← Unit 1: How can the diversity of materials be explained?
How can knowledge of elements explain the properties of matter?
the nuclear model of the atom (protons, neutrons, electrons), the use of nuclear notation, isotopes, and the calculation of relative atomic mass from isotopic composition determined by mass spectrometry
A focused VCE Chemistry Unit 1 answer on atomic structure. Covers the nuclear model of the atom, nuclear notation, isotopes, the relative atomic mass calculation from isotopic abundances, and how a mass spectrometer determines that abundance.
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What this dot point is asking
VCAA wants you to know the nuclear model of the atom (a dense positive nucleus of protons and neutrons surrounded by electrons), to read nuclear notation (mass number, atomic number), to distinguish isotopes of an element, and to use a mass spectrum to calculate a relative atomic mass from isotopic composition.
The answer
The nuclear model
An atom has a tiny, dense nucleus containing protons (positive) and neutrons (neutral), surrounded by electrons (negative) in shells. The nucleus holds almost all the mass; the electrons occupy almost all the volume.
| Particle | Relative charge | Relative mass | Location |
|---|---|---|---|
| Proton | +1 | 1 | Nucleus |
| Neutron | 0 | 1 | Nucleus |
| Electron | -1 | 1/1836 (about 0) | Shells around nucleus |
The atomic number Z is the number of protons (and defines the element). The mass number A is protons + neutrons. The number of neutrons is A - Z. In a neutral atom, electrons = protons.
Nuclear notation
An atom is written as ^A_Z X, for example ^12_6 C for carbon-12. Reading it: Z = 6 protons, A = 12 nucleons, so 6 neutrons and (in a neutral atom) 6 electrons.
For an ion, the charge is added: ^23_11 Na^+ has 11 protons, 12 neutrons and 10 electrons.
Isotopes
Isotopes are atoms of the same element (same Z) with different mass numbers (different numbers of neutrons). Examples:
- Carbon: ^12_6 C, ^13_6 C, ^14_6 C
- Chlorine: ^35_17 Cl, ^37_17 Cl
- Hydrogen: ^1_1 H (protium), ^2_1 H (deuterium), ^3_1 H (tritium)
Isotopes of an element have identical chemistry (same electron configuration) but slightly different physical properties (mass, density, rate of diffusion).
Relative isotopic mass and relative atomic mass
The relative isotopic mass of an isotope is its mass relative to one-twelfth the mass of a ^12_6 C atom. By definition, ^12_6 C has a relative isotopic mass of exactly 12.
The relative atomic mass Ar of an element is the weighted mean of the relative isotopic masses of its naturally occurring isotopes:
Ar = sum over isotopes of (relative isotopic mass x fractional abundance)
If abundances are given as percentages, divide each by 100 before multiplying. The result is dimensionless.
Mass spectrometry
A mass spectrometer separates ions by mass-to-charge ratio (m/z). The simplified workflow:
- Vaporisation: the sample is heated to a gas.
- Ionisation: high-energy electrons knock an electron off each atom or molecule, producing positive ions (usually +1).
- Acceleration: an electric field accelerates the ions.
- Deflection: a magnetic field deflects the ions; lighter ions and more highly charged ions deflect more.
- Detection: ions hit a detector and the relative numbers at each m/z are recorded.
The output is a mass spectrum: a bar chart of relative abundance (y) against m/z (x). For singly charged atomic ions (the usual case for elements), m/z equals the relative isotopic mass, and the bar heights are proportional to the natural abundance of each isotope. Plug those into the weighted-mean formula and you have Ar.
Worked example
A sample of copper gives two peaks in a mass spectrum:
m/z = 63, relative height 69.2
m/z = 65, relative height 30.8
Fractional abundances: 0.692 and 0.308 (these are already out of 100).
Ar(Cu) = (63 x 0.692) + (65 x 0.308) = 43.60 + 20.02 = 63.6
The accepted value for copper is 63.55, so this is consistent. Copper has only two stable isotopes and Cu-63 is more than twice as abundant as Cu-65, pulling the average closer to 63.
Common traps
Confusing mass number with relative atomic mass. Mass number is an integer (protons + neutrons) for a single isotope. Relative atomic mass is the weighted mean across all isotopes and is rarely a whole number (Cl is 35.45, not 35 or 36).
Forgetting to divide percentages by 100. If abundances are given as 75.78% and 24.22%, use 0.7578 and 0.2422, or divide the final answer by 100. Doing neither inflates the answer by 100x.
Adding electrons to the mass. Electrons are about 1/1836 the mass of a proton. For VCE you treat their contribution as zero. The relative isotopic mass is set by protons + neutrons.
Calling isotopes different elements. Isotopes share the same Z (same number of protons) and therefore the same element identity. Only the neutron count changes.
Misreading the y-axis as the mass. The y-axis of a mass spectrum is relative abundance. The mass (or m/z) is on the x-axis.
In one sentence
An atom is a nucleus of protons (defining the element) and neutrons surrounded by electrons, isotopes of an element differ only in neutron count, and a mass spectrometer separates ions by m/z so the weighted-mean of the isotope masses gives the relative atomic mass.
Past exam questions, worked
Real questions from past VCAA papers on this dot point, with our answer explainer.
2024 VCE3 marksChlorine has two naturally occurring isotopes, Cl-35 (relative isotopic mass 34.97, abundance 75.78%) and Cl-37 (relative isotopic mass 36.97, abundance 24.22%). Calculate the relative atomic mass of chlorine to 4 significant figures.Show worked answer →
A 3-mark answer needs the weighted-mean formula, the substitution, and a sensibly rounded answer.
Ar(Cl) = (34.97 x 0.7578) + (36.97 x 0.2422)
= 26.498 + 8.954
= 35.45
The result sits much closer to 35 than to 37 because Cl-35 is more than three times as abundant. Markers expect the answer to be reported to a sensible number of significant figures (usually matching the input data, here 4 s.f.), and they accept either fractional abundances (0.7578) or percentages (75.78), as long as you divide by 100 in the latter case.
2025 VCE4 marksA mass spectrum of magnesium shows three peaks at m/z = 24, 25 and 26 with relative heights 79, 10 and 11. (a) Identify the species responsible for each peak. (b) Calculate the relative atomic mass of magnesium.Show worked answer →
A 4-mark answer needs the identification of the cations and the weighted-mean calculation.
(a) Each peak corresponds to a singly charged positive ion of a magnesium isotope:
m/z = 24: Mg-24 cation (Mg^+, 12 protons, 12 neutrons)
m/z = 25: Mg-25 cation (12 protons, 13 neutrons)
m/z = 26: Mg-26 cation (12 protons, 14 neutrons)
(b) Convert peak heights to fractional abundances by dividing by the total (79 + 10 + 11 = 100):
Ar(Mg) = (24 x 0.79) + (25 x 0.10) + (26 x 0.11)
= 18.96 + 2.50 + 2.86
= 24.32
Mass spectrometry measures mass-to-charge ratio m/z, not mass directly. For singly charged ions (the usual case in this course) m/z equals the relative isotopic mass.