Exercises — Chromatography — TLC, column, GC, HPLC (principles)

• Stationary phase = the silica gel layer on the plate. It stays fixed; it grabs molecules.
• Mobile phase = the solvent that climbs the plate by capillary action. The whole separation is a tug-of-war between these two — nothing moves without the mobile phase.

$K=C_s/C_m$ is the ratio of "how much sits stuck" to "how much rides along." $K_Q=8>K_P=2$, so Q spends a larger share stuck on the stationary phase. Q clings harder.

Apply $R_f=\frac{\text{distance moved by spot}}{\text{distance moved by solvent front}}$. $R_f(X)=\frac{7.5}{10.0}=0.75,R_f(Y)=\frac{3.0}{10.0}=0.30$ On normal-phase (polar silica), polar sticks more → travels less → small $R_f$. Y has the smaller $R_f$, so Y is more polar. See the figure: Y (pink) lags near the bottom, X (blue) races ahead. This depends on Polarity and Intermolecular Forces.

Use $v_{\text{analyte}}=\frac{v_{\text{mobile}}}{1+k}$. $v_{\text{analyte}}=\frac{2.0}{1+3}=\frac{2.0}{4}=0.50\text{ mm/s}$ Reading it physically: with $k=3$ the molecule is stuck three units of time for every one unit riding, so it travels at one-quarter of solvent speed. Matches the hopping picture.

$k=K\frac{V_s}{V_m}$ just rescales $K$ by the physical amounts of each phase. $k=12\times \frac{0.5}{2.0}=12\times 0.25=3.0$ So even though $K=12$ (loves the stationary phase), there is so little stationary phase around that the effective clinginess is only $k=3$.

Smaller $k$ ⇒ larger $v_{\text{analyte}}=v_{\text{mobile}}/(1+k)$ ⇒ exits first. Compute the "slowdown factor" $1+k$: A gives $2$, B gives $5$, C gives $1.5$. Speeds are $\propto 1/(1+k)$, so ordering fastest→slowest is by smallest $1+k$: C ($1.5$) then A ($2$) then B ($5$). $\boxed{\text{Elution order: }C,A,B}$

Each molecule hops randomly between stuck and moving. Two identical molecules do not hop at exactly the same instants — one may cling a bit longer, the other ride a bit longer, purely by chance. The average speed is the same ($v_{\text{mobile}}/(1+k)$), but the spread around that average grows with distance travelled, like a random walk. So a band that starts as a tight dot spreads into a Gaussian-ish smear — the longer the path, the wider and more dilute. This is exactly why we don't run a column forever: resolution improves with separation of centres but degrades as each band fattens.

"Like sticks to like." Now the stationary phase is non-polar, so non-polar molecules (the oil) stick to it and lag; polar molecules (the sugar) prefer the polar mobile phase and shoot through. ⇒ The polar sugar elutes first, the non-polar oil last. This flips the normal-phase TLC rule (where polar = slow) because we swapped which phase is polar. The rule "polar sticks more" was never universal — it always meant "sticks to whichever phase matches its polarity." Related: UV-Vis Spectroscopy is the usual HPLC detector reading these bands.

Spots near the baseline mean both compounds are stuck hard to the polar silica — the mobile phase is too weak to pull them off. Make the solvent more polar. A more polar mobile phase competes with silica for the analytes' sticky sites (it can hydrogen-bond / dipole-interact with them too), so molecules spend more time riding ⇒ lower effective $k$ ⇒ larger $R_f$. The spots rise off the baseline into the useful $0.2$–$0.8$ window where their small $K$ differences can spread them apart. (Rule of thumb: increase mobile-phase polarity to raise $R_f$ on normal phase.)

The deciding question is always: can the sample be vaporised without decomposing?

• (a) → GC. Light hydrocarbons are volatile and stable, so they evaporate cleanly and travel as a gas — fast diffusion gives sharp peaks and excellent resolution.
• (b) → HPLC. Proteins are non-volatile and heat-sensitive; heating chars/denatures them before they vaporise. HPLC keeps them dissolved in liquid at room temperature. Confirm identity later with Mass Spectrometry. This is a core purification decision.

(a) $k=K{V}_s/V_m$ with $V_s/V_m=1.0/4.0=0.25$: $k_P=8\times 0.25=2.0,k_Q=20\times 0.25=5.0$ (b) $v=v_{\text{mobile}}/(1+k)$: $v_P=\frac{5.0}{1+2.0}=\frac{5.0}{3.0}=1.667\text{ mm/s}$ $v_Q=\frac{5.0}{1+5.0}=\frac{5.0}{6.0}=0.833\text{ mm/s}$ (c) time $=L/v$ with $L=100\text{ mm}$: $t_P=\frac{100}{1.667}=60.0\text{ s},t_Q=\frac{100}{0.833}=120.0\text{ s}$ (d) Time gap $=t_Q-t_P=120.0-60.0=\boxed{60.0\text{ s}}$. P (smaller $K$, smaller $k$) elutes first at $60$ s; Q clings twice as much and arrives a full minute later — cleanly separated.

With equal $K$: $k=12\times 0.25=3.0$ for both. $v=\frac{5.0}{1+3.0}=1.25\text{ mm/s}(\text{same for both})$ $t=\frac{100}{1.25}=80.0\text{ s}(\text{same for both})$ Both peaks arrive at $80$ s — they overlap into one peak; no separation occurs. Lesson: chromatography separates only when the compounds have different $K$ (hence different $k$). The whole method is "a race rigged by differential affinity" — take away the difference and the race is a tie. All the fancy hardware is just a way to amplify a difference in $K$ into a difference in arrival time.

$K\to k\to v\to t$. Different $K$ makes different $t$; equal $K$ makes one peak. Everything else — TLC, GC, HPLC — is choosing phases so the $K$ values you care about differ as much as possible.