Atomic Structure & Mass Spectrometry (OCR AS Chemistry A): Revision Note
Exam code: H032
Isotopes & relative atomic mass
Isotopes are different atoms of the same element that contain:
The same number of protons (and electrons)
A different number of neutrons
Isotopes are also described as atoms of the same element but with different mass numbers
The mass of an element is given as relative atomic mass (Ar)
This is the weighted average mass of its isotopes
The relative atomic mass of an element can be calculated from its relative isotopic abundances, using the following steps:
Multiply each isotope’s mass by its abundance
Add the results together
Divide by the total abundance (usually 100)
The equation for this is:
Relative abundance values are either provided or can be read from a mass spectrum
Worked Example
Calculating the relative atomic mass of oxygen
A sample of oxygen contains the following isotopes:
Isotope | Percentage abundance |
---|---|
16O | 99.76 |
17O | 0.04 |
18O | 0.20 |
What is the relative atomic mass of oxygen in this sample, to 2 decimal places?
A. 16.00
B. 17.18
C. 16.09
D. 17.00
Answer:
The correct answer option is A:
Ar =
Ar = 16.0044
Ar = 16.00
Mr from mass spectra
The percentage abundance of the isotopes in an element can be found using a mass spectrometer
In A-Level Chemistry, mass spectra are interpreted assuming all ions are singly charged (+1)
The basic processes of mass spectrometry are:
The sample is vapourised
The sample is ionised to form positive ions
The ions are accelerated
Heavy ions move more slowly and are less deflected
The ions are detected as a mass-to-charge ratio, written as
Each ion produces a signal proportional to its abundance
So, a larger signal means a greater abundance
A mass spectrum can be used to calculate the relative atomic mass of an element
Worked Example
Calculating the relative atomic mass of boron
Calculate the relative atomic mass of boron using its mass spectrum, to 2dp:

Answer:
The relative atomic mass of boron, to 2dp, is 10.80
Ar =
Ar = 10.801
Ar = 10.80
Relative molecular and formula mass
These terms are closely related and both involve calculating mass from atomic contributions:
Relative molecular mass refers to substances with simple molecules
For example, ethanol (C2H5OH) contains:
2 carbon atoms
6 hydrogen atoms
1 oxygen atom
Relative formula mass refers to substances with giant structures
For example, calcium chloride (CaCl2) is an ionic lattice with a repeating ratio of:1 calcium ion to 2 chloride ions
These terms are often misused
Relative molecular mass is commonly and incorrectly applied to all compounds
Both the relative molecular mass and relative formula mass are calculated by adding up the relative atomic masses (Aᵣ) of all the atoms present
Worked Example
Calculate:
The relative molecular mass of ethanol, C2H5OH
The relative formula mass of calcium chloride, CaCl2
(Ar values: C = 12.0, H = 1.0, O = 16.0, Ca = 40.1, Cl = 35.5)
Answers :
The relative molecular mass of ethanol, C2H5OH, is 46.0
Mr = (2 x C) + (5 x H) + O + H
Mr = (2 x 12.0) + (5 x 1.0) + 16.0 + 1.0
Mr = 46.0
The relative formula mass of calcium chloride, CaCl2, is 111.1
Mr = Ca + (2 x Cl)
Mr = 40.1 + (2 x 35.5)
Mr = 111.1
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