chem 12 freezing pt depression the real thing

8/3/2019 Chem 12 Freezing Pt Depression the REAL THING

1/5

Capinpuyan | Page 1 of 4

Aimee Lorraine C. Capinpuyan #7A

Experiment #3 Freezing Point Depression 12/09/10

I. AbstractThe main objective of this experiment was to determine the identity of an unknown fatty acid

using the data derived from the freezing point depression. This objective was realized through a

series of procedures, beginning with the determination of the freezing point of naphthalene

(C10H8) and followed by the determination of the new freezing point when 1.0 grams of an

unknown substance was added in increments of 0.5 grams. The resulting data was then averaged

and then plotted. Finally, after plugging in the derived values into the equation for the freezing-

point depression constant, it was found that the unknown substance was sulfur, or S8.

II. IntroductionThis experiment relied heavily on the properties that a substance takes on when a solute is

mixed in with

a pure solvent.Th

ese are called colligative properties, and they rely only on t

he

proportions of solute to solvent instead of the identities of the substances being mixed.

In this experiment, we measured the freezing point of naphthalene, and then subtracted it

from the value of the freezing point of a mixture containing naphthalene and an unknown fatty

acid. In this way, we will be obtained , or change in temperature, in order to satisfy the

following equation:

where Kf is the molal freezing-point depression constant for the pure solvent (which, n our

case, is naphthalene and has a value of 6.9 ), and m is the

concentration expressed as the number of moles of solute dissolved in 1 kg of solvent, or

molality.

III. MethodsIn determining the freezing point of naphthalene, 10 grams of the white powdery substance

was heated in water above 85inside a test tube apparatus with a two-holed stopper, with one

hole for the thermometer and the other for the metal stirrer. When the temperature of the

naphthalene reached above 85, it was removed from the water and cooled. Starting from 85,

the mixture was stirred vigorously, and its temperature was noted every thirty seconds, including

the point at which the naphthalene began to form white crystals. The crystals were melted back

into a transparent liquid, and a second trial was performed. The process was repeated with the

same amount of naphthalene plus 1.0 grams of an unknown bright yellow powder in increments

of 0.5 grams.

IV. ResultsThe results of the experiment are as follows.

Figure 1. Experiment Data Table


2/5


Pure Naphthalene Pure Naphthalene + 0.5g unknownPure Naphthalene + 1.0g

unknown

Time

(sec)

Temperature (C) Time

(sec)

Temperature (C) Time

(sec)

Temperature

(C)Trial 1 Trial 2 Ave. Trial 1 Trial 2 Ave.

0 85.0 85.0 85.0 0 85.0 85.0 85.0 0 85.0

15 - - - 15 81.5 83.8 82.7 15 83.030 80.5 82.0 81.3 30 80.0 81.5 80.8 30 82.9

45 - - - 45 79.0 79.5 79.3 45 80.0

60 77.5 78.0 77.8 60 77.0 78.8 77.9 60 79.0

75 - - - 75 76.0 76.8 76.4 75 77.0

90 76.5 76.2 76.4 90 75.5* 75.7 75.7 90 76.0

105 - - - 105 75.3 75.0* 75.3 105 75.0

120 76.0 76.0 76.0 120 75.0 75.0 75.0 120 74.5

135 - - - 135 75.0 75.0 75.0 135 74.3

150 75.9* 75.8* 75.9 150 75.0 74.9 75.0 150 74.0

165 - - - 165 75.0 74.8 74.9 165 74.0*

180 75.8 75.8 75.8 180 74.9 74.8 74.9 180 74.0195 - - - 195 74.5 74.5 74.5 195 74.0

