The shape of the lesson
By the end of the lesson, learners can
- The molecular ion and its fragments: Build the mass spectrum of a molecule in the simulator.
- Counting carbons with M+1: Every carbon atom has a 1.1% chance of being 13 C, so the more carbon atoms in the molecule, the taller the M+1 peak relative to M.
- Chlorine and bromine: the M+2 peak: Chlorine and bromine each have two common isotopes two mass units apart, so a compound containing them shows a second molecular ion peak at M+2.
- M+: last major peak, gives Mr.
- Fragments: 15 CH3+, 29 C2H5+, 43 C3H7+ or CH3CO+.
- M+1: n = 100 × (M+1) / (1.1 × M).
- M+2: 3 : 1 is Cl, 1 : 1 is Br.
One picture every idea returns to
In the spectrometer a molecule loses an electron to form the molecular ion, M+. The M+ peak, at the highest m/z apart from its isotope peaks, gives the Mr. Some molecular ions break into a fragment ion and a neutral radical; only the ion is detected, so peaks such as 15 (CH3+), 29 (C2H5+ or CHO+) and 43 (C3H7+ or CH3CO+) reveal pieces of the structure. About 1.1% of carbon is 13C, so a small M+1 peak appears and the number of carbon atoms is n = 100 × (M+1 height) / (1.1 × M height). An M+2 peak one third the height of M shows one Cl atom (35Cl : 37Cl = 3 : 1); M and M+2 of equal height show one Br atom (79Br : 81Br = 1 : 1).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Break the molecule, weigh the pieces.. A mass spectrometer ionises a molecule and records the masses of the ions. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The molecular ion and its fragments | Build the mass spectrum of a molecule in the simulator. Project tab 4 of the lab (Analytical techniques lab, a mass spectrum) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Counting carbons with M+1 | Every carbon atom has a 1.1% chance of being 13 C, so the more carbon atoms in the molecule, the taller the M+1 peak relative to M. Project tab 4 of the lab (Analytical techniques lab, a mass spectrum) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Chlorine and bromine: the M+2 peak | Chlorine and bromine each have two common isotopes two mass units apart, so a compound containing them shows a second molecular ion peak at M+2. Project tab 5 of the lab (Analytical techniques lab, chlorine and bromine patterns) and let a student drive it while the class predicts. | Cooperative Concept Mapping |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 14 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Chlorine and bromine: the M+2 peak, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The M + peak is the last significant peak before its own small isotope peaks, not the tallest (the tallest is the base peak). | In the M+1 formula use peak heights, not m/z values, and put the M+1 height on top. Only ions are detected, never the radical lost. A 3: 1 pair two units apart means chlorine; 1: 1 means bromine, so do not swap them. |
Mass spectrometry: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
In the mass spectrum of a hydrocarbon the M+ peak at m/z 86 has a relative abundance of 20.0 and the M+1 peak 1.32. Calculate the number of carbon atoms and give the molecular formula.
Write an equation for the formation of the molecular ion of propanone, and an equation for its fragmentation to give the peak at m/z 43.
A compound contains carbon, hydrogen and one halogen. Its spectrum shows molecular ion peaks at m/z 78 and 80 with heights in the ratio 3 : 1, and a large peak at 43. Identify the halogen and suggest a structure.
The M+ peak of an ester has height 24.0 and the M+1 peak 2.11. Mr = 144. Find the number of carbon atoms and suggest a molecular formula with two oxygen atoms.
Explain why the mass spectrum of 1-bromopropane shows two peaks of almost equal height at m/z 122 and 124.
Compound Q, C3H8O, is an alcohol. Its mass spectrum has M+ at 60 and a large peak at m/z 31 but only a small peak at 45. Suggest the structure of Q and explain the peak at 31.
Answer key and marking notes
- n = 100 × 1.32 / (1.1 × 20.0) = 132 / 22.0 = 6.00, so 6 carbon atoms.
