If seawater has 40 nL/L of methane, what is the equivalent molarity, please? Any help with this basic conversion is most appreciated.
Cheers,
FR

If seawater has 40 nL/L of methane, what is the equivalent molarity, please? Any help with this basic conversion is most appreciated.
Cheers,
FR
Can I assume the methane is in liquid phase? And that the density of liquid methane is 0.423 g/mL.
With those assumptions my calculations show:
40nL x (1 mL/10^6 nL) x (0.423 g/1 mL) x (1 mol/16.043g) = 1.05 x 10^6 moles
... in a liter, equivalent to 1 micromolar (1000 nM).
This is clearly wrong, the answer should be in the nanomolar range.
Any help is appreciated, I believe my basic assumptions may be off.
It is most unusual to state the quantity of dissolved methane in units of volume. It is customary in similar cases to use ppm, or some other such weight/weight measure.
If you are certain that the units you have been given are not a mistake, then assume the volume of methane is that of a gas at STP.
Weird.
*
Unless I am mistaken nL/L is ppb [parts per billion]
You may be right. I have not seen it used this way.
The reported value of nL/L is for surface seawater in the subtropical atlantic gyre, Seifert et al 1999 Biogeochemistry volume 44.
Your feedback is much appreciated, but I fear I am still at a loss on how to convert this to moles/liter.
Perhaps we assume that a liter of water is 1 kg, and that 40 ppb of one kilogram equals (40 x 10^9) kg, or (40 x 10^6)g. With a molecular weight of 16 grams/mol (or thereabouts,) there will be (40 x 10^6)/16 = ~2.5 x 10^6 moles in that same liter of seawater 
this still yields an answer in the micromolar range, and seawater, particularly in the gyres, should have methane concentrations in the nanomolar range.
...any additional help?
I found this set of conversions; what are your assumptions based on? Don't all gases have the same molar volume at standard temperature and pressure?
http://www.lenntech.com/calculators/...ermillion.htm
Sorry, my knowledge of analytical chemistry is very rudimentary
As is mine, I am not certain whether dissolved methane is a gas or a liquid or how that changes the equation.Originally Posted by samcdkey
Your link indictes an answer of 3e13 mg/m^3, or 3e17mg/liter,
... I am suspecting this is wrong as well.
I will continue to work on it.
Edit: Hmmm, using the weight option instead of the volume option at the link leads again to an estimate of 2.5 micromolar.
I am beginning to wonder if the original measurements are reliable.
What exactly is the question? Is it just to give the molarity of methane in seawater assuming the concentration is 40ppb? I may be oversimplying the question, but it seems to me that you would start by calculating the number of moles of H2O, (methane's effect on the volume should be negligible I just used 1 litre=1 kg). I didn't work this out, but apparently it is 55mol/L. Multiply this by avogadro's constant to give the number of atoms in 1 litre. I don't have a calculator handy, but that equation is 55*6.023×10^23. You divide that number by a billion, then multipy it by 40 and that should give you the number of particles of methane in 1 litre of sea water. Divide this by avogadros constant (you could probably leave avogadros constant out altogether, but I just did it this way to understand what I was doing) and you should get the number of moles of methane per litre sea water. This is apparently 2.2x10^6 mol/L.
Which is 2.2 micromolar, which is 1000 fold higher than it should be..... Should be nanomolar. Only at methane vents does methane reach micromolar concentrations, in surface seawater it is in close equilibrium with the atmosphere and shoul be around 2 nanomolar.
aaaarggghhh.
Alright. A colleague has pointed me towards the ideal gas law for this conversion:
Basic refresher please: Does this mean that an 'ideal gas'  ANY ideal gas  will have about 24.4 L/mol at atnmospheric pressure and standard temps (~22C)?... use the Ideal Gas Law  PVRT; at sea surface conditions, there are about 24.4 L/moles,
Can I also draw from this that the methane in the original measurement (40nL/L) is still considered to be in a gaseous phase? This makes little sense to me as it is seawater, but I am happy to be educated.so divide your nL numbers by 24.4, and you'll get reasonable nM values, between 2 to 4, which is atmospheric saturation, or very slightly above.
mmm... perhaps the method involves outgassing the methane from the seawater and the measurements are then made in air...... will have to check M&M.
Gases can also dissolve in water. Haven't you ever had a soda?
But when truly dissolved is it a gas? I realise these are basic misunderstandings on my part.
The gas in sodas bubbles out. Soda is carbonated under pressure, which I believe converts gas to liquid.
There is flux of methane from the ocean to the atmosphere and back again, is it considered a gas at every point in this flux.
I really don't think the true gas law or any of that stuff is relevant. You have the value in ppb, and assuming that is correct it doesn't matter about any of that other stuff. That would only be relevant if you were trying to calculate how much methane was present given the conditons etc. You already know how much is present, and are just converting from one measurement (ppb) to another (moles/litre). I've never done this before, but my reasoning seems sound, and my answer seems about right.
Check to make sure you're values are correct, then also check to make sure you are reading the units correctly. Is it definitely moles per litre?
EDIT
Actually the ideal gas law will make some difference, the 55 in my equation will change, though not by enough to make any major difference in the value.
An answer of micromolar cannot be correct, that would make the seawater 100000% supersaturated with methane relative to the atmosphere. Although I admit I do not fully understand the reasoning for using the ideal gas law,  it seems certain that there is some detail in the methods that I am not aware of, and when I do get around to having a look I'll be sure to make a note of it.
Have you taken into consideration the fact that liquid is much denser than air? Since you are looking for an answer at constant volume, molarity isn't really a good way to measure it since the number of particles in that litre could vary significantly. Unless you have taken this into account you can't really make a comparison like the one you have above. Either way, the values you are getting for the molarity of methane in sea water are accurate assuming the 40ppb is accurate.Originally Posted by free radical
« Alkaloid  Element » 