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4.9.1Unit 4. Rates, Equilibria and Further Organic Chemistry

High Resolution Mass Spectrometry — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 5 parts Lesson 51 of 56 Unit quiz

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Curriculum Equation Key معادلة تفاعل تكوين الإستر (الأسترة) - Esterification Reaction EquationFast Exam Solution
R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O
Word Equation & Conditions: حمض كربوكسيلي + كحول (في وجود حمض الكبريتيك المركز كعامل حفاز ونازع للماء) ⇌ إستر + ماء
Location & Specifications: الكيمياء العضوية للثانوية العامة والتحصيلي و Edexcel Chemistry.
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Section 1 of 5Precise Masses

Low resolution instruments give masses to the nearest whole number, whereas high resolution instruments measure to several decimal places.

Mass defect
  • Mass defect Isotopic masses are not whole numbers, except for carbon-12 which is defined as exactly 12.0000.
Different values
  • Different values Hydrogen is 1.0078, nitrogen-14 is 14.0031 and oxygen-16 is 15.9949, so each combination of atoms has a unique precise mass.
Resolution
  • Resolution High resolution mass spectrometry can distinguish ions with the same nominal mass but different molecular formulae.
Output
  • Output The molecular ion peak is quoted to four decimal places and compared with calculated masses for candidate formulae.

Section 2 of 5Working Out the Formula

The molecular ion peak gives the relative molecular mass of the compound.

1
Identify the molecular ion, M⁺, which is the peak at the highest m/z value with a significant intensity.
2
List possible molecular formulae that have the same nominal mass.
3
Calculate the precise mass of each candidate by adding the isotopic masses of its atoms.
4
Choose the formula whose calculated mass is closest to the measured value from the spectrum.

Section 3 of 5Fragmentation and Practical Points

Fragmentation and Practical Points
Fragment ions
  • Fragment ions High energy bombardment breaks the molecular ion into fragments, and their m/z values suggest the structure.
Common losses
  • Common losses A loss of 15 suggests CH₃, a loss of 29 suggests C₂H₅ or CHO, and a loss of 17 suggests OH.
Sample
  • Sample Only a very small sample is needed, and the substance is vaporised and ionised in the instrument.
Limits
  • Limits Isomers have identical precise masses, so the technique alone cannot tell apart different structures with the same formula.

Section 4 of 5Worked Example

Question

  • A compound has a molecular ion at m/z 60.0210.
  • Which formula fits: C₂H₄O₂, C₃H₈O or CH₄N₂O?
  • Use C = 12.0000, H = 1.0078, N = 14.0031, O = 15.9949.

  1. C₂H₄O₂2 × 12.0000 + 4 × 1.0078 + 2 × 15.9949 = 24.0000 + 4.0312 + 31.9898 = 60.0210.
  2. C₃H₈O36.0000 + 8.0624 + 15.9949 = 60.0573.
  3. CH₄N₂O12.0000 + 4.0312 + 28.0062 + 15.9949 = 60.0323.

Answer C₂H₄O₂, whose calculated precise mass of 60.0210 matches the measured value.

Section 5 of 5Exam Checklist & Tips

4.9.1 – Key Points for the Exam

  • Molecular ion Ion formed when the whole molecule loses one electron.
  • Nominal mass Relative mass rounded to a whole number.
  • Precise mass Mass measured to several decimal places.
  • m/z Mass-to-charge ratio of an ion.

Common Mistakes

  • Using whole number masses to distinguish between formulae with the same nominal mass.
  • Taking the largest peak rather than the highest m/z peak as the molecular ion.
  • Forgetting that isomers give identical precise masses.
  • Using average relative atomic masses from the periodic table for precise mass.

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