---
title: "Chromatography"
book: "University Chemistry — Year 2"
subject: chemistry
language: en
chapter: 31
exercises: 12
source: https://one-course.com/books/chemistry/3/en/chapter/31-chromatography
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 31 — Chromatography

A drop of an energy drink, diluted and filtered, is injected into a steel tube no longer than a pen and packed with silica grains a few micrometres across. A few minutes later a detector has drawn a line with a handful of peaks, and from the area of one of them the laboratory reports how much caffeine the can holds, to a few per cent. [Chromatography](#def-b2-chromatography-principle) separates the components of a mixture by letting them race through a column at different speeds; this chapter explains why they travel at different speeds, why their peaks have the width they have, how to separate two neighbours, and how to turn a peak into a quantity.

![An analytical laboratory: liquid chromatographs with their solvent bottles and an autosampler tray of vials; the chromatograms appear on the screen (illustration).](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/img-aebaaaa0dee7.jpg)

*An analytical laboratory: liquid chromatographs with their solvent bottles and an autosampler tray of vials; the chromatograms appear on the screen (illustration).*

**You already know.**

The school volume: paper and thin-layer [chromatography](#def-b2-chromatography-principle), chromatogram, eluent. The Year 1 volume: the partition coefficient, liquid–liquid extraction, the retention factor $R_f$ of a thin-layer plate. [Chapter 7](https://one-course.com/books/chemistry/3/en/chapter/7-binary-liquidvapour-diagrams-and-distillation#ch-b2-liquid-vapour-diagrams): the [theoretical plate](https://one-course.com/books/chemistry/3/en/chapter/7-binary-liquidvapour-diagrams-and-distillation#def-b2-liquid-vapour-diagrams-fractional-distillation) of a distillation column.

## 31.1 Principle

**Definition 31.1 (Chromatography).**

*Chromatography* separates the components of a mixture by distributing them between a *stationary phase*, fixed in a column or on a plate, and a *mobile phase*, a gas or a liquid that flows past it. Carrying the components through and out of the column with the mobile phase is *elution*.

A molecule moves only while it is in the [mobile phase](#def-b2-chromatography-principle); while it is held by the [stationary phase](#def-b2-chromatography-principle) it waits. It switches between the two phases millions of times on its way, so its average speed is set by the fraction of time it spends in each, that is by its partition equilibrium.

**Definition 31.2 (Retention).**

The *retention time* $t_R$ of a compound is the time from injection to the maximum of its peak; the *hold-up time* $t_M$ is the retention time of an unretained compound, which never enters the [stationary phase](#def-b2-chromatography-principle). The *retention factor* $k$ of a compound on a column is

$$
k = \frac{t_R - t_M}{t_M}.
$$

It is not the $R_f$ of a thin-layer plate, which measures a distance travelled, though both describe the same partition.

**Proposition 31.3 (Retention time).**

If, at equilibrium, the amounts of a compound in the stationary and [mobile phases](#def-b2-chromatography-principle) of a slice of column are $n_s$ and $n_m$, the compound moves at $u/(1 + n_s/n_m)$, where $u$ is the speed of the [mobile phase](#def-b2-chromatography-principle), and $t_R = t_M(1 + n_s/n_m)$.

**Proof.** A given molecule spends the fraction $n_m/(n_s + n_m)$ of its time in the [mobile phase](#def-b2-chromatography-principle), where it moves at $u$, and the rest at rest. Its average speed is $u\,n_m/(n_s + n_m) = u/(1 + n_s/n_m)$. The column of length $L$ is crossed in $t_R = (L/u)(1 + n_s/n_m)$, and $L/u = t_M$. ∎

**Proposition 31.4 (Retention and partition).**

The [retention factor](#def-b2-chromatography-retention) is the ratio of the amounts in the two phases, $k = n_s/n_m = KV_s/V_m$, where $K$ is the partition coefficient (concentration in the [stationary phase](#def-b2-chromatography-principle) over concentration in the [mobile phase](#def-b2-chromatography-principle)) and $V_s$, $V_m$ the volumes of the phases in the column.