- 6 × 12 = 72; 86 − 72 = 14 H, so C6H14.
- CH3COCH3 + e− → [CH3COCH3]+ + 2e− (m/z 58).
- [CH3COCH3]+ → CH3CO+ + CH3•; CH3CO+ is detected at m/z 43.
- 3 : 1 pair two units apart: one chlorine atom (35Cl and 37Cl).
- 78 − 35 = 43, so the rest of the molecule is C3H7; the peak at 43 is C3H7+.
- C3H7Cl: 1-chloropropane or 2-chloropropane.
- n = 100 × 2.11 / (1.1 × 24.0) = 211 / 26.4 = 7.99, so 8 carbon atoms.
- 8 × 12 + 2 × 16 = 128; 144 − 128 = 16 H, so C8H16O2.
- Bromine exists as 79Br and 81Br in roughly equal amounts.
- C3H779Br+ has m/z 43 + 79 = 122 and C3H781Br+ has 43 + 81 = 124, so the two molecular ions are equally common.
- The alcohol is propan-1-ol or propan-2-ol.
- m/z 31 is CH2OH+, formed when the C1–C2 bond breaks; this needs an OH on a CH2 at the end of the chain.
- So Q is propan-1-ol, CH3CH2CH2OH. Propan-2-ol would give a large peak at 45, CH3CHOH+, from loss of CH3.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The simplified mass spectrum of compound C is shown. The peaks at m/z 64 and 66 are both molecular ion peaks. Which formula fits C?
- AC2H5Cl
- BC2H5Br
- CC3H7Cl
- DC5H4
In the mass spectrum of an alkane the molecular ion peak has a relative abundance of 50.0 and the M+1 peak 2.75. How many carbon atoms are in each molecule?
- A5
- B6
- C2.5
- D55
The mass spectrum of propanal, CH3CH2CHO, has a large peak at m/z 29. Which ion or ions could cause this peak?
- AC2H5+ or CHO+
- BCH3+ only
- CCH3CO+ or C3H7+
- Dthe molecular ion, M+
How is the relative molecular mass of an organic compound found from its mass spectrum?
- Afrom the m/z of the molecular ion peak, the last major peak at the high-mass end before its small isotope peaks
- Bfrom the m/z of the tallest peak, the base peak
- Cfrom the m/z of the M+1 peak
- Dby adding the m/z values of all the fragment peaks
An organic compound has two molecular ion peaks, at m/z 136 and 138, of almost equal height. What does this show?
- Athe molecule contains one bromine atom
- Bthe molecule contains one chlorine atom
- Cthe molecule contains two chlorine atoms
- Dthe molecule contains 13 carbon atoms
Compound D has Mr 58. In its mass spectrum, shown, the M+ peak at 58 has height 30.0 and the M+1 peak at 59 has height 0.99. Which compound is D?
- Apropanone, CH3COCH3
- Bpropanal, CH3CH2CHO
- Cbutane, CH3CH2CH2CH3
- D2-methylpropane, (CH3)3CH
Dichloromethane, CH2Cl2, contains two chlorine atoms. Which molecular ion peaks, with their relative heights, does it show?
- Am/z 84, 86 and 88 in the ratio 9 : 6 : 1
- Bm/z 84 and 86 in the ratio 3 : 1
- Cm/z 84, 86 and 88 in the ratio 1 : 2 : 1
- Dm/z 84 and 88 in the ratio 3 : 1
Compound E, Mr 74, contains C, H and O. Its M+ peak has height 40.0 and its M+1 peak 1.76. The spectrum has a large peak at m/z 31 and a peak at 43, but none at 45. What is E?
