The Nitrogen % in 0.200 g of dried leaf sample is 6.2%. The Nitrate-N (measured by dissolving 0.500 g of dried leaf material in 25 ml water) is 136.51 mg/L. I need to convert both values in a similar concentration unit (Total N & NO3-N) so that I can determine the reduced N (by subtracting nitrate-N from Total N). Please help, and also need suggestions about in what concentrations usually these three plant N components are expressed in a thesis writing.
Thank you
Interesting. Total nitrogen will always be in g/kg ( For example 6.2% means 62 g/kg dry matter ), while nitrates and ammonium forms of nitrogen are mostly expressed in mg/kg ( For example 6.2 ppm means 6.2 mg/kg dry matter) ...therefore , nitrates and ammonium both forms hardly constitute 1% of total nitrogen registered in leaves.
Even though it is not my direct discipline but I can suggest u that technically dry the leaf matter in the laboratory and following the standard manuals of nitrogen analysis equipment you can solve the question I think...This very little help
Dear Samra,
In my thesis work, I have used the Jeschke-Pate procedure for determining reduced N. First, the concentrations of total N in plant dry matter are assayed using a CHN analyser and levels of NO3- ions, Cl- ions, SO4-- ions, and HP2O by anion chromatography of aqueous extracts of dry matter. Then, the total reduced N (soluble plus insoluble) is assessed as the difference between total N and NO3-N. Increments or losses of all three categories of N are estimated using the above concentration values and data for dry matter contents of plant parts at initial and final harvests.
You could have a look at their original papers, attached both here.
Hope this may help you.
Ronald
If the extract has 136.51 mg/L then this equates to
0.025 * 136.51 = 3.413 mg in your 25 ml of extract .
As this came from 500mg of dried leaf the %DW of nitrate is
100* 3.413/500 = 0.683%.
both nitrate N and total N, as %, can be given as mg.100mg-1 DW
So, nitrate N is 100 * 0.683 / 6.2 = 11% of total N
This seems very high - maybe check your weights and volumes used in the assays
The calculation could be:
136.5x25/0.5=6825 mg/kg=0.683%
the total N (6.2%)-NO3-N (0.683%)= 5.517%
The total nitrogen in your sample is quite high. this sample should be a legume species. otherwise it is likely to be to high.
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