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Problem calculating the atomic weight of an isotope.

  1. Jun 9, 2012 #1
    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?
     
  2. jcsd
  3. Jun 9, 2012 #2
    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...
     
  4. Jun 9, 2012 #3
    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;

    [itex]\frac{11.009\;amu\;×\;80.22\;+\;m\;amu\;×\;19.78}{100}[/itex] = 10.81 amu

    11.009 × 80.22 = 883.1

    [itex]\frac{(883.1\;amu\;+\;m19.78\;amu)\;×\;100}{100}[/itex] = 1081 amu

    m19.78 amu = (1081 - 883.1) amu

    ∴m = [itex]\frac{(1081 - 883.1)\;amu}{19.78}[/itex]

    So m comes as 10.005 ≈ 10.01 amu
     
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