Mass spectra & calculations
Mass spectra & calculations
Reading elemental mass spectra
🔓 CRACK IT
Use the information on the spectrum. You do not need TOF instrument stages or kinetic-energy calculations here.
- Peak position: read m/z from the x-axis. For a singly charged monatomic ion, this is numerically equal to its relative isotopic mass.
- Peak height: read relative abundance or intensity from the y-axis. The tallest isotope peak represents the most abundant isotope.
- Check whether abundances are percentages, ratios or raw intensities before calculating the weighted mean.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
- State the quantity on the horizontal axis of a mass spectrum. [1]
- State the quantity on the vertical axis. [1]
- Explain why m/z is numerically equal to relative isotopic mass in a simple elemental spectrum of singly charged ions. [1]
- State what the relative abundance of a peak represents. [1]
Total: 4 marks
Show answer & marking points
- Mass-to-charge ratio, m/z. [1]
- Relative abundance or relative intensity. [1]
- The charge number z = 1, so dividing by z does not alter the numerical mass value. [1]
- The abundance of that isotope ion relative to the others detected. [1]
| m/z | Relative abundance |
|---|---|
| 24 | 79 |
| 25 | 10 |
| 26 | 11 |
Calculate Aᵣ and identify the element. Reference Aᵣ values: Na = 23.0, Mg = 24.3, Al = 27.0. [4]
Total: 4 marks
Show answer & marking points
- Weighted numerator = 24(79) + 25(10) + 26(11) = 2432. [1]
- Total abundance = 79 + 10 + 11 = 100. [1]
- Aᵣ = Σ(isotopic mass × abundance)/Σ(abundance)
Aᵣ = 2432/100 = 24.32. [1] - Magnesium. [1]
Calculating relative atomic mass
🔓 CRACK IT
Relative atomic mass is an average, but the isotopes do not all contribute equally. A common isotope must count more than a rare one. That is what weighted mean means.
- Multiply each isotope's relative mass by its abundance.
- Add those contributions.
- Divide by the total abundance. Use 100 for percentages, but add the values yourself for a ratio or raw intensities.
Your answer must lie between the lightest and heaviest isotope masses. It should be closer to the more abundant isotope. Relative atomic mass has no unit.
Ar relative atomic mass; no unit; Σ sum of all the indicated contributions.
All masses and abundances known
Masses 28, 29 and 30 have abundances 92%, 5% and 3%.
Calculate the relative atomic mass of the element.
Worked answer
Aᵣ = Σ(isotopic mass × abundance)/Σ(abundance)
Aᵣ = [28(92) + 29(5) + 30(3)]/100 = 2811/100.
An unknown isotope mass
Mass 63 has abundance 70.0%. The other isotope has abundance 30.0%. Aᵣ = 63.60.
Calculate the relative isotopic mass of the second isotope.
Worked answer
- 63.60 = [63(70) + 30x]/100.
- 6360 = 4410 + 30x.
- x = (6360 − 4410)/30.
Two unknown abundances
Isotope masses are 50, 52 and 54. Mass 50 has abundance 20.0%, and Aᵣ = 52.20.
Calculate the percentage abundances of isotopes 52 and 54.
Worked answer
- Let mass 52 have abundance x%. Mass 54 has (80 − x)%.
- 52.20 = [50(20) + 52x + 54(80 − x)]/100.
- 5220 = 5320 − 2x, so x = 50.0.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Two isotopes have accurate relative isotopic masses 78.918 and 80.916. The lighter : heavier abundance ratio is 4.35 : 1.00. Calculate Aᵣ to three decimal places. [4]
Total: 4 marks
Show answer & marking points
- Use the mass–ratio products: 78.918(4.35) + 80.916(1.00). [1]
- Divide by 4.35 + 1.00 = 5.35. [1]
- Aᵣ = Σ(isotopic mass × abundance)/Σ(abundance)
Aᵣ = 424.2093/5.35 = 79.2914579… [1] - Aᵣ = 79.291. [1]
An isotope of relative isotopic mass 63 has abundance 69.0%. A second isotope has relative isotopic mass x and abundance 31.0%. Aᵣ = 63.62. Calculate x. [4]
Total: 4 marks
Show answer & marking points
- 63.62 = [63(69.0) + 31.0x]/100. [1]
- 6362 = 4347 + 31x. [1]
- 31x = 2015. [1]
- x = 65. [1]
An element has relative isotopic masses 24, 25 and 26. Mass 24 has abundance 79.0%, and Aᵣ = 24.32. Let the abundance of mass 25 be x%. Calculate the abundances of masses 25 and 26. [5]
Total: 5 marks
Show answer & marking points
- Mass 26 has abundance (21 − x)%. [1]
- 24.32 = [24(79) + 25x + 26(21 − x)]/100. [1]
- 2432 = 2442 − x. [1]
- Mass 25: x = 10.0%. [1]
- Mass 26: 11.0%. [1]
An element has isotopes of relative isotopic mass 10 and 11. The mass-10 abundance is x%, and Aᵣ = 10.81. Calculate each percentage abundance. [4]
Total: 4 marks
Show answer & marking points
- Mass 11 has abundance (100 − x)%. [1]
- 10.81 = [10x + 11(100 − x)]/100. [1]
- 1081 = 1100 − x, so x = 19. [1]
- Mass 10: 19%; mass 11: 81%. [1]
A student says the following data cannot be used because the intensities do not add to 100.
| m/z | Relative intensity |
|---|---|
| 90 | 74.0 |
| 91 | 15.0 |
| 92 | 24.5 |
| 94 | 25.0 |
| 96 | 4.0 |
- Explain why the student is incorrect. [1]
- Use the data to show that Aᵣ = 91.32. [4]
Total: 5 marks
Show answer & marking points
- Relative intensities are a comparative scale. They need not total 100; use their actual total. [1]
- Total intensity 142.5 [1]; weighted numerator 13013.0 [1]; divide 13013.0/142.5 = 91.319298… [1]; Aᵣ = 91.32 [1].