**Proof.** From [Proposition 31.3](#prop-b2-chromatography-retention-time), $(t_R - t_M)/t_M = n_s/n_m$. With $n_s = c_sV_s$, $n_m = c_mV_m$ and $K = c_s/c_m$, $n_s/n_m = KV_s/V_m$. ∎

Two compounds separate if their partition coefficients differ; the column geometry ($V_s/V_m$) multiplies all retentions alike. Changing the [mobile phase](#def-b2-chromatography-principle), which changes $K$, is the chemist’s main lever on retention.

![Two compounds injected together as one thin band move down a column (three snapshots, schematic). The less retained one (smaller k) runs ahead; both bands widen as they travel, the one that has gone further the more.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-6b4b6e20b675.svg)

*Two compounds injected together as one thin band move down a column (three snapshots, schematic). The less retained one (smaller $k$) runs ahead; both bands widen as they travel, the one that has gone further the more.*

## 31.2 Peak width and plate number

A band does not stay thin. Molecules of the same compound take different paths between the grains, diffuse along the column and lag behind in the [stationary phase](#def-b2-chromatography-principle) by random amounts. The peak recorded at the outlet is very nearly a Gaussian of standard deviation $\sigma_t$ (in time), and the narrower it is for a given $t_R$, the better the column.

**Definition 31.5 (Plate number).**

The *plate number* of a column for a compound is $N = (t_R/\sigma_t)^2$, where $\sigma_t$ is the standard deviation of its peak in time. The *plate height* is $H = L/N$, for a column of length $L$.

The names come from the distillation column of [Chapter 7](https://one-course.com/books/chemistry/3/en/chapter/7-binary-liquidvapour-diagrams-and-distillation#ch-b2-liquid-vapour-diagrams): a chromatographic column behaves as if it were a stack of $N$ equilibrium stages, each $H$ high. The analogy is a way of counting, not a picture of the column.

**Proposition 31.6 (Plate number from a peak).**

With $w$ the width at the base between the tangents at the inflection points and $w_{1/2}$ the width at half height,

$$
N = 16\left(\frac{t_R}{w}\right)^2 = 5.54\left(\frac{t_R}{w_{1/2}}\right)^2.
$$

**Proof.** For a Gaussian of standard deviation $\sigma$ and height $h$, the inflection points are at $\pm\sigma$, where the height is $h\mathrm e^{-1/2}$ and the slope $\mp h\mathrm e^{-1/2}/\sigma$; the tangent there reaches zero after $\sigma$ more, at $\pm2\sigma$: $w = 4\sigma$. Half height is reached where $\mathrm e^{-x^2/2\sigma^2} = 1/2$, $x = \sigma\sqrt{2\ln2}$: $w_{1/2} = 2\sqrt{2\ln2}\,\sigma \approx
2.355\,\sigma$. Substituting $\sigma = w/4$ or $\sigma = w_{1/2}/2.355$ in $N = (t_R/\sigma)^2$ gives $16$ and $2.355^2 = 5.545$, written 5.54. ∎

**Proposition 31.7 (Van Deemter equation).**

The [plate height](#def-b2-chromatography-plates) depends on the linear speed $u$ of the [mobile phase](#def-b2-chromatography-principle) as

$$
H = A + \frac{B}{u} + Cu,
$$

where $A$ accounts for the different paths through the packing, $B/u$ for diffusion along the column (worse when the molecules stay longer) and $Cu$ for the slow exchange with the [stationary phase](#def-b2-chromatography-principle) (worse when the [mobile phase](#def-b2-chromatography-principle) hurries past). $H$ is smallest at $u_{\mathrm{opt}} = \sqrt{B/C}$, where $H_{\min} =
A + 2\sqrt{BC}$.

**Proof.** *Admitted at this level.* ∎

The form of the equation is admitted; its minimum is not. $\mathrm dH/\mathrm du = -B/u^2 + C$ vanishes at $u = \sqrt{B/C}$, and there $B/u = Cu = \sqrt{BC}$, so $H_{\min} = A + 2\sqrt{BC}$; the second derivative $2B/u^3$ is positive, so this is a minimum. Small grains reduce $A$ and $C$: that is why modern liquid [chromatography](#def-b2-chromatography-principle) uses particles of a few micrometres and high pressures to push the [mobile phase](#def-b2-chromatography-principle) through them.

