Separation / Purification

Question 1

Define:

separation of a chemical mixture

Answer 1

A separation is any process that transfers or converts a mixture of substances into at least two distinct product mixtures. At least one chemical constituent must be enriched in one product and depleted in the other(s).

In some cases, a separation may fully divide the mixture into its pure constituents, but this is not necessary. Separations are classified based on the technique/apparatus as well as the specific chemistry of the mixture.

Question 2

What is necessary for a separation process to be successful?

Answer 2

Separations require differences in chemical properties such as size, shape, mass, intermolecular forces, or chemical affinity between the constituents of a mixture.

If no single difference can be used to accomplish a desired separation, multiple processes can be performed in combination.

Question 3

Define:

extraction of a chemical mixture

Answer 3

In a chemical system, extraction refers to the process of moving a solute from one phase to another for easier separation.

There are many forms of extraction, but the two forms most commonly tested on the MCAT are solid-liquid and liquid-liquid (or aqueous-organic).

Question 4

Describe the general process for aqueous-organic extraction?

Answer 4

Aqueous-organic extraction is a special type of liquid-liquid extraction, used to separate dissolved compounds based on relative solubility.

A dissolved miscible solution is exposed to an immiscible solvent which dissolves at least one constituent and leaves at least one constituent in the original solution. Since the solutions are immiscible, they can then be titrated off, and the dissolved constituent separately reconstituted through a work-up step.

Question 5

Describe the general process for solid-liquid extraction?

Answer 5

Solid-liquid extraction is used to separate solid mixtures based on relative solubility in a liquid solvent.

A solid mixture is exposed to a solvent which dissolves at least one constituent and leaves at least one constituent in the solid phase. The solid phase can then be filtered, and the dissolved mixture separately reconstituted.

Question 6

How could a mixture of white sand and table salt be nondestructively separated?

Answer 6

Via solid-liquid extraction, by placing the mixture in water. The salt will dissolve (move into the aqueous phase), while the sand will not. This allows the sand to be easily filtered. The table salt can then be recovered through evaporation.

Question 7

What kinds of organic compounds are soluble in water?

Answer 7

Any compound with chemical properties similar to water will dissolve readily in water.

Specifically: molecules with O, N, or F bonded to H that can hydrogen bond, such as ethanol, ammonia, or hydrogen flouride. Additionally: charged ions (such as Na⁺, Cl^-), and molecules with a large dipole moment (such as acetone, dimethylsulfoxide).

Question 8

What kinds of organic compounds are soluble in nonpolar solvents such as CCl₄?

Answer 8

Nonpolar compounds (which don’t dissolve well in water) tend to dissolve well in nonpolar solvents.

Specifically: hydrocarbons such as cyclohexane and benzene, ethers, and very large aldehydes and ketones.

Question 9

How does a separation funnel help with organic extractions?

Answer 9

A separation funnel helps to precisely separate aqueous and organic layers. It is shaped like an inverted water droplet, with a stopcock at the narrow bottom.

When two immiscible liquids are present in the funnel, they settle to two disctinct layers. The denser layer, on the bottom, can be poured off with the stockcock. The narrow shape of the bottom allows for maximal separation of the layers.

Question 10

How could a mixture of ethanol and benzene be nondestructively separated?

Answer 10

Through aqueous-organic extraction, by placing them in a mixture of water and a suitable organic solvent such as carbon tetrachloride in a separation funnel.

The alcohol will dissolve well in the water, while the benzene will dissolve in the carbon tetrachloride, due to polarity effects. The water and carbon tetrachloride will settle in two layers in a separation funnel and can be physically separated. The ethanol and benzene can be recovered later in a work-up step.

Question 11

Define:

acid extraction of a chemical mixture

Answer 11

Acid extraction of a chemical mixture is the aqueous-organic extraction of a mixture of solutes, after at least one species has been protonated (and at least one species remains not protonated) by the addition of an acid to the mixture.

The species which is more basic will be protonated by the acid addition, and can be extracted in the aqueous layer since it is now a cation.

Question 12

Benzene and phenylamine (C₆H₅-NH₂) are both dissolved in a sample of diethyl ether. What should be added to the mixture to allow the two compounds to be separated by extraction? Which compound will end up in the aqueous layer?

