Since the dilution is 2 mL of pre-diluted sample in 8 mL of water, it should be 1:4, and then 0,03/4=0,0075, right?
But the question is how should I include that in the original formula I was provided for the calculation of % of nitrogen? The only factor I might change after dilution in that equation is V, I suppose, as I get the c values automatically from the UV/vis... Or I should add a multiplication/division factor somewhere?