Merridith GawExperiment 1: Charles’s Law

Processed Data:

Mark Temperature (± uncertainty) Volume (± 0.05) Changes in

Temperature Changes in Volume

1 30.0 227.5 (± 0.05) - -2 46.0 250.0 (± 0.05) 16 22.53 64.0 267.0 (± 0.05) 18 174 76.0 279.0 (± 0.05) 12 125 92.0 288.0 (± 0.05) 16 11At 0 mL -202.67 0.0 (± 0.05)At -273 °C -273.15 0.0 (± 0.05)

*Graph at last Page

Discussion:

This experiment demonstrates the concept of Charle’s Law which describes the relationship between the VOLUME and TEMPERATURE of a gas while the pressure and moles remains constant. Factors affecting this experiment is the pressure, temperature, volume and amount of gas. The type of gas is not a factor, despite the fact that different gases have different partial pressure. The reason is because the total pressure will exerted would remain constant, meaning even if the gas contained is a composed of the different types of gases, this will not affect the experiment. 3 i

This experiment required the use of a levelling bulb to keep pressure constant, while the water level keeps the pressure inside and outside the apparatus to remain equal. The amount of gas is must remain constant as to observe the volume the gas occupies as temperature is heated, this is done by checking for leaks were gas can slip. It is also crucial to see that the flask containing gas should not contain water. The reason is that heating the flask can cause the water within to evaporate which will contribute to the volume of gas, there is also the possibility that the vapor condenses which increase the amount of water. Both of which causes deviation to the data.2

This experiment is conducted in atmospheric pressure or 1 atm. This low pressure allows us to assume that gas to behave ideally, and that difference can be negligible. The result of the experiment allowed us to create graph where we can observe that there is a linear relationship between temperature and volume. And theoretically through the graph, we can extrapolate the temperature at which volume would be 0, and it was found to -273°C otherwise known as the absolute zero. However, error cannot be prevented as the temperature when volume reaches 0 is found to be different from -273°C.ii Such as the rubber tubing, Rubber is known to expand when heated. The rubber tubing is used to allow hot air to travel from one container to another. However the calculating the volume of gas, the heated gas could have expanded the tubing allowing it to increase in space, decreasing the actual volume of gas shown in the burette. Another would be small leaks, which were not noticed. 1

Conclusions:

Despite not being able to obtain the extrapolated absolute temperature, the method in this experiment is still useful as this still presented the relationship between volume and temperature of gas while keeping constant amount and pressure. The errors made could have been prevented and minimize if one were to follow the instructions in extreme details, creating a better setup.

Calculation:

y=0.789 x+214.06

if y=0mL; x=−202.6664

Reference:

Blauch, David. “Charles’s Law” n.d., n.p. Accessed on June 6, 14 <http://www.chm.davidson.edu/vce/gaslaws/charleslaw.html>

ii “Excerpt From Encyclopedia Britannica” Accessed on June 6, 14<http://abyss.uoregon.edu/~js/glossary/ideal_gas_law.html>

Merridith GawExperiment 2: Freezing Point Depression

Processed Data:

Stea

ric A

cid:

Trial 1 Trial 2Temp (°C)

Second Temp (°C)

Second Temp (°C)

Second Temp (°C)

Second

85.1 30.3 62.3 331.0 80.2 31.0 57.3 330.482.4 60.1 61.7 360.3 74.9 60.3 55.8 360.980.0 91.1 56.8 392.4 71.1 90.7 54.9 391.978.2 120.4 55.0 420.3 68.3 120.5 54.2 422.375.6 150.4 54.3 450.5 66.5 150.4 53.7 450.473.6 180.4 53.8 480.8 64.2 180.8 53.5 480.770.8 210.4 53.0 512.3 62.1 211.0 53.1 510.669.1 240.3 53.0 540.7 61.4 240.4 52.9 540.667.1 271.0 52.6 570.5 59.5 271.1 52.7 570.865.6 300.4 52.5 600.9 57.9 301.5 52.4 601.3

Stea

ric A

cid

with

0.5

Aliq

uot Trial1

Stea

ric A

cid

with

1.0

Aliq

uot

Trial 1Temp (°C)

Second Temp (°C)

Second Temp (°C)

Second Temp (°C)

