Question

A) Two stable isotopes of lithium $\begin{array}{c}6\\ 3\end{array}Li$ and $\begin{array}{c}7\\ 3\end{array}Li$ have respective abundances of $7.5$ and $92.5\%$. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

B) Boron has two stable isotopes, $\begin{array}{c}10\\ 5\end{array}B$ and $\begin{array}{c}11\\ 5\end{array}B$. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of $\begin{array}{c}10\\ 5\end{array}B$ and $\begin{array}{c}11\\ 5\end{array}B$.

Answer 1

A) Given,
Mass of $\begin{array}{c}6\\ 3\end{array}Li$ lithium isotope,
$m_1=6.01515u$
Mass of $\begin{array}{c}7\\ 3\end{array}Li$ lithium isotope,
$m_2=7.01600u$
Abundance of $\begin{array}{c}6\\ 3\end{array}Li,n_1=7.5\%$
Abundance of $\begin{array}{c}7\\ 3\end{array}Li,n_2=92.5\%$
The atomic mass of lithium atom is:
$m=\frac{m_1{n}_1+m_2{n}_2}{n_1+n_2}$
Putting values, we get
$m=\frac{6.01515\times 7.5+7.01600\times 92.5}{7.5+92.5}$
$=6.940934u$
Final Answer : 6.941 u

B) Given,
Mass of $\begin{array}{c}10\\ 5\end{array}B$ Boron isotope,
$m_1=10.01294u$
Mass of $\begin{array}{c}11\\ 5\end{array}B$ Boron isotope,
$m_2=11.00931u$
The atomic mass of Boron is $10.811u$
Let us assume abundance of $\begin{array}{c}10\\ 5\end{array}B$
$n_1=x\%$
Abundance of $\begin{array}{c}11\\ 5\end{array}B,n_2=(100-x)\%$
The atomic mass of Boron atom is:
$m=\frac{m_1{n}_1+m_2{n}_2}{n_1+n_2}$
Putting values, we get
$10.811=\frac{10.01294\times x+11.00931\times (100-x)}{x+100-x}$
$1081.11=10.01294x+1100.931-11.00931x$
$x=\frac{19.821}{0.99637}=19.89\%$
Abundance of $\begin{array}{c}10\\ 5\end{array}B,n_1=19.89\%$
Abundance of $\begin{array}{c}11\\ 5\end{array}B,n_2=(100-x)\%$
$n_2=(100-19.89)\%$
$n_2=80.11\%$
Final Answer : $19.9\%,80.1\%$