Isotopy

John Dalton in one of his postulations of Atomic Theory stated that atoms of the same element are alike and different from atoms of other elements. This was, however, faulted with the discovery of nucleon mass (mass number) and neutrons by Rutherford and Chadwick respectively.

Studies showed that there are atoms of the same element that have the same atomic number, but different mass numbers as a result of the difference in their number of neutrons. These atoms are called isotopes and the phenomenon is known as isotopy. Examples of elements that exhibit isotopy include chlorine (Cl-35 and Cl-37), carbon (C-12, C-13 and C-14), oxygen (O-16, O-17 and O-18) etc.

In reality, all elements exhibit isotopy because this explains why the relative atomic masses of elements are not whole numbers.

Relative Abundance of Isotopes
In a particular element, say chlorine, the two isotopes - Cl-35 (chlorine-35) and Cl-37 (chlorine-37) are present in different quantities of 75% and 25% respectively. These are their relative abundances.

Therefore, relative abundance is said to be the amount of an isotope present in a given sample of an element. It can be expressed as a percentage, fraction or decimal. The sum of the relative abundances of all the isotopes in an element is always equal to 100 (if expressed as %) or 1 (if expressed as fractions or decimals)

Still using chlorine as an example, the relative abundance of its isotopes can be expressed as shown:

Isotopes        Relative Abundances         Sum
Cl-35                        75 %
Cl-37                        25%
                                                                     100%

Cl-35                        3/4
Cl-37                        1/4
                                                                     4/4 = 1

Cl-35                        0.75
Cl-37                        0.25
                                                                     1.00

The relative abundances can also be expressed as a ratio as stated below:
                             Cl-35 : Cl-37
                              75%  :   25%
Ratio                        3    :     1

Calculation of Relative Atomic Mass Using Relative Abundance

Since the isotopes present in a given element are in different proportions and possess different masses, it will be wrong to assume that the atomic mass of the element will be the simple addition of the masses of all the isotopes in the element.

Thus, it is always calculated as a weighted average of the mass numbers of all the isotopes using the formula:

Relative Atomic Mass (R.A.M.)
= (Relative Abundance of Isotope A x Mass No. of Isotope A) + (Relative Abundance of Isotope B x Mass No. of Isotope B) + …

Examples

Question 1
Given that the relative abundances of the two isotopes of chlorine, Cl-35 and Cl-37 are in the ratio of 3:1 respectively. Calculate the relative atomic mass of chlorine.

Answer
Isotopes           Cl-35           Cl-37
Mass number      35                37
Ratio                     3         :         1
Sum of ratio:            3 + 1= 4
Proportion (Relative Abundance) of Cl-35 = 3/4
Proportion (Relative Abundance) of Cl-37 = 1/4

Using,

Relative Atomic Mass
= (Relative Abundance of Isotope-35 x Mass No. of Isotope-35) + (Relative Abundance of Isotope-37 x Mass No. of Isotope-37)

Therefore,

Relative Atomic Mass of Chlorine = (3/4 x 35) + (1/4 x 37)
                                                            = 26.25 + 9.25
                                                            = 35.5

Question 2
If the relative atomic mass of copper is 63.54, calculate the proportions of the isotopes, Cu-65 and Cu-63 in the metal.

Answer
Relative atomic mass of Cu = 63.54
Let the relative abundance of Cu-65 = x
Let the relative abundance of Cu-63 = y

but,

x + y = 1……(i) [sum of relative abundances must be equal to 1]

from equation (i),

y = 1 - x……(ii)

Using,

Relative Atomic Mass
= (Relative Abundance of Isotope-65 x Mass No. of Isotope-65) + (Relative Abundance of Isotope-63 x Mass No. of Isotope-63)

Therefore,

63.54 = (x × 65) + (y x 63)
63.54 = 65x + 63y………(iii)

Substitute y = 1 - x into equation (iii)

63.54 = 65x + 63(1 - x)
63.54 = 65x + 63 - 63x
63.54 - 63 = 65x - 63x
  0.54 = 2x
        x = 0.54/2
           = 0.27 or 27%

Substitute the value of x into equation (ii)
       
        y = 1 - x
           = 1 - 0.27
           = 0.73 or 73%

Therefore, the proportions of Cu-65 and Cu-63 are 27% and 73% respectively.

