SME
Analytical Chemistry
Chromatography
Use this for “An unretained peak comes out at 0.85 min. Two compounds elute at 7.20 and 7.90 min with base widths 0.40 and 0.44 min. Find the resolution, both retention factors, the selectivity and the plate count.”
Diagram
R
=
2Δ
t
w₁ + w₂
=
2
×
0.6
0.3 + 0.32
=
1.94:
separated
to
the
baseline.
k
=
t − tM
tM
=
4
and
4.6;
α
=
k₂
k₁
=
1.15.
N
=
16
(
t₂
w₂
)
2
=
4,900
plates.
The dashed tangents of each peak meet the baseline one base width w apart.
This is an example. Type your own values to start a new problem.
R
=
2(
t₂ (min)
−
t₁ (min)
)
w₁ (min)
+
w₂ (min)
tM
Dead time
example · min
k₁
Retention factor of peak 1
calculated
k₂
Retention factor of peak 2
calculated
α
Selectivity
calculated
N
Plates (from peak 2)
calculated
Formulas
R
=
2
×
t₂
−
t₁
w₁
+
w₂
1.94
=
2
×
5.6 − 5
0.3 + 0.32
k₁
=
t₁
− tM
tM
4
=
5 − 1
1
k₂
=
t₂
− tM
tM
4.6
=
5.6 − 1
1
α
=
k₂
k₁
1.15
=
4.6
4
N
=
16
×
(
t₂
w₂
)
2
4,900
=
16
×
(
5.6
0.32
)
2
Assumptions
•
Widths are taken at the base, where tangents to the peak’s sides meet the baseline (four standard deviations).
•
The dead time tM is how long an unretained compound takes to cross the column.
•
A resolution of 1.5 or more separates two peaks of similar size to the baseline.
Step-by-step
WE KNOW
tM = 1 min, t₁ = 5 min, t₂ = 5.6 min, w₁ = 0.3 min, w₂ = 0.32 min
FIND
resolution (R), retention factor of peak 1 (k₁), retention factor of peak 2 (k₂), selectivity (α), plates (from peak 2) (N)
Step 1 · Find resolution (R)
Use R = 2 × (t₂ − t₁) ÷ (w₁ + w₂)
The
gap
between
the
peaks
over
their
average
base
width,
w₁
+
w₂
2
.
R
=
2
×
5.6 − 5
0.3 + 0.32
R
=
2
×
0.6
0.62
R
=
1.2
0.62
R
=
1.94
→
R
≥
1.5:
the
two
peaks
are
separated
to
the
baseline.
Step 2 · Find retention factor of peak 1 (k₁)
Use k₁ = (t₁ − tM) ÷ tM
The time the column holds the compound, over the time an unretained one takes.
k₁
=
5 − 1
1
k₁
=
4
Step 3 · Find retention factor of peak 2 (k₂)
Use k₂ = (t₂ − tM) ÷ tM
The time the column holds the compound, over the time an unretained one takes.
k₂
=
5.6 − 1
1
k₂
=
4.6
Step 4 · Find selectivity (α)
Use α = k₂ ÷ k₁
How many times longer the column holds the second compound than the first.
α
=
4.6
4
α
=
1.15
Step 5 · Find plates (from peak 2): N
Use N = 16 × (t₂ ÷ w₂)²
A
base
width
is
4σ,
so
N
=
(
t
σ
)
2
=
16
(
t
w
)
2
.
N
=
16
×
(
5.6
0.32
)
2
N
=
16
×
17.5
2
N
=
16
×
306.25
N
=
4,900
Check
Put the numbers back in. Both sides should match.
1.94
=
2
×
5.6 − 5
0.3 + 0.32
✓
4
=
5 − 1
1
✓
4.6
=
5.6 − 1
1
✓
1.15
=
4.6
4
✓
4,900
=
16
×
(
5.6
0.32
)
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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