What you are given
- Wavelength (nm): 340
- Extinction coefficient: 6220
- Absorbance (A): 0.622
- Show result in: µM
Formula: A = ε · c · l
A = epsilon x c x l, so c = A / (epsilon x l). Keep the extinction coefficient and the path length in matching units, and stay inside the linear range of the instrument.
Worked example
Formula: A = ε · c · l
c (mol/L) = A · DF / (ε · l)c = 0.622 · 1 / (6220 · 1) = 0.0001 MYour sample is 100 µM at 340 nm.
A printable one-page version is on the formula sheet.
Try it yourself
Work out the concentration from the absorbance.
Your sample is 1 mg/mL at 400 nm.
c (mg/mL) = A · DF / (E0.1% · l)c = 0.667 · 1 / (0.667 · 1) = 1 mg/mLOpen this in the calculator to change the numbers.
Work out the concentration from the absorbance.
Your sample is 40.19 µM at 750 nm.
c (mol/L) = A · DF / (ε · l)c = 0.25 · 1 / (6220 · 1) = 0.00004019 MOpen this in the calculator to change the numbers.
Ten more problems on Beer-Lambert (absorbance to concentration)
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