Freezing point of substance X (0.50 to 0.75g + 0.15g to 0.25g): 76oC
Table 1: Freezing point/melting point of naphthalene, naphthalene + (0.50 to 0.75g X) and naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X) from 0 - 150 seconds
Table 2: Freezing point/melting point of naphthalene, naphthalene + (0.50 to 0.75g X) and naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X) from 180 - 480 seconds
Table 3: Freezing point/melting point of naphthalene, naphthalene + (0.50 to 0.75g X) and naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X) from 510 - 810 seconds
Table 4: Freezing point/melting point of naphthalene, naphthalene + (0.50 to 0.75g X) and naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X) from 840 - 1080 seconds
Table 5: Freezing point/melting point of naphthalene, naphthalene + (0.50 to 0.75g X) and naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X) from 1110 - 1350 seconds
* Since the melting point and freezing of naphthalene, naphthalene + (0.50 to 0.75g X)
and naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X) are the same, the graph of
temperature versus time for the freezing point is plotted.
Analysis & Calculation:
- Calculate molecular weight of substance X in
- Naphthalene + (0.50 to 0.75g X)
Using the formula:
Mass of materials
∆T = Kf m = Kf x ------------------------------------------
Molecular weight X Kg solvent
∆T = 78 oC -77 oC
= 1 oC
Kf = 6.8 g/mol
Mass of materials = 0.50g
Kg solvent = 0.005g
(0.50g)
1 oC = (6.8 g/mol) x ------------------------------------
Molecular weight X (0.005g)
0.50 X 6.8
Molecular weight = --------------
0.005
= 680 g/mol
- Naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X)
Using the formula:
Mass of materials
∆T = Kf m = Kf X ------------------------------------------
Molecular weight X Kg solvent
∆T = 78 oC -76 oC
= 2 oC
Kf = 6.8 g/mol
Mass of materials = 0.65g
Kg solvent = 0.005g
(0.65 g)
2 oC = (6.8 g/mol) x ------------------------------------
Molecular weight X (0.005 g)
0.65 X 6.8
Molecular weight = ----------------
2 X 0.005
= 442 g/mol
- The average reading of (a) & (b)
680 + 442
Average = ----------------
2
= 561 g/mol
- Calculate the molality of substance X in
- Naphthalene + (0.50 to 0.75g X)
0.50
Molality = --------------------
680 X 0.005
= 0.1471 M
- Naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X)
0.65
Molality = --------------------
442 X 0.005
= 0.2941 M
Discussion:
In the experiment, we used naphthalene as the product which will be determined its freezing point and melting point. The characteristic of freezing point and melting point of naphthalene is study in the experiment by adding different mass of foreign substance which is indicating as substance X. By determining the freezing point of the naphthalene, ∆T, we can use it to find out the molecular weight of substance X using the given formula:
Mass of materials
∆T = Kf m = Kf X ------------------------------------------
Molecular weight X Kg solvent
Where;
∆T = freezing point depression
Kf = freezing point molar constant of solvent (where for naphthalene is 6.8 g/mol)
m = (Mass of materials / Molecular weight X Kg solvent)
Naphthalene is a white solid with a strong smell, crystalline, solid aromatic hydrocarbon with a pungent odor. It melts at 80°C, boils at 218°C, and sublimes upon heating. It is insoluble in water, somewhat soluble in ethanol, soluble in benzene, and very soluble in ether, chloroform, or carbon disulfide. Naphthalene is a natural component of fossil fuels such as petroleum and coal; it is also formed when natural products such as wood or tobacco are burned. It is used in mothballs and gives them their characteristic odor. From it are prepared derivatives that are used in the preparation of dyes and as insecticides and organic solvents. The molecular structure of naphthalene is that of two benzene rings fused together with two adjacent carbon atoms common to both rings.
When weighing the mass of naphthalene, two-decimal balance is use because the range of mass we need of substance naphthalene and substance X is in two-decimal point. The air movement in the lab is decrease to the minimum to prevent affecting the mass reading of the two-decimal balance. Naphthalene is quite irritating of its smell as it may cause some dizziness to certain people. Exposure to a large amount of naphthalene can cause red blood cells to be damaged or destroyed, a condition called hemolytic anemia, which leads to fatigue, lack of appetite, restlessness, and a pale appearance. So, when handling the substance, use mouth piece to cover the nose if dizziness is found.
The test tube contains naphthalene will then heat using water bath. The water level from outside of test tube must cover over the level of naphthalene inside the test tube. This is to ensure that all the heat is transferred from the water surrounding into the test tube evenly. Using mouth piece when there is any dizziness, because the naphthalene gas is more irritate to the nose.
