Falling ball experiment

Objective

To investigate how the height a ball is thrown from affects the penetration distance in a flosorb mixture it reaches

Hypothesis

If the height the marble is thrown from increases, then the penetration distance in flosorb will increase as well because according to this formula, d=v^2/2μg  where v= velocity, μ= friction coefficient, and g= acceleration due to gravity”  (Formulas.tutorvista.com, 2014), the distance will be longer because the velocity will be faster (we know that the final velocity will be higher because since there is more distance to go through, it will have more time to accelerate as indicated by the following formula; "v= v0  + at , where v0= initial velocity, v=final velocity, a= acceleration and t= time” (Furey, 2014)) and the friction coefficient will stay the same because the substance is the same and so will gravity. 

Variables

Materials 

-Ruler

-Meter stick

-Sodium polyacrylate

-Plastic beaker 

-Marble

-Tap water

- Spoon

-Scale

Method

-Gather all the equipment

-Take a 250 ml beaker or bigger and fill it up with water.

-Mix 7 grams of sodium polyacrylate , until the mixture looks like jelly but enough transparent.

-With your ruler or meter stick, measure 50 cm, and put the marble at that height approximately

-Throw the ball from one of the heights

-Measure the penetration distance with a ruler. If you can't because the mixture isn't as transparent as it should, push the marble to the side of the beaker and measure it.

-Don’t forget to collect all your results. You need to do each measurement 5 times.

-Repeat this procedure changing the height to 50,100, 150, 200,  and 250 cm.

-Clean up and tidy up

Data process 

References

Formulas.tutorvista.com,. (2014). Stopping Distance Formula | Formula for Stopping Distance | Formulas@TutorVista.com. Retrieved 1 June 2014, from http://formulas.tutorvista.com/physics/stopping-distance-formula.html


Furey, E. (2014). Velocity as a function of Acceleration and Time. CalculatorSoup Online Calculator Resource. Retrieved 1 June 2014, from http://www.calculatorsoup.com/calculators/physics/velocity_a_t.php

Colligative properties (Vapour pressure and Raolt's Law)

Let's take two solutions to explain this.

The first solution is one of a pure liquid.  Let's keep in mind that the vapour pressure is caused by some molecules turning from liquid to gas.

On the other hand, the second solution is one made of water (solvent) and a solute that is not volatile, meaning that it will not turn into a gas, i.e. salt. If more space is taken by the salt molecules, then less water molecules can turn to gas. So vapour pressure is lower.

In other words, the more solute or the more concentration of the solution, the lower the vapour pressure because less molecules will be able to turn into a gas.

This is not only a rule though. Later on, François Marie Raolt discovered that the...

Vapour pressure of a solution (P)= Vapour pressure of the pure solvent (P_O) x Molar fraction of the pure solvent (x_O)

Solutions Lab session

Solutions lab session

During this session you will visit a number of areas of knowledge that you have used before – precision measurements, properties of substances, molarity, molality. You will also be introduced to ideas such as the conservation of mass and additive/non-additive volumes. As you will be measuring a number of different masses, you should try to use the same set of scales for each measurement.

Firstly, complete the blank spaces in the questions below. I have provided enough information to complete all answers.

Secondly, make a blog page for this practical that uses the information to discuss the following points (explain and give examples to support your discussion):

• Is mass always conserved?
• Is volume always conserved?
• What are molality and molarity?

Firstly, complete the blank spaces in the questions below. I have provided enough information to complete all answers.

Secondly, make a blog page for this practical that uses the information to discuss the following points (explain and give examples to support your discussion):

• Is mass always conserved?

Mass is not conserved in chemical reactions. The fundamental conservation law of the universe is the conservation of mass-energy. This means that the total mass and energy before a reaction in a closed system equals the total mass and energy after the reaction. According to Einstein’s famous equation, E = mc², mass can be transformed into energy and energy can be transformed into mass.

• Is volume always conserved?

Volume can be approximately conserved in systems that have approximately constant density fluids. These include gasses at almost constant temperature and pressure or most liquids with not very extreme pressure ranges and constant temperature (since many liquids are almost incompressible over small pressure ranges. But, in reality, everything is compressible and changes density with temperature changes, so the assumption of conservation of volume is always a simplifying approximation of reality.

• What are molality and molarity?

Molarity is defined as the number of moles of solute per liter of solution. This means that if you have a 1 M solution of some compound, evaporating one liter will cause one mole of the solute to precipitate.

