A stock solution of my plant extract was prepared at a concentration of 10 mg/mL. From this stock, a working solution of 0.2 mg/mL was prepared. Then, 150 µL of the 0.2 mg/mL solution was mixed with 750 µL of Folin–Ciocalteu reagent and 600 µL of sodium carbonate solution. The mixture was incubated, and the absorbance was measured. For the 0.2 mg/mL solution, the measured absorbance was 0.13.
The total phenolic content was determined by comparison with a standard curve of gallic acid, which was expressed by the equation:
Y = 0.004X + 0.0929
where Y is the absorbance and X is the concentration of gallic acid (µg/mL).
Please calculate it for me and tell me the value.
Thank you very much.
Got it — let’s carefully calculate step by step.
We have:
Absorbance of sample = 0.13
Standard curve: Y = 0.004X + 0.0929
(where Y = absorbance, X = gallic acid concentration in µg/mL).
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Step 1: Rearrange the equation
X = \frac{Y - 0.0929}{0.004}
Step 2: Plug in the absorbance (Y = 0.13)
X = \frac{0.13 - 0.0929}{0.004}
X = \frac{0.0371}{0.004}
X = 9.275 \, \mu g/mL
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✅ So, the equivalent gallic acid concentration = 9.28 µg/mL (approx).
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Step 3: Express as Total Phenolic Content (TPC)
Since your extract solution tested was 0.2 mg/mL, and this gave 9.28 µg gallic acid equivalents per mL:
\text{TPC} = \frac{9.28 \, \mu g \, GAE/mL}{0.2 \, mg \, extract/mL}
\text{TPC} = 46.4 \, \text{µg GAE/mg extract}
Or equivalently:
\text{TPC} = 46.4 \, \text{mg GAE/g extract}
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✅ Final Answer:
The total phenolic content of your extract = ≈ 46.4 mg GAE/g extract.
Kindly go through this paper published for total Phenols and Flavanoids estimations using Gallic Acid and Quercetin, respectively.
in Oriental Journal of Chemistry June 2012 28 :993-998 by Madhu Aggarwal et al
The total phenolic content of the plant extract working solution is 9.275 µg gallic acid equivalent per ml.
H.A Adewuyi Thanks a lot, please Iwould also like an answer to this question (flavonoids calculation) The total flavonoid contents of the extract were determined by the colorimetric aluminum chloride method (Chang et al., 2002). 0.5 ml of the extract (200 μg/ml) in 80% methanol was prepared and mixed with 1.5 ml of 80% methanol, 0.1 ml of 10% aluminum chloride, and 0.1 ml of 1M potassium acetate, and diluted to 5 ml with distilled water. Blank was prepared, containing 2 ml of 80% methanol, 0.1 ml of 10% aluminum chloride, and 0.1 ml of 1M potassium acetate, and diluted to 5 ml with distilled water. Then the mixtures are allowed to stand for 30 minutes in the dark at room temperature. Absorbance was measured at 430 nm using a UV/visible spectrophotometer. The measurement was compared to a standard curve of prepared rutin solutions (25, 50, 75, 100, and 200 μg/ml). The experiment was conducted in triplicate and expressed as rutin equivalent (mg/g of dry mass) using the calibration curve equation. The absorbance in 200µg per ml is, on average, 0.04, the equation Y=0.0017X+0.025
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