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On Thu, 20 Jan 2000, Kevin Buckley wrote:

> make the throwaway comment "... 1.6 mg/liter N.  That would be about 7 mg/l
> of nitrate."  How do you (is it possible to?) calculate the amount of
> Nitrogen in xxxNO3?  Is it calculated using the atomic weights of Nitrogen &
> Oxygen in some way?  What's the general principle?

The idea is to determine the amount of nitrate that would be needed to
contain the amount of nitrogen that's present.

Nitrate has a molecular weight of 62 grams/mole (from 3 oxyen atoms at 16
grams/mole each and one nitrogen at at 14 grams/mole).  The ratio between
the weight of one mole of nitrate and one mole of nitrogen is 62/14, or

So to convert nitrogen to the equivalent amount of nitrate, mulitply the
nitrogen concentration by 4.4.

It's common in water analyses to report all of the nitrogen species as the
equivalent amount of nitrogen; that is, they often report NO3-N, NO2-N and
NH3-N instead of NO3, NO2 and NH3.  That lets you compare results without
doing any conversions.

Roger Miller

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