Second

82.5 30.5 55.3 330.4 82.1 30.2 55.5 330.678.1 60.9 53.8 360.8 77.0 60.1 54.9 361.274.2 90.1 52.8 391.2 72.8 90.5 53.2 391.370.5 120.4 51.8 421.2 69.7 120.2 52.0 421.568.6 150.1 51.0 451.5 66.8 150.3 51.0 451.765.0 180.3 50.5 482.1 64.1 180.6 50.6 482.062.9 210.8 50.1 512.6 62.3 210.6 50.1 512.160.7 241.0 50.0 542.7 59.8 240.4 50.0 542.458.6 271.2 49.9 573.0 58.5 270.4 49.7 572.557.1 300.0 49.8 603.6 57.4 300.4 49.6 602.9

Solution Mass of Substance

Freezing Points from Extrapolation (°C)

Molality (mol/Kg)

Moles of Unknown Added

Molecular Massg/mol

Stearic Acid 4.5328 58.2181 - - -

Stearic Acid + one Aliquot 0.5002 54.3058 0.8631 0.003912 127.8532

Stearic + 2 Aliquots 1.000 55.1899 0.6681 0.003028 330.2292

Average Molar Mass: 229.04118

Standard Deviation: 143.1044

*Graph at last Page

Discussion:

This experiment allowed us to observe the concept of freezing point depression. Wherein the addition of another compound in a liquid lowers its freezing point. The freezing point is the temperature at which vapor pressure of liquid and solid state of compound is equal. Meaning in a cooling curve, the intersection represents the freezing point. When a pure substance freezes, the molecules rearrange themselves into a composition that prevents flowing. In other words, they come together and cluster to form the composition of a solid. When another substance is added, the intermolecular forces that allow the molecules to freeze are disrupted. Meaning more energy is required to be removed from the system before the particles can reorder. This means that a solution will always freeze at a lower temperature than the pure solvent. It is important to note that the amount that the freezing point is lowered does not depend upon the identity of the solute; it’s just dependent upon the number of solute particles in the solution. 1 i

Stearic acid was used as the pure solvent and was mixed with an unknown fatty acid twice. All which were boiled and cooled. During the cooling process, the temperature was taken every 30 seconds for 10 minutes. This data is used to graph the cooling curve which depicts the relationship between the temperature and time to the cooling process. In the graph, there comes a certain point where the temperature remains constant or at least the change in temperature is minimal. This point is where the changes in state occurs. The two states are in equilibrium within the temperature and pressure. The heat being removed allows the phase change to occur, in this case liquid to solid. The temperature remains constant until all of the energy is removed. It can be seen in the graph the trend change upon adding the unknown aliquot. The graph showed that as more moles of unknown added the lower the freezing point is. 2

With the data gathered, a calculated average molar mass of the unknown fatty acid was seen to be 229.0411 g/mol. Given the choice of unknown fatty acid, Myristic Acid is found to have a molar mass of 228g/mol, which closest to the result. The slight deviation of mass can be caused by possibility of inaccurate reading of the thermometer leading to an inaccurate graphing and calculation. 3

Conclusions:

It can be concluded that this experiment is successful, such that the unknown was identified through the use of the molar mass calculated, although there are slight errors. The concept of freezing point depression was also observed as the graph of the cooling curve showed that the freezing point lowers as more solute are added.

Reference:

Minnesota State University. “Molar Mass from Freezing depression” , Accessed on June 6, 2014 <http://web.mnstate.edu/marasing/CHEM210L_2013_Summer/11Molar%20Mass%20from%20fp%20depression.pdf>

Calculation:

Freezing Point Depression:

y=−0.0735 x+86.889 y=−0.0359x+72.222y=−0.0735 x+86.889=−0.0359 x+72.222

x=390.0797 ; y=58.2181℃Molar Mass:

Molality=T(Solvent)−T (Mixture )

Mass(Solvent)

58.218°C−54.306 °C4.5328 g

=0.2956/ g

Molesof unknown=(molality of unknown×Kgof solvent )0.2956molg

× 4.5328gmol

=0.003912mol

Molar Mass= Mass of UnknownMolesof Unknown

0.5002g0.003912mol

=229.0411 gmol

Graphs:

i