Saturday, 9 March 2013

Sixth Lab Session - Propanol

Objective: 
To determine and measure how the different types of alcohols and molecules arrangements (size and number of Carbon atoms) affect their vapour pressure. 

Background information:
The vapor pressure of a liquid is the equilibrium pressure of a vapor above its liquid; that is, the pressure of the vapor resulting from evaporation of a liquid above a sample of the liquid  in a closed container. (This can be also applied in solids). Examples:

Substance
Vapor pressure at 25oC
diethyl ether
0.7 atm
bromine
0.3 atm
ethyl alcohol
0.08 atm
water
0.03 atm

The vapor pressure of a liquid varies with its temperature, as the following graph shows for water. The line on the graph shows the boiling temperature for water.

Graph of vapor pressure vs temperature for water
As the temperature of a liquid or solid increases its vapor pressure also increases. Conversely, vapor pressure decreases as the temperature decreases.
(MERCIS, Elissa (2010))

During this lab session, we are going to work with Propanol:

"Propanol is a colourless, mobile, neutral liquid of average volatility and has a characteristic alcoholic smell. It  is freely miscible with water." Furthermore, its molecular weight is 60.095.

Propanol has many applications
n-Propanol is used as a solvent and an intermediate. It shows less tendency to absorb water than lower alcohols, and has a considerably milder and more pleasant odour than higher alcohols.

In the coatings industry, n-Propanol is a useful medium-volatility alcohol for improving the drying characteristics, e.g. of alkyd resins, electrodeposition paints and baking finishes.

It is added to cleaners, floor polishes and metal degreasing fluids and is used as an additional solvents in the manufacture of adhesives.
n-Propanol is also present in de-icing fluids and is used as extractant and entrainer in azeotropic distillation.
n-Propanol is a feedstock in the manufacture of insecticides, herbicides and pharmaceuticals. It reacts with acetic acid to form propyl acetate, with ammonia to form propylamines and with halogens to form the corresponding propyl halides

(BASF, The Chemical Company. (5-3-2008))





Materials:


                                   
            Propanol                                                                       Schlenk tube

                                                                                       Vacuum line

                                  

                        2 rubber bands                                                              Cover



                                                
                               1 laptop                                                                         Logger Pro


                            

                          Stand                                                                          Clamp

                                                    

                                      Vaseline                                                            Gas pressure Sensor

                
                                 Rubber Tube

If you want to carry out this experiment, you must install "Logger Pro" in you computers, using this website: 


Method:
1.-Take the Schlenk tube and attach it to the stand using a clamp.


2.-Fill the Schlenk tube with 20 mL propanol (or any other alcohol you can find below).

3.-Spread vaseline in the upper part of the Schlenk tube. This will prevent our materials sticking together while creating the vacuum.  


                                                                                                                     

4.-Place the cover of the Schlenk tube on top of it (where you have already spread the vaseline). Use two rubber bands in order to ensure it does not move nor fall.
                                                   
                                                            
5.-Connect the rubber tube to the Schlenk tube so you can connect the gas pressure sensor to your laptop.



6.- Go to the vacuum line and create a vacuum (the tap should look upwards). Some propanol will start boiling, don't worry.

                                                          

7.-Collect the data and wait until the pressure collected reaches an equilibrium.



Results and calculations:
Once we finished our experiment, we waited until the pressure remained constant, and we obtained that the pressure of propanol was 9,8 kPa. We then shared the data with our classmates, and we created a table:

Table 1: Table showing the pressure in kPa of the different alcohols we have been working with

Alcohol
Pressure (kPa)
Propanol
9,8
Methanol
16,9
Hexanol
4,27
Butanol
5,25
Pentanol
4,10
Ethanol
11,7
Octanol
2,04
Heptanol
2,02

As we can see, there is an inverse proportion between the pressure and the number of carbons. For instance, the higher the number of carbons, the lower the pressure measured. 

Graph 1: Graph showing how the vapour pressure changes with the number of carbons the alcohol has.

In the graph above we can observe how the vapout pressure decreases at the same time the number of carbons increase. In other words, the pressure is inversely proportional to the number of carbons the alcohol contains. We can determine that is a logaritmic function because is the equation that is better adjusted to the points obtained. The co-relation coefficient is 0,9764,  this means that the results are not perfect. However is a number very near to 1 (which would mean that the results are perfect) so we can say that it has been an almos perfect experiment. 

Graph 2: Graph showing the inverse of the data we have obtained in this experiment (number of carbons against pressure) 
Comments about the graph:
In the inverse of the first graph (second graph) we have included the best fitting line (slope), which is a straight line (constant slope).
We can see that the co-relation coefficient (R2), which indicates us how well the line fits the points, is not as good as possible (0,8801). If this coefficient approaches to 1, this means that the experiment has been very well done, with very little mistakes.



