# How to calculate peak ratios in mass spectometry for a polyatomic species from given isotope ratios? [closed]

Element Z has 2 isotopes, 55Z and 57Z with the abundance ratio of 2:1. It exists in nature as Z3. The ratio of the peaks at m/e ratios 165:167:169:171 is 8:x:y:1. What are the values of x and y?

I tried to sketch the graph..shouldn't the graph have 100% and 50% abundance of element Z? Ive been doing this for 3 hours now...need help! thanks!

## closed as off-topic by bon, Todd Minehardt, Jannis Andreska, Freddy, Jon CusterMay 21 '16 at 19:15

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This is a permutation and combination problem.

If the element existed as Z, and not $Z_3$, then yes you are right the peaks should be 2:1, but more correctly abundance would be 66%:33% rather than 100%:50%. Keep this in mind, it will be helpful when you do the calculation.

But now we are told that it exists as $Z_3$, which means we have to have 3 atoms of Z.

1. There is only one "way" to get 171, and that is to have all three atoms as 57Z. So from probability, this would look like $$P(M=171)=(1/3)^3 = 1/27$$

2. Similarly, we consider the case of getting 165, which is all three atoms have to be 55Z. Mathematically, this would then be $$P(M=165)=(2/3)^3 = 8/27$$

3. In the case of M=167, we need two 55Z's and one 57Z. So after doing the math we find that it is $4/27$.

Since this is homework, I'm sure you can do the case of M=169 by yourself now.