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Questions on time-of-flight mass spectrometry and relative atomic mass

The four stages of a time-of-flight mass spectrometer, with electron impact and electrospray ionisation, acceleration to the same kinetic energy, ion drift in the flight tube and detection; flight times from KE = ½mv² and t = d/v; reading a mass spectrum; relative isotopic mass and relative atomic mass calculated from abundances; the molecular ion; the spectra of magnesium, chlorine and bromine; and the M+1 peak from carbon-13 used to count carbon atoms, with the M+2 peaks of organic chlorine and bromine compounds.

19 original questions · 65 marks · the time-of-flight mass spectrometry and relative atomic mass notes · Atomic structure · Flashcards for this unit

To print: the written questions (PDF, 73 KB) · the mark schemes (PDF, 75 KB)

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• 6 marks · Explain · TFM-Q12

  Explain the five peaks in the mass spectrum of chlorine gas, including the ratios of their heights.

  Mark scheme

  Indicative content

  1. chlorine atoms are chlorine-35 and chlorine-37 in a ratio of about 3 : 1, from abundances of about 75% and 25%
  2. the peaks at m/z 35 and 37 are ³⁵Cl⁺ and ³⁷Cl⁺ ions, formed when some Cl₂⁺ molecular ions break apart
  3. the heights at 35 and 37 are in the ratio 100 : 33, which is 3 : 1, the ratio of the two isotopes
  4. the peaks at 70, 72 and 74 are the molecular ions ³⁵Cl–³⁵Cl⁺, ³⁵Cl–³⁷Cl⁺ and ³⁷Cl–³⁷Cl⁺
  5. the chance of two chlorine-35 atoms is ¾ × ¾ = 9/16, and of two chlorine-37 atoms ¼ × ¼ = 1/16
  6. the mixed molecule can form two ways, 2 × ¾ × ¼ = 6/16, so the heights 45 : 30 : 5 are in the ratio 9 : 6 : 1

  Levels

  • Level 3 (5–6 marks) all five peaks are assigned to the correct ions, the 3 : 1 ratio at 35 and 37 is linked to the isotope abundances, and the 9 : 6 : 1 ratio of the molecular ion peaks is derived from the probabilities including the two ways of forming the mixed molecule; the working is shown and the ions carry their charges
  • Level 2 (3–4 marks) the atomic and molecular ion peaks are assigned and the 3 : 1 ratio explained; 9 : 6 : 1 is stated without derivation, or derived without the factor of two for ³⁵Cl³⁷Cl⁺
  • Level 1 (1–2 marks) 35 and 37 are named as the isotopes or 70 to 74 as Cl₂⁺; the molecular ions are usually given the same 3 : 1 ratio, and the ratios go unexplained

• 4 marks · State · TFM-Q1 AQA

  State, in order, the four stages of a time-of-flight mass spectrometer, and explain why the instrument is kept under a high vacuum.

  Mark scheme

  1. B1 ionisation first, then acceleration
  2. B1 ion drift in the flight tube, then detection
  3. B1 without a vacuum the ions would collide with molecules in the air
  4. B1 collisions would slow or deflect the ions, so they would arrive at the wrong time or not reach the detector

• 4 marks · Calculate · TFM-Q10 AQA

  Calculate the flight time of a ²⁴Mg⁺ ion in a spectrometer that gives every ion 2.00 × 10⁻¹⁶ J of kinetic energy and has a flight tube 1.50 m long. L = 6.022 × 10²³ mol⁻¹.

  Mark scheme

  1. M1 mass of one ion = 0.0240 kg mol⁻¹ ÷ 6.022 × 10²³ mol⁻¹ = 3.985 × 10⁻²⁶ kg
  2. M1 v = √(2KE/m) = √(2 × 2.00 × 10⁻¹⁶ ÷ 3.985 × 10⁻²⁶) = 1.002 × 10⁵ m s⁻¹
  3. M1 t = d ÷ v = 1.50 ÷ 1.002 × 10⁵; error carried forward from the speed
  4. A1 t = 1.50 × 10⁻⁵ s, to three significant figures

• 4 marks · Deduce · TFM-Q13

  Deduce the identity of an element from its mass spectrum, by calculating its relative atomic mass.

  Mark scheme

  1. M1 sum of m/z × abundance: 379.50 + 349.68 + 3538.56 + 265.09 + 259.00 = 4791.83
  2. M1 divide by the total abundance, 100.00
  3. A1 Ar = 47.9 to one decimal place
  4. B1 the element is titanium, whose Ar on the periodic table is 47.9

• 4 marks · Compare · TFM-Q16 AQA

  Compare electron impact ionisation with electrospray ionisation.

  Mark scheme

  1. B1 in electron impact the sample is vaporised, while in electrospray it is dissolved in a volatile solvent
  2. B1 electron impact knocks an electron off to form X⁺, while electrospray adds a proton to form XH⁺
  3. B1 the ion from electron impact has the mass of the particle, Mr, while the ion from electrospray has a mass of Mr + 1
  4. B1 electron impact often fragments molecules, while electrospray causes little fragmentation, so it is used for large molecules

• 4 marks · Calculate · TFM-Q19 CAIE

  Calculate the number of carbon atoms in one molecule of a compound whose mass spectrum has a molecular ion peak of relative height 62.0 and an M+1 peak of relative height 3.41, and explain why the M+1 peak exists.

  Mark scheme

  1. B1 about 1.1% of carbon atoms are carbon-13, so a few molecules contain one carbon-13 atom and are one unit heavier
  2. M1 n = (100 × height of M+1 peak) ÷ (1.1 × height of M peak)
  3. M1 n = (100 × 3.41) ÷ (1.1 × 62.0) = 341 ÷ 68.2 = 5.0
  4. A1 5 carbon atoms, a whole number, from 5.0

• 3 marks · Describe · TFM-Q2 AQA

  Describe how a sample of krypton is ionised by electron impact, and give an equation for the ionisation.

  Mark scheme

  1. B1 the sample is vaporised, if it is not already a gas
  2. B1 high-energy electrons from an electron gun knock one electron off each atom, forming 1+ ions
  3. B1 Kr(g) + e⁻ → Kr⁺(g) + 2e⁻; allow Kr(g) → Kr⁺(g) + e⁻

• 3 marks · Describe · TFM-Q3 AQA

  Describe how a sample is ionised by electrospray ionisation, and give the equation for the ionisation of a molecule X.

  Mark scheme

  1. B1 the sample is dissolved in a volatile solvent
  2. B1 the solution is forced through a fine hollow needle held at a high positive voltage, so each molecule gains a proton
  3. B1 X(g) + H⁺ → XH⁺(g); reject an equation that removes an electron

• 3 marks · Explain · TFM-Q4 AQA

  Explain why all the 1+ ions leave the acceleration stage with the same kinetic energy but reach the detector at different times.

  Mark scheme

  1. B1 the electric field gives every ion with the same charge the same kinetic energy
  2. B1 since KE = ½mv², an ion of smaller mass has a greater speed
  3. B1 every ion crosses the same length of flight tube, so the lighter ions arrive first; the flight time is proportional to √m

• 3 marks · Explain · TFM-Q5 AQA

  Explain how the detector of a time-of-flight mass spectrometer produces a signal that measures the abundance of each ion.

  Mark scheme

  1. B1 the positive ions hit a negatively charged plate and each gains an electron
  2. B1 the flow of electrons to the plate produces a current
  3. B1 the size of the current is proportional to the number of ions arriving, and so to the abundance

• 3 marks · Calculate · TFM-Q7

  Calculate the relative atomic mass of silicon, to one decimal place, from the abundances given in the table.

  Isotope	Abundance / %
  silicon-28	92.23
  silicon-29	4.67
  silicon-30	3.10

  Mark scheme

  1. M1 sum of mass × abundance: 28 × 92.23 + 29 × 4.67 + 30 × 3.10 = 2810.87
  2. M1 divide by the total abundance, 100.00
  3. A1 Ar = 28.1, to one decimal place, with no unit

• 3 marks · Calculate · TFM-Q8

  Calculate the relative atomic mass of neon, to one decimal place, from the peak heights in its mass spectrum given in the table.

  m/z	Relative height
  20	100
  21	0.3
  22	10.2

  Mark scheme

  1. M1 sum of m/z × height: 20 × 100 + 21 × 0.3 + 22 × 10.2 = 2230.7
  2. M1 divide by the sum of the heights, 100 + 0.3 + 10.2 = 110.5; error carried forward is not allowed for dividing by 100
  3. A1 Ar = 20.2, to one decimal place, with no unit

• 3 marks · Calculate · TFM-Q9

  Calculate the percentage abundance of boron-10 in boron, given that Ar of boron is 10.8 and its only isotopes are boron-10 and boron-11.

  Mark scheme

  1. M1 with x as the percentage of boron-10, 10x + 11(100 − x) = 10.8 × 100
  2. M1 1100 − x = 1080
  3. A1 boron-10 is 20% of boron, to two significant figures

• 3 marks · Calculate · TFM-Q11 AQA

  Calculate the flight time of an ⁸¹Br⁺ ion in a spectrometer in which a ⁷⁹Br⁺ ion takes 1.54 × 10⁻⁵ s, all ions having the same kinetic energy.

  Mark scheme

  1. M1 at the same kinetic energy and distance, t is proportional to √m
  2. M1 t = 1.54 × 10⁻⁵ × √(81 ÷ 79) = 1.54 × 10⁻⁵ × 1.0126
  3. A1 t = 1.56 × 10⁻⁵ s, to three significant figures

• 3 marks · Suggest · TFM-Q14 AQA

  Suggest the relative molecular mass of a protein whose largest peak in an electrospray spectrum is at m/z = 14306, and suggest why electrospray is used for it rather than electron impact.

  Mark scheme

  1. B1 the ion detected is MH⁺, so the protein's Mr is 14306 − 1 = 14305
  2. B1 electron impact would break a molecule this large into many fragments
  3. B1 electrospray is gentle, so the molecule stays whole and its molecular ion peak is seen

• 3 marks · Suggest · TFM-Q15

  Suggest why the mass spectrum of sulfur shows a small peak at m/z = 16, although no isotope of sulfur has a mass number of 16.

  Mark scheme

  1. B1 m/z is the mass of an ion divided by its charge
  2. B1 a ³²S²⁺ ion, formed when two electrons are removed, has m/z = 32 ÷ 2 = 16
  3. B1 a 2+ ion gains twice the kinetic energy in the field, so it arrives at the same time as a 1+ ion of mass 16

• 3 marks · Evaluate · TFM-Q17

  Evaluate the statement that the relative atomic mass of bromine, 79.9, is the mass of a bromine atom.

  Mark scheme

  1. B1 no bromine atom has a relative mass of 79.9: the isotopes have masses close to 79 and 81
  2. B1 Ar is the weighted mean mass of the atoms of the element, relative to one twelfth of the mass of a carbon-12 atom
  3. B1 the two isotopes are almost equally abundant, so the mean lies near 80; the statement is wrong because Ar is an average, not the mass of any one atom

• 3 marks · Suggest · TFM-Q21 CAIE

  Suggest which halogen is present in a halogenoalkane whose mass spectrum shows two molecular ion peaks of almost equal height, at m/z = 122 and 124, and suggest its molecular formula.

  Mark scheme

  1. B1 two peaks two units apart in a 1 : 1 ratio show one bromine atom, because bromine-79 and bromine-81 are almost equally abundant
  2. B1 122 − 79 = 43 is left for the alkyl group, which is C₃H₇
  3. B1 the molecular formula is C₃H₇Br

• 3 marks · Show · TFM-Q22

  Show that the relative atomic mass of copper is 63.6, to one decimal place, from the isotopic abundances given in the table.

  Isotope	Abundance / %
  ⁶³Cu	69.2
  ⁶⁵Cu	30.8

  Mark scheme

  1. M1 Ar = Σ(mass number × abundance) ÷ 100 = (63 × 69.2 + 65 × 30.8) ÷ 100
  2. M1 = (4359.6 + 2002.0) ÷ 100 = 6361.6 ÷ 100 = 63.616, with the working written out in full because the answer is given
  3. A1 Ar = 63.6 to one decimal place; the value lies nearer 63 than 65 because copper-63 is the more abundant isotope

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