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Dela Cruz, John Carlo F. Chem 157.1

Tulod, Jhune Karryl M. Experiment 2: Steam Distillation

Results and Discussion:

Graph 1: Theoretical Plot of Temperature and Vapor Pressure of Hexane, Water and Hexane-

Water Mixture

Table 1:Experimental Data for Steam Distillation

Volume of Distillate

(mL)

Temperature

(Celsius)

5 81

10 93

15 71

20 8025 78

30 84

35 87

40 80

45 87

50 78

Ave: 81.9

Table 2: Theoretical Data for Steam Distillation

Source:T

(Kelvin)

Hexane

(atm)

Water

(atm)

Mixture

(atm)

313.15 0.36659 0.07262 0.43936

323.15 0.53152 0.12145 0.65322

333.15 0.75134 0.19610 0.94792

343.15 1.03809 0.30681 1.34559

353.15 1.40494 0.46649 1.87218

363.15 1.86615 0.69114 2.55799

0

0.5

1

1.5

2

2.5

3

300 320 340 360 380 V a p o r p r e s s u r e ( a t m )

Temperature (Kelvin)

Temperature vs Vapor Pressure

Hexane

Water

Mixture



Calculations:

Using the Clausius-Clapeyron Equation for both hexane and water, the theoretical boiling point can

be calculated:

ln2,1, = −∆ 1

2

− 1

1

At same temperature T1, the equation can be manipulated into:

ln2, − ln1, 8.314 / ∙∆ , = ln1 − 2, − ln1, 8.314 / ∙

∆ ,

Where ∆ , = 40700 / and∆ , = 28850 /. Solving the equation using the

theoretical data:

ln2, − ln0.072622 8.314

40700 = ln1 − 2, − ln0.366593 8.314

28850

2 = 0.213029

Substituting the obtained value for P2 into the first equation of Hexane:

2,

= 1

−2 = 1

−0.213029

= 0.786971

ln0.786971

0.366593 = −28850

8.314 1

2

− 1

313.15

The theoretical boiling point, T2 is:

2 = 336.34 = 63.19 ℃

The percent deviation of the Experimental Boiling Point to the Theoretical Boiling Point is calculated

by:

% = −ℎℎ × 100% =81.9℃− 63.19℃63.19℃ × 100% = +29.61%

%Weight of hexane and water in the mixture:

%ℎ =( × )

( × ) + ( × ) × 100%

%ℎ =0.6548 / × 38

0.6548 / × 381 / × 12 × 100% = 65.44%

%ℎ = 100% − 65.44% = 34.56%



Discussion

Steam distillation uses the principle that the total pressure exerted by the vapor of a system

composed of two immiscible liquids is equal to the sum of the partial pressures of the individual

liquids (PT=Po

A + Po

B , where Po

A and Po

B are the individual vapor pressures of the liquids, and PT

=1atm). With this, the final pressure of 1 atm can be reached faster with the mixture than with theindividual liquids thus lowering the boiling point. This technique is mostly used for organic

compounds that may decompose at high temperatures.

In this experiment, hexane and water are used. Throughout the whole distillation process,

hexane and water molecules co-distill in proportion to their respective vapor pressures at the

distilling temperature. Since hexane has lower boiling point (higher vapor pressure) than water, a

greater amount of hexane is expected to exist in the distillate. This can be demonstrated by the

equation

mA/mB =(MWA)( Po

A) / (MWB)( Po

B)

where mA/mB is the mass ratio of the two liquids in the distillate and MWA and MWB are their

respective molecular weights.

The results of the experiment show a +29.61% error between the theoretical andexperimental boiling point values for the mixture of water and hexane. This could be a result of

contaminants in the system, or that the thermometer is placed very close to the tube where steam

from the steam generator enters the flask thus affecting the temperature readings. The mass ratio

(weight percent) of the liquids on the other hand shows that hexane is in greater proportion than

water in the distillate, which is the same as the expected result.

References:

Paula., & Atkins. (2010). Physical Chemistry ninth edition.

http://www.brynmawr.edu/chemistry/Chem/mnerzsto/steam_distillation.htmhttp://web.mst.edu/~tbone/Subjects/TBone/chem226/distmethod.pdf

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