# How to relate the transmittance to the absorbance, concentration, and molar absorptivity?

##### 1 Answer

The absorbance

#bb(A = -log T)#

*absorbance* is therefore:

#color(blue)(A) = -log(0.762) = color(blue)(0.118)#

What would the absorbance be if the percent transmittance was

#color(white)("The substance absorbs no incoming light at all.")#

**Beer's Law**:

#bb(A = epsilonbc)# ,where

#epsilon# is themolar absorptivityin#"L/mol"cdot"cm"# ,#b# is thepath lengthof the cuvette (usually#"1 cm"# ), and#c# is theconcentrationin#"mol/L"# . In this case,#b = "1.5 cm"# !

Therefore, the *molar absorptivity* is:

#color(blue)(epsilon) = A/(bc)#

#= 0.118/("1.5 cm" cdot "0.0802 M")#

#= color(blue)("0.981 L/mol"cdot"cm")#

This makes sense, because if the absorbance is low, the substance reflects most of the light that approaches it. Since absorbed light is not seen, the higher the absorbance, the darker the substance.

The darker the substance, the higher its

*very low*, since it is expected to absorb a lot of light, but it absorbs only a little since

Let's see if that's true.

#color(blue)(c) = A/(epsilonb)#

#= 0.118/("10000 L/mol"cdot"cm"cdot"1.5 cm")#

#= color(blue)(7.87 xx 10^(-6) "M")#

Indeed, the concentration is very low.