## Chemistry 10th Edition

Naturally occurring Rubidium has the following percent of each isotope: % of $^{85}Rb$ = $72.16\%$ % of $^{87}Rb$ = $27.84\%$
Atomic weight of Rubidium = $85.4678 amu$ Mass of $^{85}Rb = 84.9118 amu$ Mass of $^{87}Rb = 86.9092 amu$ Let the fraction of $^{85}Rb$ = $x$ The fraction of $^{87}Rb$ = $(1-x)$ Now, the atomic mass of an element is the weighted average of the masses of its constituent isotopes. $x(84.9118 amu) + (1-x)(86.9092 amu) = 85.4678 amu$ $84.9118x + 86.9092 - 86.9092x = 85.4678$ $-1.9974x = -1.4414$ $x = 0.7216$ Therefore, fraction of $^{85}Rb = 0.7216$ And, the fraction of $^{87}Rb = (1- 0.7216) = 0.2784$ % of $^{85}Rb$ = $72.16\%$ % of $^{87}Rb$ = $27.84\%$