Question

rubidium has two common isotopes $^{85}rb$ and $^{87}rb$. if the abundance of $^{85}rb$ is 72.20% and the abundance of $^{87}rb$ is 27.8%, what is the average atomic mass of rubidium?

Explanation:

Step1: Convert percentages to decimals

$72.20\% = 0.722$ and $27.8\%=0.278$

Step2: Use average - atomic - mass formula

The average atomic mass $A$ of an element with isotopes of masses $m_1$ and $m_2$ and abundances $x_1$ and $x_2$ is $A = m_1x_1 + m_2x_2$. For $^{85}Rb$ with mass approximately 85 amu and $^{87}Rb$ with mass approximately 87 amu, we have $A=85\times0.722 + 87\times0.278$.

Step3: Calculate the result

$85\times0.722=61.37$, $87\times0.278 = 24.186$. Then $A=61.37+24.186 = 85.556$ amu.

Answer:

$85.556$ amu