Freezing Point Depression

Topics: Ethanol, Mass, Chemistry Pages: 6 (1588 words) Published: October 10, 2012
For this experiment, the osmolality given by the machine was equal to the molality. 1) Finding the identity of a unknown substance

Molar Mass of Possible unknown substances (g/mol):
* Glycine: 75.067
* Glucose: 180.155
* Mannitol: 182.171
* Sucrose: 342.26
* Lactose: 360.312
Table 1.1: Table showing the results for the osmolality. 200mg of the unknown was dissolved with 10mL of doubly deionized water. A 0.2mL sample was taken and placed in the Osmometer. Trial Number| Osmolality (mOsm/kg H2O)|

1| 294|
2| 291|
3| 296|
Average| 293.67|

Equation used:
Molality= n1000g solvent
Molality= mM1000g solvent
Molality= mM*1000g solvent

Mass of unknown substance:
0.226 g
Mass of solvent:
9.876 g = 9.876 * 10-3 kg
0.29367OsmkgH2O=0.226gM*9.876*10-3kg solvent
0.29367OsmkgH2O=22.88 gM*kgH2O
Based on the molar mass of calculated, the identity of the unknown substance D is Glycine. 2) Finding the concentration of a known solution
Table 1.2: Table used for the preparation of the calibration curve used to find the concentration of alcohol in a wine. 0.2mL of each sample were taken and placed in the Osometer. Ethanol sample (g EtOH/100mL)| Osmolality (mOsm/kg H2O)|

5.117| 109|
7.984| 181|
11.13| 242|
14.01| 307|
18.31| 404|

Wine tested: Piesporter (White) 9.0% by volume.
Table 1.3: Table showing the results for the osmolality of a sample of wine. 1mL of wine was diluted with 10 mL of water. A 0.2mL sample was taken and put in the osometer. Trial Number| Osmolality (mOsm/kg H2O)|

1| 204|
2| 204|
Average| 204|

Figure 1.1: Calibration curve for used for the identification of the concentration of ethanol present in a sample of wine. 5 data points were taken at 5.117g EtOH/100mL solvent, 7.984g EtOH/100mL solvent, 11.13g EtOH/100mL solvent, 14.01g EtOH/100mL solvent and 18.31g EtOH/100mL solvent. Equation for line: y =22.09x-1.274

Concentration based on curve:
Osmolality = 204 mOSm/kgH2O
204 = 22.09x-1.274
x = 8.96g EtOH/100mL
x = 8.96mg EtOH/mL
Theoretical molality of ethanol:
9% by volume:
1mL * 0.09 = 0.09mL of ethanol/1mL solution
Add 9 mL of water:
0.09 mL of ethanol/10mL solution = 9 * 10-3 mL of ethanol/mL solution Using density of ethanol: 0.7893g/cm3 = 0.7893g/mL
9*10-3mL of ethanolmL solution*0.7893gmL of ethanol=7.10*10-3gmL of solution=7.10mgmL solution Percent difference:
8.96-7.10mgmL solution7.10mgmL solution*100%=26.2% difference There s a 26.2% difference, which is quite significant. The alcohol content in the solution is lower than the label of the bottom.

Osmometry is useful in determining the identity or concentration of a solution. It can be used to detect compounds in drug samples as well as determining the concentration of blood alcohol. To do this, a small sample of the solution with either unknown identity or unknown concentration is put in the osmometer. The freeze button is pressed when the number reaches 3000. From the results, we determined that the unknown substance labeled D is glycine and that the alcoholic content by volume for Piresporter is less than the labeled 9.0% by volume. The conclusion is that osmometry has many uses and is a key tool is determining traits of a compound.

This experiment consisted of two parts. Firstly to determine the identity of a provided unknown and secondly to determine the amount of alcohol present in a wine sample. In the case of determining the identity of a unknown in this case unknown D, 5 different possibilities were presented which simplified the task immensely. By using the osmolality obtained from the Advance Wide-range osmometer 3WII the molecular mass could be calculated and by comparing the result of 77.91 g/mol to the given molecular masses the deduction can be made that the unknown D is Glycine. The reason why there is a slight deviation from the obtained value may be...
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