Relative Atomic Mass Calculations Chemistry Tutorial
Key Concepts
 The relative atomic mass of an element is also known as the relative atomic weight of an element, or, the atomic weight of an element.
 Relative atomic mass is often abbreviated as r.a.m.
 The relative atomic mass of an element (its atomic weight) is given in the Periodic Table.
 The relative atomic mass of an element is the weighted average of the masses of the isotopes in the naturally occurring element relative to the mass of an atom of the carbon12 isotope which is taken to be exactly 12.
 The atomic mass unit (u) is defined as a mass equivalent to ^{1}/_{12} of the mass of one atom of carbon12.
1 u = 1.66 × 10^{27} kg
 We can estmate the the relative atomic mass (atomic weight) of an element E with the naturally occurring isotopes ^{a}E, ^{b}E, ^{c}E, etc, and with the respective abundances of A%, B%, C% etc,
relative atomic mass (r.a.m.) 
= ( 
A 100 
× a) 
+ ( 
B 100 
× b) 
+ ( 
C 100 
× c) 
+ etc 
 Given the relative atomic mass (r.a.m.) of an element and the estimated mass of each of its isotopes, we can then estimate the relative abundance of each isotope:
let x = %abundance of isotopea
and 100  x = %abundance of isotopeb
then, let r.a.m = relative atomic mass of the element:
r.a.m. 
= ( 
x 100 
× mass isotopea) 
+ ( 
100  x 100 
× mass isotopeb) 
and solve for x
 Note that we can measure the mass of each isotope and its abundance using Mass Spectroscopy
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Abundance of Naturally Occurring Isotopes
Most elements occur in nature as a mixture of different isotopes.
The element carbon, for example, exists in nature as a mixture of different isotopes: stable^{1} carbon12 atoms and carbon13 atoms.
If you were to take a sample of carbon atoms, for example the soot from a chimney or a lump of coal, you would find that most of the carbon atoms are the carbon12 isotope and only a few would be the carbon13 isotope.
We call this amount of each isotope found in the naturally occurring element its abundance, or its isotopic abundance to be more precise.
The abundance of the carbon12 isotope in naturally occurring bulk carbon is 98.90% while the abundance of the carbon13 isotope in nature is 1.10%
This means that if you take a lump of coal from nature, 98.90% of the carbon in the coal will be atoms of the carbon12 isotope, while only 1.10% of it will be atoms of the carbon13 isotope.
The table below gives the isotopic abundances for some elements on Earth:
Element 
Isotope 
Abundance (%) 
Element 
Isotope 
Abundance (%) 
hydrogen 
^{1}H 
99.99 
sodium 
^{23}Na 
100.00 
^{2}H 
0.01 
magnesium 
^{24}Mg 
78.90 
helium 
^{3}He 
0.0001 
^{25}Mg 
10.00 
^{4}He 
99.9999 
^{26}Mg 
11.10 
lithium 
^{6}Li 
7.42 
aluminium 
^{27}Al 
100.00 
^{7}Li 
92.58 
silicon 
^{28}Si 
92.23 
beryllium 
^{9}Be 
100.00 
^{29}Si 
4.67 
boron 
^{10}B 
19.80 
^{30}Si 
3.10 
^{11}B 
80.20 
phosphorus 
^{31}P 
100 
carbon 
^{12}C 
98.90 
sulfur 
^{32}S 
95.02 
^{13}C 
1.10 
^{33}S 
0.75 
nitrogen 
^{14}N 
99.63 
^{34}S 
4.21 
^{15}N 
0.37 
^{36}S 
0.02 
oxygen 
^{16}O 
99.76 
chlorine 
^{35}Cl 
75.77 
^{17}O 
0.038 
^{37}Cl 
24.23 
^{18}O 
0.20 
argon 
^{36}Ar 
0.34 
fluorine 
^{19}F 
100.00 
^{38}Ar 
0.063 
neon 
^{20}Ne 
90.60 
^{40}Ar 
99.60 
^{21}Ne 
0.26 



^{22}Ne 
9.20 



You can find a more complete list of isotopic abundances at the bottom of this page.