- Abutan-1-ol, CH3CH2CH2CH2OH
- Bbutan-2-ol, CH3CH(OH)CH2CH3
- Cpropanoic acid, CH3CH2COOH
- Dmethyl ethanoate, CH3COOCH3
Quiz key
Two molecular ion peaks two units apart with heights about 100 : 33, a 3 : 1 ratio, show one chlorine atom: 35Cl gives m/z 64 and 37Cl gives 66. 64 − 35 = 29, which is C2H5, so C is chloroethane; the peak at 29 is C2H5+ and the pair at 49 and 51 is CH2Cl+. Bromoethane would give peaks at 108 and 110 of equal height. C3H7Cl has Mr 78. A hydrocarbon of Mr 64 would have no 3 : 1 M+2 peak.
n = 100 × 2.75 / (1.1 × 50.0) = 275 / 55.0 = 5.00, so five carbon atoms: the alkane is C5H12 (Mr 72). The M+1 peak exists because 1.1% of carbon atoms are 13C. 6 comes from forgetting the 1.1 (275 / 50.0 = 5.5, rounded up). 2.5 is 2.75 / 1.1, which ignores the height of the M peak. 55 is the denominator, 1.1 × 50.0, not the answer.
C2H5+ = 2(12) + 5(1) = 29 and CHO+ = 12 + 1 + 16 = 29. Breaking the bond between the ethyl group and the CHO group of propanal can give either ion, depending on which piece keeps the charge. CH3+ gives m/z 15. CH3CO+ and C3H7+ both give 43. The molecular ion of propanal is at 58, its Mr.
The molecular ion M+ has lost only an electron, so with a charge of +1 its m/z equals Mr. It is the major peak at the highest m/z, apart from the small M+1 (and any M+2) isotope peaks just after it. The base peak is simply the most abundant ion and is often a stable fragment, such as CH3CO+ at 43. The M+1 peak is one unit too high; it is used to count carbon atoms. Fragment masses overlap and are not simply additive, because each fragment comes from a different break.
Bromine is about 50% 79Br and 50% 81Br, so a compound with one Br atom gives M and M+2 peaks of equal height. Here 136 − 79 = 57, which is C4H9: a bromobutane. One chlorine gives a 3 : 1 pair. Two chlorines give three peaks, M, M+2 and M+4, in the ratio 9 : 6 : 1. A 13C effect shows as an M+1 peak, one unit above M, and is small unless the molecule has very many carbons.
n = 100 × 0.99 / (1.1 × 30.0) = 99 / 33.0 = 3.00, so D has three carbons: C3H6O (Mr 58). The base peak at 43 is CH3CO+ (58 − 15, loss of CH3) and 15 is CH3+; there is no peak at 29, so there is no ethyl or CHO group. That is propanone. Propanal would give strong peaks at 29 (C2H5+ and CHO+). Butane and 2-methylpropane also have Mr 58 and a peak at 43 (C3H7+), but they have four carbons, so their M+1 peak would be 1.32, not 0.99.
CH2 is 14. Both 35Cl: 14 + 70 = 84, probability (3/4)2 = 9/16. One of each: 86, probability 2 × 3/4 × 1/4 = 6/16. Both 37Cl: 88, probability (1/4)2 = 1/16. So M : M+2 : M+4 = 9 : 6 : 1. A 3 : 1 pair would mean only one chlorine. 1 : 2 : 1 is the pattern for two bromine atoms, whose isotopes are 1 : 1. Leaving out the mixed-isotope ion at 86 ignores that there are two ways to have one of each isotope.
n = 100 × 1.76 / (1.1 × 40.0) = 176 / 44.0 = 4.00, so four carbons: C4H10O (Mr 74). The peak at 31 is CH2OH+, which forms when the OH is on a terminal CH2, and 43 is C3H7+, the rest of the chain: butan-1-ol. Butan-2-ol has the same Mr and carbon count, but its CH3CH(OH)+ fragment gives a peak at 45, which is absent. Propanoic acid and methyl ethanoate (C3H6O2) also have Mr 74, but only three carbons, so their M+1 peak would be 1.32.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.