Answer 12

Aqueous acid should be added to the mixture, and the phenylamine will end up in the aqueous layer.

The NH₂ group of phenylamine is basic, and will be protonated by the acid. The C₆H₅NH₃⁺ that this creates is an ionic species, and will preferentially dissolve in water over ether. Benzene will not be protonated, and will remain in the ether (organic) layer.

Question 13

Define:

base extraction of a chemical mixture

Answer 13

Base extraction of a chemical mixture is the aqueous-organic extraction of a mixture of solutes, after at least one species has been deprotonated (and at least one species remains not deprotonated) by the addition of a base to the mixture.

The species which is more acidic is deprotonated by the base addition, and can be extracted in the aqueous layer since it is now an anion.

Question 14

Benzene and phenol (C₆H₅-OH) are both dissolved in a sample of diethyl ether. What should be added to the mixture to allow the two compounds to be separated by extraction, and which compound will end up in the aqueous layer?

Answer 14

Aqueous base should be added to the mixture, and the phenol will end up in the aqueous layer.

The OH group of phenylamine is weakly acidic, and will be deprotonated by the base. The C₆H₅O^- that this creates is an ionic species, and will preferentially dissolve in water over ether. Benzene will not be deprotonated and will remain in the ether (organic) layer.

Question 15

What properties of a compound determine its boiling point?

Answer 15

A compound’s boiling point is primarily determined by two properties:

1. The size and branching of the molecule
2. The strength of the intermolecular attractive forces between the molecules

Typically, size trumps intermolecular attractive forces: very large nonpolar compounds have higher boiling points than small polar ones.

Question 16

Which has a higher boiling point, water or ethanol?

Answer 16

Water has a higher boiling point than ethanol.

The OH groups in water are capable of twice as much hydrogen bonding, compared to the single OH group in ethanol.

Question 17

Which has a higher boiling point, water or octane?

Answer 17

Octane has a higher boiling point than water.

The enormous size of octane trumps even the intermolecular hydrogen bonding strength of water.

Question 18

Which has a higher boiling point, water or acetone?

Answer 18

Water has a higher boiling point than acetone.

The OH groups in water are capable of hydrogen bonding, while the C=O group in acetone only have dipole-dipole interactions between the molecules. Hydrogen bonding is stronger than dipole-dipole.

Question 19

Which has a higher boiling point, ethanol or propane?

Answer 19

ethanol has a higher boiling point than propane.

The OH group in ethanol is capable of hydrogen bonding, compared to only London dispersion forces in propane. Hydrogen bonding is stronger than London dispersion forces.

Question 20

Which has a higher boiling point, acetone or methane?

Answer 20

Acetone has a higher boiling point than methane.

The C=O group in acetone creates dipole-dipole interactions between molecules, compared to only London dispersion forces in methane. Dipole-dipole interactions are stronger than London dispersion forces.

Question 21

Define:

distillation of a chemical mixture

Answer 21

Distillation is the act of separating a mixture of liquids by heating the mixture until the more volatile component boils away.

The boiled vapor is condensed back to liquid phase in a separate container to complete the separation.

Question 22

Under what primary condition can two liquids be separated by simple distillation?

Answer 22

The two liquids must have a boiling difference of at least 25°C to be effectively separated using simple distillation.

If the liquids’ boiling points are too close, the vapor will have a significant component of both compounds, and they won’t be effectively separated.

Question 23

What separation process can be used if the boiling points of two miscible liquids are less than 25°C apart?

Answer 23

The two liquids can be separated using fractional distillation.

In fractional distillation, a separation column is added to the distillation apparatus. The column effectively lengthens the distance over which the vapor travels, allowing the more volatile component to separate off as gas while the less volatile component recondenses.

Question 24

What separation process can be used to separate miscible liquids with very high boiling points.

Answer 24

Vacuum distillation can be used to separate liquids with very high boiling points.

By attached a vacuum pump to the heating portion of the distillation apparatus, the liquids’ boiling points can be lowered, allowing distillation at lower temperatures.

Remember: a liquid’s boiling temperature decreases as atmospheric pressure above the liquid decreases.

