Skip to main content
EnglandChemistrySyllabus dot point

How does the arrangement of sub-atomic particles and isotopes explain atomic mass and chemical identity?

Sub-atomic particles and their relative masses and charges, atomic number and mass number, isotopes and their identical chemical properties, and the determination of relative atomic mass from mass spectra.

An OCR H432 module 2 answer covering protons, neutrons and electrons, atomic and mass number, isotopes, and calculating relative atomic and isotopic mass from mass spectrometry data.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this topic is asking
  2. Sub-atomic particles
  3. Isotopes
  4. Relative masses
  5. Mass spectrometry and relative atomic mass
  6. Examples in context
  7. Try this

What this topic is asking

OCR specification point 2.1.1 wants you to describe the atom in terms of protons, neutrons and electrons with their relative masses and charges, use atomic number and mass number, define and recognise isotopes, and calculate relative atomic mass and relative isotopic mass from mass spectrometry data.

Sub-atomic particles

Protons and neutrons sit in the nucleus; electrons occupy shells around it. Their relative masses and charges are:

Particle Relative mass Relative charge
proton 11 +1+1
neutron 11 00
electron 11836\frac{1}{1836} 1-1

The proton and neutron dominate the mass; the electron is negligible by mass. A neutral atom has equal numbers of protons and electrons, so the overall charge is zero.

Isotopes

Isotopes of an element have identical chemical properties because chemical behaviour depends on the number and arrangement of electrons, which is the same for every isotope. They differ only in physical properties that depend on mass, such as density and rate of diffusion.

Relative masses

OCR expects precise definitions of the relative-mass terms, each measured against carbon-12.

  • Relative isotopic mass: the mass of an atom of an isotope compared with one twelfth the mass of an atom of carbon-12.
  • Relative atomic mass (ArA_r): the weighted mean mass of an atom of an element compared with one twelfth the mass of an atom of carbon-12.
  • Relative molecular mass (MrM_r): the weighted mean mass of a molecule compared with one twelfth the mass of an atom of carbon-12 (use relative formula mass for ionic compounds).

Mass spectrometry and relative atomic mass

A mass spectrometer ionises a sample, separates the ions by their mass-to-charge ratio (m/zm/z), and records the relative abundance of each. For singly charged ions the m/zm/z value equals the relative isotopic mass. The relative atomic mass is the weighted mean:

Examples in context

Example 1. Chlorine in mass spectra. A molecule of chlorine, Cl2\text{Cl}_2, gives three molecular-ion peaks at m/z=70m/z = 70, 7272 and 7474 because each molecule can contain two 35Cl^{35}\text{Cl}, one of each, or two 37Cl^{37}\text{Cl}. The 70:72:7470 : 72 : 74 peak ratio of roughly 9:6:19 : 6 : 1 comes straight from the 3:13 : 1 abundance of the two isotopes, a recurring exam pattern that shows isotope abundance controlling the spectrum.

Example 2. Carbon dating. The ratio of radioactive 14C^{14}\text{C} to stable 12C^{12}\text{C} in a once-living sample falls predictably as the 14C^{14}\text{C} decays. A mass spectrometer can measure this ratio, the same instrument used to find relative atomic masses in the laboratory, linking the abstract definition of isotopes to a real dating technique.

Try this

Q1. State the relative mass and relative charge of a neutron. [1 mark]

  • Cue. Relative mass 11, relative charge 00.

Q2. Explain why the two isotopes of chlorine react in the same way with sodium. [2 marks]

  • Cue. They have the same number of electrons and the same electron configuration, and chemical reactions depend on electrons, not neutrons.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20193 marksA sample of the element boron contains two isotopes, 10B^{10}\text{B} and 11B^{11}\text{B}. The relative atomic mass of this sample is 10.810.8. Calculate the percentage abundance of each isotope.
Show worked answer →

Let the abundance of 11B^{11}\text{B} be x%x\%, so the abundance of 10B^{10}\text{B} is (100x)%(100 - x)\%.

Ar=10(100x)+11x100=10.8A_r = \dfrac{10(100 - x) + 11x}{100} = 10.8 (1, sets up the weighted mean).

100010x+11x=10801000 - 10x + 11x = 1080, so x=80x = 80 (1). Therefore 11B^{11}\text{B} is 80%80\% and 10B^{10}\text{B} is 20%20\% (1).

Markers reward the weighted-mean equation, the correct algebra, and both abundances.

OCR 20214 marks(a) Define the term relative isotopic mass. (b) An element has isotopes of mass 63.063.0 (abundance 69.2%69.2\%) and 65.065.0 (abundance 30.8%30.8\%). Calculate its relative atomic mass to one decimal place and identify the element.
Show worked answer →

(a) The relative isotopic mass is the mass of an atom of an isotope compared with one twelfth the mass of an atom of carbon-12 (1).

(b) Ar=(63.0×69.2)+(65.0×30.8)100=4359.6+2002.0100=6361.6100=63.6A_r = \dfrac{(63.0 \times 69.2) + (65.0 \times 30.8)}{100} = \dfrac{4359.6 + 2002.0}{100} = \dfrac{6361.6}{100} = 63.6 (2).

Ar63.6A_r \approx 63.6 corresponds to copper (1). Markers reward the standard definition, the weighted mean, and the correct element.

Related dot points

Sources & how we know this