \) → mass = 12 (C) + 3×1 (H) + 79 (Br) = 94 amu
- \( \text{CH}_3^{81}\text{Br} \) → mass = 12 + 3 + 81 = 96 amu
- The existence of two peaks matches the near-equal abundance of the two Br isotopes (50.7% and 49.3%).
Conclusion: Both the hard and soft mass spectra support the molecular formula:
\( \text{CH}_3\text{Br} \)