Equipment:
- Butane lighter
- Marking pen
- Large trough
- Retort stand, boss head and clamp
- Thermometer
- Barometer
- Graduated cylinder (250 mL)
- Blow dryer
Procedure:
We started by almost filling the trough entirely with water. Then we filled the whole graduated cylinder with water and covered the end while we turned it upside down and put it under water. Then we placed it in the clamp on the retort stand. We weighed the butane lighter and placed it under the opening of the graduated cylinder. Then we opened the lighter and released gas into the cylinder. The gas pushed the air out and we had a gas column in the tube. When the gas column reached such a length that the water level was the same inside and outside the tube we closed the lighter and read how much gas was in the graduated cylinder. Then we measured the pressure in the room and the temperature in the water. Then we dried the lighter with a blow dryer and measured the weight.
Observations and calculations:
Temperature: 22˚C +/- 0.5˚
Pressure (in the air): 1014 mb +/- 1 mb= 101.4 kPa +/- 0.1 kPa
Vapor pressure: 2.64 kPa (found in table and therefore no uncertainty)
Partial pressure of the gas: 101.4 kPa – 2.64 kPa = 98.76 kPa
Volume of gas released:
Volume of air in the cylinder: 1.3 ml +/- 0.1 ml
Volume of gas in the cylinder when releasing of gas was done: 100 ml +/- 0.1 ml
Volume of butane gas released: Vmin: 99.9 ml – 1.4 ml = 98.5 ml
Vmax: 100.1 ml – 1.2 ml = 98.9 ml
V : 100.0 ml – 1.3 ml = 98.7 ml +/- 0.2 ml
From these measurements we can make an equation:
To make the equations easier, I will convert some of the units with a unit-conversion program.
(P1V1)/T1 = (P2V2)/T2
P1= 98.76 kPa = 0.974685 atm
V1= 98.7 mL = 0.0987 L
T1= 22˚C = 295˚K
P2= 101.3 kPa = 1 atm
V2= ?
T2= 0˚C = 273˚K
(0.974685 x 0.0987)/295 = (1 x V2)/273
V2= 0.0902 L
Then we use the fact that at STP the molar volume of an ideal gas is 22.4 L.
0.0902 L/22.4 L mol-1 = 0.00403 mol
M = m/n
M = 0.230 g / 0.00403 mol = 57.1 g/mol
Conclusion and Evaluation:
Through this experiment we have learned that it is in fact possible to calculate the molar mass of a gas through calculations and measurements of volume, temperature and pressure. It is perhaps not very useful to do this if the identity of the gas is know, but if we do not know what gas we are dealing with, this method could to certain extent be used to find out what gas it is. We can draw this conclusion because we found a value close, but not exactly, the literature value.
When we performed this practical, we did everything according to the instructions, at least as far as I can remember. However, it was a quite complicated procedure with many things that had to be dealt with carefully and I do not think it impossible that we neglected some of this. The part where we should fill a graduated test tube to the top with water and turn it upside down in the tub with out letting any gas in, was a difficult maneuver.
But all in all, the procedure was good, when performed properly.
