Redfield ratio? Red herring!

Discussion in 'Water Parameters and Additives' started by RocketRooster, 10 Jan 2013.

  1. RocketRooster

    RocketRooster

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    Well, that depends on your test kit. I had a discussion with @Ridwaan about this earlier today and something about it was bothering me. I've realised what it is.

    The MOLAR mass of nitrogen is around 14g/mole. Phosphorous is double that at around 31g/mole.

    So, point is, one atom of P is twice as heavy as one atom of N.

    Ignoring for now the fact that phosphates and nitrates have oxygen atoms (you can calculate those in if you like), if your test kit give readings in mg/l instead of ppm, your 'redfield ratio' is actually 16*14:31, or about 7:1. If your kits show ppm then 16:1 is fine. If say, your nitrate kit shows ppm and the phosphate kit mg/l, you will need to convert one or the other to calculate the RR of your water. Ideally you should be using the molar mass of the molecules, not just the N or P atoms, it actually makes a big diff to the result.

    That is, if it reaaally bothers you. :)
     
    Last edited by a moderator: 26 Nov 2015
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  3. carlosdeandrade

    carlosdeandrade

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  4. Visser

    Visser MASA Contributor

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    @RocketRooster, a very simple formula to calculate redfields ratio from NO3 & PO4 is:
    (NO3/PO4)*1.53

    with this formula you get the correct ratio without having the do all the conversions from molecular mass to ppm.

    Say you have 1 ppm nitrates & 0.1ppm phosphates

    (1/0.1) * 1.53 = 15.3 redfields ratio
     
    Last edited by a moderator: 26 Nov 2015
  5. RocketRooster

    RocketRooster Thread Starter

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    Thanks Visser. The point I'm getting at is that mg/l units need conversion, since it's a ratio of mass per volume, whereas ppm is a molecule count.

    If your nitrate test kit gives mg/l, you need to convert it to ppm, otherwise your nice and simple formula won't work though.
     
  6. RocketRooster

    RocketRooster Thread Starter

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    To wit: the standard rr is 16:1 when your units are all ppm, and 7:1 when mg/l
     
  7. Nemos Janitor

    Nemos Janitor

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    ppm can refer to a measurement of weight/weight or mass/mass or volume/volume. It is dimensionless.

    Most reef test kits refer to mg/L and ppm measuring as the same thing and no conversion is necessary. If one wants to get technical then mg/L will have also be converted for temperature, salinity, altitude etc etc. :whistling:
     
  8. dallasg

    dallasg Moderator MASA Contributor

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  9. RocketRooster

    RocketRooster Thread Starter

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    I respectfully beg to differ, Keith. Mg is a measure of mass, not weight. And since water is incompressible, a liter is a liter regardless of depth or altitude.

    The redfield ratio refers to the number of N atoms versus P atoms in the test sample. If we are just counting atoms or molecules, then 16:1 holds for ppm, since as you say its dimensionless. However, a mole of P has twice the mass of a mole of N. So 1mg of N has twice the number of atoms that 1mg of P has. The moment you have measurements in mg, then the picture changes completely and you have to take the relative mass of a nitr*te molecule and a phosphate molecule into account. 1mg/l of nitrate is emphatically not the same ppm as 1mg/of phosphate, since it isnt a measure of counts, it's a measure of mass.

    I've got 5 zim $ that says, if you take the corresponding mg/l and ppm measures shown on a test kit and go and graph them, you won't get parallel lines...
     
  10. Nemos Janitor

    Nemos Janitor

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    Yup i agree in pure water. We are talking salt water. Does the mass not alter with the amount of salt added, temperature, pressure etc?
     
  11. RocketRooster

    RocketRooster Thread Starter

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    The mass will increase as you add solutes, but the volume will remain pretty much the same at a given temperature. There is a density difference between salt and freshwater, but thats moot really since we're doing apples and apples every time we test (unless of course, and I'm sure you'll agree, if the aquarist can't keep his salinity in check then he has no urgent business with his n/p ratios :))
     
  12. Nemos Janitor

    Nemos Janitor

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    Something else to consider is most nitrate test kits measure NO3. Some give reference to N-NO3 the same applies to Phosphate test kits which measure PO4 and P-PO4. I would expect the in the case of your redfield red herring one would have to use total N & P.

    The point I am confused about is that we are measuring PO4 (phosphate) and not P (total Phosphorus). Compound to this, most Phosphate aquarium test kits only measure orthophosphate and ignore organic phosphates.

    A 0.045 ppm of PO4 (Orthophosphate) is equivelant to 0.015 ppm P-PO4 (phosphorus - Phosphate)
    A 0.5 ppm NO3 (nitrate) is aprox equivalent to 0.11 ppm N-NO3 (nitrogen - Nitrate)

    The conversion rates are multiply P-PO4 by 3.07 and N-NO3 by 3.57. How will this difference affect the ppm and mg/L conversion?

    Do I understand correct that this is what is referd to in the first post?
     
    Last edited: 11 Jan 2013
  13. RocketRooster

    RocketRooster Thread Starter

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    I agree, one does not really know, unless you speak to the manufacturer of the test kit, what exactly is being tested or how the result is derived. For example, if your test kit chemically only measures orthophosphate, is the reference card (or electronic result) calibrated (ie contains a conversion factor) to include organic phosphates or not?

    Be that as it may, let's assume for the sake of argument that your test kit does what it says on the tin and gives you an accurate overall value of plain old P04, either in mg/l or in ppm.

