Experiment 6: Fractional Distillation

• Reading Assignment– Experiment 6 (pp. 58 -64)– Operation 29

Key Point!

• When conducting a distillation, the vapor should be richer in the lower boiling component than what you started with.

Simple Distillation: Apparatus

Put in boilingstone!

Temperature Behavior During Distillation

A. Single pure componentB. Two components of similar boiling pointsC. Two components with widely different boiling points

Phase Diagram: Two Component Mixture of Liquids

Questions based upon the previous slide:

a) What is the bp of pure A?b) What is the bp of pure B?c) What is the bp of a solution with the composition of 30 % B, assuming a simple distilllation apparatus?d) What is the composition of the vapor assuming a

simple distillation apparatus?e) What is the composition of the distillate collected

assuming a simple distillation apparatus?f) What does the “tie-line,” x-y represent? Hint: the upper curve is the vapor curve and the lower curve is the

liquid curve. “Composition of the vapor and liquid that are in

equilibriuim with each other at 130 oC.”

Vapor-Liquid Composition Curve (Benzene vs. Toluene)

Vaporliquid

Questions based upon the previous slide:

a) What is the bp of pure toluene?b) What is the bp of pure benzene?c) What is the bp of a solution with the composition of 50 % benzene, assuming a simple distilllation

apparatus?d) What is the composition of the distillate assuming a simple distillation apparatus?e) How many theoretical plates would be necessary for a fractional distillation starting with a 50 % benzene

solution?

When will simple distillation do a reasonable job of separating a mixture?

1) When the difference in boiling points is over 100o

2) When the there is a fairly small amount of impurity, say less than 10 %.3) When one of the components will not distil because of a lack of volatility (i.e. sugar dissolved in water).

Raoult’s Law

PTOTAL = PANA + PBNB

NA = Mole Fraction of A = Moles A

Moles A + Moles B

Fractional Distillation: Apparatus

Put in boiling stone

Temperature vs. Volume: Fractional Distillation

Fractional Distillation Phase Diagram

How many theoretical plates are need to separate a mixture

starting at L?

• Looks like about 5 plates are needed to separate the mixture on the previous slide!

• Count the “tie-lines” (horizontal lines) to come up with the 5 plates (labelled with arrows on the next slide)!

Fractional Distillation Phase Diagram. The arrows indicate

a theoretical plate!

Theoretical Plates Required to Separate Mixtures based on

BPBoiling Point Difference Theoretical Plates

108 172 254 343 436 520 1010 207 304 502 100

Azeotrope

• Some mixtures of liquids, because of attractions or repulsions between the molecules, do not behave ideally

• These mixtures do not obey Raoult’s Law• An azeotrope is a mixture with a fixed

composition that cannot be altered by either simple or fractional distillation

• An azeotrope behaves as if it were a pure compound, and it distills from beginning to end at a constant temperature.

Types of Azeotropes

• There are two types of non-ideal behavior:– Minimum-boiling-point

• Boiling point of the mixture is lower than the boiling point of either pure component

– Maximum-boiling-point• Boiling point of the mixture is higher than

the boiling point of either pure component

Maximum Boiling-Point Azeotrope

Observations with maximum boiling azeotrope

On the right side of the diagram:Compound B will distill (lowest bp). Once B has been removed, the azeotrope will distill (highest bp).

On the left side of the diagram: Compound A will distill (lowest bp) Once A has been removed, the azeotrope will distill. (highest bp)

The azeotrope acts like a pure “compound”

Minimum Boiling-Point Azeotrope

Observations with minimum boiling azeotrope

On the right side of the diagram:The azeotrope is the lower boiling “compound,” and it will be removed first. Pure ethanol will distill oncethe azeotrope has distilled.

On the left side of the diagram: the azeotrope is the lower boiling “compound,” and it will distill first. Once the azeotrope has been removed,then pure water will distill.

The azeotrope acts like a pure “compound”

The Gas Chromatograph

Gas Chromatography: Separation of a Mixture

Gas Chromatogram

Lowestb.p.

Highestb.p. Retention

time

Triangulation of a Peak

Sample Percentage Composition Calculation

Gas Chromatography: Results

In a modern gas chromatography instrument, the results are displayed and analyzed using a computerized data station. It is no longer necessary to calculate peak areas by triangulation; this determination is made electronically.

Our analysis will be conducted on a modern data station.

Compounds in mixture: boiling points.

Cyclohexane 80 oC

Toluene 110 oC

Mixture separates by distillation according to the boilingpoint. Compounds with the lower bp come off first! The same is true on the gas chromatographic column; the lower boiling compound comes off first!

How to identify the components in your unknown

mixture

Use the retention time information from your gas chromatograms to provide a positive identification of each of the components in the mixture.

Don’t rely on the distillation plot to determine the composition of your mixture!

First Fraction: Cyclohexane/TolueneChromatogram

Solventscyclohexane

toluene

Data: Cyclohexane/Toluene First Fraction

solvents

cyclohexanetoluene

?

Calculation of percentages from the data for fraction 2

area counts/response factor = adjusted area

Cyclohexane area = 42795/1.133 = 32104Toluene area = 18129/1.381 = 13127Total area 45231

Note: this calculated area is different than thatshown on the data sheet! Use this calculated area!

Percent cyclohexane = 32104/45231 x 100 = 71.0%Percent toluene = 13127/45231 x 100 = 29.0 %

Round off numbers so that the total equals 100%

Second Fraction: Cyclohexane/TolueneChromatogram

solvents

cyclohexane

toluene

Data: Cyclohexane/Toluene Second Fraction

solvents

cyclohexane toluene

?

Calculation of percentages from the data for fraction 4

area counts/ response factor = adjusted area

Cyclohexane area = 57546/1.133 = 43170Toluene area = 191934/1.381 = 138981Total area 182151

Note: this calculated area is different than thatshown on the data sheet!

Percent cyclohexane = 43170/182151 x 100 = 23.7 %Percent toluene =138981/182151 x 100 = 76.3 %

Round off numbers so percentage = 100%