Calculating Natural Abundance of Lithium Isotopes

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To calculate the natural abundance of lithium isotopes, the average atomic mass of lithium (6.941 amu) is set equal to the weighted sum of the isotopes' masses. The equation involves two variables, X and Y, representing the proportions of lithium-6 and lithium-7, respectively. Since X and Y are not independent, an additional equation is needed, specifically that their sum equals 1 (X + Y = 1). This allows for solving the two-variable equation. Resources are available online to assist with the calculations.
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AMU Too many variables...?

Homework Statement



the atomic mass of lithium6 is 6.0151 and the amu of lithium-7 is 7.0160. calculate the natural abuncance of hese two isotopes. use the value of 6.941 amu as the value fr the average atomic mass of lithium


Homework Equations



ave atomic weight = (% of isotope 1)(amu of isotope 1)+(% of isotope 2)*(amu of isotope 2)


The Attempt at a Solution



Okay.. So the equation I set up ends up being

6.941 amu =X(7.0160)+Y(6.0151)

I have 2 variables, so here I get stuck... I have no idea how to solve when I have 2 variables...
 
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You need a second equation. X and Y are not independent. What is their sum?
 
This link should help.
http://dbhs.wvusd.k12.ca.us/webdocs/Mole/AvgAtomicWt-Reverse.html
 
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