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Effective atomic number for a mixture (with components concentration)

  1. May 20, 2013 #1
    Assume we have a concentration of any specific material (in mg/M). Let's say for example: Ag (Z=47) with concentration 10 mg/M in water (H2O, Z=7.42).

    How can I calculate the effective atomic number (Zeff.) for the silver-water mixture if the concentration of silver was 10 mg/M for example?

    The only formula I know to calculate Zeff. is the following:

    efe25af14b6a9e88727a42c54d8a1711.png

    but it uses (fn) the fraction of the total number of electrons associated with each element, not the concentration in mg/M.

    Is there any possible way to convert the concentration (in mg/M) into a fraction of total number of electrons for silver then calculate Zeff for the silver-water mixture mentioned above as an example?


    Thanks in advance.

    source: http://en.wikipedia.org/wiki/Effective_atomic_number (example for H2O provided)

     
  2. jcsd
  3. May 20, 2013 #2

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    Based on the atomic masses of silver, hydrogen, and oxygen, you can convert their masses to atom counts.

    Silver has a molar mass of 107.8682 grams.
    So for instance 10 mg of Silver is 0.010 / 107.8682 moles.
     
  4. May 20, 2013 #3
    Thanks for the response. that's for my second question..
    but how can I apply this to calculate the Zeff for the silver-water mixture using the formula in the original post above (or any other formula) ?
     
  5. May 20, 2013 #4

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    I'm a bit confused about your unit mg/M.
    What does it mean?
    Afaik 1 M = 1 mol/L.

    Anyway, if you have 10 mg silver in 1 kilogram water, you have:
    0.010 / 107.8682 moles silver
    1000 / 18 moles water, consisting of 2000 / 18 moles hydrogen, and 1000 / 18 moles oxygen.

    Therefore the number of silver electrons is 47 x 0.010 / 107.8682.
    The number of hydrogen electrons is 1 x 2000 / 18.
    And the number of oxygen electrons is 8 x 1000 / 18.

    From this you can calculate first the fractions, and then the Zeff.
     
  6. May 23, 2013 #5
    sorry, I've been busy with exams.

    Thanks a lot. your last post was very helpful.
     
  7. May 23, 2013 #6

    I like Serena

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    I take it you've already taken the exam on this subject... successfully?
     
  8. May 24, 2013 #7
    not on this subject specifically but it helped me to figure things out... I was not convinced that mg/M is even a concentration unit, it confused me but I think I've got it all clear now. Thank you <3
     
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