![A van Deemter curve (exercise parameters: A = 10\, µ m, B = 20\, µ m\, mm/ s, C = 5\, µ m\, s/ mm). The minimum, H = 30\, µ m at u = 2\, mm/ s, sits where the diffusion and exchange terms are equal.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-7d040ed2984b.svg)

*A van Deemter curve (exercise parameters: $A = 10\,\text{µ}\mathrm{m}$, $B = 20\,\text{µ}\mathrm{m}\,\mathrm{mm}/\mathrm{s}$, $C = 5\,\text{µ}\mathrm{m}\,\mathrm{s}/\mathrm{mm}$). The minimum, $H = 30\,\text{µ}\mathrm{m}$ at $u = 2\,\mathrm{mm}/\mathrm{s}$, sits where the diffusion and exchange terms are equal.*

## 31.3 Separating two compounds

**Definition 31.8 (Selectivity and resolution).**

For two neighbouring peaks with $k_2 > k_1$, the *selectivity factor* is $\alpha = k_2/k_1$, and the *resolution* is

$$
R_s = \frac{2(t_{R2} - t_{R1})}{w_1 + w_2},
$$

the distance between the peaks divided by their mean base width.

At $R_s = 1$ the peaks still overlap visibly; at $R_s = 1.5$ the signal returns to the baseline between them (baseline separation), which is the usual target for quantitative work.

![Two Gaussian peaks of equal size at three resolutions (model). At 0.75 they merge into a doublet; at 1.0 the valley is still at a quarter of the height; at 1.5 the signal is back at the baseline.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-5302fa46bd36.svg)

![Two Gaussian peaks of equal size at three resolutions (model). At 0.75 they merge into a doublet; at 1.0 the valley is still at a quarter of the height; at 1.5 the signal is back at the baseline.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-58fe184fe0dd.svg)

![Two Gaussian peaks of equal size at three resolutions (model). At 0.75 they merge into a doublet; at 1.0 the valley is still at a quarter of the height; at 1.5 the signal is back at the baseline.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-95c25e23dc27.svg)

*Two Gaussian peaks of equal size at three [resolutions](#def-b2-chromatography-separation) (model). At 0.75 they merge into a doublet; at 1.0 the valley is still at a quarter of the height; at 1.5 the signal is back at the baseline.*

**Theorem 31.9 (Resolution equation).**

For two neighbouring peaks with the same [plate number](#def-b2-chromatography-plates) $N$,

$$
R_s = \frac{\sqrt N}{4}\,\frac{\alpha - 1}{\alpha}\,\frac{k_2}{1 + k_2}.
$$

**Proof.** For close peaks the two base widths are nearly equal; take both equal to that of the second peak, $w = 4\sigma_t = 4t_{R2}/\sqrt N$ ([Definition 31.5](#def-b2-chromatography-plates)). Then $R_s = (t_{R2} -
t_{R1})/w = \sqrt N\,(t_{R2} - t_{R1})/(4t_{R2})$. With $t_R = t_M(1 + k)$ ([Proposition 31.3](#prop-b2-chromatography-retention-time)), $t_{R2} - t_{R1} = t_M(k_2 - k_1)$ and $t_{R2} =
t_M(1 + k_2)$, so $R_s = (\sqrt N/4)(k_2 - k_1)/(1 + k_2)$. Finally $k_2 - k_1 = k_2(1 - 1/\alpha) =
k_2(\alpha - 1)/\alpha$. ∎

**Method 31.10 (Improving a resolution).**

1. Efficiency: $R_s \propto \sqrt N$ . Doubling the column length doubles $N$ and the analysis time but multiplies $R_s$ by only $\sqrt2 \approx 1.41$ ; smaller particles or the optimum flow rate raise $N$ at constant length.
2. Selectivity: the factor $(\alpha - 1)/\alpha$ is the most sensitive. Change the [mobile phase](#def-b2-chromatography-principle) (solvent, pH), the [stationary phase](#def-b2-chromatography-principle) or, in [gas chromatography](#def-b2-chromatography-techniques) , the temperature.
3. Retention: $k_2/(1+k_2)$ rises steeply up to $k \approx 2$ and slowly beyond, while the time keeps growing as $1 + k$ ; aim for $k$ between about 2 and 10.

