# Folin–Ciocalteu formula

I am searching a long time now. I have a question about the Folin–Ciocalteu formula for quantification of polyphenol. How is the $62.5$ in the equation explained? Where does it come from?

This is the protocol [1, p. 38]:

$\pu{5.0 mL}$ of extract were placed in calibrated flask and filled up to $\pu{25 mL}$ by distilled water. $\pu{1.0 mL}$ of Folin-Ciocalteu reagent, $\pu{10.0 mL}$ of distilled water were added to $\pu{2.0 mL}$ of diluted extract (from $\pu{25 mL}$ calibrated flask) and filled up to $\pu{25 mL}$ by sodium carbonate solution ($\pu{290 g/L}$). An absorbance of prepared sample was measured by means of UV spectrophotometer at $\pu{760 nm}$ after $\pu{30 min}$ of incubation in darkness. The same liquid (with pure water instead of plant extract) was used as a blind test. All determinations were performed in triplicate. The content of polyphenols was calculated from the formula:

$$X = 62.5 \cdot \frac{A_1 \cdot m_2}{A_2 \cdot m_1}$$

where $A_1$ - absorbance of pyrogallol solution, $A_2$ - absorbance of investigated samples, $m_1$ - mass of investigated sample [$\pu{g}$], $m_2$ - mass of pyrogallol [$\pu{g}$].

### Reference

1. Modnicki, D.; Balcerek, M. Herba Polonica 2009, 55 (1), 35–42.

$62.5$ is just a dilution factor for the herb extract probe. From the protocol, first time the extract was diluted $5$ times (from $\pu{5 mL}$ to $\pu{25 mL}$), second time – $12.5$ times (from $\pu{2 mL}$ to $\pu{25 mL}$), resulting in total dilution factor of $5 \cdot 12.5 = 62.5$.