Estimating Isotopic Mass
The relative atomic mass of a carbon12 atom is defined as 12.00
The relative atomic mass of an atom of carbon13 is found to be 1.08333 times the mass of a carbon12 atom, that is 1.083 × 12 = 13.00
We can estimate the mass of any isotope of an element, its isotopic mass, using its mass number (A).
The mass number (A) of an isotope tells us how many protons and neutrons are in the nucleus of an atom of this isotope.
Nucleon is the term used to describe both protons and neutrons.
So, the mass number (A) tells us the number of nucleons in the nucleus of an atom of the isotope of the element.
For example:
 nucleus of an atom of the carbon12 isotope contains 12 nucleons
 nucleus of an atom of the carbon13 isotope contains 13 nucleons
The mass of a proton is almost exactly the same as the mass of a neutron.
The mass of a proton is about 1 u (1 atomic mass unit), so the mass of a neutron is also about 1 u.
The mass of an electron is so small compared to the mass of a proton or neutron that it can be ignored when estimating the mass of an isotope of an element.^{2}
We can estimate the mass of an atom of an isotope of an element by adding together the mass of its nucleons:^{3}
isotopic mass = number of nucleons × mass of nucleon
isotopic mass = number of nucleons × 1 u
For example:
 isotopic mass of carbon12 atom: 12
nucleons × 1 u/nucleon = 12 u
 isotopic mass of carbon13 atom: 13
nucleons × 1 u/nucleon = 13 u
Isotopic masses can be measured using mass spectroscopy. You will find a discussion of calculating relative atomic mass (atomic weight) using these measured isotopic masses in the Mass Spectroscopy tutorial.
Calculating the Relative Atomic Mass (Atomic Weight) of an Element
The relative atomic mass of an element is the weighted average of the masses of the isotopes in the naturally occurring element relative to the mass of an atom of the carbon12 isotope which is taken to be exactly 12.
What is a "weighted average" ?
First lets look at what the "average weight" of carbon would be:
mass of carbon12 isotope is 12 u
mass of carbon13 isotope is 13 u
So we can calculate the average mass (average weight) of carbon as:
average weight = 
12 + 13 2 
= 12.5 u 
If we look up the atomic weight of carbon in the Periodic Table we find that it is 12.01 NOT 12.5
This is because most of the atoms found in naturally occurring carbon are atoms of the carbon12 isotope while very few of the atoms will be of the carbon13 isotope.
We need to take this into account when we calculate our "average", and when we do this we refer to the result as a "weighted average".
We need to know the abundance of each isotope, that is we need to know how much of the mass (weight) of the naturally occuring bulk carbon is due to each of the isotopes (carbon12 and carbon13).
From the table in the section above we find that the isotopic abundance of carbon12 is 98.90% and the isotopic abundance of carbon13 is 1.10%
This means that if I had 100 atoms of bulk carbon (like in coal or soot), then:
 98.90% of the 100 carbon atoms are carbon12 atoms with a mass of 12 u
That is, 98.90 of the carbon atoms have a mass of 12 u
 1.10% of the 100 carbon atoms are carbon13 atoms with a mass of 13 u
That is, 1.10 of the carbon atoms have a mass of 13 u
So, the total mass of 100 naturally occurring carbon atoms is:
total mass of 100 atoms 
= 
mass of all carbon12 atoms 
+ 
mass of all carbon13 atoms 

= 
98.90 × 12 
+ 
1.10 × 13 

= 
1186.8 
+ 
14.3 

= 
1201.1 


So the mass of 100 naturally occurring carbon atoms is 1201.1 u
Therefore the "weighted average mass" of 1 carbon atom is 1201.1 ÷ 100 = 12.011 u
This value for the atomic weight of carbon agrees with the value in the Periodic Table.
Note that this does NOT mean that the mass of 1 atom of carbon is 12.011 u
It DOES mean that if we take a sample of naturally occurrring carbon we would find that the average mass of a carbon atom would be 12.011 u
Let's review how we calculated the "weighted average" mass of bulk carbon atoms:
 abundance of the first isotope multiplied by the mass of this isotope given by the mass number
 abundance of the second isotope multiplied by the mass of this isotope given by the mass number
 add these two values together then divide by 100
In general, to calculate the "weighted average" mass of an element that occurs naturally as two different isotopes, isotope 1 and isotope 2, then:
"weighted average" mass of element 
= 
(abundance isotope 1 × mass isotope 1) + (abundance isotope 2 × mass isotope 2) 100 
If we estimate the mass of each isotope by using its mass number (A), then we can rewrite the expression as:
"weighted average" mass of element 
= 
(abundance isotope 1 × A_{(isotope 1)}) + (abundance isotope 2 × A_{(isotope 2)}) 100 
Or, put another way:
"weighted average" mass of element 
= ( 
%abundance isotope 1 100 
× A_{(isotope 1)}) 
+ ( 
%abundance isotope 2 100 
× A_{(isotope 2)}) 
Calculating Isotopic Abundance from Atomic Weight
The Periodic Table gives us the weighted average for the mass of an element, referred to as the element's atomic weight (or relative atomic mass).
If we know the mass of each isotope making up this naturally occurring element (estimated by its mass number), then we can calculate the abundance of each isotope in nature.
We wrote a general mathematical expression (mathematical equation) above for calculating the "weighted average" mass, also known as the relative atomic mass or atomic weight, of an element, which was:
"weighted average" mass of element 
= 
(abundance isotope 1 × A_{(isotope 1)}) + (abundance isotope 2 × A_{(isotope 2)}) 100 
Let's say I wanted to find the abundance (%) of each isotope of nitrogen.
Nitrogen has two naturally occurring stable isotopes: nitrogen14 and nitrogen15.
So, substituting these into the mathematical equation above I get:
"weighted average" mass of nitrogen 
= 
(abundance nitrogen14 × A_{(nitrogen14)}) + (abundance nitrogen15 × A_{(nitrogen15)}) 100 
I can look up the atomic weight (relative atomic mass, or, "weighted average" mass) of nitrogen in the Periodic Table:
atomic weight of nitrogen is 14.01
I can estimate the mass of an atom of each isotope by using its mass number (A):
The mass of an atom of nitrogen14 = its mass number = 14 u
The mass of an atom of nitrogen15 = its mass number = 15 u
I can substitute these values into the mathematical equation above:
14.01 
= 
(abundance nitrogen14 × 14) + (abundance nitrogen15 × 15) 100 
I can multiply both sides of the mathematical equation by 100:
100 × 14.01 
= 100 × 
(abundance nitrogen14 × 14) + (abundance nitrogen15 × 15) 100 
1401 
= 
(abundance nitrogen14 × 14) + (abundance nitrogen15 × 15) 
But how can I solve this equation when there are 2 unknowns, the abundance of nitrogen14 is unknown and the abundance of nitrogen15 is unknown.
The trick is to remember that we are talking about percentage abundance! Which means that:
%abundance of nitrogen14 + %abundance of nitrogen15 = 100
Or, put a different way:
%abundance of nitrogen14 = 100  %abundance of nitrogen15
So, if I let the %abundance of nitrogen14 be equal to x then:
%abundance of nitrogen14 = x
%abundance of nitrogen15 = 100  x
If I substitute these into the mathematical equation above I will have only 1 unknown value, x:
1401 
= 
(abundance nitrogen14 × 14) 
+ 
(abundance nitrogen15 × 15) 
1401 
= 
(x × 14) 
+ 
([100  x] × 15) 
To solve for x I will clear the brackets first:
1401 
= 
(x × 14) 
+ 
([100  x] × 15) 
1401 
= 
14x 
+ 
(15 × 100)  15x 
1401 
= 
14x 
+ 
1500  15x 
Next I collect like terms, starting with x:
1401 
= 
14x 
+ 
1500  15x 
1401 
= 
14x  15x 
+ 
1500 
1401 
= 
x 
+ 
1500 
Then subtract 1500 from both sides of the equation:
1401  1500 
= 
x 
+ 
1500  1500 
99 
= 
x 


Note that the abundance of an isotope must be a positive number (not a negative number), so I divide both sides of the equation by 1 to find the value of x :
Then substitute this value for x back into the expressions we wrote for the abundance of each isotope:
%abundance of nitrogen14 = x = 99 %
%abundance of nitrogen15 = 100  x = 100  99 = 1%
In general, if an element occurs in nature in 2 isotopic forms, isotope 1 and isotope 2, then we can estimate the percentage abundance of each isotope using the mass number (A) of each isotope because the %abundance of isotope 2 equals 100  %abundance of isotope 1:
100 × atomic weight of element 
= 
(abundance isotope 1 × A_{(isotope 1)}) + ([100  abundance isotope 1] × A_{(isotope 2)}) 
let abundance isotope 1 = x
100 × atomic weight of element 
= 
(x × A_{(isotope 1)}) + ([100  x] × A_{(isotope 2)}) 
100 × atomic weight of element 
= 
x A_{(isotope 1)} + 100A_{(isotope 2)}  A_{(isotope 2)}x 
(100 × atomic weight of element)  (100 × A_{(isotope 2)}) 
= 
x A_{(isotope 1)}  xA_{(isotope 2)} 
100 × (atomic weight of element  A_{(isotope 2)}) 
= 
x(A_{(isotope 1)}  A_{(isotope 2)}) 
100 × (atomic weight of element  A_{(isotope 2)}) (A_{(isotope 1)}  A_{(isotope 2)}) 
= 
x 
So,
 %abundance isotope 1 = x
 %abundance isotope 2 = 100  x
Worked Example 1: Calculating the Atomic Weight of an Element
Question:
Naturally occurring silver is 51.84% silver107 and 48.16% silver109.
