S4 Chemistry · Unit 1

Structure of an Atom &
Mass Spectrum

Discovery of subatomic particles, atomic notation, isotopes, relative atomic mass, the mass spectrometer, and interpretation of mass spectra.

1.1 Discovery 1.2 Atomic Numbers 1.3 Relative Atomic Mass 1.4 Mass Spectrometer 1.5 Mass Spectra 1.6 Uses & Calculations Exercises Quiz Unit Test
1.1

Discovery of Atomic Constituents & Their Properties

The Story of the Atom

The modern atom was revealed through a series of landmark experiments. Each discovery overturned an earlier model and added a new layer of structure.

Dalton's Atomic Theory (1803)

John Dalton proposed that matter is made of indivisible, indestructible atoms. All atoms of a given element are identical in mass and properties; atoms of different elements differ in mass. Compounds form when atoms combine in fixed whole-number ratios. This model could not explain electrical phenomena, and later experiments proved atoms contain smaller particles.

Discovery of the Electron — J.J. Thomson (1897)

Using cathode ray tubes, Thomson showed that cathode rays consist of negatively charged particles deflected by electric and magnetic fields. He measured the charge-to-mass ratio: e/me = 1.76 × 1011 C kg−1. This proved atoms contain sub-atomic particles and are divisible.

Thomson's Plum Pudding Model: the atom is a uniform sphere of positive charge with electrons embedded within it — like plums in a pudding.

Rutherford's Gold Foil Experiment (1909–1911)

Rutherford, Geiger, and Marsden fired alpha particles at thin gold foil. Results:

This disproved Thomson's model. If charge were diffuse (plum pudding), only tiny deflections would occur — back-scattering is impossible without a concentrated mass.

SOURCE α foil most pass through small deflection back-scatter (rare)

Discovery of the Proton — Rutherford (1919)

Rutherford bombarded nitrogen with alpha particles and detected hydrogen nuclei — named protons. Protons carry charge +1 (1.6 × 10−19 C) and mass ≈ 1 u, and reside in the nucleus.

Discovery of the Neutron — Chadwick (1932)

Chadwick bombarded beryllium with alpha particles and detected neutral, penetrating radiation that ejected protons from paraffin wax. He identified these as neutrons — neutral particles of mass ≈ 1 u in the nucleus. Neutrons explain why atomic mass is roughly double atomic number for lighter elements.

Properties of Subatomic Particles

ParticleSymbolRelative MassRelative ChargeLocation
Protonp⁺1+1Nucleus
Neutronn⁰10Nucleus
Electrone⁻1/1836 ≈ 0−1Shells outside nucleus
ℹ️
Neutral atomIn a neutral atom: number of protons = number of electrons. The nucleus is positive because protons outweigh the effect of neutral neutrons.
Section 1.1 Quick Quiz
Atomic Constituents & Their Properties
10 Questions
Q1
Which subatomic particle was discovered first by J.J. Thomson in 1897?
Q2
The Rutherford gold foil experiment proved that:
Q3
The relative mass of a proton compared to an electron is approximately:
Q4
Which subatomic particle has no electric charge?
Q5
An atom is electrically neutral because:
Q6
James Chadwick discovered the neutron in 1932 by:
Q7
Which model of the atom did Rutherford's experiment disprove?
Q8
The relative charge of an electron is:
Q9
In Rutherford's model, why is the atom mostly empty space?
Q10
The proton was identified by Rutherford when he:
🎯

Section 1.1 — Atomic Constituents

10 Questions
Q1 of 10

Which particle has relative charge +1?

Proton: charge +1, mass 1 u. Electron: charge −1. Neutron: charge 0. Protons are in the nucleus.
Q2 of 10

Relative mass of an electron compared to a proton is:

Electron mass ≈ 1/1836 of proton. Treated as 0 for mass number purposes. Only protons + neutrons count toward mass number.
Q3 of 10

The nucleus of an atom contains:

Nucleus: protons (positive) + neutrons (neutral). Electrons orbit outside. Nucleus is tiny but very dense.
Q4 of 10

Who proposed the nuclear model from the gold foil experiment?

