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IB DP Chemistry – S1.2.3 Mass Spectrometry – Study Notes

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?

  1. The height of each peak represents the mass of the isotope.
  2. The x-axis shows the charge of each ion.
  3. The mass spectrum shows both chemical and physical properties.
  4. 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:

\( \text{CH}_3\text{Br} \)

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