# Need help on Isotopes

1. Apr 2, 2010

### allie2032

1. The problem statement, all variables and given/known data
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 text books 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.

2. Relevant equations

3. The attempt at a solution

2. Apr 3, 2010

### phyzguy

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?

3. Apr 5, 2010

### epenguin

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.

4. Apr 6, 2010

### Cilabitaon

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 $$\frac{(121 \times 57.3) + (M \times (100-57.3))}{100} = 121.84$$