SME
Analytical Chemistry
Spectrophotometry (Beer–Lambert)
Use this for “A dye has ε = 8,500 L/(mol·cm) at 520 nm. A 2.50 × 10⁻⁵ M solution sits in a 1.00 cm cell. Find its absorbance and percent transmittance.”
Diagram
A
=
εbc
=
12,000
×
1
×
4 ×
10
-5
A = 0.48
%T = 100 × 10−A = 33.1%: the beam leaves with 0.331 of its light, drawn as its width.
A
=
−log(
I
I₀
);
the
more
strongly
the
solution
absorbs,
the
deeper
its
colour.
This is an example. Type your own values to start a new problem.
A
=
ε (L/(mol·cm))
×
b (cm)
×
c (M)
%T
Percent transmittance
calculated · %
Formulas
A
=
ε
×
b
×
c
0.48
=
1.2 ×
10
4
×
1
×
(4
×
10
-5
)
A
=
−log₁₀(%
T
100
)
0.48
=
−log₁₀(
33.11
100
)
Assumptions
•
Beer’s law, A = εbc: light of one wavelength, a dilute solution (A up to about 1 stays on the straight line), and nothing else in the cell absorbs.
•
A has no unit: A = −log(I ÷ I₀), where I₀ goes into the cell and I comes out, and %T = 100 × I ÷ I₀.
•
ε is in L/(mol·cm), b in cm and c in mol/L, so the units cancel.
Step-by-step
WE KNOW
ε = 1.2 × 10⁴ L/(mol·cm), b = 1 cm, c = 4 × 10⁻⁵ M
FIND
absorbance (A), percent transmittance (%T)
Step 1 · Find absorbance (A)
Use A = ε × b × c
Beer’s
law:
multiply
ε
by
the
path
length
and
the
concentration.
A
=
1.2 ×
10
4
×
1
×
(4
×
10
-5
)
A
=
0.48
Step 2 · Find percent transmittance (%T)
Use A = −log₁₀(%T ÷ 100)
Undo
the
log:
the
light
that
gets
through
is
1
10
A
of
what
went
in,
as
a
percent.
%T
=
100
10
A
%T
=
100
10
0.48
%T
=
100
3.02
%T = 33.11%
Check
Put the numbers back in. Both sides should match.
0.48
=
1.2 ×
10
4
×
1
×
(4
×
10
-5
)
✓
0.48
=
−log₁₀(
33.11
100
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
An unknown from a calibration line
Use this for “Iron standards give the line A = 0.0820c + 0.0030 (c in mg/L). A sample diluted 5 times reads A = 0.415. How much iron was in the sample?”
Two absorbers at two wavelengths
Use this for “Co²⁺ and Ni²⁺ both absorb at 510 nm and 395 nm. From their ε at each wavelength and the mixture’s two absorbances in a 1 cm cell, find both concentrations.”
Standard addition
Use this for “A 5.00 mL water sample gives a lead signal of 0.150. With 0.50 mL of 50 mg/L lead standard added, both made up to 25.00 mL, it reads 0.270. Find the lead concentration.”