Calculate the atomic weight of silver.
Solution:
(Based on the StoPGoPS approach to problem solving.)
 What is the question asking you to do?
Calculate the "weighted average" mass (relative atomic mass, or, atomic weight) of silver
 What data (information) have you been given in the question?
Extract the data from the question:
isotope name 
mass number (A) 
isotopic abundance (%) 
silver107 
107 
51.84 
silver109 
109 
48.16 
 What is the relationship between what you know and what you need to find out?
(a) mass of an isotope can be estimated using its mass number (A) expressed in atomic mass units, u
(b)
"weighted average" mass of silver 
= ( 
%abundance silver107 100 
× A_{(silver107)}) 
+ ( 
%abundance silver109 100 
× A_{(silver109)}) 
 Substitute in the values and solve:
"weighted average" mass of silver 
= ( 
51.84 100 
× 107) 
+ ( 
48.16 100 
× 109) 

= 
55.469 
+ 
52.494 

= 
107.96 


 Is your answer plausible?
Do a rough calculation:
Since about half (≈50%) of the atoms of silver are silver107 and the other half (≈50%) are silver109, the weighted average mass for silver will lie about halfway between the mass of silver107 (107) and silver109 (109):
atomic weight ≈ ½ × (107 + 109) = 108
Since our calculated value of 107.96 is close to 108 we are reasonably confident that our answer is correct.
You can also look up the atomic weight of silver in the periodic table and find that it is 107.9.
Since our calculated value of 107.96 is about the same as the value in the periodic table we are reasonably confident that our answer is plausible.
 State your solution to the problem "atomic weight of silver":
atomic weight of silver is 107.96 u
Worked Example 2: Calculating an Element's Isotopic Abundance
Question:
Copper consists of two isotopes, copper63 and copper65.
Its atomic weight (relative atomic mass) is given as 63.62 u
Find the % abundance of each isotope.
Solution:
(Based on the StoPGoPS approach to problem solving.)
 What is the question asking you to do?
(a) Calculate the % abundance of copper63
(b) Calculate the % abundance of copper65
 What data (information) have you been given in the question?
Extract the data from the question:
isotope name 
mass number (A) 
approximate mass of isotope (u) 
copper63 
63 
63 
copper65 
65 
65 
atomic weight of copper = 63.62 u
 What is the relationship between what you know and what you need to find out?
(a)
"weighted average" mass of copper 
= ( 
%abundance copper63 100 
× A_{(copper63)}) 
+ ( 
%abundance copper65 100 
× A_{(copper65)}) 
(b) let %abundance copper63 = x
and %abundance copper65 = 100  x
 Substitute the values into the equation and solve:
(a)
63.62 
= ( 
x 100 
× 63) 
+ ( 
100  x 100 
× 65) 
63.62 
= 
63x 100 
+ 
(65 × 100)  65x 100 
63.62 
= 
63x 100 
+ 
6500  65x 100 
63.62 
= 
63x + 6500  65x 100 
100 × 63.62 
= 
100 × 63x + 6500  65x
100 
6362 
= 
63x + 6500  65x 
6362 
= 
2x + 6500 
6362  6500 
= 
2x 
138 
= 
2x 
138 2 
= 
2x 2 
69 
= 
x 
(b) %abundance copper63 = x = 69 %
%abundance copper65 = 100  x = 100  69 = 31 %
 Is your answer plausible?