Question 25

Can two liquids be separated with distillation if their boiling points differ by 15°C?

Answer 25

Yes. The two liquids can be separated if fractional distillation is used.

Only simple distillation requires the liquids to have greater than 25°C difference in boiling points.

Question 26

What type of distillation would be used to separate a water and acetone mixture?

Which substance will be the separated condensate?

Note: b.p. of water = 100°C, b.p. of acetone = -95°C.

Answer 26

Simple distillation.

Acetone will boil first at a lower temperature and will end up as the condensate.

Simple distillation is ideal here, since the difference in boiling temperature between acetone and water is ~195°C.

Question 27

What type of distillation would be used to separate a water and isoheptane mixture?

Which substance will be the separated condensate?

Note: b.p. of water = 100°C, b.p. of isoheptane= 90°C.

Answer 27

Fractional distillation.

Isoheptane will boil first at a lower temperature and will end up as the condensate.

Fractional distillation is ideal here, since the difference in boiling temperature is less than 25°C.

Question 28

What type of distillation would be used to separate a triacontane and glycerine mixture?

Which substance will be the separated condensate?

Note: b.p. of triacontane = 450°C, b.p. of glycerine = 290°C.

Answer 28

Vacuum distillation.

Glycerine will boil first at a lower temperature and will end up as the condensate.

Vacuum distillation is ideal here, since the boiling temperatures are both very high.

Question 29

Define:

Chromatography

Answer 29

Chromatography is the separation and/or identification of complex mixtures of chemicals using their varying affinities for differing **stationary phase **and mobile phase materials.

Chromatography can be used to separate a mixture, to identify the components of the mixture, or both.

Question 30

Describe the difference between a stationary and mobile phase in chromotography.

Answer 30

The stationary phase does not change position relative to the aparatus. Common nonpolar stationary phase materials are silica beads and alumina fibers.

The mobile phase flows through the aparatus. Common polar mobile phase materials are water and acetone.

Question 31

What is the basic scheme of any form of chromatography?

Answer 31

All forms of chromatography consist of a chemical mixture being introduced to a stationary phase, across which a mobile phase flows. The phases should have differing properties, such as polar vs nonpolar.

Compounds within the mixture will have relative affinities towards either the stationary or mobile phase and move at consequently move at different characteristic rates.

Question 32

Define:

R_f value

Answer 32

The Rf value, or retardation factor, is a measure of how great an affinity the sample has towards the mobile phase.

R_f = d_substance/d_solvent

R_f = S_{in mobile phase}/S_total

Where: d = distance moved, and S = quantity of substance.

Question 33

If a substance has an R_f value of 1, what does that indicate about it?

Answer 33

An R_f value of 1 indicates that the substance has complete affinity for the mobile phase, and no affinity for the stationary phase. The substance moves with the solvent front.

R_f = d_substance/d_solvent

R_f = S_{in mobile phase}/S_total

Where: d= distance moved, and S = quantity of substance.

Question 34

If a substance has an R_f value of 0, what does that indicate about it?

Answer 34

An R_f value of 0 indicates that the substance has complete affinity for the stationary phase, and no affinity for the mobile phase. The substance does not move at all.

R_f = d_substance/d_solvent

R_f = S_{in mobile phase}/S_total

Where: d= distance moved, and S = quantity of substance.

Question 35

Which substance will move farther along the plate, if A has an R_f = .75 and B has an R_f = .6? How do you know?

Answer 35

Substance A will move farther than substance B. A higher value of R_f always indicates greater affinity for the mobile phase and will always more farther.

R_f = d_substance/d_solvent

Where: d= distance moved.

Question 36

What are the necessary properties of a compound, if it is to be analyzed via gas chromatography?

Answer 36

For any compound to be analyzed via gas chromatography, it must be sufficiently volatile that it can be vaporized at reasonable temperatures and pressures.

This prevents gas chromatography from being useful for analyzing extremely large molecules, or biological ones that would denature.

Question 37

In gas chromatography, what are the properties of the mobile phase?

Answer 37

In gas chromatography, the mobile phase is an inert nonpolar gas which flows through the flow column.

Vapor which is less polar will have higher affinity for the mobile phase, and elute through the column more quickly.

Question 38

In gas chromatography, what are the properties of the stationary phase?

Answer 38

In gas chromatography, the stationary phase is a polar liquid which coats the inside of a flow column.

Vapor which is more polar will have higher affinity for the stationary phase, and elute through the column more slowly.

Question 39

In thin-layer chromatography, what are the properties of the sample to be analyzed?

Answer 39

In thin-layer chromatography the sample (either dissolved in a solvent or in a fluid form) must be able to move across and deposit on the stationary phase.

As the mobile phase wicks along the stationary phase, different sample molecules will move at different characteristic rates, depending on their polarity.

Question 40

In thin-layer chromatography, what are the properties of the stationary phase?

Answer 40

In thin-layer chromatography (TLC), the stationary phase is a thin layer covering a solid plate. The thin layer is capable of wicking a liquid via capillary action.

The most common stationary phase in TLC is a polar silica layer covering an impermeable ceramic plate; though paper chromatography is also common in labs.

Question 41

In thin-layer chromatography, what are the properties of the mobile phase?

Answer 41

In thin-layer chromatography (TLC), the mobile phase is a liquid which wicks up the stationary phase via capillary action.

The most common materials used as mobile phase in TLC are nonpolar organic solvents such as CCl₄ or large ethers.

Question 42

What are the main differences between paper chromatography and the main category of thin layer chromatography (TLC) that it falls under?

Answer 42

In paper chromatography the stationary phase is a single substance: high quality filter paper; whereas TLC is often a solid with a special stationary phase coating.

In both types of chromatography, the mobile phase is a liquid; though in paper chromatography polar solvents such as water and acetone are much more common.

Question 43

Which type of chromatography is best for analyzing a small amount of a natural organic ink for purity? Why?

Answer 43

Paper chromatography (or generally TLC) is best suited for this experiment.

While gas chromatograhy is often used for very small samples, the organic compounds in the ink might denature and skew the results.

Question 44

Which type of chromatography is best for quickly analyzing a small sample of a dirt for inorganic compounds? Why?

Answer 44

Gas chromatography is best suited for this experiment, especially since speed is a requirement.

While it may be possible to find suitable solvents and run TLC, gas chromatograhy is often preferred for very small samples. Since the compounds are inorganic, there is not a risk of denaturing them when vaporizing.

Question 45

Define:

recrystallization

Answer 45

Recrystallization is a separation process for solid mixtures.

There are many different types of apparatus and specific processes. The most common is single-solvent recrystallization, since the other types: hot-filtration, multi-solvent, and seeding are all additional subsets within single-solvent.

Question 46

Describe the general process for single-solvent recrystallization.

Answer 46

A mixture of solids are introduced to the least amount of a hot solvent necessary to completely dissolve. The solution is cooled just until the point where the pure solid recrystallizes.

More consistent temperature and longer time length of crystallization results in purer solids.

Question 47

Describe the requirements and differences for hot-filtration recrystallization.

Answer 47

Hot-filtration recrystallization is used if there are additional insoluble impurities. The initial step is to dissolve the crystals and filter the insoluble impurities, then proceed to do standard recrystallization.

It’s important that the filtering apparatus remain hot, to prevent crystals from forming during filtration.

Question 48

Describe the requirements and differences for multi-solvent recrystallization.

Answer 48

Multi-solvent recrystallization is used if the solubilities of the solids are too similar. A new solvent is introduced and causes the less soluble compound to recrystallize.

Heating and cooling are not required, but are often used to achieve more pure results.

Question 49

Describe the requirements and differences for seeding.

Answer 49

Seeding is common in all recrystallizations as an additional step. A small crystal of pure solid is added to the saturated solution to initiate crystal formation.

This can also be done by providing any crystal surface for growth, such as scratching the inside of the glass beaker.

Question 50

What separation process would be used to purify a contaminated sample of calcium chloride?

Answer 50

Recrystallization.

Most likely, hot-fitration or multi-solvent recrystallization, since CaCl₂ has relatively high solubility. Either the impurities won’t dissolve as readily (hot-filtration), or a second solvent could be added to precipitate the impurities and leave pure CaCl₂ in solution (multi-solvent).