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ihaveabutt
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Boron has two naturally occurring isotopes, lOB and 11B. We know that 80.22% of its atoms are 11B, atomic weight 11.009 amu. From the natural atomic weight given on the inside back cover, calculate the atomic weight of the lOB isotope.
Solution
If 80.22% of all boron atoms are 11B, then 100.00 - 80.22, or 19.78%, are the unknown isotope. We can use W to represent the unknown atomic weight in our calculation:
(0.8022 x 11.009) + (0.1978 x W) = 10.81 amu (natural atomic weight)
W = (10.81-8.831)/0.1978 = 10.01 amu
On the part “(0.8022 x 11.009) + (0.1978 x W) = 10.81 amu (natural atomic weight),” how do I know what to multiply 0.1978 by if W is unknown? How did they arrive at 10.81?