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Chemistry 18 Online
OpenStudy (anonymous):

The Element rhenium has two naturally occurring isotopes, 185 Re and 187 Re, with an average atomic mass of 186.207 amu. Rhenium is 62.60% 187 Re, and the atomic mass of 187 Re is 186.956 amu. Calculate the mass of 185 Re. How do you do this?

OpenStudy (aaronq):

write the expression as if you were to calculate the average atomic mass, i.e. \(\Sigma (mass_{isotope\;A}*\%abundace_{isotope\;A})\) since: (abundance % of isotope 185) + (abundance % of isotope 187) = 1 you can substitute in: (abundance % of isotope 187)= 1-(abundance % of isotope 185)

OpenStudy (anonymous):

I don't get it, where did the one come from?

OpenStudy (aaronq):

they have to equal to 100%, 1 is the same as 100% without the percent.

OpenStudy (anonymous):

So is the equation 62.60=1-x?

OpenStudy (aaronq):

it would be 0.6260=1-x or 62.60=100-x both are equally valid. but that's what you have to substitute into the first equation

OpenStudy (anonymous):

Thank you, I understand now!

OpenStudy (aaronq):

good stuff. no problem, dude

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