Experiment n6: Pressure of alcohol vapours

Aim:

To determine the pressure exerted by heptanol vapours, and compare it to the one in different alcohols.

Background information:

Vapour pressure is the pressure exerted by a vapour in equilibrium with the solid or liquid phase of the same substance.
In a closed container, the vapour pressure is the vapour that is exerted by the evaporation of a liquid above a sample of that same liquid. The pressure of the vapour will vary on the temperature that is found in.

The vapor pressure of a liquid depends on the molecular structure of the liquid and on the temperature. Increasing the temperature increases the average kinetic energy of the liquid molecules, and the fraction of molecules get the sufficient energy to escape from the liquid phase. So, the vapor pressure increases with temperature.

We notice that vapour pressure was produced by the gas of the alcohol. Also, the vapour pressure depends on the temperature that it is found in. For example, if we had an alcohol that was found in a container in room temperature, the vapour pressure released will depend on the 27ºC of the room (or 300ºK) when using the ideal gas equation to calculate de pressure of the molecule:

PV = nRT

Alcohols are substances containing an OH group attached to a hydrocarbon group.

Alcohols have similar characteristics since they are composed of the same atoms: Carbon, Hydrogen and Oxygen. The alcohols we have worked with, always follow a same pattern: They start with a CH3, end with an OH and are followed with the amount of Carbons in the middle that will determine its name.

Meth - 1 Carbon

Eth - 2 Carbons

 Prop - 3 Carbons

 But - 4 Carbons

Pent - 5 Carbons

Hex - 6 Carbons

Hept - 7 Carbons

Oct - 8 Carbons

Materials:

- Heptanol

- Schlenk tube

- Stopper

- Elastic band

- Vaseline

- Vacuum line (in the image)

- Pressure sensor

- Laptop

- Logger Pro software

Method:

1- Set up the equipment with all its components. The schlenk tube should be arranged with rubber bands as shown in the picture.

2- Pour the heptanol inside the schlenk tube (more or less 20 mL) and close it using a rubber tube. Connect the schlenk tube to the vacuum schlenk line.

3- The stopcock allows a vacuum to enter the apparatus. Turn it. You will observe that after some time it will start to boil since the pressure is lowered and the temperature increases. 

4- Connect the pressure sensor to the schlenk tube by attaching the small insertion cable to one of the exits of the schlenk tube.

5- Take down the results of the pressure released on the logger pro software and draw it on a table.

6- Make a graph the results of the pressures released from the other alcohols from other classmates. 

Results (from our classmates):

The pressure exerted by octanol's vapour was of 2,04 kPa.

The pressure exerted by heptanol's vapour was of 2,02 kPa.

The pressure exerted by hexanol's vapour was of 4,27 kPa.

The pressure exerted by pentanol's vapour was of 4,10 kPa.

The pressure exerted by butanol's vapour was of 5,25 kPa.

The pressure exerted by propanol's vapour was of 9,80 kPa.

The pressure exerted by ethanol's vapour was of 11,7 kPa.

The pressure exerted by methanol's vapour was of 16,90 kPa.

Calculations:

In order to elaborate a table and a graph we decided to show the relationship between the pressures of the different alcohols and their number of carbons. We calculated the inverse on the pressures due to be able to analyze a linear function and not a polynomial one.

Methanol: Carbon 1 = 1/2,02 kPa = 0,495049504505
Ethanol: Carbons 2 = 1/11,7 kPa = 0,08547008547009

Propanol: Carbons 3 = 1/4,10 kPa = 0,24390243902439
Butanol: Carbons 4 = 1/5,25 kPa = 0,19047619047619

Pentanol: Carbons 5 = 1/9,80 kPa = 0,10204081632653

Hexanol: Carbons 6 = 1/4,27 kPa = 0,21186440677966

Heptanol: Carbons 7 = 1/2,04 kPa = 0,49019607843137

Octanol: Carbons 8 = 1/16,90 kPa = 0,05917159763314

Table and graph: Showing the relationship between the pressure exerted by vapours of different alcohols and their numbers of carbons.

In this graph we chose not to put a straight best fitting line, as we thought a curved line could show in a better way this type of growth and proportion of this data, as from it we can deduce that the number of carbon atoms is indirectly proportional to the pressure of that compound.

Table and graph: Showing the relationship between the inverse of the pressure exerted by vapours of different alcohols and their numbers of carbons.

The best fitting linear used was linear because a straight linear could express better the increase on this type of function. 

Conclusions from the data:

- We notice that the pressure changes with the type of alcohol.

- All alcohols contains the same elements, carbon, oxygen and hydrogen.

- The shape of the molecule could vary in many different ways (shown in pictures below).

- So we got the conclusion that the bigger the molecule, the smaller the pressure, observing the results we could see how when the molecular weight of alcohol increase, the pressure exerted by the gas decreases,

Curro´s video:

Conclusions:

We have reached to a conclusion, which is that the bigger the molecule, the lower the pressure, but why does this occur:

Paco´s video:

Jaime´s video:

Structure of heptanol:

To represent alcohols we decide to used color balls representing a heptanol molecule, the red one is oxygen, the white one is hydrogen and the black one is carbon.

The following images show the representation of the different shapes that an heptanol molecule can assume:

We could observe that a heptanol molecule could adopt many different shapes, and observing how the other molecules were we notice that heptanol could adopt much more shapes than the other alcohols except octanol, hence, the bigger the molecular mass, the opportunity of the molecule of adopting more different positions.

From observing other classmates' results with other alcohols we find that depending on the amount of carbon atoms that an alcohol possesses, the more or less shapes it can take up. From the images we can observe that the bonds between the carbons and the oxygen are the ones that provide the molecule a bending point, meaning that the molecule would be able to bend in that point, for example octanol, that has 8 carbons, is going to be able to bend in 8 points while methanol will only be able to bend in one point, as it has only one carbon (heptanol is able to bend in 7 points, as it has 7 carbons)

 Bibliography:

Chem.purdue.edu (n.d.). Vapor Pressure. [online] Retrieved from: http://www.chem.purdue.edu/gchelp/liquids/vpress.html [Accessed: 8 Jun 2013].

Sal.M (2011) Vapour pressures/states of matter. Retrieved from: https://www.khanacademy.org/science/chemistry/states-of-matter/v/vapor-pressure 

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