## 31.4 Techniques

**Definition 31.11 (Chromatographic techniques).**

In *gas chromatography* (GC), the [mobile phase](#def-b2-chromatography-principle) is a gas (helium, hydrogen or nitrogen) and the [stationary phase](#def-b2-chromatography-principle) a liquid film coating the inside of a long capillary column, kept in an oven. In *high-performance liquid chromatography* (HPLC), a pump pushes a liquid [mobile phase](#def-b2-chromatography-principle) through a short column packed with small particles. In *reversed phase* HPLC, the [stationary phase](#def-b2-chromatography-principle) is nonpolar (silica carrying long alkyl chains) and the [mobile phase](#def-b2-chromatography-principle) a polar mixture of water with methanol or acetonitrile. In *gradient elution*, the composition of the [mobile phase](#def-b2-chromatography-principle) is changed during the run, in liquid [chromatography](#def-b2-chromatography-principle), to elute strongly retained compounds faster.

![Block diagrams. Top: a gas chromatograph; the sample is vaporised in the injector, carried by the gas through a capillary column tens of metres long coiled in an oven, and detected at the outlet. Bottom: a liquid chromatograph; the pump mixes the solvents (gradient elution) and pushes them at high pressure through a short packed column to a UV absorbance cell.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-d8b66a3d202c.svg)

![Block diagrams. Top: a gas chromatograph; the sample is vaporised in the injector, carried by the gas through a capillary column tens of metres long coiled in an oven, and detected at the outlet. Bottom: a liquid chromatograph; the pump mixes the solvents (gradient elution) and pushes them at high pressure through a short packed column to a UV absorbance cell.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-c6a45144f4e3.svg)

*Block diagrams. Top: a gas chromatograph; the sample is vaporised in the injector, carried by the gas through a capillary column tens of metres long coiled in an oven, and detected at the outlet. Bottom: a liquid chromatograph; the pump mixes the solvents ([gradient elution](#def-b2-chromatography-techniques)) and pushes them at high pressure through a short packed column to a UV absorbance cell.*

The choice between them follows from volatility. GC needs compounds that vaporise without decomposing, roughly below $400{}^{\circ}\mathrm{C}$: solvents, fuels, fragrances, small organic molecules. Raising the oven temperature during the run (a temperature ramp) elutes heavier compounds sooner, as a gradient does in HPLC. HPLC handles everything that dissolves, including salts, sugars, drugs and proteins. In [reversed phase](#def-b2-chromatography-techniques) the most polar compounds leave first and the most nonpolar last; adding more organic solvent to the [mobile phase](#def-b2-chromatography-principle) lowers every retention. Preparative column [chromatography](#def-b2-chromatography-principle), and its faster version, flash [chromatography](#def-b2-chromatography-principle), apply the same principle on a large scale to purify grams of product, usually on polar silica with mixtures of hexane and ethyl acetate.

## 31.5 Quantitative analysis

The area of a peak is proportional to the amount of compound that reached the detector, but the proportionality constant depends on the compound and on the detector. Injected volumes, of the order of a microlitre, are not perfectly reproducible. An [internal standard](#def-b2-chromatography-quantitative) removes both difficulties.

**Definition 31.12 (Response factor and internal standard).**

An *internal standard* is a compound, absent from the sample, added in a known amount to every solution analysed; it must be resolved from all other peaks and behave like the analyte. The *response factor* $F$ of an analyte relative to the internal standard is defined by

$$
\frac{A_X}{A_{IS}} = F\,\frac{c_X}{c_{IS}},
$$

where $A$ are peak areas and $c$ concentrations in the injected solution.

