
#1
Jul2311, 04:35 PM

P: 425

1. The problem statement, all variables and given/known data
1) "The relative atomic mass of antimony is 121.8. Antimony exists as two isotopes; antimony121 and antimony123. Calculate the relative abundances of the two isotopes." 2) "The relative atomic mass of rubidium is 85.5. Rubidium consists of two isotopes, rubidium85 and rubidium87. Calculate the percentage of rubidium87 in naturally occurring rubidium." 2. Relevant equations N/A 3. The attempt at a solution I thought these questions should be relatively easy but the solutions at the back of the book don't seem to agree with my answers. For question 1, I did [121*x + 123*(100x)]/100 = 121.8 and solved for x. Likewise for question 2. For question 1, I get that antimony121 is 60% and antimony123 is 40%, but the textbook says the answers should be 37.5% and 62.5% respectively. For question 2, I get Rb85 occurring 75% and Rb87 occurring 25%, yet the textbook says Rb87 occurs 23.5%. However I checked my answers and I do get the average relative atomic masses that they state. Where have I gone wrong? Thanks. 



#2
Jul2311, 06:29 PM

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RGV 



#3
Jul2311, 06:49 PM

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PF Gold
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You need to know the atomic mass of each of the isotopes: antimony121, antimony123, rubidium85, and rubidium87. You can't assume that they're 121, 123, 85, and 87.
Wikipedia page on isotopes of Antimony. Wikipedia page on isotopes of Rubidium. 



#4
Jul2411, 04:52 AM

P: 425

Abundance of Isotopes
Hello,
Thank you for your responses. I'm just a little confused and would appreciate if someone could explain to me why my method is incorrect though. For example, for chlorine, Cl35 occurs 75% of the time and Cl37 occurs 25% of the time. So the average, according to the textbook, is; (35*0.75)+(37*0.25) = 35.5 Which is why 35.5 amu is the relative atomic mass of chlorine. In the questions I did, I followed the same method, but since the percentages are unknown I let the percentage be x, and the other percentage be (100x), since percentages must total to 100. I don't understand why the atomic mass A_{r} of Rb85 wouldn't just be 85? All questions are supposed to be doable using only the textbook. The question before this I got correct using this method, and the method working they've shown looks identical to mine... 



#5
Jul2411, 09:34 AM

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In my first reply I asked you a questionfor a good reason. So far you have not answered it, but had you done so you would see right away where your problem lies.
RGV 



#6
Jul2411, 11:01 AM

P: 425

Hello,
I saw your post but I still don't see the error, solving for x I get x = 50% (or 0.5) which is how often each of your numbers (5 and 5) occur... I don't understand. I'm not sure how your example is relevant because it doesn't involve weighted averages... For chlorine, the approximate relative atomic mass is 35.5, because the textbook explains that Cl35 occurs 75% of the time and Cl37 25% of the time, so by weighted averages, 35*0.75 + 37*0.25 = 35.5 This is exactly what I have done for my questions, so why are they incorrect? 



#7
Jul2411, 11:46 AM

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P: 9,818

Find and use more accurate data for the average atomic masses and isotope masses. The links SammyS gave you are excellent.
ehild 



#8
Jul2411, 01:19 PM

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I believe that the 60% you calculated is the percent mass of Sb121 in a sample of antimony. Since each atom of Sb121 has less mass than each atom of Sb123, the number of atoms of Sb121 is more than 60% of the total number of antimony atoms in the sample.




#9
Jul2411, 01:40 PM

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I was responding to the unedited post and had not noticed that you had edited it. That said, you got good advice from Sammy S. The actual weight of Antimony 121 is not exactly 121 times the weight of one nucleon, because of binding energy, etc. Of course, there is the possibility that the book's answer is wrong; that happens frequently.
RGV RGV 



#10
Jul2511, 02:35 AM

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P: 9,818

Yes, the book mixed the abundances, that of Sb121 is greater. You can find accurate abundances here: http://presolar.wustl.edu/ref/IsotopeAbundanceTable.pdf
ehild 


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