Rutherford (1911): most alpha particles passed through gold foil, some deflected sharply → small dense positive nucleus. Thomson had proposed the 'plum pudding' model earlier.
Q5 of 10

A neutron has relative charge of:

Neutron: mass = 1 u, charge = 0. Neutral — no charge. Contributes to mass number but not atomic number.
Q6 of 10

Mass number A equals:

A = protons + neutrons (nucleons). Example: ²³Na: 11 protons + 12 neutrons = A 23. Electrons have negligible mass.
Q7 of 10

In a neutral atom, electrons equal:

Neutral: no net charge → electrons = protons. Ions have gained/lost electrons so e⁻ ≠ protons.
Q8 of 10

J.J. Thomson's cathode rays were identified as:

Thomson (1897): same charge/mass ratio regardless of cathode material → electrons are universal. Led to discovery of the electron.
Q9 of 10

Atomic number determines:

Z = protons = chemical identity. All atoms of the same element have the same Z. Changing Z changes the element.
Q10 of 10

Chemical bonding involves primarily:

Bonding = sharing (covalent) or transfer (ionic) of outer/valence electrons. Protons and neutrons are locked in the nucleus.
1.2

Atomic Number, Mass Number & Isotopic Mass

Atomic Number (Z)The number of protons in the nucleus. Defines the element — unique to each element. In a neutral atom, Z also equals the number of electrons.
Mass Number (A)The total number of protons + neutrons (nucleons) in the nucleus.
A = Z + N  →  N = A − Z

Standard Nuclear Notation

ᴬX e.g. ²³Na → A=23, Z=11 ᶻ protons=11, neutrons=23−11=12

The superscript is the mass number; the subscript is the atomic number.

IsotopesAtoms of the same element (same Z) with different mass numbers (different number of neutrons). Same chemical properties (same electron config), different physical properties (different mass).
ElementIsotopeProtons (Z)NeutronsMass No. (A)
Hydrogen¹H (protium)101
Hydrogen²H (deuterium)112
Hydrogen³H (tritium)123
Carbon¹²C6612
Carbon¹³C6713
Carbon¹⁴C6814
Chlorine³⁵Cl171835
Chlorine³⁷Cl172037
Isotopic MassThe exact mass of one atom of a specific isotope, measured relative to 1/12 the mass of ¹²C. Isotopic masses are close to (but not exactly) whole numbers due to nuclear binding energy (mass defect).
💡
Why isotopes have the same chemistryChemical behaviour is governed by electron configuration, which depends only on the number of protons (Z). Neutrons sit in the nucleus and play no role in bonding.
Section 1.2 Quick Quiz
Atomic Number, Mass Number & Isotopic Mass
10 Questions
Q1
The atomic number (Z) of an element is defined as:
Q2
The mass number (A) of an atom is:
Q3
An atom of ²⁷Al (Z=13) contains how many neutrons?
Q4
Isotopes of an element differ in:
Q5
Which of the following are isotopes of oxygen?
Q6
⁵⁶₂₆Fe tells us this iron atom has:
Q7
Why do isotopes have identical chemical properties?
Q8
A neutral atom of an element has 19 electrons. Its ion X⁺ would have:
Q9
Chlorine has atomic number 17 and mass number 35. How many neutrons does ³⁵₁₇Cl have?
Q10
The term 'nuclide' refers to:
🎯

Section 1.2 — Atomic Number & Isotopes

10 Questions
Q1 of 10

Isotopes are atoms of the same element with:

Isotopes: same Z (same element) but different neutron number → different A. Example: ¹²C, ¹³C, ¹⁴C all have 6 protons.
Q2 of 10

²³Na has how many neutrons? (Z=11)

Neutrons = A − Z = 23 − 11 = 12.
Q3 of 10

²³⁵U and ²³⁸U (Z=92) differ in:

Both uranium: same Z=92. Neutrons: ²³⁵U → 143; ²³⁸U → 146. Same chemistry, different nuclear properties.
Q4 of 10

Isotopes have the same:

Same Z → same electron configuration → same valence electrons → same chemical properties. Different neutrons → different physical properties.
Q5 of 10

³⁵Cl (Z=17) has how many neutrons?

Neutrons = 35 − 17 = 18. The other common isotope ³⁷Cl has 37−17 = 20 neutrons.
Q6 of 10

Mass number is always:

A = protons + neutrons — both integers → A is always an integer. Relative atomic mass IS decimal (weighted average).
Q7 of 10

Which property do isotopes NOT share?

Isotopes differ in NEUTRON number. They share: protons, electrons, chemical properties, electron configuration.
Q8 of 10

¹⁴C (Z=6) has how many neutrons?

Neutrons = 14 − 6 = 8. ¹⁴C is radioactive — used in carbon dating. ¹²C (most common) has 6 neutrons.
Q9 of 10

Mg²⁺ ion (Z=12) has how many electrons?

