Loav · VCE Chemistry · Unit 4 / Area of Study 2
U4A2-05 · Expanded original 2026 teaching guide · Study guide
Why a mixture separates
Chromatography separates components because they interact differently with a stationary phase and a moving mobile phase. A component spends time associated with each phase. Under fixed conditions, stronger retention by the stationary phase relative to transport in the mobile phase generally means slower movement through the system. Separation depends on the relative balance of interactions, not a universal rule that “the more polar substance comes first”. Changing either phase can change the order.
High performance liquid chromatography, HPLC, uses a liquid mobile phase. In a chromatogram the horizontal axis records time and the vertical axis records detector response. Each resolved peak represents a detected component emerging over a time interval. Retention time is measured from injection to the peak's characteristic maximum under the stated method. Features of instrument construction and operation are outside this guide's data-analysis scope.
Compare an unknown with authenticated standards run under the same conditions: mobile-phase composition, stationary phase and other relevant conditions must match. A retention-time match supports an identity but does not prove uniqueness, because different substances can have the same retention time or co-elute. A single peak does not guarantee absolute purity: two components might overlap or a contaminant might not be detected. Conversely, several genuine sample peaks provide evidence for multiple detected components after considering blanks and solvent signals.
Area answers a different question from retention time
Retention time supports identification. Peak area, with a suitable calibration, supports quantification. Prepare a calibration graph from known concentrations and their measured responses under consistent conditions. The vertical axis might be integrated peak area in arbitrary response units; the horizontal axis might be concentration in mg L⁻¹. Keep the same response measure for standards and unknown. A peak's height cannot be substituted for its area without a stated validated relationship.
Fit the specified relationship and retain the intercept unless evidence justifies a zero intercept. Blank/background response can produce a non-zero intercept. Use interpolation within the tested concentration range rather than assuming a straight line continues indefinitely. If the unknown is above range, an appropriately planned dilution and remeasurement can place it within range. Standards and sample preparation must suit the matrix and analyte. A beautifully straight graph does not remove contamination or dilution errors.
Worked example 1: calibrate, then undo dilution
An original simulated HPLC calibration uses standards between 1.00 and 8.00 mg L⁻¹. The fitted relationship is A = 250c + 100, where A is peak area in response units and c is concentration in mg L⁻¹. A diluted unknown gives A = 1350. Rearranging, c = (1350 − 100)/250 = 5.00 mg L⁻¹, within the calibration range.
The solution was prepared by transferring 10.00 mL of original sample and diluting to 100.0 mL. Dilution factor = 100.0/10.00 = 10.00, so original concentration = 5.00 × 10.00 = 50.0 mg L⁻¹. Using 1350/250 would incorrectly ignore the intercept and give 5.40 mg L⁻¹. Using the 100 mL flask volume as an analyte amount would confuse volume with concentration.
Combining evidence is a constraint problem
Begin with the molecular formula or identified molecular-ion mass, then classify functional groups with IR. Use carbon-13 environment counts and shifts to constrain carbon types and symmetry. Use proton shifts, integral ratios and compatible splitting to join fragments. Finally check every atom and every significant observation. The best structure is one that explains all the evidence, not just the most visually memorable peak.
A useful working table has four columns: observation, inference, candidate feature, and remaining uncertainty. “IR 1740 cm⁻¹” is an observation; “compatible with ester C=O” is an inference; “contains –COO–” is a candidate feature. It still does not tell you which alkyl group is on which side. NMR can address that question. If several structures survive, report the ambiguity and specify additional evidence instead of claiming false certainty.
Worked example 2: original schematic structure dossier
A pure compound has formula C₄H₈O₂ and identified molecular ion m/z = 88. IR shows an absorption at 1740 cm⁻¹ and no broad OH feature. Carbon-13 data show four environments, including an ester-carbonyl signal at 170 ppm. Proton data show a 3H singlet at 2.0 ppm, a 2H quartet at 4.1 ppm and a 3H triplet at 1.0 ppm. These values are original simplified teaching data within the relevant 2026 ranges, not a measured spectrum.
First, 4(12) + 8 + 2(16) = 88 agrees with the formula and molecular ion. The IR and carbonyl shift support an ester. The 2H quartet/3H triplet pair suggests an ethyl group; the CH₂ at 4.1 ppm is consistent with the oxygen-linked ethyl group of an ester. The 3H singlet at 2.0 ppm supports CH₃ directly attached to the carbonyl with no neighbouring proton set splitting it. Together these give ethyl ethanoate, CH₃C(=O)OCH₂CH₃.
Audit: four distinct carbon positions explain four carbon-13 environments; proton counts sum to eight; the formula has two oxygen atoms; there is no OH group. A rival ester, methyl propanoate, would place the 3H oxygen-linked group in a singlet downfield and its CH₂ beside the carbonyl rather than the ester oxygen. It does not explain this complete proton pattern as well. A lone m/z 88 or lone carbonyl band would not have been sufficient.
Choose the method for the question
To screen a possible functional group cheaply, a suitable chemical test may help. To compare crystalline-product purity, melting range supplies useful evidence. To quantify a particular compound within a separable mixture, calibrated chromatography can be appropriate. For structural identity of a reasonably pure organic compound, complementary MS, IR and NMR data are powerful. A mixture's overlapping NMR signals may make interpretation difficult, so separation can be necessary first. Explain why the method suits the requested information, not merely what equipment is available.
