Table : Re-185 = 184.95 amu
Re-187 = 186.96 amu
Let X = fraction abundance of 185, then
1-X = fraction abundance of 187
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X(184.95) + (1-X)*186.96 = 186.21
Solve for X and 1-X. then convert to % by multiplying X*100 and (1-X)*100 to give percent abundance.
Post your work if you get stuck.
Q: the average atomic mass of naturally occurring rhenium is 186.21 amu. There are two common isotopes of naturally occuring rhenium, using the information given in the table below, calculate the % abundance of naturally occuring rhenium.
Table : Re-185 = 184.95 amu
Re-187 = 186.96 amu
How do I do this?
1 answer