I am embarrassed that I cannot figure this out.
Lithium has only two naturally occurring isotopes. The mass of lithium6 is 6.01512 (amu) and the mass of lithium7 is 7.01601 (amu). Use the atomic mass of lithium (6.941) to calculate the relative abundances of the two isotopes.
I know the answers but I just cannot figure out how to get them. I have tried using the equation
atomic mass=(%isotope1 x mass of isotope 1) + (%isotope 2 + mass of isotope 2)
but this does not seem to get me anywhere. Can someone please help me out? You do not even have to do the whole problem, just give me a little direction.
Lithium has only two naturally occurring isotopes. The mass of lithium6 is 6.01512 (amu) and the mass of lithium7 is 7.01601 (amu). Use the atomic mass of lithium (6.941) to calculate the relative abundances of the two isotopes.
I know the answers but I just cannot figure out how to get them. I have tried using the equation
atomic mass=(%isotope1 x mass of isotope 1) + (%isotope 2 + mass of isotope 2)
but this does not seem to get me anywhere. Can someone please help me out? You do not even have to do the whole problem, just give me a little direction.

Hi John Q,
I am french (Boulogne sur mer 62200  FRANCE)
x % of Lithium6
y % of Lithium7
x + y = 100
y = 100  x
M(Li) x 100 = x • M(Li6) + y • M(Li7)
M(Li) x 100 = x • M(Li6) + (100  x) • M(Li7)
. . . then :
M(Li) x 100 = x • (M(Li6)  M(Li7)) + 100 M(Li7)
. . . and :
x = ( 100 M(Li)  100 M(Li7) ) / ( M(Li6)  M(Li7) )
x = ( 6,941  7,01601 ) x 100 / ( 6,01512  7,01601 )
x = 7,49 % of Li6
. . . and y = 100  x
. . . . . . .y = 92,51 % of Li7
I hope to have answered your question.
.
I am french (Boulogne sur mer 62200  FRANCE)
x % of Lithium6
y % of Lithium7
x + y = 100
y = 100  x
M(Li) x 100 = x • M(Li6) + y • M(Li7)
M(Li) x 100 = x • M(Li6) + (100  x) • M(Li7)
. . . then :
M(Li) x 100 = x • (M(Li6)  M(Li7)) + 100 M(Li7)
. . . and :
x = ( 100 M(Li)  100 M(Li7) ) / ( M(Li6)  M(Li7) )
x = ( 6,941  7,01601 ) x 100 / ( 6,01512  7,01601 )
x = 7,49 % of Li6
. . . and y = 100  x
. . . . . . .y = 92,51 % of Li7
I hope to have answered your question.
.