After the naphthalene was melted and ready to taken out to record its freezing point, apparatus must ready setup and the time and temperature is taken immediately after the test tube contain naphthalene is taken out from the hot water. This is because the freezing of naphthalene is happen immediately. If the temperature is taken slow for first few seconds it may cause the data result become incorrect and it is hard to determine the freezing point. So the few first minutes is decided to record the temperature in every 15 seconds as in short period. Then after that every 30 seconds per record of temperature to be apply as in longer period where the changes in temperature is not that fast.
While taking the temperature records, the solution is stirred constantly to transfer the heat evenly in the solution. Using a rod glass to stir it or simply just use the thermometer to stir while we can also easily record the temperature faster. As the mercury bulb located bottom of the thermometer is easily to be break because it is only make up by a thin layer of glass, so more care was taken while stirring. Error while recording the reading of temperature would be minimized as the meniscus level of mercury is corrected 90 degree to the observer’s eyes. This is to avoid the inaccuracy of data recorded.
The temperature taken in given time was recorded in a proper table. A graph of temperature against time was plotted to show the freezing point for naphthalene only, naphthalene + (0.50 to 0.75g X) and naphthalene + (0.50 to 0.75g X) + (0.15 to 0.25g X). The freezing point for naphthalene we get in this experiment was 78 oC which is two degrees lower than the theoretical value which is 80 oC. This may due to the improper heating or naphthalene that was used was contaminated. The graph we get was not a smooth stair case like line graph. The curve in between the constant temperature (straight line) is sometimes either too high or too low from the original curve. This may due to the improper recording of the temperature. Constant stirring is needed when cooling of the solution from the water bath to give the smooth and proper graph.
Precaution Step:
- When finish the experiment, DO NOT pour the waste of naphthalene or the mixture substance into sink. Throw them into a bottle.
- Do not inhale too much gas released when heating naphthalene as it is toxic and can cause eye irritation, confusion, excitement, malaise, abdominal pain, irritation to the bladder, profuse sweating, jaundice, hematopoietic, hemoglobinuria, renal shutdown, and dermatitis.
- Wear safety gloves and safety goggles when handling naphthalene as it is toxic to us.
- Use tong when handle test tube to prevent burns to hand when test tube are hot or affecting result by moisture of hand when test tube is cold.
- Switch off the fan when weighing the substance or material. Air movements are easily caused the reading of balance to be run off.
Conclusion:
- The average molecular weight for substances X is 561 g/mol.
-
Freezing point for naphthalene is 78 oC; naphthalene with 0.50g substance X is 77 oC; naphthalene with 0.65g substance X is 76 oC.
- The molarity of substance X is 0.1471 M in naphthalene with 0.50g substance X and 0.2941 M in naphthalene with 0.65g substance X.
References:
- www.nv.cc.va.us/alexandria/science/tableofcontents.htm
- www.chemicool.com
- http://www.tarleton.edu/~alow/1084exp1.htm
- http://dbhs.wvusd.k12.ca.us/webdocs/Solutions/BP-Elev-and-FP-Lower.html
- http://en.wikipedia.org/wiki/Naphthalene
- http://icn2.umeche.maine.edu/genchemlabs/Freezing_Point/freezing4.htm
- http://chemed.chem.purdue.edu/genchem/probsolv/colligative/kf1.3.html
Questions:
1) How can the Kf value for a solvent be determined?
The same method as the experiment can be carried out by using a known volatile substance instead of the unknown substance X.
∆T = Kf x Mass of materials 1
------------------------------------------ x -----------------------
Molecular weight of solvent kg of solvent
According to the equation above, the data that should be collected from the experiment is the depressed temperature ∆Tf , the mass of solute that is used (in grams) and the mass of solvent (in kg). If the solute that being used is known, then the molecular weight of the solute can be known. For example, if glucose is the solute that being used, the molecular mass of glucose would be 180. From the equation above, the kf value can be determined.
kf = ( ∆Tf x kg of solvent x molecular weight of solute )
--------------------------------------------------------------
grams of solute
2) Compound A, C8H8N2, melts at 80.5oC and compound B, C10H8, melts at 80.6oC. Compare Z which is given is the same as A or B. Suggest a simple method to determine the identity of Z.
The same method of freezing point depressing method can be carried out to determine the molecular weight of the substance Z. If the molecular weight is 132, then the substance is a compound Z. If the molecular weight is 128, then the substance is compound B. From the experiment, the depression of the temperature, the mass of solute Z (gram) and the mass of solvent (kg) should be recorded. Then the molecular weight of the substance Z can be calculated by using the following equation:
Molecular weight of the solute = kf (grams of solute)
-----------------------
∆Tf (kg of solvent)
Since the melting point of substance A is 80.5°C and B is 80.6°C, a solvent with a higher melting point should be used. For example, sodium chloride (NaCl) can be used as NaCl has a high melting point which is 323 °C.