Mass spectra & relative masses

9 practice questions

Reading elemental mass spectra

Learning goalInterpret peak positions and intensities, and use spectra to identify elements.

🔓 CRACK IT

Read a mass spectrum using its axes. The x-axis shows m/z; the y-axis shows the relative abundance of each ion.

The x-axis is mass-to-charge ratio, m/z. For singly charged mononuclear ions, it is numerically equal to relative isotopic mass.
The y-axis is relative abundance or intensity. A taller peak means more ions of that isotope.
Use the isotope pattern and calculated weighted mean to identify the element. Check against its Periodic Table value.

EXAM TRAPPeak heights are relative quantities. They need not add to 100 unless they are given as percentages.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 16

m/z	Relative abundance / %
19.9924	90.48
20.9938	0.27
21.9914	9.25

State the quantity shown on each axis of an elemental mass spectrum. [2]
Calculate the relative atomic mass to four significant figures. [3]
Identify the element. [1]

Total: 6 marks

Show answer & marking points

x-axis: m/z. y-axis: relative abundance or intensity. [2]
Aᵣ = [(19.9924 × 90.48) + (20.9938 × 0.27) + (21.9914 × 9.25)] / 100 = 20.18001128 = 20.18. Correct weighted sum [1], denominator [1], answer [1].
Neon. [1]

Exam-style question 17

A singly charged ion has KE = 7.50 × 10⁻¹⁶ J and travels 1.25 m in 10.92 µs. Use L = 6.022 × 10²³ mol⁻¹.

Calculate its mass number. [5]
The element has exactly two isotopes. The other mass number is two greater. Currents are 3.20 nA for the lighter isotope and 1.10 nA for the heavier. Calculate their percentage abundances. [2]
Calculate Aᵣ to three decimal places. [2]
Explain why the heavier isotope gives the smaller current. [1]

Total: 10 marks

Show answer & marking points

v = d/t.
v = 1.25/(10.92 × 10⁻⁶) = 1.1447 × 10⁵ m s⁻¹ [1].
m = 2KE/v² [1].
m = (2 × 7.50 × 10⁻¹⁶)/(1.1446886… × 10⁵)² = 1.14476544 × 10⁻²⁵ kg [1].
Relative mass = m × L × 1000.
Relative mass = (1.14476544 × 10⁻²⁵) × (6.022 × 10²³) × 1000 = 68.94 [1].
Mass number = 69 [1].
Total current = 4.30 nA. Isotope 69: 3.20/4.30 × 100 = 74.4% [1]. Isotope 71: 25.6% [1].
Aᵣ = Σ(isotopic mass × abundance)/Σ(abundance)
Aᵣ = [69(3.20) + 71(1.10)]/4.30 [1] = 69.512 [1].
It is less abundant, so fewer ions of that isotope arrive. [1]

Exam-style question 18

Stretch: this molecular-isotope probability calculation goes beyond the usual mononuclear-ion Aᵣ calculation requirement.

Element X has isotopes ²⁴X and ²⁶X with abundances 79.0% and 21.0%. X₂ is analysed by electron impact. Assume molecular ions form without fragmentation.

State the three m/z values for X₂⁺. [2]
Calculate the expected percentage ratio of the peaks. [3]
Explain why the middle peak can form in two ways. [1]

Total: 6 marks

Show answer & marking points

m/z = 48, 50, 52. [2]
P(48) = 0.790² = 62.41% [1]; P(50) = 2(0.790)(0.210) = 33.18% [1]; P(52) = 0.210² = 4.41% [1]. Ratio = 62.41 : 33.18 : 4.41.
Mixed pairs can be ²⁴X–²⁶X or ²⁶X–²⁴X, so double the mixed-isotope probability. [1]

✓ DONE WHENYou can interpret both axes, identify an element and distinguish abundance from flight time.

Calculating relative atomic mass

Learning goalCalculate weighted means and solve missing-mass and missing-abundance problems.

🔓 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 = Σ(isotopic mass × abundance) / Σ(abundances)

A_r relative atomic mass; no unit; Σ sum of all the indicated contributions.

Worked example

Type 1: all masses and abundances known

Isotopes 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.

Aᵣ = 28.11

Worked example

Type 2: one mass unknown

An element has Aᵣ = 63.62. Isotope 63 has abundance 69%; the other isotope has abundance 31%.

Calculate the relative isotopic mass of the second isotope.

Worked answer

63.62 = [63(69) + 31x] / 100.
6362 = 4347 + 31x.
x = (6362 − 4347) / 31 = 65.

The second mass number is 65.

Worked example

Type 3: two abundances unknown

Isotopes 24, 25 and 26 give Aᵣ = 24.32. Isotope 24 has abundance 79.0%.

Calculate the percentage abundances of isotopes 25 and 26.

Worked answer

Let isotope 25 have abundance x%. Isotope 26 then has abundance (21 − x)%.
24.32 = [24(79) + 25x + 26(21 − x)] / 100.
2432 = 2442 − x, so x = 10.

Isotope 25: 10%; isotope 26: 11%.

EXAM TRAPFor three isotopes, the last abundance is 100 − known abundance − x.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 19

An element has isotopes with 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.

Total: 4 marks

Show answer & marking points

Use the ratio as weighting factors. [1]
Aᵣ = [(78.918 × 4.35) + (80.916 × 1.00)] / 5.35. [1]
Aᵣ = 79.2914579… [1]
Aᵣ = 79.291. [1]

Exam-style question 20

An element has Aᵣ = 58.573. One isotope has accurate relative isotopic mass 57.935 and abundance 68.0%. The second isotope has abundance 32.0%.

Calculate the accurate relative isotopic mass of the second isotope to three decimal places. [3]
State its mass number. [1]

Total: 4 marks

Show answer & marking points

58.573 = 57.935(0.680) + 0.320x [1]. x = [58.573 − 57.935(0.680)] / 0.320 [1] = 59.92875 = 59.929 [1].
Mass number = 60. [1]

Exam-style question 21

Two isotopes have relative isotopic masses 10 and 11. Aᵣ = 10.8. Calculate their percentage abundances.

Total: 4 marks

Show answer & marking points

Let mass 10 have abundance x%; mass 11 has (100 − x)%.
10.8 = [10x + 11(100 − x)]/100.
1080 = 1100 − x, so x = 20.
Mass 10: 20%; mass 11: 80%.

✓ DONE WHENYou can solve all three weighted-mean question types and check the result.

Relative molecular mass from mass spectra

Learning goalIdentify molecular and protonated molecular ions and distinguish them from the base peak.

🔓 CRACK IT

First identify which ion the question describes. A molecular ion formed by electron impact has lost one electron; its mass is effectively the mass of the original molecule.

The molecular-ion peak is usually the peak furthest to the right, because the intact molecule is heavier than its fragments. Isotope peaks, such as a small M+1 peak, can appear just beyond it.

For a singly charged molecular ion, use its m/z as Mᵣ. An electrospray [MH]⁺ ion includes an extra proton, so subtract 1 from its m/z to find Mᵣ of M.

The base peak is simply the most intense peak. It may be a fragment, so it does not automatically give the mass of the intact molecule. These shortcuts assume singly charged ions; check the charge and ion label first.

Molecular-ion peak, M⁺: The intact molecule after one electron is removed. For a singly charged molecular ion, its m/z gives the relative mass of that molecular ion.
Base peak: The tallest peak, representing the most abundant ion. It is not automatically the molecular ion.

Electron impact: Mᵣ = m/z of the identified M⁺ ion

M⁺ the singly charged molecular ion; M a neutral molecule; M⁺ is its positive molecular ion and [MH]⁺ is its protonated ion; M_r 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.

Electrospray: Mᵣ = m/z of [MH]⁺ − 1

[MH]⁺ the protonated molecule: M with an extra H⁺ attached; M a neutral molecule; M⁺ is its positive molecular ion and [MH]⁺ is its protonated ion; M_r 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.

Worked example

Read the right peak

A [MH]⁺ ion appears at 151. In another spectrum, the base peak is 43 and the identified M⁺ peak is 72.

Determine the relative molecular mass of the compound in each spectrum.

Worked answer

For the protonated ion, subtract 1. For the identified molecular ion, use its m/z.

Mᵣ = 150 and 72 respectively.

EXAM TRAPDo not automatically choose the tallest or the highest-m/z peak. Identify the intact molecular ion and account for charge/protonation.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 22

An electron-impact spectrum has peaks at m/z 41, 55 and 70. The base peak is at 41; the molecular-ion peak is at 70.

State the relative molecular mass. [1]
Explain the term base peak and why it does not necessarily give Mᵣ. [2]

Total: 3 marks

Show answer & marking points

Mᵣ = 70. [1]
The base peak is the most intense peak, assigned relative intensity 100 [1]. It may be a fragment ion, so it need not give Mᵣ [1].

Exam-style question 23

A reference ion of relative isotopic mass 125.0 takes 18.6 µs. A singly charged [MH]⁺ ion takes 23.3 µs in the same instrument. Both have the same kinetic energy and travel the same distance.

Calculate the relative mass of [MH]⁺ on the carbon-12 scale. [4]
Determine Mᵣ of M. [1]
Why is electrospray often used for a large molecule? [1]
An electron-impact spectrum of M has a base peak at 43 and molecular-ion peak at 195. Which confirms Mᵣ, and why? [2]

Total: 8 marks

Show answer & marking points

Equal KE and distance give m/t² = constant [1]. m₂/m₁ = (t₂/t₁)² [1]. m₂ = m₁(t₂/t₁)²
m₂ = 125.0(23.3/18.6)² [1] = 196.153… ≈ 196 to 3 s.f. [1].
Mᵣ ≈ 196.153… − 1 = 195.153… ≈ 195. [1]
Less fragmentation, so the large molecule is more likely to remain intact. [1]
Use m/z 195 because it is the intact molecular ion [1]. The intense fragment at 43 does not give Mᵣ [1].

✓ DONE WHENYou can extract Mᵣ from the identified M⁺ or [MH]⁺ peak.

Relative masses: definitions and language

Learning goalDefine relative atomic and molecular mass using carbon-12 and use relative formula mass for ionic compounds.

🔓 CRACK IT

A relative mass compares a particle's mass with a common reference: one-twelfth of the mass of one carbon-12 atom. It is a ratio of two masses, so it has no unit.

Read the particle named in the definition. Relative isotopic mass refers to one isotope. Relative atomic mass is a weighted mean that allows for an element's isotope abundances.

Relative molecular mass refers to molecules. For a substance without separate molecules, such as an ionic compound, use relative formula mass. Molar mass is a different quantity: the mass of one mole, usually in g mol⁻¹.

Relative isotopic mass: The mass of an atom of one isotope compared with one-twelfth of the mass of a carbon-12 atom.
Relative atomic mass, Aᵣ: The weighted mean mass of an atom of an element compared with one-twelfth of the mass of a carbon-12 atom.
Relative molecular mass, Mᵣ: The mean mass of a molecule compared with one-twelfth of the mass of a carbon-12 atom.
Relative formula mass: The sum of Aᵣ values in one formula unit. Use this term for ionic compounds.

Mᵣ or relative formula mass = Σ(Aᵣ × number of each atom)

A_r relative atomic mass; no unit; Σ sum of all the indicated contributions.

Worked example

Molecules or ionic compounds?

Calculate the relative molecular mass of CO₂ and the relative formula mass of NaCl. Use Aᵣ values: C = 12.0, O = 16.0, Na = 23.0 and Cl = 35.5.

Worked answer

CO₂ is molecular: Mᵣ = sum of the Aᵣ values of all atoms in the formula
Mᵣ = 12.0 + 2(16.0) = 44.0.
NaCl is ionic: relative formula mass = 23.0 + 35.5 = 58.5.

44.0 and 58.5; neither has units.

EXAM TRAPInclude “weighted mean” in Aᵣ and the one-twelfth carbon-12 reference in both definitions.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 24

Define relative atomic mass. [2]
Define relative molecular mass. [2]
State the term used instead of relative molecular mass for an ionic substance. [1]

Total: 5 marks

Show answer & marking points

The weighted mean mass of an atom of an element compared with one-twelfth of the mass of a carbon-12 atom. [2]
The mean mass of a molecule compared with one-twelfth of the mass of a carbon-12 atom. [2]
Relative formula mass. [1]

✓ DONE WHENYou can give precise definitions and choose the correct term without adding units.

Mass spectra & relative masses: AQA A-level Chemistry