The atomic weight of copper is given as 63.62.
Because this weight is closer to that of isotopic mass of copper63 than it is to that of the isotopic mass of copper65 we know that copper63 must be more abundant (greater than 50%).
Our calculated value for the abundance of copper63 is 69% which is greater than 50% so we are reasonably confident that our answer is plausible.
Note that the %abundance of both isotopes must add up to 100%.
69% + 31% = 100%
So we are reasonably confident that our calculated values for the abundance of each isotope is plausible.
 State your solution to the problem "abundance of each isotope of copper":
abundance of copper63 = 69%
abundance of copper65 = 31%
Footnotes
1. A stable isotope is one that does not undergo radioactive decay (nuclear decay).
An unstable isotope is one that does undergo radioactive decay, and therefore, the abundance of a naturally occurring unstable isotope will decrease over time ....
Isotopic abundances can change over time, if a radioactive isotope decays to produce a stable isotope of a different element then the isotopic abundance of this stable isotope will increase over time.
Manmade nuclear reactions will also change the isotopic abundance.
There is a discussion of the variation in isotopic abundance in the Carbon14 Dating tutorial.
2. Mass of a proton = 1.673 × 10^{27} kg = 1.01 u
Mass of a neutron = 1.675 × 10^{27} kg = 1.01 u
Mass of an electron = 9.109 × 10^{31} kg = 0.000549 u
3. In effect we are estimating the mass of the nucleus rather than the atom (since we are ignoring the mass of the electrons which contribute very little to the mass of an atom).
When we measure the mass of a nucleus we find the observed mass is less than the sum of the masses of all the nucleons.
This is called the mass defect.
Isotopic Abundance Data
Element 
Isotope 
Abundance (%) 
hydrogen 
^{1}H 
99.99 
^{2}H 
0.01 
helium 
^{3}He 
0.0001 
^{4}He 
99.9999 
lithium 
^{6}Li 
7.42 
^{7}Li 
92.58 
beryllium 
^{9}Be 
100.00 
boron 
^{10}B 
19.80 
^{11}B 
80.20 
carbon 
^{12}C 
98.90 
^{13}C 
1.10 
nitrogen 
^{14}N 
99.63 
^{15}N 
0.37 
oxygen 
^{16}O 
99.76 
^{17}O 
0.038 
^{18}O 
0.20 
fluorine 
^{19}F 
100.00 
neon 
^{20}Ne 
90.60 
^{21}Ne 
0.26 
^{22}Ne 
9.20 
sodium 
^{23}Na 
100 

^{} 

magnesium 
^{24}Mg 
78.90 
^{25}Mg 
10.00 
^{26}Mg 
11.10 
aluminium 
^{27}Al 
100 
silicon 
^{28}Si 
