The Dumas method is one of the simplest procedures for determining the molar mas

Question

The Dumas method is one of the simplest procedures for determining the molar mass of an with a unknown volatile liquid. In the Dumas method, volatile liquid sample is heated in a flask at of tiny the entire sample vaporizes. Because the volume occupied by the vapor at atmospheric pressure is much larger than the volume occupied by the liquid, some of the vapor will escape from the flask. However, the vapor remaining in flask will contain the number of moles of the substance that fills the volume of the flask at the experimental pressure and vapor temperature. Why is the barometric (i.e., atmospheric) pressure considered to be the vapor, i.e., how does the experimental procedure ensure this? Page reference ___ Explain carefully. Why isn't it necessary to weigh the of liquid initially put into the flask? Explain. A certain volatile hydrocarbon (a binary compound of carbon and hydrogen) is found to be 92.3% carbon, by mass. In a separate experiment, utilizing the Dumas method, a 4.00 mL pure liquid sample of this hydrocarbon is vaporized in a 125 mL Florence flask when the barometric pressure is 768.0 torr. After the excess gas escapes, the temperature is measured as 98.0 degree C. The flask and contents are subsequently cooled to 25 degree C (density water at 25 degree C = 0.997044 g/mL) and the vapor to a liquid. The flask is then emptied, cleaned, and filled with water. When weighed on a balance, the difference in weight between the flask filled to the brim with water and the dry empty flask at 25 degree C is 128.12 g. The empty flask - fitted with a foil cap pierced with a pinhole - weighs 25.3478 g The weight of the flask and contents is found to be 25.6803 g Please determine the following, The empirical formula of this hydrocarbon. The volume that the vapor occupied in the Erlenmeyer flask (in L). The molecular formula of this hydrocarbon.

Explanation / Answer

Dumas method for estimating molecular mass of an unknown volatile liquid

1. The pressure of the vapor is equal to the atmospheric pressure. The boiling point of the volatile liquid is defined as the point where the vapor pressure of the liquid equals to that of the atmospheric pressure and thus, the pressure of vapor is taken as the atmospheric pressure at which the liquid exists only in vapor state.

2. It is not necessary to weigh the liquid put initially in the tube as in the experiment we consider the weight of the vapor formed after heating the system which should directly imply how much liquid was there initially.

3. mass of unknown liquid (m) = 25.6803 - 25.3478

                           = 0.3325 g

Pressure (P) = 768/760

               = 1.0105 atm

Volume (V) = 128.12 g/0.997044 g/ml

             = 128.50 ml

             = 0.1285 L

Temperature (T) = 98 + 273

                            = 371 K

R = gas constant

Using,

molar mass = mRT/PV

                    = 0.3325 x 0.08205 x 371/1.0105 x 0.1285

                    = 77.95 g/mol of unknown liquid

92.3% is carbon = 77.95 x 0.923 = 71.95 g

Hydrogen = 77.95 - 71.95 = 6 g

Empirical formula of unknown liquid = CH (for C6H6)

Volume that vapor occupies in the erlenmeyer flask = 0.1285 L

molecular formula = C6H6

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