**Proposition 31.13 (Internal-standard quantification).**

$F$ measured on a calibration solution of known $c_X$ and $c_{IS}$ gives, for a sample spiked with the [internal standard](#def-b2-chromatography-quantitative) at $c_{IS}$, $c_X = (A_X/A_{IS})\,c_{IS}/F$, whatever the volume injected.

**Proof.** Each area is proportional to the amount injected: $A_X = s_X c_X v$, $A_{IS} = s_{IS} c_{IS} v$, with detector sensitivities $s$ and injected volume $v$. The ratio $A_X/A_{IS} = (s_X/s_{IS})(c_X/c_{IS})$ no longer contains $v$, and $F = s_X/s_{IS}$ is a constant of the method, measured once on the calibration solution. Solving for $c_X$ gives the result. ∎

**Method 31.14 (Quantifying with an internal standard).**

1. Choose a standard close in structure to the analyte, absent from the sample, eluting near it but resolved ( $R_s \ge 1.5$ ).
2. Inject a calibration solution with known $c_X$ and $c_{IS}$ ; compute $F$ from the area ratio.
3. Add the standard to the sample solution at the same $c_{IS}$ ; inject; read $A_X/A_{IS}$ .
4. Compute $c_X = (A_X/A_{IS})\,c_{IS}/F$ , then go back through every dilution to the original sample.

An external calibration (a series of standards of the analyte alone, then the sample, all injected the same way) is simpler but relies on the injected volume and the detector staying constant between runs.

**History — Colour writing, 1906.**

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/img-9930c3f5f945.jpg)

The botanist Mikhail Tsvet poured an extract of green leaves onto a glass tube filled with powdered chalk and washed it through with a solvent: the pigments separated into coloured bands, green chlorophylls and yellow carotenoids. In 1906 he named the method [chromatography](#def-b2-chromatography-principle), “colour writing”. It was largely ignored for a quarter of a century, until it was taken up again for natural products in the 1930s. (Portrait: unknown author, public domain; Wikimedia Commons.)

**Safety.**

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-4b496db0d6c3.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-afc03563fb34.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-58320b3f229b.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-b3bf7bb196be.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-748906458fb6.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-76745c6dd418.svg)

Acetonitrile and methanol, the usual organic components of HPLC [mobile phases](#def-b2-chromatography-principle), are flammable and toxic; hexane, used in column [chromatography](#def-b2-chromatography-principle), is flammable, a health hazard and toxic to aquatic life. Solvent waste is collected, never poured down the sink.

## 31.6 Exercises

**Exercise 31.1 ★.**

A compound has $t_R = 5.0\,\mathrm{min}$ on a column whose [hold-up time](#def-b2-chromatography-retention) is $1.0\,\mathrm{min}$. Compute its [retention factor](#def-b2-chromatography-retention) and the fraction of its time spent in the [mobile phase](#def-b2-chromatography-principle).

**Solution of Exercise 31.1.**

$k = (5.0 - 1.0)/1.0 = 4.0$; fraction of time in the [mobile phase](#def-b2-chromatography-principle) $1/(1 + k) = 0.20$.

**Exercise 31.2 ★.**

A peak at $t_R = 6.0\,\mathrm{min}$ has a base width of $0.30\,\mathrm{min}$. Compute the [plate number](#def-b2-chromatography-plates).

**Solution of Exercise 31.2.**

$N = 16(6.0/0.30)^2 = 16 \times 400 = 6400$.

**Exercise 31.3 ★.**

A $250\,\mathrm{mm}$ column gives $N = 10\,000$ for a compound. Compute the [plate height](#def-b2-chromatography-plates).

**Solution of Exercise 31.3.**

$H = 250\,\mathrm{mm}/10\,000 = 0.025\,\mathrm{mm} = 25\,\text{µ}\mathrm{m}$.

**Exercise 31.4 ★.**

GC or HPLC? Ethanol in blood; a protein; caffeine in a drink; benzene in petrol; sugars in fruit juice.

