University:
University of Alberta
Course:
CHEM 101 | Introductory University Chemistry I
Academic year:

2024
Views:

367

Pages:

4
Author:

Neena N.

Relative atomic mass A naturally occurring sample of an element has isotopes present in the same proportions, no matter where the sample has come from. The average relative mass of these mixtures of isotopes is known as the relative atomic mass (Ar). The relative atomic mass is a weighted average of the isotopes as it takes into account the relative amount of each isotope in a sample of the element. An instrument called a mass spectrometer can be used to obtain the relative atomic masses of elements, as it separates isotopes of the element being measured and determines 1. the mass of each isotope being measured 12C relative to and 2. the relative abundance of the isotopes. The diagram below shows the type of image produced by a mass spectrometer, which is known as a mass spectrum. The mass spectrum tells us: 1. The number of peaks indicates the number of isotopes. 2. The position of the peak on the horizontal axis indicated the relative isotopic mass 3. The relative heights of the peaks correspond to the relative abundance of the isotopes. Example: Chlorine From the mass spectrum above, it can be seen that chlorine as two isotopes, one with a weight of 35 12 and the other 37, relative to C. The abundance can also be observed. This information can be summarised in the table below Isotope Relative isotopic mass Relative abundance (%) 35Cl 34.969 75.8 37Cl 36.966 24.2 This information can be used to calculate the relative atomic mass. Calculating relative atomic mass For each isotope you multiply its relative isotopic mass by its abundance. Example: Chlorine Total relative mass of atoms of chlorine = 34.969 x 75.80 + 36.966 × 24.20 Average relative atomic mass of one chlorine atom 34.969 x 75.80 + 36.966 × 24.20 100 2650.6 + 894.58 100 = 35.452 Therefore the relative atomic mass of chlorine is 35.45 or Ar (CI) = 35.45 = = Example: The relative atomic mass of rubidium is 85.47. The relative isotopic masses of its two isotopes are 84.95 and 86.94. Step Calculation State the relative abundances of the isotopes in Abundance of 84.95 isotope = x terms of x, where x is the abundance of the Abundance of 86.94 isotope = 100 - x lighter isotope Substitute the relative isotopic masses, relative 85.47 = \frac{(84.95x) + (86.94 \times (100 abundances and relative atomic mass into the x))}{100} formula: A_r = \frac{\text{relative isotopic mass} \times \% \text{abundance}}{100} Expand the top line of the equation 85.47 = \frac{84.95x + (8694 - 86.94x)}{100} Solve the equation to find x, the relative abundance of lightest isotope. 8547 = 84.95x + 8694 - 86.94x 8547 - 8694 = 84.95x - 86.94x -147 = -1.99x x = \frac{-147}{-1.99} = 73.87\% Determine the abundance of the heavier isotope. Abundance of 86.94 isotope = 100 - x = 100 - 73.87 = 26.13\%