Molality is defined as the number of moles of solute per kilogram of solvent. To make a 1 m solution, you'd take one mole of a substance and add it to 1 L of solvent. As a result, the final volume of a 1 m solution will be somewhat more than 1 L.

Notice the subtle difference between the definition of molarity and the definition of molality. The former

expresses the concentration of a solution as a ratio of solute to solution, while the other expresses the

concentration of a solution as a ratio of solute to solvent.

Molality = moles of solute per kilogram of solvent.

Molarity = moles of solute per liter of solution.

1.    Working out the volume of 2.5 g sodium chloride using cyclohexane.

a.     Measure 3 mL of cyclohexane with a pipette and pour it into a dry measuring cylinder.

Weigh the cylinder with the cyclohexane:                                ___73.50g__.

Weigh 2.50 g of sodium chloride and place it in the cylinder as well. 

Weigh the whole apparatus:                                                      ___76.00g_.

Does the total mass equal the masses of the different parts?   ____yes____.

A French scientist named Lavoisier stated that “matter cannot be created or destroyed, so mass is always conserved”. Does your data agree (approximately) with this statement?                                       ____yes____.

b.    Why does sodium chloride not dissolve in cyclohexane (Hint: which kind of substance are they – ionic, covalent (organic) or metallic)?

Cyclohexane is a very non polar solvent and doesn’t have strong enough dipole. And sodium chloride has a very strong dipole because it is an ionic compound.

Water can dissolve NaCl because oxygen has a strong dipole moment and can stabilize/coordinate the Na+ and Cl- ions.

As it does not dissolve, we can work out the volume of the salt by measuring the change in volume of the mixture:

What was the initial volume of cyclohexane?                                  3ml         .

What is the final volume (after adding the salt)?                        __4.5mL_.

What is the volume of the sodium chloride?                             _____1.5ml___.

2.    Is mass conserved when 2.5 g of salt is dissolved in water?

Weigh a clean, dry 25 mL measuring cylinder:                                __70.00g_.

Take 10 mL of water with a pipette and pour it in the cylinder.

Weigh it again, now with the water:                                                           __80.00g_.

            What is the mass of the water?                                                        ___10.00g_____.

           

What should the mass of water be per gram? (use the internet)      ___1g/ml_____.

Weigh 2.50 g of sodium chloride. Add it to the water and dissolve it.

Weigh the whole apparatus:                                                            __82.50g_.

            Does the total mass equal the masses of the different parts?        ____Yes____.

Is mass conserved?                                                                            ____Yes____.

What is the final volume of the solution?                                        __11.0mL__.

3.    Is volume ´additive´ (can we just add the individual volumes to get the final volume) when 2.5 g sodium chloride is dissolved in water?

What was the initial volume of water in part 2?                             __10mL_.

What volume should be taken up by the salt solution?                   ___12.5ml_____.

What is the actual final volume of your sodium chloride solution?            __11.mL__.

Is there a difference between your answer and what you predicted? Explain why there is or might be:

Yes it changes because it dissolves in water, and when it dissolves the volume changes.

4.    Work out the molarity and molality of the sodium chloride solution:

Molarity, M (mol/L) = number of moles of solute ÷ volume of solution (L)

Calculate the molarity of your sodium chloride solution (in water):

Volume= 11ml

Molarity/Concentration= 2.50 mole per liter.

Molarity:

2.5/58=0,043g/mol

0,043/0,011=3.9mol/L

Molarity in water:

12,5g/76=0,164g/mol

0.164/0.011=14.9mol/L

            Molality, m (mol/kg) = number of moles ÷ mass of solvent (kg)

            Calculate the molality of your sodium chloride solution (in water):

0.043/0.01=4.3mol/kg

5. Bonus questions

·         Why is it suggested to use the same set of scales for each measurement?

It is suggested to use the same set of scales for each measurement to compare the results and prove hypothesis.

·         What are “colligative properties”?

Colligative properties are properties that can only be measured for solutions and it depend on the ratio of the number of particles of solute and solvent in the solution, not the identity of the solute.