**Solution of Exercise 31.4.**

Ethanol in blood: GC (volatile). Protein: HPLC. Caffeine in a drink: HPLC (in water, not volatile enough without preparation). Benzene in petrol: GC. Sugars: HPLC (they decompose before they boil).

**Exercise 31.5 ★★.**

Two peaks: $t_{R1} = 4.0\,\mathrm{min}$, $w_1 = 0.40\,\mathrm{min}$; $t_{R2} = 4.6\,\mathrm{min}$, $w_2 =
0.50\,\mathrm{min}$. Compute the [resolution](#def-b2-chromatography-separation). Are they baseline separated?

**Solution of Exercise 31.5.**

$R_s = 2(4.6 - 4.0)/(0.40 + 0.50) = 1.2/0.90 \approx 1.3$: not quite baseline separated (below 1.5).

**Exercise 31.6 ★★.**

How many plates are needed to separate two compounds with $\alpha = 1.10$ and $k_2 = 4.0$ at $R_s = 1.5$?

**Solution of Exercise 31.6.**

$\sqrt N = 4R_s\,\dfrac{\alpha}{\alpha - 1}\,\dfrac{1 + k_2}{k_2} = 4 \times 1.5 \times 11 \times 1.25 =
82.5$, so $N \approx 6.8 \times 10^3$.

**Exercise 31.7 ★★.**

A column has $A = 5\,\text{µ}\mathrm{m}$, $B = 8\,\text{µ}\mathrm{m}\,\mathrm{mm}/\mathrm{s}$, $C = 2\,\text{µ}\mathrm{m}\,\mathrm{s}/\mathrm{mm}$ (exercise data). Find the optimum speed and the smallest [plate height](#def-b2-chromatography-plates).

**Solution of Exercise 31.7.**

$u_{\mathrm{opt}} = \sqrt{8/2} = 2\,\mathrm{mm}/\mathrm{s}$; $H_{\min} = 5 + 2\sqrt{16} = 13\,\text{µ}\mathrm{m}$.

**Exercise 31.8 ★★.**

Predict the order of [elution](#def-b2-chromatography-principle) of benzene, phenol and toluene in reversed-phase HPLC with a water–methanol [mobile phase](#def-b2-chromatography-principle), and what happens when the methanol fraction is raised.

**Solution of Exercise 31.8.**

Phenol (polar $\ce{OH}$, hydrogen bonds with water) first, then benzene, then toluene (one more $\ce{CH3}$, more nonpolar). More methanol makes the [mobile phase](#def-b2-chromatography-principle) less polar: all three elute earlier, in the same order.

**Exercise 31.9 ★★.**

A calibration solution of an analyte at $20.0\,\mathrm{mg}/\mathrm{L}$ with the [internal standard](#def-b2-chromatography-quantitative) at $10.0\,\mathrm{mg}/\mathrm{L}$ gives areas 860 and 400. A sample spiked with the standard at $10.0\,\mathrm{mg}/\mathrm{L}$ gives areas 645 and 410. Compute the [response factor](#def-b2-chromatography-quantitative) and the concentration of the analyte (exercise data).

**Solution of Exercise 31.9.**

$F = (860/400)/(20.0/10.0) = 2.15/2.00 = 1.075$. Sample: $c_X = (645/410) \times 10.0/1.075 \approx
14.6\,\mathrm{mg}/\mathrm{L}$.

**Exercise 31.10 ★★★.**

A separation gives $R_s = 1.06$. What column length, relative to the present one, gives $R_s = 1.5$, and what does it cost in time? What other levers are there?

**Solution of Exercise 31.10.**

$R_s \propto \sqrt N \propto \sqrt L$: $L$ must be multiplied by $(1.5/1.06)^2 \approx 2.0$, which doubles the analysis time and the pressure. Other levers: the selectivity ([mobile phase](#def-b2-chromatography-principle), [stationary phase](#def-b2-chromatography-principle), temperature), smaller particles, the optimum flow rate.

**Exercise 31.11 ★★★.**

For two Gaussian peaks of equal height and width with base width $w = 4\sigma$, compute the signal midway between them, relative to the peak height, at $R_s = 1$ and at $R_s = 1.5$ (neglect the far peak’s contribution at each maximum).

