What is the relative abundance of Lithium 6 & Lithium 7 isotopes with RAM of 6.944 if their masses are 6.0167 & 7.0179 respectively?
To find the relative abundance of Lithium-6 and Lithium-7 isotopes, we can use the equation:
Relative abundance = (Fraction of isotope x Mass of isotope)/Average atomic mass
The average atomic mass is given as 6.944. Let's assume the fraction of Lithium-6 is x, then the fraction of Lithium-7 would be (1-x) since the sum of the fractions of all isotopes must equal 1.
For Lithium-6:
Relative abundance of Lithium-6 = (x * Mass of Lithium-6)/Average atomic mass
For Lithium-7:
Relative abundance of Lithium-7 = ((1-x) * Mass of Lithium-7)/Average atomic mass
We are given the masses of Lithium-6 and Lithium-7 as 6.0167 and 7.0179, respectively. Plugging in these values and the average atomic mass, we can set up the following equations:
(x * 6.0167)/6.944 = Relative abundance of Lithium-6
((1-x) * 7.0179)/6.944 = Relative abundance of Lithium-7
Simplifying these equations, we get:
0.8661x = Relative abundance of Lithium-6
1.0107 - 1.0107x = Relative abundance of Lithium-7
To solve for x, we can set these two equations equal to each other:
0.8661x = 1.0107 - 1.0107x
Adding 1.0107x to both sides:
1.8768x = 1.0107
Dividing both sides by 1.8768:
x = 1.0107/1.8768
x ≈ 0.5383
The fraction of Lithium-6 is approximately 0.5383. Therefore, the fraction of Lithium-7 would be (1 - 0.5383), or approximately 0.4617.
Hence, the relative abundance of Lithium-6 is approximately 53.83% and the relative abundance of Lithium-7 is approximately 46.17%.