Practice with worked answers
1. Two compounds have retention times of 4.2 min in separate methods with different mobile phases. Are they the same?
No reliable identity comparison follows. Retention depends on conditions; standards and unknown must be compared under the same method. Even a same-condition match supports rather than uniquely proves identity.
2. Use A = 250c + 100 for area 850. The analysed sample was diluted fivefold. Find original concentration.
c = (850 − 100)/250 = 3.00 mg L⁻¹ for the analysed dilution, inside the 1.00–8.00 range. Original concentration = 3.00 × 5 = 15.0 mg L⁻¹, treating the stated factor as exact.
3. Area 2600 is measured with the same 1.00–8.00 mg L⁻¹ calibration. Is reporting 10.0 mg L⁻¹ securely justified?
Algebra gives 10.0 mg L⁻¹, but it lies outside the calibrated range. This is extrapolation. Dilute appropriately and remeasure within range before relying on the result, then apply the known dilution factor.
4. A sample gives one HPLC peak and an ester IR band. Does that prove pure ethyl ethanoate?
No. One detected peak does not exclude co-elution or undetected impurities, and an ester band does not identify which ester. Same-condition standards and complementary MS/NMR would strengthen the identity and purity assessment.
Quality and safety: retain raw chromatograms, calibration points and dilution records, including anomalous results. Repetition assesses scatter but cannot automatically correct systematic calibration bias. These examples are simulated. Real solvents, sample extraction and instrument use require school supervision and approved chemical-specific controls.
Go deeper: identification and quantification have separate assumptions
A chromatographic peak must be assigned correctly before its area can give the desired analyte concentration. If two substances co-elute, the integrated response may include both. Applying the target's calibration to the combined area can overestimate the target, even when the calibration line is excellent. Different substances can also have different response factors, so two equal areas do not automatically mean equal molar concentrations or equal masses.
Compare a sample with a blank and with the target standard under the same conditions. A blank signal helps identify contamination or solvent-related contributions. If co-elution is suspected, a suitably changed separation method may resolve the peak, but the standards and calibration must then be checked under that method. Do not compare a new sample retention time against an old-method standard and call the difference an identity change. Complementary spectra of a separated fraction help determine whether a peak assignment is credible.
Worked example 3: from a sample extract to an amount, with a selectivity check
An original simulated investigation transfers a 0.750 g solid sample quantitatively into 150.0 mL solution. A 5.00 mL portion is diluted to 50.00 mL. For a target marker, a validated same-condition calibration is A = 400c + 80 with c in mg L⁻¹, over 0.500–6.00 mg L⁻¹. The assigned peak has area 1280. First solve c = (1280 − 80)/400 = 3.00 mg L⁻¹. This is inside the calibration range. Undo the tenfold dilution to obtain 30.0 mg L⁻¹ in the initial extract.
Multiply by the extract volume in litres: m = 30.0 × 0.1500 = 4.50 mg. Convert sample mass to 750 mg, giving target mass fraction (4.50/750) × 100 = 0.600%. This conclusion assumes complete transfer and extraction and a selective response for the target. The 5.00 mL portion is not the whole original extract; multiplying 3.00 mg L⁻¹ directly by 0.1500 L would undercount by ten.
Now challenge selectivity: the fraction's molecular ion matches the target, but its NMR contains additional non-solvent resonances not present in the reference. Matching mass alone does not rescue the calculation; another component may co-elute. Report 0.600% as the result conditional on the peak assignment and selectivity assumptions, not as a confirmed target percentage. Resolve the interference or validate a selective measurement before asserting the amount. A numerical answer can be arithmetically correct while its chemical interpretation remains unverified.
A disciplined structure decision
When evidence conflicts, do not average incompatible structures or ignore the awkward signal. Check whether the sample is pure, whether the reference and unknown were measured comparably, whether a peak belongs to solvent, and whether the molecular ion was correctly assigned. Then revisit the candidate. For example, an apparent ester plus a strong acid OH could indicate an acid candidate, water contamination, or a mixture; it is not permission to draw an ester with an unexplained extra OH while keeping the old formula.
Choose additional evidence by its expected information gain. Another mass measurement at the same nominal resolution does little to distinguish structural isomers with the same formula. A proton pattern that distinguishes oxygen-linked methyl from oxygen-linked ethyl may decide between two ester isomers. Conversely, a complex mixture may need improved separation before any additional NMR interpretation is meaningful. The next method should address the unresolved ambiguity, not simply add another instrument name to the answer.
Transfer practice
5. Two mixture peaks have equal areas. Can you conclude that each contributes 50% of the mixture by mass?
No. Detector response per unit mass can differ, and components outside the detected peaks may be present. Separate suitable calibrations and complete sample accounting are needed. Equal area is a response comparison, not automatically a composition comparison.
6. Use A = 400c + 80 for an area of 880 from a fourfold diluted solution. The original extract volume is 250.0 mL. Calculate marker mass in the extract.
Analysed concentration = (880 − 80)/400 = 2.00 mg L⁻¹. Original extract concentration = 8.00 mg L⁻¹. Mass = 8.00 × 0.2500 = 2.00 mg. Keep concentration units and litres explicit; the total volume is needed only when converting concentration to amount.
7. An unknown and standard share retention time and molecular-ion mass. Give a plausible alternative to identical compounds, and a useful next test.
They may be structural isomers that co-elute under this method. Compare diagnostic proton/carbon NMR or an authenticated IR fingerprint under appropriate conditions; changing the chromatographic separation can also test co-elution. Agreement on two non-unique features strengthens a hypothesis but does not logically eliminate all alternatives.