**Solution of Exercise 31.11.**

The peaks are $4\sigma R_s$ apart, so the midpoint is $2\sigma R_s$ from each. $R_s = 1$: each peak contributes $\mathrm e^{-(2)^2/2} = \mathrm e^{-2} = 0.135$; the valley is at $0.27$ of the peak height. $R_s = 1.5$: $2\mathrm e^{-(3)^2/2} = 2\mathrm e^{-4.5} \approx 0.022$, about 2 %: back to the baseline for practical purposes.

**Exercise 31.12 ★★★.**

At constant $N$ and $\alpha$, compare the [resolution](#def-b2-chromatography-separation) factor $k_2/(1+k_2)$ and the relative analysis time $1 + k_2$ for $k_2 = 2$, 5 and 10. Which range of $k$ would you choose?

**Solution of Exercise 31.12.**

$k_2/(1+k_2)$: 0.67, 0.83, 0.91; time in units of $t_M$: 3, 6, 11. Going from 2 to 5 gains 25 % in [resolution](#def-b2-chromatography-separation) for twice the time; from 5 to 10, 9 % for nearly twice again. A $k$ from about 2 to 5 is the usual compromise.

## 31.7 Problem: Caffeine in an Energy Drink

**Problem 31.1.**

Weekend problem — reversed-phase elution order, column performance from a chromatogram, improving the separation, and quantification with an internal standard

Exercise data (invented, not a real method). An energy drink is diluted tenfold ($5.00\,\mathrm{mL}$ to $50.0\,\mathrm{mL}$) and theophylline is added as [internal standard](#def-b2-chromatography-quantitative) at $40.0\,\mathrm{mg}/\mathrm{L}$ in the diluted solution. Column: $150\,\mathrm{mm}$, [reversed phase](#def-b2-chromatography-techniques); [mobile phase](#def-b2-chromatography-principle) water–methanol; UV detection. The unretained matrix (sugars and salts) elutes at $t_M = 1.0\,\mathrm{min}$; theophylline at $3.0\,\mathrm{min}$ (base width $0.18\,\mathrm{min}$); caffeine at $3.4\,\mathrm{min}$ (base width $0.20\,\mathrm{min}$). Calibration: caffeine $50.0\,\mathrm{mg}/\mathrm{L}$ and theophylline $40.0\,\mathrm{mg}/\mathrm{L}$ give areas 1500 and 1000. Sample: areas 970 (caffeine) and 1010 (theophylline). A can holds $250\,\mathrm{mL}$.

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-chromatography/fig-ed0f0be80a7e.svg)

**Part I — [Reversed phase](#def-b2-chromatography-techniques).**

1. Describe the stationary and [mobile phases](#def-b2-chromatography-principle) of a reversed-phase column.
2. Why do the sugars elute at the [hold-up time](#def-b2-chromatography-retention) ?
3. Caffeine is 1,3,7-trimethylxanthine, theophylline 1,3-dimethylxanthine. Explain their order of [elution](#def-b2-chromatography-principle) .
4. Why is theophylline a good [internal standard](#def-b2-chromatography-quantitative) here?
5. What would happen to both [retention times](#def-b2-chromatography-retention) if the methanol fraction were raised?
6. What does the [hold-up time](#def-b2-chromatography-retention) measure?

**Part II — Column performance.**

7. Compute the [retention factors](#def-b2-chromatography-retention) of theophylline and caffeine.
8. Compute the [selectivity factor](#def-b2-chromatography-separation) .
9. Compute the [plate number](#def-b2-chromatography-plates) for caffeine and for theophylline.
10. Compute the [plate height](#def-b2-chromatography-plates) for caffeine.
11. Compute the [resolution](#def-b2-chromatography-separation) between the two peaks.
12. Check it with the [resolution](#def-b2-chromatography-separation) equation.
13. Are the peaks baseline separated?

