Help with easy problem
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# Help with easy problem

[From: ] [author: ] [Date: 11-06-21] [Hit: ]
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.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,......
I am embarrassed that I cannot figure this out.

Lithium has only two naturally occurring isotopes. The mass of lithium-6 is 6.01512 (amu) and the mass of lithium-7 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.

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Hi John Q,

I am french (Boulogne sur mer 62200 - FRANCE)

x % of Lithium-6

y % of Lithium-7

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