Friday, October 14, 2011

Determination of the Molar Entropy of Camphor


1. To study the molar entropy of fusion of camphor

2. To determine the molar entropy of fusion ( ) of camphor


Camphor, C10H16O is an aromatic crystalline compound which is obtained from naturally from the wood or leaves of the camphor tree or synthesized. The camphor is used in this experiment to determine its molar entropy of fusion. Molar entropy of fusion is denoted as ( ), which represents the entropy increases when a particular substance is melting. The entropy is always depends on the state of the substance. The disorder in entropy of a substance is the lowest in solid state, whereas liquid is more disorder and followed by gas state with highest disorder in entropy. In another words, the degree of disorder increases in the transition from a closely packed molecule arrangement solid to the disorganized molecules arrangement of liquid state. But, in this experiment, the entropy of camphor is the energy absorbed by the camphor is disturbed by the adding of naphthalene in the system.

The molar entropy of fusion ( ) of camphor can be determined cryoscopically. Entropy, S is a thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that system can contain energy. When the energy of a thermodynamic system is concentrated in a relatively few energy states, the entropy of the system is low. When the same energy is dispersed or spread out over a great many energy states, the entropy of the system is high. Cryoscopy is a technique used for determining the molecular weight of a solute by dissolving a known quantity of the solute into a solvent and records the drops of freezing point of the solvent. For dilute solutions of naphthalene in camphor is described as below


Colligative properties are defined as those properties of solutions that depend on the number of dissolved particles in solution, but not on the natures of the solutes. The freezing point and boiling point of the particular solutions will be affected. This is because a liquid solution containing a solvent with an impurity is more disordered than a pure solvent. Therefore more energy must be removed to take the solution from its highly disordered liquid state to its ordered solid state than to do the same for a pure solvent. Hence, solutions of solvents with impurities freeze at a lower temperature than pure solvents due to colligative effect. In this experiment, colligative property is related to the freezing point depression and molality of a liquid.

The molality of a liquid is calculated by using the number of mole of solut is divided by the mass of solvent used in kg


The freezing point depression is calculated by

Apparatus: test tube, beaker (250cm3), Bunsen burner, thermometer (0-200 )

Materials: camphor (M=152g/mol), naphthalene (M=128g/mol)


1. 2g of camphor is being melted and stirred in a test tube.

2. The test tube in an air jacket and is allowed it to solidify.

3. The air jacket is removed and the test tube is heated gently until melting commerce.

4. The content is stirred with the thermometer and the temperature at the last crystal just disappears.

5. 0.05g of naphthalene is added into this test tube and the process is repeated.

6. The procedure is repeated for further 0.05g of naphthalene up to a total weight of 0.20g of naphthalene.

Results and calculation:

Table 1 Weight of naphthalene and camphor, no of mole for naphthalene and camphor, mole fraction, freezing point and T

Weight of naphthalene (g)

Number of moles of naphthalene (mol×10-4)

Weight of camphor (g)

Number of moles of camphor (mol)


Freezing point (K)

∆ T










































*Molecular mass of naphthalene and camphor are 128.06 g/mol and 152.13g/mol

ΔT= depression of freezing point of the mixture in K

= Kfp x m

Kf p= freezing point depression constant (Kfp of camphor= 37.7 /m)

m = molality of the solute= No. of solute mole (mol)/ mass of solvent (kg)


A graph of Kf versus X was plotted.

∆T = Kf X

= 12 / 0.0288

= 417

= 16 / 0.0559

= 286

= 24 / 0.0815

= 294

= 36 / 0.1059

= 340



f = RTo2/ Hf Equation 1

f = Hf / To Equation 2

thus substitute Equation 2 Equation 1

f = RTo/ Sf

Sf = RT0/ f

From the graph plotted,

Lim Kf = RT0 / ∆Sf

X 0

RT0 / ∆Sf = 222.5

222.5 = 8.314 * (172 + 273) / ∆Sf

∆Sf = 16.628 J K­-1 mol-1

∆Sf = ∆Hf / T0

∆Hf = 16.628 J K­-1 mol-1 * (172+273) K

= 7.399 kJ mol-1



In this experiment, the molar entropy, Sf of the camphor is obtained which has the value of 15 J/mol. Before obtain the molar entropy, we have to measure the temperatures change in each of the different amount of naphthalene added into the camphor. After the temperature changes are obtained, the cryoscopic constant for each amount of naphthalene added are calculated, the value of cryoscopic costant for each are shown in table 2. By using the value obtained, a graph of cryoscopic constant against mole fraction of naphthalene in the mixture is plotted and the y-intercept represents the cryoscopic constant of pure camphor which is 246.5.

According to the Thermodynamics Second Law, the entropy of a substance is the measure of the degree of disorder in a particular system. The value of the entropy of a distribution of atoms and molecules in a thermodynamic system is a measure of the disorder in the arrangements of its particles. State of the substance can determine the entropy of its substance. The degree of disorder is increase from solid state to liquid state and finally to gaseous state. In another word, the entropy of the same substance has higher entropy when it is in gaseous state compared with solid or liquid state. Molar entropy of fusion of camphor is denoted as ( ), which represents the entropy increases when a pure camphor is melting.

The different amounts of naphthalene are added into camphor as the pure camphor act as the control experiment. The pure freezing point of camphor is 172 in this experiment. After added the naphthalene, the freezing temperature are decrease as more amount of naphthalene is added into the same amount of camphor. The more the naphthalene added, the lower the freezing temperature of the particular mixture. This is showed that the system of camphor is being disturbed by naphthalene. The naphthalene molecules are rearranged with the camphor naphthalene in order to form bond in between each other. For a substance in liquid state to freeze, the naphthalene and camphor molecules begin to form cluster of molecules to convert to solid state. The molecule moves in high speed are not able to form a cluster of molecule with each other, so the molecules prefer move slower by reducing their kinetic energy. The kinetic energy is converted into heat energy that is being released to the surrounding as the temperature of the mixture decrease. Thus, the freezing point of the solvent with different solute is lower than the freezing temperature of pure solvent.

Precaution steps:

In this experiment, there is some precaution steps should be taken. Firstly, the zero error should be avoided when measuring the weight of naphthalene and camphor. This can be avoided by press the tare button when use it. Secondly, the reading of temperature on thermometer should be taken parallel to the eyes to avoid parallax error.


  1. how does the graph looks like actually? how do we get the lim Kf from the graph? TQ.