Q.(a) Two stable isotopes of lithium $^{6}_{3}\text{Li}$ and $^{7}_{3}\text{Li}$ 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.

The listed atomic mass of an element is the abundance-weighted average of its isotopic masses; running the same weighting equation in reverse recovers an unknown abundance from a known average mass.

1. Atomic mass of lithium The periodic-table atomic mass of an element is the abundance-weighted average over its naturally occurring isotopes:

   $M = f_1 m_1 + f_2 m_2$

   where $f_1, f_2$ are fractional abundances and $m_1, m_2$ the isotopic masses.

   $M(\text{Li}) = (0.075)(6.01512\ \text{u}) + (0.925)(7.01600\ \text{u})$

   $= 0.45113\ \text{u} + 6.48980\ \text{u} = 6.94093\ \text{u} \approx 6.941\ \text{u}$

   This matches the periodic-table value for lithium, a good sanity check.
2. Abundances of boron isotopes Let $x$ be the fractional abundance of $^{10}_{5}\text{B}$; then $(1-x)$ is the abundance of $^{11}_{5}\text{B}$. The weighted average must equal boron's known atomic mass:

   $10.811 = x(10.01294) + (1-x)(11.00931)$

   $10.811 = 11.00931 - x(11.00931 - 10.01294)$

   $10.811 - 11.00931 = -x(0.99637)$

   $-0.19831 = -0.99637\,x \implies x = \frac{0.19831}{0.99637} = 0.1990$

   So $^{10}_{5}\text{B}$ is present at about $19.9\%$, and $^{11}_{5}\text{B}$ makes up the remaining $\approx 80.1\%$ — close to boron's real isotopic ratio of roughly 1:4.
   ✓Final answer

   1. $M(\text{Li}) \approx \boxed{6.941\ \text{u}}$
   2. $^{10}_{5}\text{B} \approx \boxed{19.9\%}$, $^{11}_{5}\text{B} \approx \boxed{80.1\%}$