Question

Suppose that a 5.0£10¡5 M solution of a compound had a %T equal to 70.80 when measured in a
cuvette with a 1.0 cm path length. Find themolar extinction coefficient of this compound.

Answer 1

The following expression shows the relationship between the absorbance and percent transmittance;

$A=2-log(\%T)$

Percent transmittance $(\%T)$ $= 70.80$

Plug in the value into the formula

$A=2-log(70.80)\newline A=2-1.85=0.15$

Beer Lambert Law states that the amount of energy absorbed or transmitted by a solvent is directly proportional to the concentration of the solute and molar absorptivity of the solution i.e

$A=\epsilon lc \ where,$

$A=Absorbance=0.15\newline \epsilon=molar\ extinction\ co efficient\newline l=path \ length=1cm\newline C= concentration=5.0*10^{5} M$ Rearrange the formula and make $\epsilon$ the subject.

$\epsilon=\dfrac{A}{l*c}$

Substituting the values into the formula;

$\epsilon=\dfrac{0.15}{5*10^5M*1cm}=3000M^{-1}cm^{-1}$

Therefore $\epsilon$$=3000M^{-1}cm^{-1}$