A naturally occurring sample of the element boron has a relative atomic mass of 10.8 In this sample, boron exists as two isotopes. Calculate the percentage abundance of 10B in this naturally occurring sample of boron.
The answer is 20%. This is a failry common question and stomps many students.
The easiest way to answer this question, in my opinion, is to see that this question is asking you to figure out the ratio of isotopes. Looking at the number 10.8 we can see that for every 8 11Bs we have 2 10Bs and so (2/8+2)*100 = 20%
Another way of looking at it is by showing an average of the percentage abundaces and making them equate to 10.8.
10x/100 + 11(100–x)/100 = 10.8
10x + 1100 – 11x = 1080
∴ x = 1100 – 1080 = 20%
TA
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