Bibliography:

• Chem.purdue.edu. 2014. Vapor Pressure. [online] Available at: http://www.chem.purdue.edu/gchelp/liquids/vpress.html [Accessed: 12 Mar 2014].
• Environmentalchemistry.com. 2014. Molarity, Molality and Normality (EnvironmentalChemistry.com). [online] Available at: http://environmentalchemistry.com/yogi/chemistry/MolarityMolalityNormality.html [Accessed: 12 Mar 2014].
• Princeton.edu. 2014. Vapor pressure. [online] Available at: http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Vapor_pressure.html [Accessed: 12 Mar 2014].
• Science.uwaterloo.ca. 2014. Intermolecular Forces. [online] Available at: http://www.science.uwaterloo.ca/~cchieh/cact/c123/intermol.html [Accessed: 12 Mar 2014].
• Answers.yahoo.com. (2014). Is volume conserved or can it be? - yahoo answers. [online] Retrieved from: http://answers.yahoo.com/question/index?qid=20090923032810AA2TAZW [Accessed: 11 Mar 2014].
• 
  
• Environmentalchemistry.com. (2014). Molarity, molality and normality (environmentalchemistry.com). [online] Retrieved from: http://environmentalchemistry.com/yogi/chemistry/MolarityMolalityNormality.html [Accessed: 11 Mar 2014].
• 
  
• Lightandmatter.com. (2014). Conservation of mass and energy. [online] Retrieved from: http://www.lightandmatter.com/html_books/7cp/ch01/ch01.html [Accessed: 11 Mar 2014].

Vapor pressure and intermolecular forces

Introduction

The vapor pressure of a liquid is the equilibrium pressure of a vapor above its liquid (or solid); that is, the pressure of the vapor resulting from evaporation of a liquid (or solid) above a sample of the liquid (or solid) in a closed container. The vapor pressure of a liquid varies with its temperature. As the temperature of a liquid or solid increases its vapor pressure also increases. 

Vapour pressure and intermolecular force:

When a solid or a liquid evaporates to a gas in a closed container, the molecules cannot escape.

Some of the gas molecules will eventually strike the condensed phase and condense back into it.

When the rate of condensation of the gas becomes equal to the rate of evaporation of the liquid or solid, the amount of gas, liquid and/or solid no longer changes.

The gas in the container is in equilibrium with the liquid or solid.

Some of the factors that affect vapor pressure are;

• Surface Area: the surface area of the solid or liquid in contact with the gas has no effect on the vapor pressure.

• Types of Molecules: the types of molecules that make up a solid or liquid determine its vapor pressure. If the intermolecular forces between molecules are:
• relatively strong, the vapor pressure will be relatively low.
• relatively weak, the vapor pressure will be relatively high.

Knowing this useful background information, we will investigate intermolecular forces further. We carried out an experiment where different compounds (each group investigated one compound) were put under different temperatures and the vapor pressure was measured. The results will be highly affected by the intermolecular forces of attraction the particles have. Let's see how this happens, but before let's state our objectives during this session. 

Objective 1. To improve practical skills – use of Schlenk tube, pressure sensors, vacuum line.

Objective 2. To investigate the structure and properties of one particular chemical.

Objective 3. To investigate the effect of temperature on vapour pressure.

Objective 4. To compare results with other groups (with other chemicals) and relate them to “intermolecular forces”.

Objective 1. These are some photos of the chemical we chose, Buthanol:

Objective 2. This video is an explanation on the molecule of buthanol:

Objective 3. Here is the graph that shows the relationship between pressure and time. Keep in mind that as time goes by, the temperature increases, the last bit of this function should be the second last, so we could see the relationship a bit clearer; this is because the temperature there would have been 25 °C, while the other two bits show what happens to the pressure when the temperature is changed to 15 °C and 35 °C. If we had plotted a graph showing pressure vs. temperature, then the order of the results would not have mattered. 

Graph of the data we obtained:

In this graph it is shown how the pressure changed with time but the reason why it changed is because of the change in temperature.When the temperature of the substance, buthanol, increased, the pressure did too because the particles got more energy to move around the container. As the particles had more energy, they collided more strongly and quickly against each other and against the walls of the container which lead to a higher pressure. Gay Lussac's law explains this relation between temperature and pressure, as this law says that temperature and pressure directly proportional. This means that as the temperature increases, the pressure does too.

Also, as buthanol was heated up, the particles had more energy to turn into a gas and more vapour pressure was created.

Improvements in the method:
As we can notice in the graph, at a certain point, we skipped one of the temperatures. When we realised that we skipped that temperature (25ºC), we went back and did it after heating buthanol 35ºC, so that is why on the right of the graph the pressure suddenly increases and at the end decreases because as we mentioned earlier, temperature and pressure are directly proportional. We should have followed the order of the temperatures to not have drastic increases of temperature, so this could be improved. 

Objective 4. We compared our results with other groups', these groups investigated different chemical compounds.

Before going through the results, let's recap the types of intermolecular forces we have studied.

