Structured Questions
Complete each question on paper, then check your answers against the mark scheme.
Question 1: TOF Mass Spectrometry
8 marks(a) Describe the processes of electrospray ionisation and electron impact ionisation in a Time of Flight (TOF) mass spectrometer. Compare when each method is typically used. [4]
(b) An ion of ⁸²Kr⁺ is accelerated in a TOF mass spectrometer. The kinetic energy of the ion is 4.82 × 10⁻¹⁵ J. The length of the drift tube is 1.20 m. Calculate the time of flight of this ion. Give your answer to 3 significant figures. (Avogadro constant \(L = 6.022 \times 10^{23}\text{ mol}^{-1}\)). [4]
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(a)
- Electron impact: Sample is vaporised and high energy electrons from an electron gun are fired at it, knocking off an electron to form 1+ ions. [1] Used for elements or low molecular mass compounds. [1]
- Electrospray: Sample is dissolved in a polar solvent and injected through a fine hypodermic needle connected to a high voltage supply. Each molecule gains a proton (H+) to form MH+ ions. [1] Used for high molecular mass compounds (e.g. proteins) to prevent fragmentation. [1]
(b)
- Mass of one ⁸²Kr⁺ ion in kg:
\(m = \frac{0.082}{6.022 \times 10^{23}} = 1.3617 \times 10^{-25}\text{ kg}\) [1] - Using \(\text{KE} = \frac{1}{2} m v^2\):
\(v = \sqrt{\frac{2 \times \text{KE}}{m}} = \sqrt{\frac{2 \times 4.82 \times 10^{-15}}{1.3617 \times 10^{-25}}} = 2.6607 \times 10^5\text{ m s}^{-1}\) [1] - Using \(t = \frac{d}{v}\):
\(t = \frac{1.20}{2.6607 \times 10^5} = 4.51 \times 10^{-6}\text{ s}\) (or 4.51 μs) [2]
Question 2: Electron Configurations and Trends
7 marks(a) Write the full electron configuration of the S²⁻ ion and the Ni atom. [2]
(b) Explain why the second ionisation energy of sodium is much higher than its first ionisation energy. [2]
(c) Explain why the first ionisation energy of sulfur is lower than that of phosphorus, despite sulfur having a higher nuclear charge. [3]
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(a)
- S²⁻: 1s²2s²2p⁶3s²3p⁶ [1]
- Ni: 1s²2s²2p⁶3s²3p⁶3d⁸4s² [1]
(b)
- The second electron of sodium is removed from the 2p sub-shell, which is in a lower energy level / closer to the nucleus than the 3s sub-shell. [1]
- This electron experiences less shielding, and a stronger electrostatic attraction from the nucleus. [1]
(c)
- Phosphorus has three unpaired electrons, one in each of the three 3p orbitals (3p³). [1]
- Sulfur has four electrons in the 3p sub-shell (3p⁴), meaning two electrons are paired in one 3p orbital. [1]
- The repulsion between these paired electrons in the same orbital makes the outer electron in sulfur easier to remove. [1]
Question 3: Mass Spectrometry and Ar
4 marks(a) Define relative atomic mass (\(A_r\)). [2]
(b) A mass spectrum of a sample of bromine contains peaks at \(m/z = 79\) and \(m/z = 81\). The peak heights are 50.5% and 49.5% respectively. Calculate the relative atomic mass of the bromine in this sample. Give your answer to 2 decimal places. [2]
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(a)
- The weighted average mass of an atom of an element [1]
- relative to 1/12th of the mass of an atom of carbon-12. [1]
(b)
- \(A_r = \frac{79 \times 50.5 + 81 \times 49.5}{100}\) [1]
- \(A_r = 79.99\) [1]
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