Need Help on Isotopes: Find Mass Number of Antimony

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SUMMARY

The discussion centers on calculating the mass number of the second isotope of naturally occurring antimony (Sb), given that one isotope, 121Sb, has an abundance of 57.3% and the average molar mass is 121.84 g/mol. The equation used to find the mass number of the other isotope is derived from the average mass formula for isotopes: \((121 \times 57.3) + (M \times (100-57.3)) = 121.84 \times 100\). This leads to the conclusion that the mass number of the second isotope must be greater than 121 to achieve the specified average. The problem-solving approach emphasizes understanding the average mass calculation for isotopes.

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Homework Statement


I have been working on this for ages and can't work it out.
Naturally occurring antimony (Sb) has a molar mass of 121.84 g/mol and contains only two isotopes. One is 121Sb which is 57.3% abundant. What is the mass number of the other isotope of naturally occurring antimony?

I have been through all my textbooks and can't find the answer, have also googled and I am totally stuck. I need to show how I came to my answer. I don't need to have the question done for me just how to go about working it out.
Thanks in advance for any help.



Homework Equations





The Attempt at a Solution

 
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The molar mass is the average mass of all of the isotopes. If 57% is 121, what must the other one isotope be in order for the average to be 121.84?
 
Yes, it's got to be a small whole number, now just over 40% if this other isotope adds 0.84 to the average, I think I can work out what it must be in my head, and if not just try 1, 2, 3,... you will never have to go very far.
 
epenguin said:
Yes, it's got to be a small whole number, now just over 40% if this other isotope adds 0.84 to the average, I think I can work out what it must be in my head, and if not just try 1, 2, 3,... you will never have to go very far.

I don't follow your reasoning; the number must be greater than 121 for the average to be 0.84 higher with a lower abundance. I may just be misunderstanding, so sorry=]

allie, I'll assume you know how to take the average mass of two isotopes, but you may not have looked at the full equation for the mass; maybe this'll help :)

in this case \frac{(121 \times 57.3) + (M \times (100-57.3))}{100} = 121.84
 

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