Conclusions:








(In 00:29 and so on, it means a lower boiling point, not a higher one)






Bibliography:
BASF, The Chemical Company. (5-3-2008) n-Propanol. Taken from: http://www.solvents.basf.com/portal/load/fid228771/n_Propanol_e_03_08.pdf

MERCIS, Elissa (2010). Vapour Pressure. Taken from: http://www.chem.purdue.edu/gchelp/liquids/vpress.html

Organization:
Objective: Mondaza
Background: Eugenia, Maite
Materials: Maite, Mondaza
Method: Mondaza
Results and calculations: Fco. Javier, Maite, Mondaza
Conclusions: Luis, Maite, Eugenia and Mondaza.

Tuesday, 12 February 2013

Fifth Lab Session - Gas law apparatus


05/02/2013

Objective: To determine how the volume of the gas affects its pressure.

Background information:
A gas law apparatus, as its name indicates, is used to study the different laws and properties of gases, such as pressure or volume and to demonstrate their relationships. It is composed of the thermometer (to indicate the temperature of the gas while changing pressure and volume), the piston with the cylinder (to change the variables), the pressure indicator and the valve that allows the compressed air to escape. (FLINN, 2011)

It has an easy mechanism that makes you able to test the relationship between pressure and volume in gases. We have to move the piston up or down so that the pressure decreases (moving it from the lower part to he top of the cylinder) or increases (compressing the air from the upper part to the lower one)


There is a directly proportional relationship between volume (V) and temperature (T), (Charle's-Gay Lussac's Law), which says that when volume increases, temperature increases. 
Even though we are not going to focus on the temperature, we have noticed that the temperature changes very little while the volume of the gas changes.




Procedure:

Spin around the handle, so when it goes downwards, the volume of the gas decreases; and when it goes upwards, the volume increases.
In order to take notes of the results, move the handle up or downwards, so it reaches the volume indicated on the tables below.

Table 1:  Table showing the relationship between the volume of air and its pressure


We have been measuring pressure above atmospheric pressure (extra pressure).

In order to calculate the atmospheric pressure (1 atm, which is equal to 101325 Pa) in hectoPascals, we have done a conversion factor: 

Then, we added this value to the pressure we obtained in the experiment.

In the fourth column we can see the volume we obtained, and in the third, the volume obtained added to the atmospheric pressure.


Graph 1 (a): Graph showing how the  air pressure decreases while the volume increases

Pressure (hPa) 
Volume (mL) 


Graph 1 (b): Graph showing the inverse of the data we have obtained in this experiment (Volume against Pressure)
   Pressure (hPa)
                                                                               Volume (mL)


Table 2: Table showing the pressure and inverse pressure of air in hectoPascals obtained while we increase the volume of it.



Graph 2 (a): Graph showing how the pressure decreases while the volume of air increases. 

Pressure (hPa)
Volume (mL)

Graph 2 (b): Graph showing  how inverse pressure of the data obtained above increases at the same time the volume increases.

Pressure (hPa)
Volume (mL)





Conclusions


As we  can see in the video, tables and graphs above, the pressure increases while the volume decreases. This experiment is just a demonstration of “Boyle’s Law”.
We’ve concluded that the volume of the gas and its pressure are inversely proportional.
We have found this experiment very interesting because we have understood this law very quickly, without any difficulties and in a practice way.


Bibliography

FLINN, S. (2011): The Gas Law apparatus. Taken from:
http://www.flinnsci.com/store/Scripts/prodView.asp?idproduct=22163















Tuesday, 5 February 2013

Fourth Lab Session - How to make a battery with redox couples

Mª Eugenia Fernández de los Ronderos, Maite Pérez Lagares and Luis Galán Ruiz
05/02/2013


Objective: To build a battery (energy emitter) with two redox couples.

Materials:

                       
     - U-tube                                      - Multimeter                                        - lead (II) nitrate

              
    - magnesium                              - magnesium sulfate (aqueous solution)                        -lead

- 2 beakers                                     -Sodium Chloride                                      -Water 

                                                         
                                                                          - Cotton

Background information:
Redox reactions, or oxidation-reduction reactions, primarily involve the transfer of electrons between two chemical species. 
The compound that loses electrons is said to be oxidized, the one that gains electrons is said to be reduced.
There are also specific terms that describe the specific chemial species. 
A compound that is oxidized is referred to as a reducting agent, while a compound that is reduced is referred to as the oxidizing agent. (UNIVERSITY OF NORTH CAROLINA, 2012) 

Due this phenomenom, the redox couples we use today, can generate electricity.

Procedure:

1.- Fill up the U tube with NaCl and water, and put cotton in both ends, so that the solution does not fall out.            

                                                     

2.- Pour aproximately 10 mL  of lead (II) nitrate inside of a beaker.
3.- Pour aproximately 10 mL of magnesium sulfate in the other beaker.
4.- Make sure the multimeter's cables are in the following position:




5.- Put a long piece of magnesium inside the beaker containing magnesium sulfate.
6.- Put the longest fragment of lead (Pb) inside the other beaker containing lead (II) nitrate.
7.- Pin the piece of magnesium which is inside the beaker with the black clamp of the multimeter.
8.- Pin the fragment of lead with the red clamp.
9.- Set the "U tube" upside down, so that each of the ends of the tube get into both beakers, in order to conduct electricity.
10.- Turn on the multimeter to the "2000 millivolts" function.
11.- Collect the data in your notebook.