Mg: 12 electrons. Mg²⁺: lost 2 → 10 electrons. Isoelectronic with Ne.
Q10 of 10

Oxygen (Z=8) with 10 neutrons has mass number:

A = Z + neutrons = 8 + 10 = 18. This is ¹⁸O — a stable heavy isotope of oxygen.
1.3

Calculation of Relative Atomic Mass

Relative Atomic Mass (Aᵣ)The weighted mean mass of an atom of an element relative to 1/12 the mass of a ¹²C atom. Accounts for the natural abundance of each isotope.
Aᵣ = Σ (isotopic mass × fractional abundance) = (m₁ × a₁) + (m₂ × a₂) + ... where aᵢ = % abundance ÷ 100
WORKED EXAMPLE

Aᵣ of Chlorine

³⁵Cl (75.0%) and ³⁷Cl (25.0%). Calculate Aᵣ(Cl).
1
Convert % → fractions: ³⁵Cl = 0.750, ³⁷Cl = 0.250
2
Aᵣ = (35 × 0.750) + (37 × 0.250) = 26.25 + 9.25
3
Aᵣ(Cl) = 35.5
WORKED EXAMPLE

Aᵣ of Boron

¹⁰B (19.9%) and ¹¹B (80.1%). Calculate Aᵣ(B).
1
Aᵣ = (10 × 0.199) + (11 × 0.801) = 1.99 + 8.811
2
Aᵣ(B) = 10.8
WORKED EXAMPLE

Finding % Abundance from Aᵣ

An element has two isotopes at m/z = 63 and m/z = 65. Aᵣ = 63.5. Find each % abundance.
1
Let x = fraction of m/z 63. Then (1−x) = fraction of m/z 65.
2
63.5 = (63 × x) + (65 × (1−x)) = 63x + 65 − 65x = 65 − 2x
3
2x = 65 − 63.5 = 1.5 → x = 0.75
4
m/z 63 = 75%, m/z 65 = 25%
Section 1.3 Quick Quiz
Calculation of Relative Atomic Mass
10 Questions
Q1
The relative atomic mass (Ar) is defined as:
Q2
Boron has ¹⁰B (19.9%) and ¹¹B (80.1%). Calculate Ar(B):
Q3
Why is Ar(Cl) = 35.5 and not a whole number?
Q4
Copper has ⁶³Cu (69.2%) and ⁶⁵Cu (30.8%). Calculate Ar(Cu):
Q5
Magnesium has: ²⁴Mg (79%), ²⁵Mg (10%), ²⁶Mg (11%). Calculate Ar(Mg):
Q6
The standard used to define relative atomic mass is:
Q7
An element has one isotope with mass 40 (96.9%) and another with mass 38 (0.06%) and 36 (0.34%). Ar ≈:
Q8
If an element has Ar = 35.5, which of these is definitely TRUE?
Q9
Neon has three isotopes: ²⁰Ne (90.48%), ²¹Ne (0.27%), ²²Ne (9.25%). Calculate Ar(Ne):
Q10
Which Ar value suggests an element has mainly ONE dominant isotope with very small amounts of others?
🎯

Section 1.3 — Relative Atomic Mass

10 Questions
Q1 of 10

Relative atomic mass is measured relative to:

1 u = 1/12 mass of ¹²C atom. This is the IUPAC standard. Carbon-12 chosen for precision in mass spectrometry.
Q2 of 10

³⁵Cl (75%) and ³⁷Cl (25%). Ar of Cl =

Ar = (35×75 + 37×25)/100 = (2625+925)/100 = 35.5. Weighted average — no atom actually has mass 35.5.
Q3 of 10

Formula for Ar from isotopic data:

Ar = Σ(isotopic mass × fractional abundance). Must weight by abundance — most abundant contributes most.
Q4 of 10

¹⁰B (20%) and ¹¹B (80%). Ar of B =

Ar = (10×20 + 11×80)/100 = (200+880)/100 = 10.8. Closer to 11 because ¹¹B dominates.
Q5 of 10

Ar of chlorine is not a whole number because:

Weighted average of 35 and 37 with abundances 75% and 25% → 35.5. Individual atoms are 35 or 37 (integers).
Q6 of 10

Ne isotopes: ²⁰Ne (90.5%), ²¹Ne (0.3%), ²²Ne (9.2%). Which contributes MOST to Ar?

²⁰Ne at 90.5% dominates. Ar(Ne) ≈ 20.18 — very close to 20.
Q7 of 10

Ar = 24.3 corresponds to which element?

Mg (Z=12) has Ar = 24.3. Isotopes: ²⁴Mg (78.6%), ²⁵Mg (10.1%), ²⁶Mg (11.3%).
Q8 of 10

50% ⁶³X + 50% ⁶⁵X. Ar =

Ar = (63×50 + 65×50)/100 = 64. This reflects copper-like composition.
Q9 of 10

Ar has no units because:

Ar = (atom mass)/(1/12 × mass ¹²C). Both masses in same units → ratio is dimensionless.
Q10 of 10

Si has Ar = 28.1. Its most abundant isotope is:

Ar close to 28 → ²⁸Si most abundant (92.2%). ²⁹Si (4.7%) and ³⁰Si (3.1%) pull Ar slightly above 28.
1.4

Description & Functioning of the Mass Spectrometer

Purpose

The mass spectrometer separates ions by their mass-to-charge ratio (m/z) to determine isotopic masses, natural abundances, relative atomic and molecular masses, and molecular structures.

