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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Cantacoxinha on April 08, 2022, 04:02:43 AM

Title: How to include a dilution step?
Post by: Cantacoxinha on April 08, 2022, 04:02:43 AM
I'm analyzing total nitrogen content in tobacco. In the procedure I'm following, I have to oxidise ground tobacco sample with concd sulfuric acid and then transfer 6 mL of digested mixture to a 200 mL flask and dilute with water. The solution is then analyzed by UV/Vis.

However, the absorbance was way too high and I had to dilute the sample, so I took 2 mL of the already diluted 200 mL solution, I diluted it up to 10 mL (2 mL sample + 8 mL water) and that was used for measurement.

Now, I have to express the total nitrogen content as % of sample dry weight. The formula I'm using is shown on the pic.

c - the concentration of total nitrogen in ppm as calculated from calibration curve

V - volume of the sample in mL (I suppose 200 mL originally?)

m - mass of tobacco sample in mg

M - moisture content in tobacco sample in %



For my specific case, how do I include the dilution step in this formula and which factor should I use?
Title: Re: How to include a dilution step?
Post by: Hunter2 on April 08, 2022, 04:32:03 AM
.
6 ml in 200 ml = 0,03 ml/ml, if you take now 2 ml and fill up to 10 ml, how much do you have then. How much is the dilution.
Title: Re: How to include a dilution step?
Post by: Cantacoxinha on April 08, 2022, 06:01:23 AM
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?
Title: Re: How to include a dilution step?
Post by: Hunter2 on April 08, 2022, 06:09:45 AM
No
6 ml in 200 ml is 0,03 ml/ml, if you take 2 ml then you have 0,06 ml and this is dissolved in 10 ml means 0,006 ml/ml new concentration.

So the dilution is from 0,03 ml /ml to 0,006 ml/ml, what  means 1:5.

So the result what you get from UV measurment you have to multiply by 5 to get value before dilution.

Better is to get a new calibration curve developed with solution 1:5.