92.23 
^{29}Si 
4.67 
^{30}Si 
3.10 
phosphorus 
^{31}P 
100 
sulfur 
^{32}S 
95.02 
^{33}S 
0.75 
^{34}S 
4.21 
^{36}S 
0.02 
chlorine 
^{35}Cl 
75.77 
^{37}Cl 
24.23 
argon 
^{36}Ar 
0.34 
^{38}Ar 
0.063 
^{40}Ar 
99.60 
potassium 
^{39}K 
93.20 
^{40}K 
0.012 
^{41}K 
6.73 
calcium 
^{40}Ca 
96.95 
^{42}Ca 
0.65 
^{43}Ca 
0.14 
^{44}Ca 
2.086 
^{46}Ca 
0.004 
^{48}Ca 
0.19 
scandium 
^{45}Sc 
100 
titanium 
^{46}Ti 
8.00 
^{47}Ti 
7.30 
^{48}Ti 
73.80 
^{49}Ti 
5.50 
^{50}Ti 
5.40 
vanadium 
^{50}V 
0.25 
^{51}V 
99.75 
chromium 
^{50}Cr 
4.35 
^{52}Cr 
83.79 
^{53}Cr 
9.50 
^{54}Cr 
2.36 
manganese 
^{55}Mn 
100 
iron 
^{54}Fe 
5.80 
^{56}Fe 
91.72 
^{57}Fe 
2.20 
^{58}Fe 
0.28 
cobalt 
^{59}Co 
100 
nickel 
^{58}Ni 
68.27 
^{60}Ni 
26.10 
^{61}Ni 
1.13 
^{62}Ni 
3.59 
^{64}Ni 
0.91 
copper 
^{63}Cu 
69.17 
^{65}Cu 
30.83 
zinc 
^{64}Zn 
48.60 
^{66}Zn 
27.90 
^{67}Zn 
4.10 
^{68}Zn 
18.80 
^{70}Zn 
0.60 
gallium 
^{69}Ga 
60.10 
^{71}Ga 
39.90 
germanium 
^{70}Ge 
20.50 
^{72}Ge 
27.40 
^{73}Ge 
7.80 
^{74}Ge 
36.50 
^{76}Ge 
7.80 
arsenic 
^{75}As 
100 
selenium 
^{74}Se 
0.90 
^{76}Se 
9.00 
^{77}Se 
7.60 
^{78}Se 
23.50 
^{80}Se 
49.60 
^{82}Se 
9.40 
bromine 
^{79}Br 
50.69 
^{81}Br 
49.31 
krypton 
^{78}Kr 
0.35 
^{80}Kr 
2.25 
^{82}Kr 
11.60 
^{83}Kr 
11.50 
^{84}Kr 
57.00 
^{86}Kr 
17.30 
rubidium 
^{85}Rb 
72.17 
^{87}Rb 
27.84 
strontium 
^{84}Sr 
0.56 
^{86}Sr 
9.86 
^{87}Sr 
7.00 
^{88}Sr 
82.58 
yttrium 
^{89}Y 
100 
zirconium 
^{90}Zr 
51.45 
^{91}Zr 
11.27 
^{92}Zr 
17.17 
^{94}Zr 
17.33 
^{96}Zr 
2.78 
niobium 
^{93}Nb 
100 
molybdenum 
^{92}Mo 
14.84 
^{94}Mo 
9.25 
^{95}Mo 
15.92 
^{96}Mo 
16.68 
^{97}Mo 
9.55 
^{98}Mo 
24.13 
^{100}Mo 
9.63 
ruthenium 
^{96}Ru 
5.52 
^{98}Ru 
1.88 
^{99}Ru 
12.70 
^{100}Ru 
12.60 
^{101}Ru 
17.00 
^{102}Ru 
31.60 
^{104}Ru 
18.70 
rhodium 
^{103}Rh 
100 
palladium 
^{102}Pd 
1.02 
^{104}Pd 
11.14 
^{105}Pd 
22.33 
^{106}Pd 
27.33 
^{108}Pd 
24.46 
^{110}Pd 
11.72 
silver 
^{107}Ag 
51.84 
^{109}Ag 
48.16 
cadmium 
^{106}Cd 
1.25 
^{108}Cd 
0.89 
^{110}Cd 
12.49 
^{111}Cd 
12.80 
^{112}Cd 
24.13 
^{113}Cd 
12.22 
^{114}Cd 
28.73 
^{116}Cd 
7.49 
indium 
^{113}In 
4.30 
^{115}In 
95.70 
tin 
^{112}Sn 
0.97 
^{114}Sn 
0.65 
^{115}Sn 
0.36 
^{116}Sn 
14.70 
^{117}Sn 
7.70 
^{118}Sn 
24.30 
^{119}Sn 
8.60 
^{120}Sn 
32.40 
^{122}Sn 
4.60 
^{124}Sn 
5.60 
antimony 
^{121}Sb 