    One mole of any entity is roughly 6 * 10^23 entities. That's Avogadro's number. It could be electrons, atoms, molecules or whatever.

    If you look at the periodic table of the elements, you will see each element has an atomic weight listed, this is the average atomic mass for the element times Avogadro's number (i.e one mole of the element). In the case of Nitrogen, one mole has a mass of 14.0067 grams. (It's obviously a huge number!)

    Hydrogen: ~1 gram
    Nitrogen: ~14 grams
    Phosphorous: ~31 grams
    Oxygen: ~16 grams

    (I've rounded these off to simplify things).

    Hence, Nitrite has a molar mass (grams) of (1*14) + (2*16) = 46g. Nitrate is (1*14) + (3*16) = 62g. Phosphate is (1*31) + (4*16) = 95g.

    It thus follows, if you have a milligram of Nitrite in a sample, you effectively have

    6 * 10^23 * ((1*14) + (2*16)) / 1000 nitrite molecules in your sample. Since each molecule has one N atom, that's the number of N atoms you have.

    Similarly, for phosphate, you have

    6 * 10^23 * ((1*31) + (4*16)) / 1000 molecules (and thus that number of P atoms).

    The two formulae yield wildly differing values, which is logical since a P04 molecule is much heavier than a NO2 molecule. 1mg of NO2 has many more N atoms in it than 1mg of P04 has P atoms in it.

    It also makes sense to me that the ratio of your N : P should be all the species of N versus all the species of P. In other words, you should add your nitrite ppm value to your nitrate ppm value before comparing it with the P04 ppm value.

    The Redfield ratio only concerns itself with the number of elemental C, N and P atoms. Hence, if your readings are in mg/l and there is no corresponding ppm scale, you have to convert the readings to ppm so that you get a reflection of the number of elemental atoms in the sample. Working with the masses alone will give you a false result. The easiest way is to simply multiply the mg/l value by the number of times that NO2 or NO3 are lighter than P04. This will convert the NO2/NO3 mg/l values into a form that is consistent to use for counting atoms. This way you don't need to concern yourself with volumes, you just want relative values that represent atom counts. (We're only interested in a ratio, after all, not how many atoms we have).

    95 / 46 = 2.065 (this is the conversion factor for Nitrite)
    95 / 62 = 1.53 (conversion factor for nitrate)

    (The conversion factor for phosphate is one, since 95 / 95 = 1 , obviously)

    So, to yield your redfield ratio having only mg/l values, you would do this:

    (NO2 mg/l value * 2.065) + (N03 mg/l value * 1.53) : P04 mg/l value


    I trust this makes sense! I'm starting to get blurry vision! :))
     
    Last edited: 11 Jan 2013
  14. Dillan

    Dillan

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    And if you are confused you must thing how lost on this topic people as stupid as me are
    :whistling:
     
  15. Nemos Janitor

    Nemos Janitor

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    Am I on the right track when I say that ...

    N represents 22.58% of NO3 (N= 14 and O= 16x3 =48)= 62. (14/62=22.58%)
    Therefore 1mg/L of NO3 = 1 x 0.2258 = 0.2258mg of N

    Likewise..

    P represents 32.63% of PO4 (P=31 and O =16x4=64)= 95 (31/95=32.63%)
    Therefore 1mg/L of PO4 = 1 x .3263 = 0.3263mg of P
     
    Last edited: 11 Jan 2013
  16. RocketRooster

    RocketRooster Thread Starter

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    Hundreds. So now you 'need' to know how many atoms of N you have relative to P.

    So 1 mg of N has 31/14 = 2.21 times as many atoms as one mg of P.

    You could go and work out how many atoms that is by using Avogadro's number, but we don't care about that - we just want the ratio.

    So easiest is, for each mg per litre, you have 0.3263 mg of P, and (2.21 x 0.2258 = 0.499mg) of N.

    So, if both readings are 1mg/l (hypothetically) your ratio of N : P is 0.499 : 0.3263 (or 1.53 : 1) :)


    This explains where @Visser's formula comes from. The caveat is that you can't apply it when you have ppm values. Everything must be either ppm (in which case you use it as is), or mg/l (in which case you apply Visser's formula).

    Short and sweet:

    If you have ppm readings--> NO2ppm + N03ppm : PO4ppm
    If you have mg/l readings--> (NO2mg/l * 2.065) + (NO3mg/l * 1.53) : PO4mg/l
     
    Last edited by a moderator: 26 Nov 2015
  17. Nemos Janitor

    Nemos Janitor

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    Great I am nearier to understanding where you are getting to.

    So if we wanted to get a more accurate mg/L we would have to work out the weight/ mass of sea water. This would entail converting Salinity to true sg (sg@ 4c) say the sample is sg=1.0265 @ 4c that is a seawater mg of 1.0265mg. That would mean deviding N & P by 1.0265 or is it multiplying.
     
  18. RocketRooster

    RocketRooster Thread Starter

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    You'd need to take the temperature and salinity into account if you wanted to know the component masses of your litre of seawater accurately.

    I am of the opininion that that's overcomplicating things though, and that the accuracy you gain is lost through statistical variance.
     
  19. dallasg

    dallasg Moderator MASA Contributor

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    Last edited by a moderator: 26 Nov 2015
  20. RocketRooster

    RocketRooster Thread Starter

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    Same as you mate. Prostitute out my coding skills to the highest bidder.

    (My tertiary education was B.Sc. Biological Sciences though. Wasted youth!)
     
    Last edited: 11 Jan 2013
  21. dallasg

    dallasg Moderator MASA Contributor

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    Nice :)
     
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