StageWhat Happens
1. VaporisationSample is heated to produce gaseous atoms or molecules. Required for ions to travel freely through the vacuum.
2. IonisationHigh-energy electrons (70 eV) bombard the gas, knocking out one electron per atom: M → M⁺ + e⁻. This is electron impact (EI) ionisation. All ions carry +1 charge.
3. AccelerationIons are accelerated through a high-voltage electric field. All ions gain the same kinetic energy (KE = qV). Lighter ions travel faster: v = √(2qV/m).
4. DeflectionIons enter a magnetic field. Lighter ions (lower m/z) are deflected more; only ions of a specific m/z reach the detector as the field is varied.
5. DetectionIons hit the detector and generate an electrical current proportional to ion abundance. The output is a mass spectrum (m/z vs relative abundance).
VAPORISE IONISE ACCELERATE DEFLECT DETECT Sample Chamber Electron Gun Electric Plates Magnet Field Detector (recorder) High vacuum throughout
⚠️
High vacuum is essentialWithout a vacuum, ions would collide with air molecules, scattering them unpredictably and preventing accurate m/z measurement.
Section 1.4 Quick Quiz
Description & Functioning of the Mass Spectrometer
10 Questions
Q1
The correct order of stages in a mass spectrometer is:
Q2
In the ionisation stage, positive ions are formed by:
Q3
The mass spectrometer must operate under high vacuum because:
Q4
Lighter ions (smaller m/z) are deflected by the magnetic field:
Q5
The purpose of accelerating ions in the electric field is:
Q6
What does the y-axis of a mass spectrum represent?
Q7
The base peak in a mass spectrum is:
Q8
In a mass spectrum of neon, peaks appear at m/z = 20, 21, and 22. This means neon has:
Q9
Why must the sample be vaporised before entering the mass spectrometer?
Q10
The detector in a mass spectrometer produces a signal proportional to:
🎯

Section 1.4 — Mass Spectrometer

10 Questions
Q1 of 10

In a mass spectrometer, the sample is first:

Step 1: vaporisation (gaseous sample). Step 2: ionisation — electron bombardment removes electrons → positive ions. Then acceleration, deflection, detection.
Q2 of 10

Ions in a mass spectrometer are deflected by:

Magnetic field deflects moving charged particles. Heavier ions (greater m/z) are deflected less than lighter ions at the same charge. This separates ions by mass.
Q3 of 10

The y-axis of a mass spectrum shows:

y-axis = relative abundance (how many ions of that m/z reach the detector). x-axis = m/z ratio. Tallest peak = base peak (most abundant fragment or isotope).
Q4 of 10

Ions are accelerated in the mass spectrometer by:

High voltage electric field accelerates positive ions from ionisation chamber toward the analyser. All ions of same charge gain same kinetic energy: KE = qV = ½mv².
Q5 of 10

The detector in a mass spectrometer measures:

Detector collects ions and measures the current — more ions → larger current → higher peak on spectrum.
Q6 of 10

m/z ratio in mass spectrometry stands for:

m/z = mass of ion / charge number. For singly charged ions (z=1), m/z = mass number. Doubly charged ions have m/z = mass/2.
Q7 of 10

In the mass spectrometer, why must a high vacuum be maintained?

High vacuum: removes air molecules so ions travel in straight paths from ionisation to detector without collisions.
Q8 of 10

Ionisation in a mass spectrometer removes:

Electron bombardment: high-energy electrons knock electrons out of atoms/molecules → M⁺ ions (for atomic samples) or molecular ions/fragments.
Q9 of 10

Which ion reaches the detector last (takes longest to deflect)?

Heavier ions (higher m/z) are deflected less by the magnetic field → require stronger field to reach detector. Heaviest ions are detected last as magnetic field is swept.
Q10 of 10

The mass spectrum of chlorine shows two molecular ion peaks at m/z 70, 72, 74. This is because:

Cl₂ molecular ions: ³⁵Cl³⁵Cl (m/z 70), ³⁵Cl³⁷Cl (m/z 72), ³⁷Cl³⁷Cl (m/z 74). Ratio of peak heights reflects isotope abundances.
1.5

Interpretation of Mass Spectra

Reading a Mass Spectrum

x-axis: m/z (mass-to-charge ratio). For singly charged ions (z=1), this equals the mass number.

y-axis: Relative abundance (%). The tallest peak = base peak = 100% — the most abundant ion.

Each peak corresponds to one isotope. Peak height ∝ abundance of that isotope.

Chlorine Mass Spectrum

Mass Spectrum of Chlorine 0 50 75 100 75% 25% 35 37 ³⁵Cl ³⁷Cl m/z Relative Abundance (%)

Calculating Aᵣ from the Spectrum

Use the formula with % abundances directly (divide by total):

Aᵣ = (35×75 + 37×25) / (75+25) = (2625 + 925) / 100 = 35.5

Neon Mass Spectrum Example

Neon has three isotopes: ²⁰Ne (90.48%), ²¹Ne (0.27%), ²²Ne (9.25%). Three peaks appear at m/z = 20, 21, 22. The base peak is m/z = 20 (most abundant).

Aᵣ(Ne) = (20×90.48 + 21×0.27 + 22×9.25) / 100 = (1809.6 + 5.67 + 203.5) / 100 = 20.2
Section 1.5 Quick Quiz
Interpretation of Mass Spectra
10 Questions
Q1
A mass spectrum shows two peaks: m/z=63 (69%) and m/z=65 (31%). The Ar is:
Q2
In the mass spectrum of Cl₂ gas, the molecular ion region shows peaks at m/z = 70, 72, 74 in approximate ratio:
Q3
The molecular ion peak (M⁺) in a mass spectrum appears at:
Q4
A compound with Mr=78 shows M+1 peak at 3.3% relative to M⁺. This suggests:
Q5
The mass spectrum of bromine gas (Br₂) shows a 1:2:1 pattern for the molecular ion region because:
Q6
In a mass spectrum, the peak at m/z = M+2 for a compound with ONE chlorine atom would be approximately what fraction of the M⁺ peak?
Q7
A mass spectrum shows the tallest peak (base peak) at m/z=28 and M⁺ at m/z=44. This could be:
Q8
Strontium's mass spectrum shows peaks at m/z = 84, 86, 87, 88 with heights 0.6%, 9.9%, 7.0%, 82.6%. The base peak is at:
Q9
An element shows 3 peaks in its mass spectrum. This definitively tells us:
Q10
From the mass spectrum of an element, which information CANNOT be obtained directly?
1.6

Uses of the Mass Spectrometer & Calculations

UseDetails
Isotope identificationDetermines the exact masses and natural abundances of all isotopes of an element.
Calculating AᵣUses peak positions (m/z) and heights (abundance) to compute weighted mean atomic mass.
Relative molecular mass (Mᵣ)The molecular ion peak M⁺ gives the Mᵣ of a compound directly.
Structural determinationFragmentation patterns reveal functional groups and molecular structure in organic chemistry.
Radiocarbon datingMeasures ¹²C/¹⁴C ratio to date ancient organic samples (archaeology, geology).
Drug testing / forensicsIdentifies substances by unique mass spectra — "molecular fingerprinting".
Quality controlUsed in pharmaceuticals, food science, and environmental monitoring.
Space explorationMars rovers carry mass spectrometers to analyse atmospheric/soil composition.
WORKED EXAMPLE

Copper — Aᵣ from Mass Spectrum

Mass spectrum of Cu: m/z = 63 (69.2%) and m/z = 65 (30.8%). Calculate Aᵣ(Cu).
1
Aᵣ = (63×69.2 + 65×30.8) / 100
2
= (4359.6 + 2002) / 100 = 6361.6 / 100
3
Aᵣ(Cu) = 63.6
WORKED EXAMPLE

Bromine — Finding % Abundance from Aᵣ

Br has two isotopes: ⁷⁹Br and ⁸¹Br. Aᵣ(Br) = 79.9. Find each % abundance.
1
Let x = fraction of ⁷⁹Br. 79.9 = (79×x) + (81×(1−x))
2
79.9 = 79x + 81 − 81x = 81 − 2x → 2x = 1.1 → x = 0.55
3
⁷⁹Br = 55.0%, ⁸¹Br = 45.0% — base peak at m/z = 79.
Section 1.6 Quick Quiz
Uses of the Mass Spectrometer & Calculations
10 Questions
Q1
Which of the following is NOT a use of the mass spectrometer?
Q2
Carbon-14 dating works by measuring the ratio of ¹⁴C to ¹²C because:
Q3
GC-MS (gas chromatography-mass spectrometry) is particularly useful because:
Q4
In sport doping analysis, mass spectrometry is preferred because:
Q5
A compound has Mr=46 and M+1 peak = 2.2% relative to M⁺. How many carbon atoms does it have?
Q6
A mass spectrum shows M⁺ at m/z=78 and a 3:1 pattern for M⁺ and M+2 peaks. This indicates:
Q7
High-resolution mass spectrometry (HRMS) can distinguish between CO (Mr=28) and N₂ (Mr=28) because:
Q8
The mass spectrum of HCl shows peaks at m/z=36 and 38 in ratio 3:1. This is because:
Q9
An element shows a single peak at m/z=27 in its mass spectrum. This element most likely has:
Q10
Environmental scientists use mass spectrometry to:
🎯

Section 1.5 — Interpreting Mass Spectra

10 Questions
Q1 of 10

The base peak in a mass spectrum is:

