Question

A certain volatile hydrocarbon (a binary compound of carbon and hydrogen) is found to be 92.3% carbon by mass. In a seperate 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 exess gas escapes the temperature is measures as 98.0 C. The flask and contents are subsequenlty cooled to 25 C (densitry water at 25 C = 0.997044 g/mL ) and the vapor condenses to a liquid. The flask is then empetied cleaned and filled with water. When weighed on the balance, the difference in weight between the flask filled to the brim with water and the dry empty flask at 25 C is 128.12 g

The empty flask - fitted with a foil cap pierced with a pinhole - weighs 25.3478g

The weight of the flask and contents is found to be 25.6803g

Determine the Emperical formula, the volume that the vapor occupied in the Florence flask and the molecular formula of this hydrocarbon.

Answer 1

Mass of flask + foil = 25.3478 g

Mass of flask + foil + contents = 25.6803 g

Therefore, mass of the hydrocarbon = (25.6803 – 25.3478) g = 0.3325 g

Also, it is given that mass of water in the flask = 128.12 g

Density of water at 25⁰C = 0.997044 g/mL.

Therefore, volume of water in the flask = volume occupied by the vapour of the hydrocarbon = mass of water/density of water = (128.12 g)/(0.99704 g/mL) = 128.500 mL = (128.500 mL)*(1 L/1000 mL) = 0.1285 L.

Pressure of the gas in the flask = 768 torr = (768 torr)*(1 atm/760 torr) = 1.0105 atm.

Temperate of the hydrocarbon in the flask = 98⁰C = 371 K.

Let M be the molar mass of the hydrocarbon so that

moles of the hydrocarbon in the flask = (weight of hydrocarbon in the flask/Molar mass of the hydrocarbon)

====> n = (0.3325 g/M) where n = moles of the hydrocarbon in the flask.

Assuming ideal behavior, we must have,

P*V = n*R*T

====> (1.0105 atm)*(0.1285 L) = (0.3325 g/M)*(0.082 L-atm/mol.K)*(371 K)

====> M = 0.3325*0.082*371/(1.0105*0.1285) g/mol = 77.90 g/mol ≈ 78.00 g/mol (ans).

Next, calculate the empirical mass of the hydrocarbon. The hydrocarbon contains 92.3% C by mass.

Percentage H by mass = (100 – 92.3) = 7.7

Find out the moles of C and H in 100 g of the hydrocarbon by dividing by the molar mass of C and H.

Moles C in 100 g hydrocarbon = 92.3/12 = 7.692

Moles H in 100 g hydrocarbon = 7.7/1.008 = 7.639

Simplest ratio of moles C:H = 7.692:7.639 = 1.006:1 ≈ 1:1

The empirical formula of the compound is CH and the empirical mass is (12*1 + 1.008*1) g =13.008 g

Let the molecular formula of the compound be (CH)_x where x is the atomicity of each element.

Therefore, (CH)_x = 78.00

====> 13.008x = 78.00

====> x = 5.996 ≈ 6.00

The molecular formula is (CH)₆ ≈ C₆H₆.

Therefore, we have the following answers:

Emperical formula = CH

Volume of the vapour in the Florence flask = 0.1285 L

Molecular formula for the hydrocarbon = C₆H₆ (ans).