Question

How do you calculate the average atomic mass of rubidium if $ 71.17\% $ of its atoms have a mass of $ 84.92amu $ and $ 27.83\% $ of its atoms have a mass of $ 86.91amu $ ?

Answer 1

Hint: We need to understand the concept of calculating the average of the given substances with respect to its percentage existence. An average is always the sum of the number divided by the number of entities. We shall first explain this concept using simple whole numbers and their percentages and then calculate the average atomic mass of the given entities.

Complete step by step answer:
Let us first calculate the average of numbers 1, 2, 3 and 4 to understand the concept of calculating average. The average of 1, 2, 3, 4 is $ \dfrac{{1 + 2 + 3 + 4}}{4} = 2.5 $ . We now consider the percentages of these numbers. Let there be $ 25\% $ of number $ 1 $ , $ 30\% $ of number $ 2 $ , $ 35\% $ of number $ 3 $ and $ 40\% $ of number $ 4 $ . The average can be calculated by multiplying the numbers with their respective percentages and then adding them up.
Hence the average is $ \left( {0.25 \times 1} \right) + \left( {0.30 \times 2} \right) + (0.35 \times 3) + (0.40 \times 4) = 2.5 $
Similarly, we can calculate the average atomic masses of rubidium given that $ 71.17\% $ of its atoms have a mass of $ 84.92amu $ and $ 27.83\% $ of its atoms have a mass of $ 86.91amu $ . Hence the average atomic mass will be:
$ (0.7217 \times 84.91) + (0.2783 \times 86.91) = 85.466amu $
Rounding this calculated value up to two decimal places, we get $ 85.47amu $ .

Therefore, the average atomic mass of rubidium if $ 71.17\% $ of its atoms have a mass of $ 84.92amu $ and $ 27.83\% $ of its atoms have a mass of $ 86.91amu $ is $ 85.47amu $ .

Note: It must be noted that the element rubidium has two stable isotopes $ ^{85}Rb $ and $ ^{87}Rb $ whose average atomic mass is $ 85.47amu $ . The more abundant an isotope is, the more its influence on the average atomic mass. The percentages given are also known as abundance which is divided by $ 100 $ in using it to calculate the average atomic mass.