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Fraction of electrons removed

  1. Aug 26, 2013 #1
    1. The problem statement, all variables and given/known data

    A 48.7 g ball of copper has a net charge of 2.2 µC. What fraction of the copper's electrons have been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5.).

    2. Relevant equations



    3. The attempt at a solution

    Number of electrons normally present ..
    (48.7/63.5)mol x 6.022^23(atoms per mol) x 29(electrons per atom) = 1.34^24 electrons

    Electrons removed = n
    ne = 2.20^-6C .. .. n = 2.20^-6C / 1.60^-19C/e .. .. n = 1.38^13 electrons removed

    Fraction removed = 1.38^13 removed / 1.34^24 total

    But the answer is wrong. I don't know what I am doing wrong?
     
  2. jcsd
  3. Aug 26, 2013 #2
    Check the result. 6.022*10^23 (note the correct notation) times 29 should be pretty close to 180*10^23 ≈ 2*10^25, and 48.7/63.5 is close to 1, so you should be in the 10^25 ballpark, not 10^24.
     
  4. Aug 26, 2013 #3
    Yup, you're correct! Thanks. Sometimes it's really hard catching your own mistake. Thanks again!
     
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