Dumas Method Lab - First Chemistry lab in Chem 112 PDF

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First Chemistry lab in Chem 112...

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Experiment 1: Dumas Method Purpose The purpose of this experiment is to find the molar mass of an unknown substance, through the use of the Dumas Method, the molar mass will then be compared to the universal molar masses of four different alcohols in order to determine which alcohol was the unknown substance.

Introduction In order to determine the molar mass of the unknown substance, the use of the ideal gas law, PV=nRT, P for pressure, V f or volume, n f or number of mole, R  is a gas constant and T for temperature, can help calculate the number of moles present in the sample. The ideal gas law is a model that demonstrates the relationship between volume, pressure, temperature and particles in a gas, it serves as a reference to the behaviour of real gases (1). As the organic liquid in this experiment is volatile, it means that it can easily be converted to gas particles, thus the use of the ideal gas law is acceptable for this experiment. After finding the number of moles of the gas, the molar mass equation, M=m/n , M  for molar mass, m for mass and n  for number of moles, can be applied in order to find the molar mass of the gas. This method is called the Dumas Method, it can determine the molar mass of an unknown substance. Dumas Method was derived from a French chemist called, Jean-Baptiste-André Dumas, who developed a method to determine vapour density of substances (2).

Procedure 1. Record the mass of one 125ml Erlenmeyer flask, one rubber band and one 2.5” square aluminum foil and repeat for a total of 3 sets 2. Add 2ml of volatile liquid into each flask and secure it with the rubber band and aluminum foil 3. Heat up liquid until boiling using a hot water bath containing a magnetic stir bar on a hot plate 4. Once the hot bath is boiling, record temperature using probe and the given atmospheric pressure 5. Remove flasks and cool for 10-15 minutes 6. Record mass of flasks and determine the difference before and after heating

Observations The colour of the unknown volatile liquid was colourless and was placed in a dark glass container, in addition, while wafting the unknown liquid, it smelled quite similar to rubbing alcohol. Once the flasks were placed into the warm water baths, condensation immediately began to occur near the bottom of the flasks. As the liquid was colourless, it was difficult to determine whether it has fully vaporised or whether it was still in liquid state. After removing the flasks from the hot water baths, there was no liquid left at the bottom, upon cooling, a small amount of liquid was collected at the bottom of the flask. After finding the difference before and after the heating, all three masses of the condensed vapour were very similar with only a ±0.01 difference.

Questions 1. Calculate the mean (molar mass) and standard error of the sample for your results.

Name: Karen Chan

Partner: Teah Persaud

Student No: 20173198

Student No: 20156595

Lab Section: 15

Bench # (on computer screen): 32

DATA SHEET

Sample number from the stock bottle you used.

M (Mass)

Trial 1

Trial 2

Trial 3

92.77 - 93.02

93.39 - 93.66

90.94 - 91.20

= 0.25

= 0.27

= 0.26

P (Pressure) V (Volume)

1

102.6 155

R (Constant Gas)

155

155

8.31451

T (Temperature)

371.2

371.2

372.2

n (number of moles)

(102.6 kPa x 0.155 L) /

(102.6 kPa) x (0.155 L) /

(102.6 kPa x 0.155 L) /

Molar Mass (Mass/moles) Mean of Molar Mass Standard Error

(8.31451 J/K·mol x 371.2 (8.31451 J/K·mol x 371.2 (8.31451 J/K·mol x 372.2 K) K) K) = 5.15 x 10-3

= 5.15 x 10-3

= 5.14 x 10-3

0.25/5.15 x 10-3

0.27/5.15 x 10-3

0.26/5.15 x 10-3

= 50.5 g/mol

= 52.4 g/mol

= 50.7 g/mol

(50.5 + 52.4 + 50.7) / 3 = 51.2 g/mol 0.852

2. Based on the molar mass you determined, identify the particular organic compound you used from the list of options presented on the datasheet. Based on the molar mass of the different compounds, I believe that my particular organic compound used was ethanol. This was determined because the average molar mass that was collected from the experiment is 51.2 g/mol, while the actual molar mass of ethanol is 46.07 g/mol. The difference between the two molar masses is 5.13 g/mol, it was the closest value to the actual molar mass out of the four volatile liquids provided. There were not other properties that could be determined from thus the smallest difference from the experimental molar mass and the actual molar mass of the different liquids were the best factor in identifying the organic compound.

3. Give three sources of experimental error that may have caused your answer to be incorrect and explain how your error would have changed your final conclusion. 1. Incomplete vaporisation of organic liquid to gas As mentioned previously, the colour of the liquid is colorless, thus determining when the liquid is completely vaporised was difficult. This could’ve affected the data due to the added mass of the liquid and assuming that the unknown substance is fully vaporised into gaseous form. Using the ideal gas law, you assume that all molecules in the flask are in gaseous state thus if the presence of a liquid would not be applicable. 2. Unsecured sealing of aluminum foil onto flask The method in ensuring the gases in the flask does not escape is not reliable as when transferring the bottle back to our work station, evaporation would’ve occurred which would affect the amount of liquid being vapourised. In addition, the sealing of the flask may not have been fully sealed, which in turn, when heating the flask, the gas may have escaped from the flask. 3. Unexpected liquid trapped under the foil After the heating process of the liquid, condensation would occur inside the flask during cooling. Liquids would form under the foil after heating thus can affect the mass of the compound as it may be overestimated....