R has isotopes of relative isotopic mass 64, 66 and 68. Mass 68 has abundance 25.0%; the mass-64 : mass-66 abundance ratio is 3 : 2.
- Calculate each percentage abundance. [3]
- Calculate Aᵣ. [3]
Total: 6 marks
Show answer & marking points
- 75.0% remains [1]. Divide this in the ratio 3 : 2, so one part is 15.0% [1]. Abundances are 45.0%, 30.0%, 25.0% for masses 64, 66, 68 respectively [1].
- Weighted numerator = 64(45.0) + 66(30.0) + 68(25.0) [1]. Divide by 100 [1] to obtain 65.60 [1].
The table gives the isotope data for element X.
| Relative isotopic mass | 58 | 60 | 61 | 62 |
|---|---|---|---|---|
| Abundance / % | 68.0 | x | 1.1 | 3.6 |
- Calculate x. [1]
- Calculate Aᵣ of X to two decimal places. [4]
- X forms ionic XCl₂. Calculate its relative formula mass using Aᵣ(Cl) = 35.45. [2]
- Explain why relative formula mass is used for XCl₂. [1]
Total: 8 marks
Show answer & marking points
- x = 100 − 68.0 − 1.1 − 3.6 = 27.3%. [1]
- Use all four weighted mass–abundance products [1]: 58(68.0) + 60(27.3) + 61(1.1) + 62(3.6) = 5872.3 [1]. Divide by 100 [1]: Aᵣ = 58.723 = 58.72 [1].
- Use the unrounded Aᵣ: 58.723 + 2(35.45) [1] = 129.623 = 129.62 [1].
- XCl₂ has a repeating ionic lattice, not discrete molecules. [1]
Molecular ions: analytical crossover
🔓 CRACK IT
This is an analytical-chemistry crossover, included to support the molecular-ion questions in your practice pack. It connects this topic to the later organic-analysis content.
A molecular ion is the intact molecule after losing an electron. For a singly charged molecular ion, its m/z gives the molecule's relative molecular mass.
The molecular-ion peak is usually the peak furthest to the right. A small M+1 isotope peak can appear just beyond it.
A molecule can also break into smaller ions called fragments. The tallest peak is the base peak; it need not be the molecular-ion peak. Identify the peak the question tells you to use.
- Molecular ion, M⁺
- The intact molecule after removal of one electron. For the identified singly charged molecular ion, m/z gives its relative mass.
- Base peak
- The most intense peak, conventionally scaled to 100. It may represent a fragment ion.
Mr relative molecular or formula mass; no unit; m/z mass-to-charge ratio: m represents the ion’s relative mass and z its charge number.
Mr relative molecular or formula mass; no unit.
Use the identified ion
A spectrum has an identified singly charged molecular ion at m/z 72 and a base peak at m/z 43.
Determine the relative molecular mass of the compound and explain which peak you use.
Worked answer
Use the intact molecular ion, not the most intense fragment.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
A compound’s spectrum has an identified singly charged molecular-ion peak at m/z 86 and a base peak at m/z 43.
- State Mᵣ. [1]
- State what is meant by the base peak. [1]
- Explain why the base peak is not necessarily used to determine Mᵣ. [1]
Total: 3 marks
Show answer & marking points
- Mᵣ = 86. [1]
- The most intense peak, conventionally assigned relative intensity 100. [1]
- It may be a fragment ion and therefore may not represent the intact molecule. [1]
Stretch / crossover: this combines mass spectra with empirical and molecular formulae. A compound contains 54.5% C, 9.1% H and 36.4% O by mass. Its identified singly charged molecular-ion peak is at m/z 88; its base peak is at m/z 43. Use Aᵣ: C 12.0, H 1.0, O 16.0.
- Determine its empirical formula. [4]
- State Mᵣ. [1]
- Determine its molecular formula. [2]
- Explain why the m/z 43 peak is not used to determine Mᵣ. [1]
Total: 8 marks
Show answer & marking points
- Divide by Aᵣ: C 54.5/12 = 4.5417…, H 9.1/1 = 9.10, O 36.4/16 = 2.275 [1]. Divide all by 2.275 [1] to obtain approximately 2 : 4 : 1 [1]. Empirical formula C₂H₄O [1].
- Mᵣ = 88. [1]
- Empirical-formula mass = 44, so multiplier = 88/44 = 2 [1]. Molecular formula C₄H₈O₂ [1].
- The m/z 43 base peak represents a fragment rather than the intact molecular ion. [1]