Question #3037d

1 Answer
Jun 17, 2017

Answer:

#""^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%#