Module 2: Foundations in Chemistry · Year 12
2.1.1 Atomic structure and isotopes
Use atomic number, mass number and ionic charge to count particles, then use isotope data and mass spectra to calculate relative masses.
What you need to know
Open a line for a quick recap. If it feels obvious, move straight to the linked practice.
2.1.1 a Describe isotopes as atoms with the same proton number but different neutron numbers, so their masses differ. Quick revision
When you compare two isotopes of the same element, look at the proton number first: it stays the same. What changes is the number of neutrons, so the mass number changes too.
Take ³⁵Cl and ³⁷Cl. Both have 17 protons, but they contain 18 and 20 neutrons respectively. Neutral atoms of both isotopes have the same electron arrangement, so you should expect them to react chemically in the same way.
2.1.1 b(i) Work out the numbers of protons, neutrons and electrons in atoms from atomic number and mass number. Quick revision
When you are given a nuclide symbol, you can get all three particle numbers from it. The atomic number, Z, tells you the number of protons. Subtract Z from the mass number, A, to get the number of neutrons.
For a neutral atom, the electron number is the same as the proton number.
For ²⁷Al, Z = 13 and A = 27.
- protons = 13
- neutrons = 27 − 13 = 14
- electrons = 13
Answer So ²⁷Al has 13 protons, 14 neutrons and 13 electrons.
2.1.1 b(ii) Work out the numbers of protons, neutrons and electrons in ions when given the ionic charge. Quick revision
For an ion, keep the nucleus fixed and change only the electron count. A positive ion has lost electrons; a negative ion has gained them.
I usually start from the atomic number and then apply the charge. For Mg²⁺, 12 protons means a neutral Mg atom would have 12 electrons, so the 2+ ion has 10.
For ²⁴Mg²⁺, Z = 12 and A = 24.
- protons = 12
- neutrons = 24 − 12 = 12
- electrons = 12 − 2 = 10
Answer ²⁴Mg²⁺ has 12 protons, 12 neutrons and 10 electrons.
2.1.1 c(i) Give an exam-ready definition of relative isotopic mass using the carbon-12 mass scale. Quick revision
For this definition, I’d learn the wording quite precisely. Relative isotopic mass is the mass of one atom of an isotope compared with one twelfth of the mass of one atom of carbon-12.
It is a ratio, so it has no unit. Also keep it separate from mass number: mass number is a whole-number count of protons and neutrons, while relative isotopic mass is a measured relative mass.
2.1.1 c(ii) Give an exam-ready definition of relative atomic mass as a weighted mean on the carbon-12 mass scale. Quick revision
If you think about chlorine carefully, there is something slightly odd about its relative atomic mass. Individual chlorine isotopes have masses close to whole numbers such as 35 and 37, yet the Aᵣ of chlorine is about 35.5. That happens because a natural sample contains a mixture of isotopes.
For the definition, I’d learn: relative atomic mass, Aᵣ, is the weighted mean mass of an atom of an element compared with one twelfth of the mass of one atom of carbon-12. “Weighted” matters because the more abundant isotopes contribute more to the mean.
Suppose a chlorine sample contains 75% ³⁵Cl and 25% ³⁷Cl.
- Aᵣ = [(35 × 75) + (37 × 25)] ÷ 100
- Aᵣ = 35.5
Answer The answer lies closer to 35 because ³⁵Cl is more abundant.
2.1.1 d(i) Read a mass spectrum to obtain relative isotopic masses and abundances; this OCR point is limited to singly charged ions and does not require knowledge of how the instrument works. Quick revision
When you read one of these elemental mass spectra, use the horizontal axis for m/z and the vertical axis for relative abundance. OCR limits this point to singly charged ions, so z = +1 and the m/z value gives the relative isotopic mass directly.
Then compare the peak heights or intensities to get the isotope abundances. You can work with percentages or with a ratio such as 3 : 1; you do not need to convert a ratio to percentages first.
You see peaks at m/z 35 and 37 with intensities 3 : 1.
- ³⁵X abundance = 3 parts
- ³⁷X abundance = 1 part
Answer The isotope ratio is 3 : 1, which is equivalent to 75% : 25%.
2.1.1 d(ii) Calculate relative atomic mass from isotopic abundances. Quick revision
For an Aᵣ calculation, multiply each isotopic mass by its abundance, add those products, then divide by the total abundance. If you are given percentages that add to 100, the denominator is 100. If you are given peak intensities, add the intensities instead.
Keep the calculation in one line where you can see the weighting clearly; it makes denominator mistakes much easier to spot.
A spectrum has peaks at 24, 25 and 26 with relative intensities 10, 4 and 1.
- Aᵣ = [(24 × 10) + (25 × 4) + (26 × 1)] ÷ (10 + 4 + 1)
- Aᵣ = 366 ÷ 15 = 24.4
Answer Aᵣ = 24.4.
2.1.1 e Calculate relative molecular mass for simple molecules and relative formula mass for giant structures from relative atomic masses. Quick revision
To find the relative mass, count every atom in the formula and add its Aᵣ contribution. For a simple molecular substance we call the result relative molecular mass, Mᵣ. For a giant structure such as NaCl, use relative formula mass because there is no separate molecule.
The maths is the same in both cases; what changes is the name you give the result.
Using H = 1.0, S = 32.1, O = 16.0, Na = 23.0 and Cl = 35.5:
- Mᵣ(H₂SO₄) = 2(1.0) + 32.1 + 4(16.0) = 98.1
- relative formula mass of NaCl = 23.0 + 35.5 = 58.5
Answer Use Mᵣ for H₂SO₄ and relative formula mass for NaCl.