Dumas Method Experiment: In this experiment, we measured a flask with an aluminu

Question

Dumas Method Experiment:

In this experiment, we measured a flask with an aluminum foil and elastic band set up then added 4mL of methanol inside the flask and sealed with the foil covering and boiled in a water bath until all the liquid vaporized (this will drive out any air in the flask and push out all excess vapour). Then we cooled it to let the vapour condense and reweighed the flask with the vapour in it. The following questions pertain to the experiment:

For each of the following experimental conditions determine whether the calculated value for molar mass would be: A) too high B) Too low C) Unaffected. In each case, explain how this result occurs.

a) After removing he flask from the water bath, the experimenter cools the flask to room temperature but does not dry it. The calculated molar mass will be: A) too high B) Too low C) Unaffected. Explain.

b) The flask is removed from the water bath containing vapour only, the experimenter cools the flask to room temperature and some vapour condenses inside the flask. The calculated molar mass will be:  A) too high B) Too low C) Unaffected. Explain.

c) The flask volume is not measured; instead the experimenter assumes the flask volume to be exactly 125.0 mL. The calculated molar mass will be:  A) too high B) Too low C) Unaffected. Explain.

d) From the time the mass of the unused flask assembly (flask, foil and elastic band) is recorded the flask is handled several times with oily fingers. The calculated molar mass will be A) too high B) Too low C) Unaffected. Explain.

Explanation / Answer

You can answer each part of the question by observing the equation: MW = (m.R.T)/(V.P) where: MW = Molecular Weight of the Unknown m = mass of the unkown in the flask R = Gas Constant T = Temperature of the boiling water bath V = Volume of the flask (may not be equal to 125 mL) P = Atmospheric Pressure or Pressure in the room

Ans (a): Too High. See how you determine the mass of the unknown, by subtracting the mass of empty flask from the mass of flask after condensation. If you will not dry the flask properly then there will be some water droplets on the outer side of the flask which will contribute to the mass of unknown and from above equation we can say that MW will be more if mass is more as there are directly proportional to each other.

Ans (b): Unaffected. Because the mass of unknown will be the mass of the condensed vapour + mass of vapour inside the flask (if any present after the condensation). The condensed mass will not lead to any extras as it is the part of actual one.

Ans (c): It may depend, if the actual calculated volume is greater than 125 mL then calculated MW will be small and if it is less then MW will be high. (MW is inversely proportional to V)

Ans (d): same as in part (a).

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