Explaining and how to calculate the relative atomic mass RAM or Ar of
- Every atom has its own unique atomic
mass based on a standard comparison or relative scale
e.g. it has been based on hydrogen H = 1 amu and oxygen O = 16 amu in the past.
The relative atomic
mass scale is now based on an isotope of carbon, carbon-12,
which is given the value of 12.0000 amu.
- In other words the relative atomic mass
of an element is now based on the arbitrary value of the carbon-12 isotope
being assigned a mass of 12.0000 by international agreement!
- Examples are shown in the Periodic Table
- Note that because of the presence of
neutrons in the nucleus, the relative atomic mass is usually at least double
the atomic/proton number because there usually at the number of neutrons as
protons in the nucleus (mass proton = 1, neutron = 1).
- Also note, that for many calculations
purposes, relative atomic masses are usually quoted and used at this
academic level to one decimal place eg.
- hydrogen H = 1.0 or 1, calcium Ca= 40.0 or
40, chlorine Cl = 35.5, copper Cu = 63.6, silver Ag 108 or 107.9 at A level
- Sometimes at A level, values of relative
atomic masses to two decimal places may be quoted.
- In using the symbol Ar for
RAM you should bear in mind that the letter A on its own usually means the mass number of a particular isotope
and amu is the acronym shorthand for atomic mass units)
- However there are complications due to isotopes and
so very accurate atomic masses
are not whole numbers.
- Isotopes are atoms of the same element with different
masses due to different numbers of neutrons. The very accurate atomic mass scale
is based on a specific isotope of carbon, carbon-12, 12C = 12.0000
units exactly, for most purposes C = 12 is used for simplicity.
the three isotopes of hydrogen, though the vast majority of hydrogen atoms have
a mass of 1. When their accurate isotopic masses, and their % abundance are
taken into account the average accurate relative mass for hydrogen =
1.008, but for most purposes H = 1 is good enough!
- The strict definition of relative
atomic mass (Ar) is that it equals average mass of all the
isotopic atoms present in the element compared to 1/12th
the mass of a carbon-12 atom.
- So, you are taking into account the
different isotopic masses of the same elements, but also their %
abundance in the element.
- Therefore you need to know the
percentage (%) of each isotope of an element in order to accurately
calculate the element's relative atomic mass.
Examples of relative atomic mass calculations
for GCSE/IGCSE/AS level students
- Example 1.1: bromine consists of
50% 79Br and 50% 81Br, calculate the Ar of bromine.
- Ar = [ (50 x 79) + (50
x 81) ] /100 = 80
- So the relative atomic mass of
bromine is 80 or RAM or Ar(Br) = 80
- Note the full working shown.
Yes, ok, you can do it in your head BUT many students ignore the %'s and
just average all the isotopic masses (mass numbers) given, in this case
bromine-79 and bromine-81.
- Example 1.2: chlorine consists of
75% chlorine-35 and 25% chlorine-37.
- Think of the data based on 100
atoms, so 75 have a mass of 35 and 25 atoms have a mass of 37.
- The average mass = [ (75 x 35) +
(25 x 37) ] / 100 = 35.5
- So the relative atomic mass of
chlorine is 35.5 or RAM or Ar(Cl) = 35.5
- Note: 35Cl and
37Cl are the most common isotopes of chlorine, but, there
are tiny percentages of other chlorine isotopes which are usually
ignored at GCSE/IGCSE and Advanced GCE AS/A2 A level.
- Example 1.3:
The mass number for any isotope
is the sum of the protons and neutrons in
the nucleus, and is always an integer i.e. a whole number.
Examples for Advanced Level Chemistry students only
Calculation of relative atomic mass
= the accurate mass of a single isotope of
an element compared to 1/12th the mass of a
carbon-12 atom e.g. the accurate mass of
is 58.9332 !
If we were to redo the chlorine example
1.1 above, which is quite adequate for GCSE purposes, more accurately at A
level, we would do ....
chlorine is 75.77% 35Cl of
isotopic mass 34.9689 and 24.23% 37Cl of isotopic mass 36.9658
so Ar(Cl) = [(75.77 x
34.9689) + (24.23 x 36.9658)] / 100 =
35.4527 (but 35.5 is usually ok in calculations pre-university!)
Mass Spectrometer and isotope analysis
on the GCSE-AS(basic) Atomic Structure Notes, with further RAM calculations.
Calculations of % composition of isotopes
It is possible to do the reverse
of a relative atomic mass calculation if you know the Ar and
which isotopes are present.
It involves a little bit of
The Ar of boron is
10.81 and consists of only two isotopes, boron-10 and boron-11
The relative atomic mass of
boron was obtained accurately in the past and
mass spectrometers can sort
out the isotopes present.
If you let X = % of boron
10, then 100-X is equal to % of boron-11
Therefore Ar(B) = (X
x 10) + [(100-X) x 11) / 100 = 10.81
so, 10X -11X +1100
=100 x 10.81
-X + 1100 = 1081, 1100 -
1081 = X (change sides change sign!)
therefore X = 19
so naturally occurring boron
consists of 19% 10B and 81% 11B (the
data books quote 18.7 and 81.3)
type in answer
Honly or multiple choice
OTHER CALCULATION PAGES
What is relative atomic mass?,
relative isotopic mass & calculating relative atomic mass
formula/molecular mass of a compound or element molecule
Law of Conservation of Mass and simple reacting mass calculations
Composition by percentage mass of elements
in a compound
Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)
Reacting mass ratio calculations of reactants and products
moles) and brief mention of actual percent % yield and theoretical yield,
and formula mass determination
Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
(relating reacting masses and formula
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
Moles and the molar volume of a gas, Avogadro's Law
Reacting gas volume
ratios, Avogadro's Law
and Gay-Lussac's Law (ratio of gaseous
Molarity, volumes and solution
concentrations (and diagrams of apparatus)
do volumetric titration calculations e.g. acid-alkali titrations
(and diagrams of apparatus)
Electrolysis products calculations (negative cathode and positive anode products)
e.g. % purity, % percentage & theoretical yield, volumetric titration
apparatus, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
Energy transfers in physical/chemical changes,
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
Radioactivity & half-life calculations including
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