Determine the molar mass of butane.

MOLAR MASS OF BUTANE

Name: Ingrid Foss Ballo

Class: 2IBJ

Team: Ingrid Foss Ballo and Maike Jule Sonnewald Petersen

Date: Tuesday 30th of January, 2004

AIMS

The purpose of this activity is to determine the molar mass of butane using gas from a butane lighter. The experiment depends on the fact that the molar volume of an ideal gas is 22,4L at standard temperature and pressure (0 °C and 101,3 kPa). By measuring the volume occupied by a known mass of butane, and knowing that its molar volume is 22,4L at S.T.P., it is possible to calculate the molar mass of butane.

METHODS AND TOOLS

Equipment

Piece of glass tubing – approximately 24 cm long and with a diameter of approximately 25 mm, closed at one end (with markings for mL).
Container for water
Butane lighter
Restort stand, boss head and clamp
Thermometer (-10-100 °C)

Procedure

We filled both the glass tubing and the container with water. Then we inverted the tubing and place the open end of it under the surface of the water in the water container, as shown in figure I:

Then we clamped the glass tubing in position and measured the temperature of the laboratory. We also obtained a reading of the atmospheric pressure in millibars. We weighed the butane ligher on a balance to at least two decimal places. We then held the lighter upright, under water in the container and placed it directly under the glass tubing, as shown in figure II:

We opened the trigger of the lighter and allowed the gas releaced to displace the water from the glass tubing. We kept the trigger open until the water levels inside and outside the tbe was about the same and then removed the butane lighter.

We adjusted the height of the glass tubing to make the water level inside and outside the tube the same, and read how many mL of butane we now had in the glass tubing.

At last, we dryed the butane lighter using a paper towel and by blowing ...

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We adjusted the height of the glass tubing to make the water level inside and outside the tube the same, and read how many mL of butane we now had in the glass tubing.

At last, we dryed the butane lighter using a paper towel and by blowing warm air over the top, using a hair dryer. Then we reweighed the lighter and calculated the mass of butane. This was done several times, to make sure that the lighter was completely dry.

Then we calculated the partial pressure of butane, the volume butane would occupy at S.T.P. and hence the number of moles of butane relased. From the measured mass, we calculated the molar mass of butane.

DATA COLLECTION

Diameter of glass tubing: 25 mm ± 1 mm (measured with a ruler)

Length of glass tubing: 24 cm ± 0,1 cm (measured with a ruler)

Weight of lighter before experiment: 24,734 g ± 0, 001 g

Temperature of laboratory: 22 °C ± 0,1 °C

Volume of butane inside the glass tubing: 100 mL ± 0, 1 mL

Vapour pressure of water at 22 °C: 2,64 kPa (according to a table in “Chemistry Laboratory Manual”)

Weight of lighter after experiment: 24,503g ± 0,001 g (First measured value was 24,519 g ± 0,001 g, so it is obvious that the lighter at that time contained some water.)

Measured atmospheric pressure: 1014,5 millibars ± 0,1 millibar.

DATA PROCESSING AND PRESENTATION

Calculating mass of butane released from the lighter:

mbutane = initial weight of lighter – final weight of lighter

mbutane = 24,734 g ± 0,001 g – 24,503 ± 0,001 g

mbutane = 0, 231 g ± 0,001 g

Calculating the partial pressure of the butane:

This is done by subtracting the vapour pressure of water from the measured atmospheric pressure. Values of the vapour pressure of water was found in Appendix 4.

Vapour pressure of water at 22 °C: 2,64 kPa (according to a table in “Chemistry Laboratory Manual”)

Measured atmospheric pressure: 1014,5 millibars.

1013 millibars = 101,3 kPa = 1 atm

1014,5 millibars = 101,45 kPa (= approximately 1 atm.)

101,45 kPa – 2,64 kPa = 98,81 kPa (This is an approximate value as our measurements of temperature and atmospheric pressure include uncertainties.)

Calculating the volume the butane occupy at S.T.P:

We use the equation:

(V1P1) / T1 = (V2P2) / T2

Temperature of the laboratory is assumed to be constant throughout the experiment.

We therefore use V1P1 = V2P2 and rearrange the formula to find V1:

V1 = (V2P2) / P1 and put in the values:

Measured volume 100 mL = 0,100 L (V2)

Standard pressure = 101,3 kPa (P1)

Measured pressure of butane = 98,81 kPa (P2)

V1 = (98,81 kPa ⋅ 0,100 L) / 101,3 kPa

V1 ≈ 0, 098 L

Calculating the number of moles of butane released at S.T.P.:

This is done by using the calculated value of the butane volume at S.T.P. and by knowing that the molar volume of the mass of butane at S.T.P. is 22,4 L.

Number of moles released at S.T.P. = Volume at S.T.P. / Molar volume at S.T.P.

Number of moles released at S.T.P. = 0, 098 L / 22, 4 L

Number of moles released at S.T.P. = 0,004375

Number of moles released at S.T.P. ≈ 0,004

Calculating the molar mass of butane:

This is done by using the measured mass. We use the formula:

Molar mass = mass / number of moles

M = m / n

mbutane = 0, 231 g ± 0,001 g

Number of moles of butane released at S.T.P. = 0,004375

M = 0,231 g / 0, 001375

M = 52, 8

CONCLUSION

We got a relatively good value for the molar mass of butane. By comparing to literature, we find that the molar mass of butane (C4H10) is 58,14. Our result was 52,8, which is quite good considering uncertainties.

Possible errors is air bubbles inside the glass tubing which might have affected the volume of butane calculated. I know that there was some air bubbles, as this was very difficult to avoid. We might have gotten a better result if we had used a stopper in the one end of the glass tubing so it would have been easier to avoid air bubbles.

Additional to this, there might also still have been some water left inside the lighter when reweighing it. We weighted it several times to get the best possible result. As we waited for some time, it was clearly shown on the weight that it had some water inside when we first weighted it. Our last result was better than the first one, but not perfect. Maybe it could have been an idea to allow the lighter to rest for about 24 hours to make sure it was perfectly dry.

The difference between our result and the literary result would mainly be caused by uncertainties.