Daisy D.

asked • 09/11/19

chemistry problem, explain because I got lost doing this and its not giving me the answer that we have

The atomic mass of chlorine is 35.45 amu. Calculate the mass of the first isotope if it has a % natural abundance of 75.77% and the second isotope has % natural abundance of 24.23% and mass of 36.97 amu.

David D.

Hi Daisy, So, 35.45 amu represents the average amu of chlorine. Chlorine has two isotopes according to the above each representing a fraction of the 35.45 amu. To calculate this, you need to do some algebra: Let X represent the amu of the first isotope. 35.45 amu = 0.7577 (x) + 0.2423 (36.97 amu)__________ 35.45 = 0.7577 (x) + 8.957____________ 35.45 - 8.957 = 0.7577 (x)______________ 26.493 = 0.7577 (x) ---> divide by 0.7577__________ x = 34.97 amu _________(mind your significant figures), double check mine if the answer requires it. Sometimes students make the mistake by using the 35.45 amu as the average and assuming there is a 50% to 50% ratio of both isotopes. In this case the ratio was 75.77% to 24.23% so these must be applied in your math. Hope that helps!
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09/11/19

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Gary B. answered • 09/11/19

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the two stable isotopes of chlorine are present at 75.77% and 24.23% respectively. Therefore the relationship between their atomic masses is as follows:


35.45 = (0.7577 x unknown amu) + (0.2423 x 36.97)


rearranging this equation gives:


unknown amu = [35.45 - (0.2423 x 36.97)]/0.7577 = 34.96 amu

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