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.

(i) The chlorine isotope ,$=\begin{array}{c}\mathrm{35}\\ \mathrm{17}\end{array}$Cl has a mass of 35 u and its abundance (or proportion) in nature is 75%.

(ii) The chlorine isotope $=\begin{array}{c}\mathrm{37}\\ \mathrm{17}\end{array}$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 = $35x\frac{75}{100}$ + $37x\frac{25}{100}$=$\frac{2625}{100}$ + $\frac{925}{100}$

= 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 $=\begin{array}{c}\mathrm{79}\\ \mathrm{35}\end{array}$Br and $=\begin{array}{c}\mathrm{81}\\ \mathrm{35}\end{array}$Br.If the abundance of $=\begin{array}{c}\mathrm{79}\\ \mathrm{35}\end{array}$Br isotope is 49.7% and that of $=\begin{array}{c}\mathrm{81}\\ \mathrm{35}\end{array}$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 $=\begin{array}{c}\mathrm{79}\\ \mathrm{35}\end{array}$Br isotope is 79 u and its abundance is 49.7%.

(ii) The mass of $=\begin{array}{c}\mathrm{81}\\ \mathrm{35}\end{array}$Br isotope is 81 u and its abundance is 50.3%.

So,Average atomic mass of bromine = $79x\frac{49.7}{100}$ + 81 x $\frac{50.3}{100}$

$\frac{3926.3}{100}$+$\frac{4074.3}{100}$

= 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 $=\begin{array}{c}\mathrm{16}\\ 8\end{array}$X and $=\begin{array}{c}\mathrm{18}\\ 8\end{array}$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 $=\begin{array}{c}\mathrm{16}\\ 8\end{array}$X isotope is 16 u.Suppose its percentage in the sample is x %.

(ii) The mass of $=\begin{array}{c}\mathrm{18}\\ 8\end{array}$X isotope is 18 u. Its percentage in the sample will be (100 - x) %.

So,Average atomic mass of X = $16x\frac{\mathrm{x}}{100}$ + 18 x$\frac{\mathrm{(100-x)}}{100}$

But the average atomic mass of X has been given to be 16.2 u.Therefore,

16.2=$16x\frac{\mathrm{x}}{100}$ + 18 x$\frac{\mathrm{(100-x)}}{100}$

or 16.2=$\frac{\mathrm{16x+1800-18x}}{100}$$$

or 16.2 x 100 = 1800 - 2x

or 2x = 1800 - 1620

or 2x = 180

And x=$\frac{180}{2}$$$

or x=90

Thus, the percentage of the isotope $=\begin{array}{c}\mathrm{16}\\ 8\end{array}$X in the sample is 90%.The percentage of the other isotope $=\begin{array}{c}\mathrm{18}\\ 8\end{array}$X in the sample will be 100 - 90 = 10%.

1. Drawback of Rutherford’s Model of the Atom

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2. Rutherford’s Experiment - Discovery of Nucleus

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3. For the symbols H, D and T,tabulate three sub-atomic particles found in each of them.

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4. 4.Isotopes of Neon.

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5. Structure of The Atom - Study Points

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6. 3.Isotopes of Oxygen.

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7. Radioactive Isotopes

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8. Characteristics of a Neutron

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9. Nucleus

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10. Arrangement Of Electrons In The Atoms

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11. Characteristics of an Electron

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12. (b) Valency of Magnesium

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13. Limitations of Rutherford's model of the atom

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14. Mass Number

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15. All about Names of the Chemical Elements

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16. How to learn naming Chemical Formulae?

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17. (d) Covalency of Nitrogen

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18. 1. Isotopes of Hydrogen.

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19. Characteristics of a Proton

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20. Comparison between Proton, Neutron and Electron

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21. Electronic Configurations of First 20 Elements

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22. (d) Valency of Chlorine

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23. Thomson's Model Of The Atom

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24. Discovery of Neutron

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25. (e) Valency of Oxygen

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26. Rules for writing of distribution of electrons in various shells for the first 18 elements

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27. Isotopes

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28. 2.Covalency

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29. Covalency of Oxygen

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30. (a) Covalency of Hydrogen

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31. Bohr's Model Of The Atom

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32. Discovery Of Electron

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33. Rutherford's Model Of The Atom

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34. 2.Isotopes of Carbon.

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35. Atomic Number

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36. The Physical Properties of the Isotopes of an Element are Different

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37. (a) Valency of Sodium

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38. Charged Particles in Matter

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39. All the Isotopes of an Element Have Identical Chemical Properties

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40. Discovery of Proton

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41. Valaency Of Elements

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42. Relationship Between Mass Number and Atomic Number

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43. (f) Valency of Nitrogen

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44. Valaence Electrons (Or Valancy Electrons)

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45. Electronic Configurations of Noble Gases (or Inert Gases)

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46. Covalency of Chlorine

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47. Cause of Chemical Combination

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48. Reason for the Fractional Atomic Masses of Elements

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49. Applications of Radioactive Isotopes

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50. Relation Between Valency and Valence Electrons

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