# Question 3037d

Jun 17, 2017

$\text{^6"Li}$: 7.59%

$\text{^7"Li}$: 92.4%

#### Explanation:

We're asked to find the relative abundance (expressed commonly as a percentage abundance of the total number of isotopes) of the two naturally-occurring isotopes of $\text{Li}$.

The relative atomic mass of $\text{Li}$ is $6.94$ $\text{amu}$ (from periodic table).

We can set up an equation representing the fractional abundance of each isotope:

overbrace((x))^(% "abundace lithium-"6)(6.01512"amu") + overbrace((1-x))^("remaining is lithium-"7)(7.01601"amu") = 6.94"amu"

Let's use algebra to solve this equation:

$\left(6.01512 \text{amu")x + 7.01601"amu" - (7.01601"amu}\right) x = 6.94$

Combining like terms:

$\left(- 1.00089\right) x = - 0.07601$

x = color(red)(0.07594

This represents the fractional abundance of $\text{^6"Li}$. The remaining quantity ($1$ minus this value) represents that of $\text{^7"Li}$:

$1 - \textcolor{red}{0.07594} = 0.92406$

Therefore, the percentage abundance of the two isotopes is

%""^6"Li" = (color(red)(0.07594))(100) = color(red)(7.59%

%""^7"Li" = (color(red)(0.92406))(100) = color(red)(92.4%#