# RELATIVE ATOMIC MASSES OF ELEMENTS

## ISOTOPE

Isotopes Definition: Isotopes are atoms with the same number of protons, but differing numbers of neutrons. Isotopes are different forms of a single element.

Examples: Carbon 12 and Carbon 14 are both isotopes of carbon, one with 6 neutrons and one with 8 neutrons (both with 6 protons).

## ATOMIC WEIGHTS AND ISOTOPIC ABUNDANCE

The atomic weight of an element is the relative atomic mass of that element. It is actually a weighted mass of the elements isotopes (if any) and their relative abundance.

You know that the sum of the percentages of the isotopes is equal to 1 (100%), so the relative abundance of the isotopes can be found using simple algebra.

Example #1:

Silver (Atomic weight 107.868) has two naturally-occurring isotopes with isotopic weights of 106.90509 and 108.90470. What is the percentage abundance of the lighter isotope?

### To avoid mistakes, use “x” as the multiplier for the isotope percentage you wish to find. In this case, you want to find the percentage of the lighter isotope, so the “x” is associated with 106.90509. Since the sum of the isotopic abundance percentages is equal to 1 (100%), the formula is:

108.90470 (1 – x) + 106.90509 (x) = 107.868

Multiplying, re-arranging and condensing the above formula results in:

108.90470 – 108.90470x + 106.90509x = 107.868

– 108.90470x + 106.90509x = – 108.90470 + 107.868

– 1.9996x = – 1.0367

x = 0.5185

Therefore, the answer is 51.85 %

Example #2:

An imaginary element (Atomic weight 93.7140) has three naturally-occurring isotopes with isotopic weights of 92.9469, 93.2923 and 94.9030. The abundance of the lightest isotope is 42.38 %. What is the percentage abundance of the heaviest isotope?

In this case, we know the abundance of one of the isotopes. We know the percentages of the lighter isotope (42.38 %) and the percentage of the heavier isotope (x), so the percentage of the middle isotope is equal to 1 (100%) minus the other two percentages (1 – 0.4238 – x).

92.9469 (0.4238) + 93.2923 [(1-0.4238)-x] + 94.9030x = 93.7140

39.3909 + 53.7550 – 93.2923x + 94.9030x = 93.7140

93.14 + 1.6107x = 93.7140

1.6107x = 0.4217

x = 26.18

## EVALUATION

1.      Define the following terms  i. Atomic number   ii. mass number  iii. Isotopes  iv. Isotopy

2.       Determine the relative atomic mass of carbon from a sample with the followingdata.98.9% of carbon -12 and 1.1% of carbon-13.