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Topic: Relative Isotope Abundance Given Percent  (Read 4503 times)

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Offline stumpedstudent

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Relative Isotope Abundance Given Percent
« on: September 06, 2010, 06:56:33 PM »
The questions I'm having trouble with is: Chlorine has two stable isotopes. The mass of one isotope is 34.97 amu. Its relative abundance is 75.53% What is the mass of the other stable isotope?

I've tried setting up this problem in several ways since I know the percent must equal 100, but when I subtract 75.53% from 100% I get 24.47%. I then try several different ways to divide it, but I never get the correct answer. However, I have figured out how to calculate given the masses and the average atomic mass, but I can't find the answer given the percent only. Does anyone know what I'm doing wrong? I need a proper formula. Thank you for your help.



Offline Jorriss

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Re: Relative Isotope Abundance Given Percent
« Reply #1 on: September 07, 2010, 12:09:41 AM »
You're leaving out a key piece of data that's necessary to solve this problem, the actual weighted atomic mass. Did you know that was needed or why?



Offline stumpedstudent

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Re: Relative Isotope Abundance Given Percent
« Reply #2 on: September 07, 2010, 08:05:27 AM »
Do you mean the actual mass of chlorine on the periodic table or the average atomic mass? I have a vague idea that the average atomic mass is needed to solve the problem. Maybe, that is why I was able to solve the other problems that give it and the other masses, but this question doesn't provide the average atomic mass. How would I calculate that?

For Example: given that indium has a mass of 112.9043 amu and a mass of 114.9041 amu. The average atomic mass is 114.82 amu.

To calculate the percent relative abundance you would have to set up two equations:

1. 112.9043(X)+114.9041(Y)=114.82                 Then you would be able to subtract 114.9041 from all numbers and you would be left: 
2.  X  +  Y  = 1                                              .1.9998(X)+0(Y)=-0.0841
                                                                    After solving the problem by dividing -0.0841 by 1.9998 which equals 0.042 then all you               
                                                                   would need to do is plug that into X in the second equation and solve.
                                                                  Thus, relative abundance of 113In=4.2%
                                                                   and relative abundance of 115In=95.8%
« Last Edit: September 07, 2010, 08:26:55 AM by stumpedstudent »

Offline Jorriss

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Re: Relative Isotope Abundance Given Percent
« Reply #3 on: September 07, 2010, 10:13:16 PM »
Without using the weighted mass from the periodic table, I don't see how it's possible.

You have the idea about setting up a system of linear equations.

x + y = 1
and

Isotope1(x)+isotope2(y)=average mass.

Here you have your second equation, with two unknowns - isotope 2 and the average mass. Without one of those, how can you solve for the other?

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