Chemistry Lab

Topics: Chemistry, Atom, Oxygen Pages: 2 (687 words) Published: March 23, 2013
Chemistry 121
Colligative Properties Lab
Demonstration of Selected Calculations from Choice I
Determination of Kf for Naphthalene
To determine the Kf for naphthalene, we need to find the difference in the freezing point of pure naphthalene and the solution of 1,4-dichlorobenzene in naphthalene. Let's say that we did this experiment, used 1.00 g 1,4-dichlorobenzene in 10.00 g naphthalene, and found that the freezing temperature of pure naphthalene was 78.2°C, while that of the solution was 75.4°C. This gives us a Tf of 78.2°C - 75.4°C = 2.8°C. Using the equation for freezing point depression and solving for Kf, we have... Tf = Kfmsolute

Kf = Tf/msolute
where msolute equals the molality of the solute. What is the molality of the solute? msolute=molality of solute = moles of solute/kg solvent
moles of solute = 1.00 g 1,4-DCB/146.9 g/mol = 6.81 x 10-3 moles 1,4-DCB kg of solvent = 10.00 g naphthalene/1000 g/kg = 0.01 kg solvent msolute = 6.81 x 10-3 moles 1,4-DCB/0.01 kg naphthalene = 0.681 m Kf = 2.8°C/0.681 m = 4.112 K·kg/mol

The actual value for Kf for naphthalene is 7.45 K·kg/mol, so we're a fair amount off the mark. This is a rather crude experiment, so that's to be expected. Determination of the Molecular Mass/Molecular Formula of Elemental Sulfur Although this experiment didn't go as well as I would have liked, there seemed to be some problems with the interpretation of the data as they were obtained, so I thought I'd demonstrate how to do this calculation. Elemental sulfur has a molecular formula of S8 (there are 8 sulfur atoms in a molecule of sulfur, just like there are 2 hydrogen atoms in a molecule of hydrogen). So, the purpose of this experiment was really just to see how close you could come to this answer. To find the molar mass (and, using the atomic mass, the molecular formula) for sulfur, we first need to find the freezing point of a solution of sulfur in naphthalene. Let's say we made such a solution with 0.500 g powdered sulfur in 10.00...

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