The atomic masses of many elements are in fractions and not whole numbers. For example, the atomic mass of chlorine is 35.5 u whereas that of copper is 63.5 u.
The ractional atomic masses of elements are due to the existence of their isotopes having different masses. Most of the elements have more than one natural isotope having different masses. Since the atomic mass of an element is the average relative mass of all the natural isotopes of that element, most elements have fractional atomic masses. For example,chlorine has two isotopes
Cl and
Cl with abundance of 75% and 25% respectively. In other words,natural chlorine consists of two types of atoms,one having a mass of 35 u and the other having a mass of 37 u in the proportion of 75% and 25% respectively. Thus, the average mass of a chlorine atom will be 75% of 35 and 25% of 37, which is 35.5 u. This gives us the atomic mass of chlorine as 35.5 u. The calculation of average atomic mass of chlorine will become more clear from the following discussion.
(i) The chlorine isotope ,
Cl has a mass of 35 u and its abundance (or proportion) in nature is 75%.
(ii) The chlorine isotope
Cl has a mass of 37 u and its abundance (or proportion) in nature is 25%.
This means that the isotope of mass 35 u will contribute 75 per cent to the average atomic mass of chlorine whereas the isotope of mass 37 u will contribute 25 per cent to the average atomic mass of chlorine.So,
Average atomic mass of chlorine =
+
=
+
= 26.25 + 9.25
= 35.5 u
Thus, the average atomic mass of chlorine is 35.5 u.
All the naturally occurring isotopes of an element are present in a fixed proportion, so the average atomic mass of an element is fixed (or constant).
We will now solve some problems based on isotopes. In order to find out whether two (or more) atoms are isotopes of the same element or not, we should look at the number of protons and neutrons in them.If they contain the same number of protons but different number of neutrons,they will be isotopes of the same element. The number of electrons in them will also be the same. This point will be helpful in solving the following problems.
Sample Problem 1. The number of protons, neutrons and electrons in species A to E are given in the following table :
Species | Protons | Neutrons | Electrons |
---|
A B C D E | 6 18 17 9 17 | 6 22 20 10 18 | 4 18 17 11 17 |
Indicate from the above table the species that represent a pair of isotopes.
Solution.Those species which contain the same number of protons but different number of neutrons will be a pair of isotopes. In the above table only two species C and E have the same number of protons (17 each) but different number of neutrons (20 and 18 respectively). Thus, C and E are a pair of isotopes.(Please note that the atomic number 17 is of chlorine, so C and E are actually the two isotopes of chlorine).
Sample Problem 2.Composition of the nuclei of two atomic species X and Y is given as under :
X Y
Protons :6 6
Neutrons :6 8
Give the mass numbers of X and Y. What is the relation between the two species and which element or elements they represent ?
Solution. We know that:
Mass number = No. of protons + No.of neutrons
So, Mass number of X = 6 + 6
= 12
Mass number of Y =6 + 8
= 14
Thus, the mass number of X is 12 and that of Y is 14.
Now,X contains 6 protons, therefore, the atomic number of X is 6. Y contains 6 protons,therefore, the atomic number of Y is also 6. Since X and Y have the same atomic number (of 6) but different mass numbers (of 12 and 14),they are a pair of isotopes. Atomic number 6 is of carbon element. So, both X and Y represent carbon element.
Sample Problem 3. Bromine occurs in nature mainly in the form of two isotopes
Br and
Br.If the abundance of
Br isotope is 49.7% and that of
Br isotope is 50.3%,calculate the average atomic mass of bromine.
Solution.We know that upper digit in the symbol of an isotope represents its mass (which is the same as its mass number).Now :
(i) The mass of
Br isotope is 79 u and its abundance is 49.7%.
(ii) The mass of
Br isotope is 81 u and its abundance is 50.3%.
So,Average atomic mass of bromine =
+ 81 x
+
= 39.263 + 40.743
= 80.006
= 80 u
Thus,the average atomic mass of bromine is 80 u.
Sample Problem 4. A sample of an element X contains two isotopes
X and
X.If the average atomic mass of this sample of the element be 16.2 u, calculate the percentage of the two isotopes in this sample.
Solution. In order to solve this problem, we will have to suppose that the percentage of one of the isotopes in the sample is x, so that the percentage of the other isotope in the sample will be (100 - x). Now :
(i) The mass of
X isotope is 16 u.Suppose its percentage in the sample is x %.
(ii) The mass of
X isotope is 18 u. Its percentage in the sample will be (100 - x) %.
So,Average atomic mass of X =
+ 18 x
But the average atomic mass of X has been given to be 16.2 u.Therefore,
16.2=
+ 18 x
or 16.2=