210 75.5 75.5 75.5 210 74.5 74.4 74.5 210 73.9

225 - - - 225 74.5 74.4 74.5 225 73.8

240 75.4 75.4 75.4 240 74.5 74.2 74.4 240 73.5

255 - - - 255 74.5 74.1 74.3 255 73.5

270 75.3 75.3 75.3 270 74.3 74.0 74.2 270 73.4

285 - - - 285 74.0 74.0 74.0 285 73.2

300 75.1 75.1 75.1 300 74.0 74.0 74.0 300 73.1

315 - - - 315 73.9 74.0 74.0 315 73.0

330 75.0 ** 75.0** 75.0 330 73.9 74.0 74.0 330 72.5

345 - - - 345 73.3 73.9 73.6 345 72.5

360 75.0 75.0 75.0 360 73.3 73.8 73.6 360 72.5

375 - - - 375 73.0 73.5 73.3 375 72.5

390 75.0 75.0 75.0 390 73.0 73.5 73.3 390 72.4

405 - - - 405 73.0 73.4 73.2 405 72.4

420 75.0 75.0 75.0 420 73.0 73.3 73.2 420 72.3

435 - - - 435 73.0 73.2 73.1 435 72.3

450 - - - 450 73.0 73.1 73.1 450 72.3

465 - - - 465 73.0 73.0 73.0 465 72.0 **

480 - - - 480 73.0 73.0 73.0 480 -

495 - - - 495 73.0** 73.0 73.0 495 -510 - - - 510 72.9 73.0** 72.9 510 -

525 - - - 525 72.9 72.9 72.9 525 -

540 - - - 540 72.8 72.8 72.8 540 -

555 - - - 555 72.5 72.5 72.5 555 -

*crystallization begins ** mixture is completely solid


3/5


The above cooling curve compares the recorded temperatures and times from the three parts

of the experiment. The blue line, describing pure naphthalene, is the most elevated, while the red

and the green lines, describing the naphthalene plus unknown substance mixture, are lower. This

shows the depression that the freezing point is subjected to when a pure solute is mixed in with a

solvent.

As evidenced by the graph shown above, the freezing point is clearly lowered when 1.0

grams of a new substance is added to the pure solution (green line). According toPrinciples of

General Chemistry by Martin Silberberg, the reason behind this phenomenon is that since in a

solution, only solvent molecules can solidify, traces of solute molecules may be left behind,

forming a slightly more concentrated solution. At t

he freezing point, t

he solid solvent and t

he

liquid solution are in equilibrium, and because the vapor pressure of the solution is lower than

that of the solvent at any given temperature, the solution freezes at a lower temperature than the

solvent, or the solution solidifies at a lower temperature.

Calculations

Kfnaph = 6.9

Solving for

freezing point of pure substance) - (freezing point of solution)

= 75.9 C 74.0 C

1.9 C or 2.0 C

Solving form:

2.0 C = 6.9

m

______________________________________________________________________________________

65

70

75

80

85

90

0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510 540

Temperature(C)

Time (seconds)

Figure 2.Temperature vs.TimeChart

Naphthalene.

Naphthalene +0.5 g unknown

Naphthalene +1.0 g unknown


4/5


m = 0.289855072

or 0.29

Solving for unknown: Let x be the molar mass of the unknown.

m =

m =

m =

=

=

x= 345 g

Computing for the percentage error

Percent error in molar mass:

Percent error in freezing point:

The value computed for the molecular weight of the unknown is 345 grams. Taking into

account the bright yellow powdery appearance of the unknown, we can infer that it is sulfur, or

S8. However, there is a big gap between the known mass of S8 (256.56 g/mol) and the calculated

one. Possibly, one source of error in this case is that not all of the sulfur was transferred into thetest tube, or there was an incorrect weighing of the sulfur. Another possible source is a

misreading of the temperature. There is always an uncertainty regarding the data for the

temperatures we gathered since the thermometer used was accurate only to a tenths of a degree,

and the readings of those small graduations were mere approximations. Lastly, while it is

possible that foreign material might have fallen into the mixture, the results would not have been

affected if this foreign material were insoluble, because this insoluble material cannot participate

in the reaction.


5/5


V. ConclusionIn conclusion, we can compute for the value of the molecular mass of an unknown substance

in a solution by first deriving the freezing point of the pure solution, followed by deriving that of

the mixture, and then lastly by plugging those values into the freezing-point depression constant

equation as shown above.

VI. References1. http://en.wikipedia.org/wiki/Naphthalene2. http://www.tarleton.edu/Faculty/alow/1084exp1.htm3. http://www.brainmass.com/homework-help/chemistry/other/2317024. Principles of General Chemistry, M. Silberberg

chem 12 freezing pt depression the real thing

Documents