Base peak: most intense peak, assigned 100% relative abundance. All other peaks expressed relative to it. Often (but not always) the most abundant isotope or most stable fragment.
Q2 of 10

The molecular ion peak (M⁺) appears at:

M⁺ peak: the unfragmented ionised molecule. m/z = molecular mass (for z=1). Used to determine molecular mass. Often the rightmost peak before m/z = 0.
Q3 of 10

In the mass spectrum of Zr (zirconium), 5 peaks are seen. This means:

Number of peaks in elemental mass spectrum = number of stable isotopes. Zr has 5 stable isotopes: ⁹⁰Zr, ⁹¹Zr, ⁹²Zr, ⁹⁴Zr, ⁹⁶Zr.
Q4 of 10

Using isotope peaks to calculate Ar: peaks at m/z 24 (79%), 25 (10%), 26 (11%). Ar ≈

Ar = (24×79 + 25×10 + 26×11)/100 = (1896+250+286)/100 = 2432/100 = 24.32 ≈ 24.3. This is magnesium.
Q5 of 10

A mass spectrum shows peaks at m/z 79 and 81 in ratio 1:1. The element is most likely:

Br has two isotopes ⁷⁹Br and ⁸¹Br in approximately 50:50 ratio. Cl isotopes at 35 and 37, ratio 3:1. Iodine is essentially monoisotopic at 127.
Q6 of 10

Fragment ions in a mass spectrum are formed by:

After ionisation, molecular ions (M⁺) may fragment: bonds break → smaller ions detected at lower m/z values. Pattern of fragments gives structural information.
Q7 of 10

The M+1 peak in organic spectra is mainly due to:

¹³C has 1.1% natural abundance. Each carbon in the molecule gives ~1.1% chance of M+1 peak. M+1/M × 100 ≈ number of carbons × 1.1. Used to count carbon atoms.
Q8 of 10

In a mass spectrum, why does the peak intensity at m/z 35 exceed that at m/z 37 for chlorine?

Abundance: ³⁵Cl 75%, ³⁷Cl 25%. Ratio of peaks = ratio of abundances = 3:1. More ³⁵Cl ions reach detector → higher peak.
Q9 of 10

A compound gives M⁺ at m/z 78 and a base peak at m/z 77. The loss of 1 unit suggests loss of:

Loss of 1 mass unit = loss of one hydrogen atom (H, mass = 1). Very common fragmentation pattern in aromatic compounds (e.g. benzene C₆H₆: M⁺ = 78, [C₆H₅]⁺ = 77).
Q10 of 10

To identify an unknown element from its mass spectrum, you would:

From mass spectrum: positions of peaks → isotopic masses; heights → relative abundances → calculate Ar = Σ(mass × abundance)/100. Compare Ar to periodic table to identify element.

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✏️

Exercises

🧪

Multiple Choice Quiz — 25 Questions

Unit 1 Quiz

Select one answer per question
Q1
Who first proposed that atoms are indivisible solid spheres?
Dalton (1803) proposed the indivisible atom. Thomson later discovered electrons, disproving this.
Q2
Thomson discovered electrons using:
Thomson used cathode ray tubes to deflect cathode rays with electric/magnetic fields, measuring e/m of electrons.
Q3
In Rutherford's gold foil experiment, what did most alpha particles do?
Most alpha particles passed straight through because the atom is mostly empty space.
Q4
Who discovered the neutron?
James Chadwick discovered the neutron in 1932 by bombarding beryllium with alpha particles.
Q5
The atomic number (Z) equals the number of:
Z = number of protons. This uniquely defines the element. In a neutral atom Z also = number of electrons.
Q6
How many electrons does ⁴⁰Ca²⁺ have?
Ca has Z=20 → 20 electrons neutral. Ca²⁺ loses 2 electrons: 20−2 = 18.
Q7
Isotopes of the same element differ in the number of:
Isotopes: same Z (protons), different A (mass number), therefore different neutrons.
Q8
The relative atomic mass is defined relative to:
By IUPAC definition, Aᵣ is on the ¹²C scale: 1 u = 1/12 × mass of one ¹²C atom.
Q9
³⁵Cl (75%) and ³⁷Cl (25%). What is Aᵣ(Cl)?
Aᵣ = (35×0.75)+(37×0.25) = 26.25+9.25 = 35.5
Q10
In the mass spectrometer, ions are separated by their:
The deflection in the magnetic field depends on m/z. Lighter ions are deflected more.
Q11
Which stage converts atoms to positive ions in the mass spectrometer?
Ionisation: high-energy electrons from an electron gun knock out one electron per atom to form M⁺.
Q12
The base peak in a mass spectrum is:
The base peak is the tallest peak, assigned 100% relative abundance — represents the most abundant ion.
Q13
How many neutrons are in ²³⁸U (Z = 92)?
N = A − Z = 238 − 92 = 146 neutrons.
Q14
Why must the mass spectrometer operate under high vacuum?
Without vacuum, ions would scatter off air molecules unpredictably, giving false m/z readings.
Q15
The relative charge of an electron is:
Electron: relative charge = −1, relative mass ≈ 0 (1/1836). Proton: +1, mass = 1.
Q16
Which is NOT a use of the mass spectrometer?
Melting points are measured with a thermometer/melting point apparatus, not a mass spectrometer.
Q17
Deuterium (²H) and protium (¹H) have the same:
Same Z → same electron config → same chemical properties. Physical properties differ due to different mass.
Q18
In the magnetic field of a mass spectrometer, heavier ions are deflected _______ lighter ions.
Heavier ions have more inertia → harder to deflect → deflected less than lighter ions.
Q19
Silicon: ²⁸Si (92%), ²⁹Si (5%), ³⁰Si (3%). Which value is closest to Aᵣ(Si)?
Aᵣ ≈ (28×0.92)+(29×0.05)+(30×0.03) = 25.76+1.45+0.90 = 28.11 ≈ 28.1
Q20
A neutral atom has mass number 39 and 20 neutrons. Its atomic number is:
Z = A − N = 39 − 20 = 19 (potassium, K).
Q21
A neutral atom has 17 protons and 18 neutrons. Its symbol is:
Z=17 → Cl. A = 17+18 = 35. → ³⁵Cl.
Q22
Which scientist proposed a dense, positive nucleus at the atom's centre?
Rutherford's nuclear model (1911) proposed the dense positive nucleus from his gold foil experiment.
Q23
The ion ⁵⁶Fe³⁺ has how many electrons?
Fe: Z=26. Fe³⁺ loses 3 electrons: 26−3 = 23 electrons.
Q24
Electron impact ionisation in the mass spectrometer produces:
EI removes one electron per atom: M → M⁺ + e⁻. Singly charged positive ions are formed.
Q25
Mass spectrum: m/z=10 (20%), m/z=11 (80%). Aᵣ = ?
Aᵣ = (10×20+11×80)/100 = (200+880)/100 = 10.8 — this is boron.
📝

Unit Test — 50 Marks

Section A — Short Answer

20 marks
Q1 [4 marks]

Name the scientist who discovered each subatomic particle and briefly describe the experiment used: (a) electron [2]; (b) nucleus [1]; (c) neutron [1].

(a) J.J. Thomson (1897) — cathode ray tube experiment. Cathode rays deflected by electric/magnetic fields; measured e/m of electrons, proving they are negatively charged sub-atomic particles. [2] (b) Rutherford (1909–1911) — gold foil experiment. Back-scatter of alpha particles proved existence of a dense, positive nucleus. [1] (c) Chadwick (1932) — beryllium bombarded with alpha particles; neutral penetrating radiation ejecting protons from paraffin identified as neutrons. [1]
Q2 [4 marks]

³¹P (phosphorus, Z=15). (a) State the protons, neutrons, and electrons in neutral ³¹P. [3] (b) State the electron count in P³⁻. [1]

(a) Protons=15, neutrons=31−15=16, electrons=15. [3] (b) P³⁻ gains 3 electrons: 15+3=18 electrons. [1]
Q3 [4 marks]

Gallium: ⁶⁹Ga (60.1%) and ⁷¹Ga (39.9%). (a) Calculate Aᵣ(Ga) to 1 d.p. [2] (b) Identify the base peak and explain why. [2]

(a) Aᵣ = (69×0.601)+(71×0.399) = 41.469+28.329 = 69.8 [2] (b) Base peak at m/z=69 (⁶⁹Ga): it is more abundant (60.1%>39.9%), so more ⁶⁹Ga⁺ ions reach the detector, producing the largest signal. [2]
Q4 [4 marks]

Explain why isotopes of an element have identical chemical properties but different physical properties. Give one example of a physical property that differs. [4]

Chemical properties depend on electrons [1], which depends on Z (proton number). Since isotopes have the same Z, they have identical electron configurations → identical chemical reactivity [1]. Physical properties depend on mass [1]. Since isotopes differ in neutron number, they have different masses → different physical properties, e.g. ¹H₂O (bp 100°C) vs ²H₂O (heavy water, bp 101.4°C); or different densities/rates of diffusion [1].
Q5 [4 marks]

Describe the acceleration and deflection stages of the mass spectrometer, stating the physical quantity that determines ion behaviour at each stage. [4]

Acceleration [2]: Ions accelerated through a high-voltage electric field; all gain the same kinetic energy (KE=qV). Since KE=½mv², lighter ions reach higher speed. Property: mass (and charge). Deflection [2]: Ions travel through a magnetic field that curves their path. The radius of curvature depends on m/z — lighter ions (lower m/z) are deflected more (smaller radius). Varying field strength brings different m/z ions to the detector in sequence. Property: mass-to-charge ratio (m/z).