Results and Conclusions

As we can see in the image below, the energy emitted is around 1259 mV (milli-Volts).

                                                             

The redox couples used are the following:


Pb2+/Pb
Its reduction potential is +1,693 V

Mg2+/Mg
Its reduction potential is -2,363 V

As we know electrons move from left to right, the same way we write it. lead has more tendency to reduce because its reduction potencial is higher than magnesium. (+1,693 > -2,363)

The Daniell cell will remain:
Mg/Mg2+// Pb2+/Pb

So when we have these redox couples, the Daniell cell produces:

 +1,693 + 2,363 = 4,056 V.

While the potencial difference is:
-(2,363) + 1,693 = - 0,67 V

The symbol has no importance, because what we want to know is the potential difference; and the symbol what indicates us is the direction of the electrons. So, the potential difference is 0,67 V. 








References:
University of North Carolina (2012). Redox Reactions. Taken from: 


  

Sunday, 3 February 2013

Third Lab Sesion - Redox titration of hydrogen peroxide and potassium permanganate

Eugenia Fernández de los Ronderos Jiménez and Fco Javier Mondaza Hernández 


Objective: To determine the volume of KMnO4 needed for 4mL of H2SO4 and the concentration of hydrogen peroxide.

Background:
Titration is a method of analysis which allow us to determine the precise endpoint of a reaction, and therefore, the precise quantity of reactant in the titration flask.
A burette is used to deliver the second reactant to the flask and an indicator or pH Meter is used to detect the endpoint of the reaction. (DARMOUTH COLLEGE, 2012)

Materials

-Stand
-Erlenmeyer flask
-Burette
-Clamp stand
-Beaker
-Pippete 
-Clamp
-KMnO (0,05 M)
H2SO(4 M)
-100 mL volumetric flask
-Comercial oxygenated water (H2O2)



                                           KMnO4                                   Beaker, Pippete and Erlenmeyer flask.
 Procedure:
1. Set up the material. (picture).
2. Fill the burette with KMnO4.




3. Adjust the level of the burette to 0.
4. Take 4 mL of  H2Owith the pippete and add it to the Erlenmeyer flask.
5. Add 4 mL (we do not need to be accurate) of the H2SO to the Erlenmeyer flask.
6. Titrate the redox solution, until no reaction is taking place. For this, you have to open the tap carefully and see how the "KMnO" behaves when it is in contact with the solution.

Image 10
Final look of the solution.

Results







Our results were the following one:

4 mL H2SO4 + 4 mL H2O2 + 29,2 mL KMnO4 

And the reaction observed is this one:

KMnO4H2SO4 + H2O2 --> MnSO4 + O2   H2O + K2SO4

Carrying out a simple balancing redox reaction, we can balance it in order to continue with our calculations:


2KMnO4 +3 H2SO4 + 5H2O2 --> 2MnSO4 + 5O2 + 8H2O + K2SO4


-Therefore, we can calculate the moles of KMnO4:

M = moles/ volume in L. // 0.05 = x/ 0.0292 //0.00146 moles KMnO4

-Taking into account the ratio between 
KMnOand H2O2, we calculate the moles of the last one:

 5 H2O2 / KMnO4   --> (0.00146 *5) :2 = moles H2O2 // 0.0073 : 2 = moles H2O//

0.00365 = moles H2O2

-Therefore, we can now calculate the molarity of  H2O2:

M = moles /volume in L // M = 0.00365 / 0,004 L // M  H2O2  = 0.9125

-According to the theoretical relation between H2Oand the O2 produced (1 volume  H2Oproduces 10 volumes of  O2) we carry out the following reasoning:

 H2O2 = n O2

V O2 (Standard conditions for gases) = 3,65 m moles * 22.4 ml/mol =  81,76 mL O2

-Finally, we calculate if the mL of O2 produced were the expected ones:

Expected mL O2: 4mL H2O2 * 10 = 40 mL

% error: (Experimental value - Theoretical value) / Theoretical value *100 = (81,76 - 40) / 40  *100= 41,76 / 40 *100 = 104, 4%


Conclusions:
In this experiment we have observed that there is always a relation between the reactants of a chemical reaction. Due to this, relations, we can calculate  different characteristics of the compounds using easy formulas.
Also, we have learned how to carry out a chemical procedure; tritation. As we have seen, this process consists of determining the precise quantity of a reactant needed to carry out a determined chemical reaction. We have learned this process could be really useful to calculate extra-characteristics of the reactants we are using.
Besides these points, we have calculated the molarity of H2O2 (0.9125 M) and we have been able to check if the volume of O2 was the correct one according to the commercial acid relation. We found out it was not (the percentage error was more than 100%), so we conclude that the hydrogen peroxide we were using was not a commercial one (so its concentration was different), or that there was not standard conditions for the O2 produced (so the results were altered).


Bibliography:

DARMOUTH COLLEGE (2012) : Titration Technique. Taken from: