Mass spectrometry — AQA A-Level Chemistry
Test yourself on Mass spectrometry with AQA A-Level practice questions.
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Mass spectrometry explained
In a mass spectrometer a molecule loses one electron to become the molecular ion, so the peak at the highest mass to charge ratio, ignoring small isotope peaks beyond it, gives the relative molecular mass directly.
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That alone leaves several possible formulae, because many combinations share the same whole-number mass. A high resolution instrument measures the mass to four or five decimal places, and because the precise masses of individual isotopes are not whole numbers, formulae that appear identical at whole-number resolution separate cleanly. What the technique cannot do is separate isomers: two isomers have exactly the same molecular formula and therefore exactly the same precise molecular mass, which is why AQA offers infrared fingerprinting instead in that situation.
Students should be able to: use precise atomic masses and the precise molecular mass to determine the molecular formula of a compound.
Work from the precise atomic masses given in the question, for example ¹H = 1.00783, ¹²C = 12.00000, ¹⁴N = 14.00307 and ¹⁶O = 15.99491. List the candidate formulae with the right whole-number mass, then add the precise masses for each. At a nominal mass of 60, C₃H₈O comes to 60.0576, CH₄N₂O to 60.0324 and C₂H₄O₂ to 60.0211, so a measured value of 60.0211 identifies the last of the three without ambiguity. Compare to the same number of decimal places the question quotes, and say which candidates you have rejected and why. Use the precise mass of the individual isotope, not the weighted average relative atomic mass in the periodic table, which already blends isotopes together.
Your focus
- Identify the molecular ion peak on a mass spectrum and state the relative molecular mass it gives.
- Explain why high resolution is needed to choose between formulae of the same whole-number mass.
- Explain why mass spectrometry alone cannot distinguish two structural isomers.
Show all 6 objectives
- Calculate the precise molecular mass of a candidate formula from precise atomic masses.
- Reject a candidate formula by quoting its calculated precise mass beside the measured value and stating the difference that rules it out.
- Explain why average relative atomic masses cannot be used in this calculation.
Mass spectrometry exam tips
Marking Points
- one mark for identifying the molecular ion peak as the peak at the highest mass to charge ratio
- one mark for equating the mass to charge ratio of that peak with the relative molecular mass
- one mark for saying precise, rather than whole-number, atomic masses must be used
- one mark for selecting the formula whose calculated precise mass matches the measured value
- one mark for calculating the precise mass of one candidate formula from precise atomic masses
- one mark for calculating a second candidate for comparison
- one mark for comparing to the same number of decimal places as the measured value
- one mark for stating the formula that matches and rejecting the others with a reason
Examiner Tips
- 💡State that the molecular ion has lost one electron, so its mass is effectively that of the molecule.
- 💡Work to as many decimal places as the data gives you, and only round at the very end.
- 💡If a question asks you to distinguish isomers, reach for the infrared fingerprint region, not the mass spectrum.
- 💡Set the arithmetic out element by element so a slip in one term is easy to find.
- 💡Keep every decimal place through the working and compare only at the end.
- 💡Always state the rejected candidate and the value that rules it out; the comparison carries a mark.
Common Mistakes
- reading the base peak, the tallest one, as the molecular ion peak
- rounding the precise masses, which destroys the discrimination the method depends on
- expecting high resolution mass spectrometry to distinguish two isomers
- saying the molecular ion is positive because a proton has been added rather than an electron removed
- using the periodic table's average relative atomic masses instead of the precise isotopic masses
- rounding to two decimal places, at which two candidates can still look identical
- assuming the nearest whole-number mass is enough to fix the formula
- giving an empirical formula when the molecular formula was asked for