57.30 
^{123}Sb 
42.70 
tellurium 
^{120}Te 
0.096 
^{122}Te 
2.60 
^{123}Te 
0.91 
^{124}Te 
4.82 
^{125}Te 
7.14 
^{126}Te 
18.95 
^{128}Te 
31.69 
^{130}Te 
33.80 
iodine 
^{127}I 
100 
xenon 
^{124}Xe 
0.10 
^{126}Xe 
0.09 
^{128}Xe 
1.91 
^{129}Xe 
26.40 
^{130}Xe 
4.10 
^{131}Xe 
21.20 
^{132}Xe 
26.90 
^{134}Xe 
10.40 
^{136}Xe 
8.90 
cesium 
^{133}Cs 
100 
barium 
^{130}Ba 
0.11 
^{132}Ba 
0.10 
^{134}Ba 
2.42 
^{135}Ba 
6.59 
^{136}Ba 
7.85 
^{137}Ba 
11.23 
^{138}Ba 
71.70 
lanthanum 
^{138}La 
0.09 
^{139}La 
99.91 
cerium 
^{136}Ce 
0.19 
^{138}Ce 
0.25 
^{140}Ce 
88.48 
^{142}Ce 
11.08 
praseodymium 
^{141}Pr 
100 
neodymium 
^{142}Nd 
27.13 
^{143}Nd 
12.18 
^{144}Nd 
23.80 
^{145}Nd 
8.30 
^{146}Nd 
17.19 
^{148}Nd 
5.76 
^{150}Nd 
5.64 
samarium 
^{144}Sm 
3.10 
^{147}Sm 
15.00 
^{148}Sm 
11.30 
^{149}Sm 
13.80 
^{150}Sm 
7.40 
^{152}Sm 
26.70 
^{154}Sm 
22.70 
europium 
^{151}Eu 
47.80 
^{153}Eu 
52.20 
gadolinium 
^{152}Gd 
0.20 
^{154}Gd 
2.18 
^{155}Gd 
14.80 
^{156}Gd 
20.47 
^{157}Gd 
15.65 
^{158}Gd 
24.84 
^{160}Gd 
21.86 
terbium 
^{159}Tb 
100 
dysprosium 
^{156}Dy 
0.06 
^{158}Dy 
0.10 
^{160}Dy 
2.34 
^{161}Dy 
18.90 
^{162}Dy 
25.50 
^{163}Dy 
24.90 
^{164}Dy 
28.20 
holmium 
^{165}Ho 
100 
erbium 
^{162}Er 
0.14 
^{164}Er 
1.61 
^{166}Er 
33.60 
^{167}Er 
22.95 
^{168}Er 
26.80 
^{170}Er 
14.90 
thulium 
^{169}Tm 
100 
ytterbium 
^{168}Yb 
0.13 
^{170}Yb 
3.05 
^{171}Yb 
14.30 
^{172}Yb 
21.90 
^{173}Yb 
16.12 
^{174}Yb 
31.80 
^{176}Yb 
12.70 
lutetium 
^{175}Lu 
97.40 
^{176}Lu 
2.60 
hafnium 
^{174}Hf 
0.16 
^{176}Hf 
5.20 
^{177}Hf 
18.60 
^{178}Hf 
27.10 
^{179}Hf 
13.74 
^{180}Hf 
35.20 
tantalum 
^{180}Ta 
0.012 
^{181}Ta 
99.99 
tungsten 
^{180}W 
0.013 
^{182}W 
26.30 
^{183}W 
14.30 
^{184}W 
30.67 
^{186}W 
28.60 
rhenium 
^{185}Re 
37.40 
^{187}Re 
62.60 
osmium 
^{184}Os 
0.02 
^{186}Os 
1.58 
^{187}Os 
1.60 
^{188}Os 
13.30 
^{189}Os 
16.10 
^{190}Os 
26.40 
^{192}Os 
41.00 
iridium 
^{191}Ir 
37.30 
^{193}Ir 
62.70 
platinum 
^{190}Pt 
0.01 
^{192}Pt 
0.79 
^{194}Pt 
32.90 
^{195}Pt 
33.80 
^{196}Pt 
25.30 
^{198}Pt 
7.20 
gold 
^{197}Au 
100 
mercury 
^{196}Hg 
0.15 
^{198}Hg 
10.10 
^{199}Hg 
17.00 
^{200}Hg 
23.10 
^{201}Hg 
13.20 
^{202}Hg 
29.65 
^{204}Hg 
6.80 
thallium 
^{203}Tl 
29.52 
^{205}Tl 
70.48 
lead 
^{204}Pb 
1.40 
^{206}Pb 
24.10 
^{207}Pb 
22.10 
^{208}Pb 
52.40 