Free MCAT-style practice, 5C
MCAT Separations Practice Questions: Extraction, Distillation and Chromatography
Ten free MCAT-style lab separation questions on acid-base extraction, distillation, recrystallization, TLC and Rf values, column and gas chromatography, and melting point, with full explanations.
Quick concept primer
Separations exploit a difference between compounds. Extraction uses solubility: a compound partitions between two immiscible solvents, and "like dissolves like." In acid-base extraction, a reagent converts one compound into a charged salt that moves into the aqueous layer. Aqueous sodium bicarbonate deprotonates carboxylic acids but not most phenols, sodium hydroxide deprotonates both, and dilute hydrochloric acid protonates amines. Which layer sits on top depends only on density.
Distillation separates liquids by boiling point. Simple distillation works when boiling points differ by more than about 25 °C; fractional distillation handles closer pairs; vacuum distillation lowers boiling points for compounds that decompose when heated. Recrystallization purifies solids with a solvent that dissolves the product well when hot and poorly when cold. A pure solid melts sharply near its known value, while impurities lower and broaden the range.
In chromatography, compounds divide their time between a stationary phase and a mobile phase. On polar silica, polar compounds stick and move slowly, so they have low Rf values (distance traveled by the spot divided by distance traveled by the solvent front) and elute late from a column. In gas chromatography, more volatile compounds elute first.
Traps: Rf can never exceed 1, and organic solvents are not always the top layer.
10 practice questions
Question 1
A student runs a silica gel TLC plate of three compounds using a fairly nonpolar solvent. Distances are measured from the baseline.
| Spot | Distance from baseline (cm) |
|---|---|
| Solvent front | 8.0 |
| Compound P | 6.4 |
| Compound Q | 2.0 |
| Compound R | 4.8 |
What is the Rf of compound R, and which compound is most polar?
- 0.60; compound Q is most polar
- 0.60; compound P is most polar
- 0.40; compound Q is most polar
- 1.67; compound P is most polar
Show answer and explanation
Answer: A. 0.60; compound Q is most polar
Rf = spot distance / solvent front distance = 4.8/8.0 = 0.60. On polar silica, the most polar compound binds the stationary phase most strongly and travels least, so compound Q (Rf 0.25) is most polar. The 1.67 choice inverts the ratio, which is impossible since Rf cannot exceed 1, and 0.40 is 1 minus the correct value.
Question 2
A diethyl ether solution contains a carboxylic acid (pKa about 4.2), a phenol (pKa about 10) and a neutral aromatic hydrocarbon. It is shaken with aqueous sodium bicarbonate, whose conjugate acid has a pKa of about 6.4. Which compound or compounds move mainly into the aqueous layer?
- The hydrocarbon only
- The carboxylic acid only
- Both the carboxylic acid and the phenol
- The phenol only
Show answer and explanation
Answer: B. The carboxylic acid only
Bicarbonate can deprotonate an acid whose pKa is well below 6.4, so the carboxylic acid becomes a water-soluble carboxylate salt. The phenol, with a pKa near 10, is too weak an acid to be deprotonated by bicarbonate and stays neutral in the ether. A stronger base such as sodium hydroxide would be needed to pull the phenol out too.
Question 3
An ether solution contains a neutral amine and a neutral ketone. Which aqueous wash would move only the amine into the aqueous layer?
- Dilute sodium hydroxide
- Saturated sodium bicarbonate
- Dilute hydrochloric acid
- Saturated sodium chloride (brine)
Show answer and explanation
Answer: C. Dilute hydrochloric acid
Acid protonates the basic amine to form an ammonium salt, which is charged and dissolves in water, while the ketone stays in the ether. Basic washes such as sodium hydroxide or bicarbonate leave the amine neutral and organic-soluble. Brine mainly removes dissolved water from the organic layer and does not react with either compound.
Question 4
A student shakes an aqueous reaction mixture with dichloromethane (density about 1.33 g/mL) to extract a nonpolar product. Where will the product mostly be found after the layers separate?
- In the upper layer, because organic solvents always float on water
- In the aqueous layer, because water is the more polar solvent
- Evenly split between the two layers
- In the lower layer, because dichloromethane is denser than water
Show answer and explanation
Answer: D. In the lower layer, because dichloromethane is denser than water
The nonpolar product prefers the organic solvent, and layer position depends only on density. Dichloromethane, at about 1.33 g/mL, is denser than water, so it forms the bottom layer. Choice A is a common trap based on solvents like diethyl ether, which is less dense than water and floats.
Question 5
For which pair of miscible liquids is fractional distillation, rather than simple distillation, most clearly needed?
