[articuno]

This is more of a mathematical question but here goes.

The Beer-Lambert equation rearranged is: ε=A/cl
Where c = concentration (mol.L^-1), A = absorbance (AU) and l = path length (cm)
Since “mol.L^-1” and “cm” are in the denominator, molar absorptivity has the units L.mol^-1.cm^-1.

Edit (Thanks to chenbeier):
The prefix milli- (m) means multiplication by 10^-3 to convert to the unit without a prefix eg. 350 mL = 350*10^-3 L = 0.35 L.
Similarly, the prefix kilo- (kg) means multiplication by 10^3 to convert to the unit without a prefix eg. 1.2 kg = 1.2*10^3 g = 1200 g
And similarly, the prefix micro- (μ) means multiplication by 10^-6 to convert to the unit without a prefix eg. 250000 μm = 250000*10^-6 m = 0.25 m

The question is, is molar absorptivity in L.mol^-1.cm^-1 numerically the same as molar absorptivity in mL.mmol^-1.cm^-1 given that 1 mol.L^-1 is equals 1 mmol.mL^-1?

Here is my attempt at answering my own question.

Suppose you have a solution that has a concentration of 0.00025 mol.L^-1. That’s 0.00025 moles per 1 litre. Converting to mmol per litre, the concentration is 0.25 mmol.L^-1. So now that’s 0.25 mmol per 1 litre which is the same is 0.25 mmol per 1000 mL. Expressing the concentration in mmol per mL (mmol.mL^-1), we return to the number 0.00025 once more (0.00025 mmol.mL^-1). Therefore, the concentration in mol.L^-1 is numerically the same as the concentration in mmol.mL^-1.
Suppose we are trying to work out molar absorptivity and A = 0.5 , c = 0.00025 mol.L^-1 and l = 1 cm.



We established earlier that mol.L^-1 = mmol.mL^-1. Redoing the calculation above but using mmol.L^-1 instead as the unit for concentration, A = 0.5 , c = 0.00025 mmol.mL^-1 and l = 1 cm.



To conclude, am I right in saying that numerically, the molar absorptivity in L.mol^-1.cm^-1 is equal to  the molar absorptivity in mL.mmol^-1.cm^-1?

[chenbeier]

This statement I would again think about, because the middle part is wrong.

The prefix milli- effectively means multiply by x10^-3 to convert from a unit with no prefix to milli-. For instance:
0.567 g = 0.567 x10^-3 mg = 567 mg
0.42 L = 0.42 x10^-3 mL = 420 mL
3.2 Sv = 3.2 x10^-3 mL = 320 mSv

If you have a/b then ma/mb is the same expression but also ka/kb.

[articuno]

This statement I would again think about, because the middle part is wrong.

The prefix milli- effectively means multiply by x10^-3 to convert from a unit with no prefix to milli-. For instance:
0.567 g = 0.567 x10^-3 mg = 567 mg
0.42 L = 0.42 x10^-3 mL = 420 mL
3.2 Sv = 3.2 x10^-3 mL = 320 mSv

If you have a/b then ma/mb is the same expression but also ka/kb.

My bad. Thanks for pointing that out. I badly phrased that and should reword to:

The prefix milli- (m) means multiplication by 10^-3 to convert to the unit without a prefix eg.

350 mL = 350*10^-3 L = 0.35 L

Similarly, the prefix kilo- (kg) means multiplication by 10^3 to convert to the unit without a prefix eg.

1.2 kg = 1.2*10^3 g = 1200 g

And similarly, the prefix micro- (μ) means multiplication by 10^-6 to convert to the unit without a prefix eg.

250000 μm = 250000*10^-6 m = 0.25 m

So based on your reply and back to the question, is L.mol^-1.cm^-1 numerically equal to mL.mmol^-1.cm^-1 ?

[chenbeier]

Yes it is.

[articuno]

Thank you so much