**Part III — Faster or better.**

14. To save time, the column is shortened to $75\,\mathrm{mm}$ (same packing and flow). What becomes of the [resolution](#def-b2-chromatography-separation) ?
15. And of the analysis time?
16. What would a $300\,\mathrm{mm}$ column give instead, and at what cost?
17. Which lever acts on $\alpha$ ?
18. Would raising $k$ help much here?

**Part IV — Quantification.**

19. Compute the [response factor](#def-b2-chromatography-quantitative) of caffeine relative to theophylline.
20. Why does the injected volume not matter?
21. Compute the concentration of caffeine in the diluted solution.
22. Compute the concentration of caffeine in the drink.
23. What would go wrong if a matrix compound co-eluted with theophylline?
24. What would an external calibration require instead?
25. State the mass of caffeine in a $250\,\mathrm{mL}$ can.

**Solution of Problem 31.1.**

**1.** Stationary: silica grains carrying long alkyl chains, nonpolar. Mobile: water with methanol, polar. **2.** They are very polar and stay in the [mobile phase](#def-b2-chromatography-principle): $k \approx 0$. **3.** Caffeine has one more methyl group than theophylline, and theophylline an $\ce{N-H}$ that hydrogen-bonds with water: theophylline is more polar and elutes first. **4.** It is close in structure (similar response and behaviour), absent from the drink, and elutes near caffeine while being resolved from it. **5.** Both would decrease: a less polar [mobile phase](#def-b2-chromatography-principle) competes better with the [stationary phase](#def-b2-chromatography-principle). **6.** The time the [mobile phase](#def-b2-chromatography-principle) takes to cross the column, the time a compound spends moving. **7.** $k_{\mathrm{theo}} = (3.0 - 1.0)/1.0 = 2.0$; $k_{\mathrm{caf}} = 2.4$. **8.** $\alpha = 2.4/2.0 = 1.2$. **9.** Caffeine $N = 16(3.4/0.20)^2 = 4624 \approx 4.6 \times 10^3$; theophylline $16(3.0/0.18)^2
\approx 4.4 \times 10^3$. **10.** $H = 150\,\mathrm{mm}/4624 \approx 0.032\,\mathrm{mm} = 32\,\text{µ}\mathrm{m}$. **11.** $R_s = 2(3.4 - 3.0)/(0.18 + 0.20) = 0.80/0.38 \approx 2.1$. **12.** $\frac{\sqrt{4624}}{4} \times \frac{0.2}{1.2} \times \frac{2.4}{3.4} = 17 \times 0.167 \times
0.706 \approx 2.0$; the small difference comes from the equal-width assumption of the equation. **13.** Yes: $R_s > 1.5$. **14.** $N$ is halved: $R_s \approx 2.1/\sqrt2 \approx 1.5$, just at baseline separation. **15.** Halved: caffeine at $1.7\,\mathrm{min}$. **16.** $R_s \approx 2.1\sqrt2 \approx 3.0$, for twice the time and twice the pressure: a waste here. **17.** The [mobile phase](#def-b2-chromatography-principle) composition (methanol fraction, pH, another organic solvent) or the [stationary phase](#def-b2-chromatography-principle). **18.** Little: $k_2/(1 + k_2) = 0.71$ already; at $k_2 = 5$ it would be 0.83 (+18 %) for an analysis time multiplied by $6/3.4 \approx 1.8$. **19.** $F = (1500/1000)/(50.0/40.0) = 1.50/1.25 = 1.20$. **20.** Both peaks come from the same injection: the volume cancels in the ratio of areas. **21.** $c = (970/1010) \times 40.0/1.20 \approx 32.0\,\mathrm{mg}/\mathrm{L}$. **22.** Tenfold dilution: $320\,\mathrm{mg}/\mathrm{L}$. **23.** $A_{IS}$ would be too large, and the caffeine result too low. **24.** A series of caffeine standards injected under exactly the same conditions as the sample, a calibration line of area against concentration, and reproducible injection volumes. **25.** $320\,\mathrm{mg}/\mathrm{L} \times 0.250\,\mathrm{L}$: $\boldsymbol{\approx 80\,\mathrm{mg}}$ of caffeine per can (exercise data).