## Problem calculating the atomic weight of an isotope.

 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?
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study
 Hi youhaveabuttandamazinglgysodoI They didn't 'arrive' at 10.81, 10.81 is a given, it was measured, presumably. W is the unknown and is multiplied by its known ratio so as to put the equation that is then solved Cheers...
 OK!!! This is done like this. 80.22% B atoms weight 11.009 amu and the other 19.78% weight lets say "m". So we can construct a statement using all the data given and that is; $\frac{11.009\;amu\;×\;80.22\;+\;m\;amu\;×\;19.78}{100}$ = 10.81 amu 11.009 × 80.22 = 883.1 $\frac{(883.1\;amu\;+\;m19.78\;amu)\;×\;100}{100}$ = 1081 amu m19.78 amu = (1081 - 883.1) amu ∴m = $\frac{(1081 - 883.1)\;amu}{19.78}$ So m comes as 10.005 ≈ 10.01 amu