chem 354 lecture 4

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Experiment 6: Simple and Fractional Distillation

• Reading Assignment– Experiment 6 (pp. 51 -57)– Technique 13, Parts A (pp. 694-702)– Technique 14 (pp. 703-715)– Technique 15 (pp. 715-732)– Technique 22 (pp. 797-818)


Experiment 6: Simple and Fractional Distillation

• Work in pairs. Each pair will conduct both the simple and fractional distillations. There are three unknowns, A, B, and C. Perform the experiment as follows:

• Day One: Working in pairs, use simple distillation to separate the unknown (Experiment 6A).

• Day Two: Again, working in pairs, repeat the experiment using fractional distillation on the same unknown (Experiment 6A)

• Do not do Experiment 6B• The products from each day’s distillation will be

analyzed by gas chromatography.


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!


Correct Thermometer Placement

Thermometer must be below this level


Your equipment has a built-in thermometer adapter, soyour equipment will look a bit different. Look at the setup in the hood before you start assembling theequipment.

Ask your instructor if you will be attaching the vacuum adapter! Some instructors will ask you to leave off thispiece of glassware!

There are wooden blocks that can be used to raise theapparatus. The wooden blocks are in the cupboard under the hood.


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


Raoult’s law calculations

See Figure 15.6 on page 720 for example calculations.


Fractional Distillation: Apparatus

Put in boiling stone


Vaporization-Condensation

bp of pure A = 51°

bp of pure B = 87°


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


Microscale distillation: Hickman Head


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”


Dean-Stark Water Separator


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 unknowns: boiling points. There will only be two components in each

unknownHexanes (mixture of isomers) 68-70 oC

Cyclohexane 80 oC

Heptane 98 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!


Gas Chromatography: Standards

Retentiontime

solvents

The x axis is in min.


Notice: 1) hexane has the lowest retention time 2) toluene has the highest retention time

The four compounds come off in the order of increasingboiling point.

Hexane is actually a mixture of three compounds. It is usually called “hexanes”

hexane cyclohexane heptane

CH3

toluene

Increasing b.p.


Preparing distillation samples for gas chromatography

After you have collected 1mL of distillate, then collect the next two drops in one of the special gas chromatography tubes. Add the solvent that is suggested by your instructor (methylene chloride or acetone). Screw on the cap and use a marking pen to put your initials on the tube.

After 4.5 mL has been distilled, repeat the process indicated above.

Charles Wandler will give a presentation in the lab on the instrument that we will use for gas chromatography. This includes a handout that tells you how to retrieve you data. The data will be available in the computer lab (CB 280). He will demonstrate where to put the tubes. He has a signup sheet and a carousel to put the samples in.


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!


Retention Times and Response Factors

Component Retention Time (min)

Response Factor

Hexanes(mixture of isomers)

3.054 1.022

Cyclohexane 3.491 1.133

Heptane 3.812 1.000

Toluene 4.331 1.381

NOTE: These values are for illustration purposes. Your actual values will be different!


First Fraction: Cyclohexane/TolueneChromatogram

Solventscyclohexane

toluene


Data: Cyclohexane/Toluene First Fraction

solvents

cyclohexanetoluene

?


Calculation of percentages from the data for fraction 1

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

?



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%


First Fraction: Hexane/HeptaneChromatogram

hexanes

heptane

solvents

?


Data: Hexane/Heptane First Fraction

solvents

?Three peaks for hexanes

heptane



area counts/response factor = adjusted area

Hexanes area = 1251 + 60375 + 8147 = 69773/1.022 = 68271Heptane area = 26374/ 1.000 = 26374 Total area =94645

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

Percent hexanes = 68271/94645 x 100 = 72.1 %Percent heptane = 26374/94645 x 100 = 27.9 %

Round off numbers so that the total equals 100%


Second Fraction: Hexane/HeptaneChromatogram

solvents

hexanes

heptane


Data: Hexane/Heptane Second Fraction

chem 354 lecture 4

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