Hello everybody
In the measurement of anthocyanins, I've calculated a concentration value of 6800 mg/L for cyanidin-3-glucoside, through the next formula: (Abs x M.W. x D.F. X 1000)/(ε x 1), where Abs: Absorbance; M.W.: Molecular weight; D.F.: Dilution factor and ε: Molar absorptivity.
Could you tell how I should convert mg/L to mg/g? Please, Thank you.
Hi Andres,
First of all you need to know the quantity of the sample which was used for the determination of anthocyanins and, optionally, its dry matter content. Let's assume that you used 1 g of some plant material and dry matter content of that plant material was 0,20 g d.m./g. You used 10 ml of some solvent. After diluting your extracts you continued the analysis and determined that there was 6800 mg of anthocyanins per 1 litre of your original extract. Now you have to carry out the following calculations:
6800 mg of anthocyanins ----------------- 1000 ml of the crude extract
x ----------------- 10 ml of the crude extract
x = 68 mg of anthocyanins
You used 1 g of the plant material for the preparation of 10 ml of the crude extract, so:
68 mg of anthocyanins ------------------- 1 g of the plant material
You can finish here (so the result is 68 mg/g) but if you want express your results as mg in dry matter then you need to do the following calculations:
68 mg of anthocyanins ------------- 1 g of the plant material ----------- 0,20 g of d.m,
x --------------------------------------------------------- 1 g of d.m.
x = 400 mg of anthocyanins/1 g d.m.
Obviously results here are odd but I picked some number just to show you calculations based on the example.
Hope this helps.
Best wishes,
Łukasz
All you need to know is density of your chemical.
Density = mass/volume.
Therefore, volume (L) = mass/density
then, mg/L= mg*density/mass.
Cheers,
Kishore
I also agree with Kishore Ragi by the fact that the density will be useful for this calculation
Hi
I have a problem to calculate total anthocyanins. I have an extract with 102444 ppm concentration. I took 1ml of extract and dilute it to 10ml. then took 1ml and added 5ml of buffers. average absorbance became 0.389 and major anthocyanin of extract is delphinidin. after calculations I obtained 362 ppm anthocyanins but in literature it is 1700ppm. can you help me, please?
Hello
Dear Mahdi
According to this analysis "Characterization and Measurement of Anthocyanins by UV-Visible Spectroscopy"
Serial dilutions are recommended to ensure accurate measurements of highly concentrated, high density, or dried samples. Perform a weight-by-volume dilution with distilled water to obtain a single-strength solution (e.g., usually around 10° Brix for fruit juices), followed by a second dilution using 0.025 M potassium chloride buffer, pH 1.0. Both dilution factors must be considered when calculating monomeric anthocyanin content.
For example, 1 g of a 75° Brix juice concentrate was diluted to a final volume of 10 ml with distilled water (dilution factor = 10; assuming a density of 1 g/ml for juice). Then, the appropriate dilution factor for the sample was determined by diluting 0.2 ml of the solution with 2.8 ml of 0.025 M potassium chloride buffer, pH 1.0 (dilution factor = 15). To calculate monomeric anthocyanin content, color density, or polymeric color, the dilution factor to use would be: DF = (10 × 15) = 150.
Please, try to perform your analysis with this information.
Regards
Hello
Dear Navina
Dilution factor is calculated through dividing the final volume by the initial one. For example, if you take 1 ml of X solution and mix it with 9 ml of any solvent, it is a dilution of 1:10. Therefore, final volume is 10 and the initial one is 1 ml. Finally, 10 / 1 = 10. This is the dilution factor.
Regards
If it was a sol, for exemple you did leaching or extraction, you have to convert the unity from mg/l to mg/Kg, so usually we use this method:
for example (leaching):
100g ->1L
the concentration of nitrite is : 2mg/l
so you do 2/100 (mg/g), after that you get 0.02*1000 to have 20mg(nitrite)/kg sol
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