Question 3
An analysis of a sample of iron using mass spectroscopy showed that the relative abundances of the four isotopes of iron are 5.8%, 91.6%, 2.3% and 0.3% for Fe-54, Fe-56, Fe-57 and Fe-58 respectively. From the result, determine the relative atomic mass of iron.

Answer
Isotopes                                  Fe-54    Fe-56    Fe-57    Fe-58
Mass Numbers                         54          56          57           58
Relative Abundances (%)       5.8         91.6       2.3          0.3

Using,

Relative Atomic Mass
= (Relative Abundance of Isotope-54 x Mass No. of Isotope-54) + (Relative Abundance of Isotope-56 x Mass No. of Isotope-56) + (Relative Abundance of Isotope-57 x Mass No. of Isotope-57) + (Relative Abundance of Isotope-58 x Mass No. of Isotope-58)

Therefore,

R.A.M. of Fe = (5.8% x 54) + (91.6% x 56) + (2.3% x 57) + (0.3% x 58)

(To convert percentage to decimal, divide the percentage value by 100)

R.A.M. of Fe = (0.058 x 54) + (0.916 x 56) + (0.023 x 57) + (0.003 x 58)

R.A.M of Fe = 3.132 + 51.296 + 1.311 + 0.174
                      = 55.913

Question 4
Given that the mean relative mass of oxygen, which contains three isotopes of mass numbers 16, 17 and 18 is 16.0044. Determine the percentage compositions of isotopes 16 and 17, if the relative abundance of the last isotope is 0.2%.

Answer
Relative atomic mass of O = 16.0044
Let the relative abundance of O-16 = x
Let the relative abundance of O-17 = y
Relative abundance of last isotope (O-18) = 0.2% = 0.002

but,

x + y + 0.002 = 1……(i)
x + y = 1 - 0.002
x + y = 0.998……(ii)

from equation (ii),

y = 0.998 - x……(iii)

Using,

Relative Atomic Mass of Oxygen
= (Relative Abundance of Isotope-16x Mass No. of Isotope-16) + (Relative Abundance of Isotope-17 x Mass No. of Isotope-17) + (Relative Abundance of Isotope-18 x Mass No. of Isotope-18)

Therefore,

16.0044 = (x × 16) + (y x 17) + (0.002 x 18)
16.0044 = 16x + 17y + 0.036………(iv)

Substitute y = 0.998 - x into equation (iv)

16.0044 = 16x + 17(0.998 - x) + 0.036
16.0044 = 16x + 16.966 - 17x + 0.036
16.0044 - 16.966 - 0.036 = 16x - 17x
-0.9976 = -x
            x = 0.9976 or 99.76%

Substitute the value of x into equation (iii)
       
        y = 0.998 - x
           = 0.998 - 0.9976
           = 0.0004 or 0.04%

Therefore, the proportions of O-16 and O-17 are 99.76% and 0.04% respectively.

Do These

Question 1
The relative atomic mass of neon is 21.5. Its two isotopes, Ne-20 and Ne-X, occur in nature in a ratio of 1:3. Find the value of X.

Question 2
What is the relative atomic mass of carbon, if the relative abundances of its three isotopes 12, 13 and 14 are 98.7%, 1.1% and 0.2% respectively?

Question 3
a) What are isotopes? Name any three elements that exhibit isotopy and give their respective isotopes.
b) Explain what you understand as relative abundance.
c) Why is the relative atomic mass of chlorine 35.5

Question 4
The abundance in nature of three isotopes of magnesium of mass numbers 24, 25, and 26 are 78.6%, x and y. Calculate the values of x and y, if the relative atomic mass of magnesium is 24.327

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