Van der Waals. It happens because electrons move randomly, and at some point, there can be more electrons on one side of the molecule. Electrons from other molecules nearby will be repelled from this negative charge, and since opposite charges attract . It is the weakest of all intermolecular forces

Permanent dipole-dipole. Occurs between polar molecules, so it is stronger than Van der Waals.

Hydrogen bonding.  The theory behind it would be the same as a permanent dipole-dipole interaction but it is a bond that occurs between hydrogen and a FON particle; that is fluorine, oxygen and nitrogen. It is the strongest of all types of bonding.

Now, let's go one by one through the results we obtained.

In the case of pentane, it is the only molecule that has Van der Waals forces of attraction as it's one and only intermolecular force. We must recap that Van der Waals  (also known as London dispersion forces) is the weakest of all forces of attraction. So, even if pentane is a very big and long molecule, when compared to the rest of molecules, it has the lowest boiling point, as well as the highest vapor pressures.

Moving onto ethyl acetate, this molecule seems to have two types of intermolecular forces; Van der Waals and permanent dipole-dipole. The molecule is quite big and it has permanent dipole-dipole forces of attraction because oxygen is quite electronegative. However even if oxygen is one of the FON elements, since it is not bonded to hydrogen but carbon, then we can only call this a dipole-dipole interaction. As a result, the boiling point becomes higher when compared to pentane's and the vapor pressures become lower as well.

If we examine butyl acetate, we will first notice that it has the highest boiling point of them all. This is because it has two types of forces present in a very strong way. It has strong Van der Waals forces because it is a big and large molecule. Moreover, since oxygen is very electronegative, this makes the Van der Waals forces of attraction stronger as electrons will concentrate in that area. Furthermore, it also has strong permanent dipole-dipole because oxygen is very electronegative.

Moreover, the case of 1-Butanol is an interesting one. It has a relatively high boiling point, yet the vapor pressure is extremely low. This could be due to the arrangement of particles. The permanent dipole-dipole interaction in butyl acetate outweighs the occurring hydrogen bonding in 1-Butanol, therefore causing a higher boiling point in butyl acetate.

Last but not least, propyl acetate. It's got Van der Waals forces of attraction and permanent dipole-dipole interaction. The vapor pressures are relatively low but not lower than butyl acetate's. This means that the attraction is weaker, also causing a lower boiling point than butyl acetate. The permanent dipole-dipole would be quite as strong one as oxygen is very electronegative.

Volume vs. Pressure experiment

Volume vs. Pressure experiment

1)    Tables

Volume (mL)	Pressure (hPa)
65	0
64	20
63	40
62	50
61	70
60	85
59	110
58	140
57	150
56	170
55	200
54	225
53	250
52	275
51	300
50	330
49	350
48	375
47	420
46	450
45	490
44	510
43	550
42	590
41	625
40	675
39	710
38	770
37	830
36	855
35	900
34	955
33	1000
32	1010
31	1150
30	1220
29	1285
28	1425
27	1500
26	1560
25	1720
24	1750
23	1930
22	2030
21	2160
20	2325

Volume (mL)	Pressure (hPa)
0.02	0.00
0.02	20.00
0.02	40.00
0.02	50.00
0.02	70.00
0.02	85.00
0.02	110.00
0.02	140.00
0.02	150.00
0.02	170.00
0.02	200.00
0.02	225.00
0.02	250.00
0.02	275.00
0.02	300.00
0.02	330.00
0.02	350.00
0.02	375.00
0.02	420.00
0.02	450.00
0.02	490.00
0.02	510.00
0.02	550.00
0.02	590.00
0.02	625.00
0.02	675.00
0.02	710.00
0.02	770.00
0.02	830.00
0.02	855.00
0.02	900.00
0.02	955.00
0.02	1000.00
0.02	1010.00
0.02	1150.00
0.02	1220.00
0.02	1285.00
0.02	1425.00
0.02	1500.00
0.02	1560.00
0.03	1720.00
0.03	1750.00
0.03	1930.00
0.03	2030.00
0.03	2160.00
0.03	2325.00

2) Graphs

3) Conclusion

While it’s true that our trend lines don’t really make sense in the context of the problem; one of them is a polynomial, we can agree that we can obtain some valuable conclusions by looking at our graphs. The first one somewhat shows us an exponential function and although, the second one does too, we can assume that our data points aren’t extremely reliable and that this function could be considered linear, hence the linear trend line. 

In other words, we see that as the volume increases, the pressure of the gas-air decreases proportionally. In fact, volume and pressure are inversely proportional.

This leads us to Boyle’s law, which says:

This equation proves the exact same theory as the conclusion above.