Question

Natural rubidium has the average mass 85.4678 u and is composed of isotopes ⁸⁵Rb (mass =84.9117 u) and ⁸⁷Rb. The ratio of atoms $\frac{{}^{\mathbf{85}}\mathbf{R}\mathbf{b}}{{}^{\mathbf{87}}\mathbf{R}\mathbf{b}}$in natural rubidium is 2.591. Calculate the mass of ⁸⁷Rb.

Short answer

Themass of ⁸⁷Rb is86.91

Step-by-step solution

Definition

The molar mass of a material is defined as the mass of the substancecontained in one moleof the substance, and it is equal to the total massof the component atoms. It may be stated mathematically as follows:
Molar mass =$\frac{Weighting}{numberofmoles}$

Determining the mass of 87Rb

From the given,$\frac{{}^{85}Rb}{{}^{87}Rb}=2.591$

$\begin{array}{rcl}N\left({}^{85}Rb\right)& =& 2.591\times N\left({}^{87}Rb\right)\\ \frac{N\left({}^{85}Rb\right)\times 84.9117+N\left({}^{87}Rb\right)\times x}{N\left({}^{85}Rb\right)+N\left({}^{87}Rb\right)}& =& 85.4678\\ \frac{N\left({}^{85}Rb\right)\times (220\times x)}{3.59\times N\left({}^{87}Rb\right)}& =& 85.4678\\ (220\times x)& =& 306.915\\ x& =& 86.91\end{array}$

Therefore the mass is86.91