#### Answer

Naturally occurring Rubidium has the following percent of each isotope:
% of $^{85}Rb$ = $72.16\%$
% of $^{87}Rb$ = $27.84\%$

#### Work Step by Step

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 $^{87}Rb$ = $27.84\%$