- Boiling points of 40 °C and 150 °C
- Boiling points of 60 °C and 190 °C
- Boiling points of 72 °C and 84 °C
- Boiling points of 95 °C and 210 °C
Show answer and explanation
Answer: C. Boiling points of 72 °C and 84 °C
Simple distillation separates liquids well only when their boiling points differ by more than about 25 °C. The 72 and 84 °C pair differs by just 12 °C, so it needs a fractionating column, which provides many vaporization and condensation cycles. The other pairs differ by more than 100 °C and separate well by simple distillation.
Question 6
A liquid product boils at about 310 °C at atmospheric pressure but decomposes above about 200 °C. Which approach best purifies it by distillation?
- Distill it under reduced pressure, which lowers its boiling point
- Use fractional distillation at atmospheric pressure, which lowers its boiling point
- Distill it at higher than atmospheric pressure to speed evaporation
- Use simple distillation at atmospheric pressure with faster heating
Show answer and explanation
Answer: A. Distill it under reduced pressure, which lowers its boiling point
A liquid boils when its vapor pressure equals the external pressure, so lowering the pressure lets it boil at a lower temperature, below the point where it decomposes. A fractionating column improves separation but does not lower the boiling point, and raising the pressure or heating faster would push the temperature even higher.
Question 7
Each of the following is a desirable property of a solvent for recrystallizing a solid product EXCEPT:
- It dissolves the product readily near the solvent's boiling point
- It dissolves the product poorly in an ice bath
- It keeps small amounts of soluble impurities dissolved when cold
- It dissolves the product readily at room temperature
Show answer and explanation
Answer: D. It dissolves the product readily at room temperature
Recrystallization depends on a steep solubility difference: the product must dissolve when hot and crystallize when cold. A solvent that dissolves the product readily at room temperature would keep much of it in solution and give a poor recovery. Keeping impurities dissolved in the cold mother liquor is exactly how the method purifies the product.
Question 8
An unknown white solid melts sharply. A chemist runs mixed melting points with two reference compounds, X and Y.
| Sample | Melting range (°C) |
|---|---|
| Unknown alone | 132 to 133 |
| Reference X alone | 132 to 133 |
| Reference Y alone | 132 to 133 |
| Unknown mixed 1:1 with X | 118 to 126 |
| Unknown mixed 1:1 with Y | 132 to 133 |
What is the best conclusion?
- The unknown is X, because mixing a compound with itself lowers its melting point
- The unknown is Y, because mixing with Y did not lower or broaden the melting range
- The unknown is a mixture of X and Y
- No conclusion is possible, because all three pure samples melt at the same temperature
Show answer and explanation
Answer: B. The unknown is Y, because mixing with Y did not lower or broaden the melting range
Mixing two different compounds acts like adding an impurity, which lowers and broadens the melting range, as seen with X. Mixing a compound with more of itself changes nothing, as seen with Y, so the unknown is Y. The sharp melting of the unknown alone argues against it being a mixture, and the mixed test solves the problem that the pure melting points match.
Question 9
A mixture of a hydrocarbon, an ester and a carboxylic acid of similar size is loaded onto a silica gel column and eluted with a mostly hexane solvent. In what order do the compounds most likely come off the column, from first to last?
- Hydrocarbon, ester, carboxylic acid
- Carboxylic acid, ester, hydrocarbon
- Ester, hydrocarbon, carboxylic acid
- All three elute together, because the eluent is nonpolar
Show answer and explanation
Answer: A. Hydrocarbon, ester, carboxylic acid
Silica is polar, so polar compounds adsorb strongly and move slowly in a nonpolar eluent. The hydrocarbon interacts least and elutes first, the moderately polar ester comes next, and the carboxylic acid, which hydrogen bonds strongly with silica, elutes last. Choice B is the order expected on a nonpolar, reverse-phase column.
Question 10
Three alkanes with boiling points of 98, 126 and 174 °C are injected into a gas chromatograph with a nonpolar column and helium carrier gas. Which statement best describes the result?
- The 174 °C alkane elutes first, because heavier molecules carry more momentum
- All three elute together, because the helium does not interact with them
- The 98 °C alkane elutes first, because it is the most volatile and travels mostly with the carrier gas
- Elution order depends only on the flow rate of the carrier gas
Show answer and explanation
Answer: C. The 98 °C alkane elutes first, because it is the most volatile and travels mostly with the carrier gas
In gas chromatography, compounds partition between the moving gas and the stationary liquid coating. With similar polarity, the most volatile compound, the one with the lowest boiling point, is carried along by the gas for the largest share of its transit and reaches the detector first. The carrier gas is inert, but the stationary phase still separates the compounds, and flow rate shifts all retention times without reversing their order.
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