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Redfield ratio

Discussion in 'General Plant Topics' started by samipedersen@hotmail.com, May 18, 2011.

  1. samipedersen@hotmail.com

    samipedersen@hotmail.com Junior Poster

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    Hello

    We are four students from the University of Copenhagen in Denmark who are having trouble figuring out the Redfield Ratio. We are doing an assignment about iron fertilization in the Oceans. We would like to convert the ratio, from an atomic ratio (106:16:1) to a mass based ratio. This is because we have values for NO3 and PO4. So far we reached these results for a mass based ratio.
    C:N:p = 41,1:7,2:1. Is this correct?

    Best regards
    Sami
     
  2. Tom Barr

    Tom Barr Founder
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    Well, you are already doing well to realize that the ratio is atomic(atoms), not based on mass.
    You got at least 10 points right there avoiding that mistake.

    Note, RR is atoms of C: N : P.

    So you simply take their mass molar equivalents to convert to mass.

    106 x 12.01 for C and 16 x 14.0067 g/mol for N and 1 x 30.9737 for P.

    1273.06: 224.1072: 30.97 for mass.
    Divide by the lowest no# in this case P(30.9737)

    Mass ratio:
    41.1 : 7.235 : 1

    If you have NO3 and PO4, you'll need to do this for those, not N and P. So include the O3/O4 in the molar calculations.

    To go back to atomic ratios, then do the reverse.
     
  3. Greg Watson

    Greg Watson Administrator
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    Here is an old post that Mr. Barr wrote that also might be relevant (note the links are dead so I've made them non-clickable). I also remember Tom writing something awhile back about the ratios varying in the dry weight analysis from different aquatic plant species.

    Best wishes,
    Greg
     
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