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Help with nanomolar calculations

  1. Sep 25, 2008 #1
    Appreciate some backup, checking my calculations to see where I messed up - - it's been 30years since college chemistry:





    Let's say I have a compound X which has a molecular weight of 320, so, 320 grams = 1 mole.

    Next, I have a 1% stock solution A of that compound X (1 gram dissolved in 100 grams water).

    We know that 1 ml = 1 gram for water, and we know that approx. 20 drops are in one ml.

    1000 mg. = 1 gram.

    20 drops X 100 ml = 2000 drops, or, 1000 milligrams divided by 2000 drops =.5 mg of the compound X in every drop of 1% stock solution A.

    Now, put that 1 drop of 1% stock solution A in 240 ml of water to create stock solution B.

    That is rather close to a 1/4800th dilution.

    Then use 1 drop from this stock dilution B to dissolve into 1000 ml of water which yields .000104 mg of compound X in a whole liter of water (final stock solution C).

    We know that 1 mg. per liter is 1 ppm, so 0.000104 mg. would be 0.000104 ppm, close to 0.104 ppb, or 104 nanomolar.

    Now, a mole = 6.022 X 10 ^23 molecules of any compound.

    A nanomole = 10^-12, essentially that is 1 trillionth, so we can calculate that 10^23 minus 10^12 yields 6.022 X 10^11 molecules in a nanomole.

    and 100 nanomoles is 6.022 X 10^13 molecules.

    Therefore, in that one liter of water there are better than 60 trillion molecules of compound X floating around in the final stock solution C - - even though the dilution is rather incredible and ultimately results in about only one ten thousandth of a milligram of compound X in the entire liter of final stock solution C.
     
  2. jcsd
  3. Sep 25, 2008 #2

    chemisttree

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    Are you sure that you messed up?
     
  4. Sep 25, 2008 #3

    Borek

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    I would check here.
     
  5. Sep 25, 2008 #4
    I don't know if I made a mistake - - that's why I'm asking for others to check out my calcs.

    I cannot see any mistakes in the line that Borek shows.


    Anybody up to the challenge ?
     
  6. Sep 26, 2008 #5

    Borek

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    So you suggest that 104 nanograms are 104 nanomoles even if the molar mass is 320 g/mole? :wink:
     
  7. Sep 28, 2008 #6
    Hmmmm,


    Yes, this phrase is wrong:

    "We know that 1 mg. per liter is 1 ppm, so 0.000104 mg. would be 0.000104 ppm, close to 0.104 ppb, or 104 nanomolar."

    I should not confuse molar mass with actual weight.

    The phrase above could end with 104 ppt (trillion).

    Hmmm, it seems therefore that 100 nanomolar would be 32000 nanograms, since 320 grams = 1 mol.

    Now, 32000 nanograms = 0.032 milligrams I believe.

    So that is about 307.7 times more compound X needed to achieve 100 nanomolar than I previously calculated - - - I was off by about 307 times . . .

    . . . and my assertion that in that one liter of water there are better than 60 trillion molecules of compound X floating around in the final stock solution C does not change - - - it should still be accurate and valid . . .

    Thank you Borek.
     
  8. Sep 29, 2008 #7

    Borek

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    I think you were off exactly by the factor of 320, and 307.7 is just effect of rounding. Your previous calculations contained hidden assumption that molar mass is 1, now you use 320.
     
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