Question

Question #3037d

Answer 1

`""^6"Li"`: `7.59%`

`""^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 `"Li"`.

The relative atomic mass of `"Li"` is `6.94` `"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:

`(6.01512"amu")x + 7.01601"amu" - (7.01601"amu")x = 6.94`

Combining like terms:

`(-1.00089)x = -0.07601`

`x = color(red)(0.07594`

This represents the fractional abundance of `""^6"Li"`. The remaining quantity (`1` minus this value) represents that of `""^7"Li"`:

`1- color(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%`