Lithium has two known stable isotopes—lithium-6 and lithium-7. Lithium-6 has an isotopic mass of 6.015 amu and percent abundance of 7.59%; lithium-7 has a mass of 7.016 amu at 92.41% abundance. Calculate the average atomic mass of lithium.
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 (fromperiodic table).
We can set up an equation representing the fractional abundance of each isotope:
% abundace lithium-6
(x) (6.01512amu)
remaining is lithium-7
+ (1 - x) (7.01601amu) = 6.94amu
Let's use algebra to solve this equation:
(6.01512amu)x + 7.01601amu
-(7.01601amu)x - 6.94
combining like terms:
(-1.00089)x= - 0.07601
x=0.07594
This represents the fractional abundance of ⁶Li.The remaining quantity ( 1minus this value) represents that of ⁷Li:
1 - 0.07594 = 0.92406
therefore, the percentage abundance of the two isotopes is:
Answers & Comments
Answer:
⁶Li: 7.59%
⁷Li: 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 (fromperiodic table).
We can set up an equation representing the fractional abundance of each isotope:
% abundace lithium-6
(x) (6.01512amu)
remaining is lithium-7
+ (1 - x) (7.01601amu) = 6.94amu
Let's use algebra to solve this equation:
(6.01512amu)x + 7.01601amu
-(7.01601amu)x - 6.94
combining like terms:
(-1.00089)x= - 0.07601
x=0.07594
This represents the fractional abundance of ⁶Li.The remaining quantity ( 1minus this value) represents that of ⁷Li:
1 - 0.07594 = 0.92406
therefore, the percentage abundance of the two isotopes is:
% ⁶Li = (0.7594) (100) = 7.59%
% ⁷Li = (0.92406) (100) = 92.4%