Section B — Extended Answer

30 marks
Q6 [8 marks]

Trace the development of atomic models from Dalton to Rutherford. For each model state (i) what it proposed and (ii) the evidence that supported or overturned it. [8]

Dalton (1803) [2]: (i) Atoms are indivisible solid spheres; all atoms of one element are identical. Explained conservation of mass and definite proportions. (ii) Disproved when Thomson discovered electrons — atoms contain smaller particles.
Thomson's plum pudding model (1904) [3]: (i) Atom is a uniform sphere of diffuse positive charge with electrons embedded within it; explains electrical neutrality. (ii) Supported by discovery of electrons (cathode rays). Disproved by Rutherford's gold foil — if charge were diffuse, back-scatter would be impossible.
Rutherford's nuclear model (1911) [3]: (i) Atom mostly empty space; tiny dense positive nucleus at centre; electrons orbit at a distance. (ii) Gold foil experiment: most α through (empty space), some deflected (nucleus repels +α), tiny fraction back-scattered (dense nucleus). Limitation: could not explain stability of orbiting electrons (addressed later by Bohr and quantum mechanics).
Q7 [8 marks]

Strontium (Sr, Z=38) has four isotopes: ⁸⁴Sr (0.56%), ⁸⁶Sr (9.86%), ⁸⁷Sr (7.00%), ⁸⁸Sr (82.58%). (a) Calculate Aᵣ(Sr) to 2 d.p. [3] (b) Describe the expected mass spectrum: axes, relative peak heights, and base peak. [3] (c) State the neutron count of each isotope. [2]

(a) Aᵣ = (84×0.0056)+(86×0.0986)+(87×0.0700)+(88×0.8258) = 0.4704+8.4796+6.090+72.6704 = 87.71 [3]
(b) x-axis: m/z (84–88); y-axis: relative abundance (%). Four peaks: m/z=84 (~0.6%, very small), m/z=86 (~10%, small), m/z=87 (~7%, small), m/z=88 (tallest — base peak at 100%); all other peaks are short relative to m/z=88. [3]
(c) N = A−Z: ⁸⁴Sr: 46n; ⁸⁶Sr: 48n; ⁸⁷Sr: 49n; ⁸⁸Sr: 50n. [2]
Q8 [6 marks]

Describe how a sample of neon gas is analysed in a mass spectrometer to produce its mass spectrum. Explain how ions are formed, separated by mass, and detected. Include why a vacuum is essential. [6]

(1) Neon enters the instrument as a gas through an inlet valve into the vacuum chamber. [1] (2) Ionisation: high-energy electrons (electron gun, ~70 eV) bombard Ne atoms, each collision removes one electron: Ne → Ne⁺ + e⁻. [1] (3) Acceleration: Ne⁺ ions accelerated through a high-voltage electric field; all ions gain the same KE = qV. Lighter ²⁰Ne⁺ ions travel faster than heavier ²²Ne⁺ ions. [1] (4) Deflection: ions enter a magnetic field. Lighter ions (²⁰Ne⁺, m/z=20) curve more sharply; heavier ions (²²Ne⁺, m/z=22) curve less. By varying field strength, each ion (²⁰Ne⁺, ²¹Ne⁺, ²²Ne⁺) is directed to the detector in turn. [2] (5) Detection: each ion type produces an electrical current at the detector; signal intensity ∝ ion abundance. Result: three peaks at m/z=20 (base peak, ~90%), 21 (~0.3%), 22 (~9%). Vacuum essential: without it, ions would collide with air molecules, scatter unpredictably, and give false m/z readings. [1]
Q9 [8 marks]

An unknown element Q shows two peaks in its mass spectrum: m/z=107 (52.0%) and m/z=109 (48.0%). (a) Calculate Aᵣ(Q). [2] (b) Identify element Q. [1] (c) State protons and neutrons in each isotope. [2] (d) Explain why both isotopes occupy the same position in the periodic table. [2] (e) Suggest one use of the mass spectrometer relevant to element Q. [1]

(a) Aᵣ = (107×52.0+109×48.0)/100 = (5564+5232)/100 = 107.96 ≈ 108 [2]
(b) Aᵣ≈108, Z=47 → Silver (Ag) [1]
(c) ¹⁰⁷Ag: p=47, n=60. ¹⁰⁹Ag: p=47, n=62. [2]
(d) Position in the periodic table is determined by atomic number Z (proton number). Both isotopes have Z=47, so identical electron configurations and identical chemical properties → same position. Neutron count affects mass only, not chemistry or periodic table placement. [2]
(e) Identifying silver isotope ratios in ancient artifacts or ore samples (provenance/authentication); determining the exact Aᵣ of silver; quality control in silver refining; forensic identification of silver compounds. [1]
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