September 29: Finishing Biodiesel lab, doing pressure-volume experiment and starting pressure-temperature experiment

Today I was involved in three experiments, firstly was finishing off the Biodiesel lab.

After previously completing the process of adding the KOH, sunflower oil and methanol into a cylinder and then stirring it with the stir plate my group now had to separate it. This process involved pouring the substance into a separating funnel and then removing the excess Glycerol from the Biodiesiel into a Juicy Water bottle. After leaving the substances in the funnel for 10 minutes, the density of the Glycerol and Biodiesiel had caused them to separate into different mixtures. Hence we were able to separate them with ease. Then, upon weighing the bottle and cap with and without the fuel, it was possible to calculate the percentage yield. Since our theoretical yield had been 111.52g (0.127*3=0.381 moles*292=111.252g) and the actual value had been 105.03g (136.88g-31.85g=105.03) we had a percentage value of 94.41%, not too shabby.

The second experiment involved the pressure-volume experiment, which consisted of observing Boyle's law. The procedure was attaching a plastic gasyringe to a pressure sensor and then observing varying volumes of gas in the syringe and how the pressure corresponded on a LoggerPro program. After experimenting with 5, 10, 15, and 20 ml of air, a pattern occurred that the smaller the volume in ml, the higher the pressure in kPa.

Finally, my last experiment involved the observing of the relationship between pressure and temperature. The procedure meant plugging a 125ml Erlenmeyer flask with a Vernier Gas Pressure Sensor and putting in varying degrees of baths to see the effects. The baths temperature in Celsius varied from a nearly boiling bath, to a warm water bath, to a room temperature bath, and an ice water bath. Although I am yet to finish the experiment, it is likely that the hotter the bath, the higher the pressure in kPa. However, this will be confirmed tomorrow to make sure that this hypothesis.

September 25: Biodiesel and Stoichiometry Lab

Biodiesel and Stoichiometry Lab

Today, we conduced another lab about finding the percentage yield of biodiesel in the lab "Biodiesel and Stoichiometry Lab", in Section 1.31.

Equation used: Glycende (veg. oil) + 3CH3OH = Biodiesel + Glycerol

Purpose of Lab: Students will learn about the process of creating biodiesel. Biodiesel is created from fat or oil. This fat or oil can sometimes be feedstock fat or oil. It is reacted with an alcohol, in our case methanol, with a catalyst (sodium hydroxide).  Biodiesel can be used to as fuel for cars. Also the Glycerol can be used to create soap. 

Today in class most students did all two steps in the lab. First was the reaction, second was the separation. 

Procedure Part 1: 

1. Measure out 25 (we halved the amounts) ml of methanol (3CH3OH).
2. Weight 1.12 grams of KOH (catalyst).
3. Combine methanol and KOH, stir until the KOH dissolves. You know have potassium methoxide.
4. Measure out 125 ml of vegetable oil and add to methodixe mixture. 
5. Heat mixture to in-between 55-60 Celsius. Stir on plate, however, keep it constantly around 55-60 Celsius. Do this for around 45 minutes. 
6. After 45 minutes pour into separation funnel and let it sit for 10-15 minutes. 

Procedure Part 2: 

1. Slowly drain the bottom layer into a small beaker. This is the layer of glycerol. When all glycerol is done stop the plug. 
2. Drain the rest of the biodiesel and measure the mass. 

Measuring the masses of the starting materials: 

1. Mass = Density x Volume
2. Mass of KOH  
1. 1.11 (±) 0.1 g
3. Mass of methanol  (3CH3OH) 
1. 0.79 (density) x 25 (volume) = 19.75 g
4. Mass of oil 
1.  0.9 (density) x 125 (volume) = 112.5 g

Finding the Limiting Reactant: 

1. Mole of the Oil is .127 g 
1. mass/ molar mass = mole
2. 112.5/ 885.4 (given) 
3. .127 
2. Mole of 3CH3OH is .616
1. Mass/ molar mass = mole
2. 19.75/ 32.05
3. .616
3. The ratio is 1:3 
1. times .127 by 3 you get .39 
2. You have more then .39 mole of 3CH3OH 
3. So 3CH3OH is in excess and the oil is the limiting factor 

Finding the theoretical yield of the biodiesel:

1. How much of the product will be created in ideal conditions
2. Multiply the amount of moles of the limiting reagent by the ration of the limiting regent 
1. .127  x 3 
2. .39
3. Multiple that by the molar mass.
1. 0.39  = mass/ 292 
1. 292 is the molar mass of the biodiesel 
2. theoretical yield= 113.8

Finding the actual yield of biodiesel:

1. How much of the product will be created 
2. 104.85 g 
1. Difference from others is we spilled some of our biodiesel so we will have less. 

Finding the Percentage Yield: 

Actual Yield/Theoretical Yield x 100 = Percentage Yield

104.85/113.8 x 100 = 92.13%

Homework: Finish lab and other questions. Quiz next class about topic 1.2.

September 23rd : Determining a Percentage Yield

Determining a Percentage Yield

Today, during this flipped unit of stoichiometry, the class worked on the experiment "Determining a Percentage Yield", in Section 1.31.

Equation used: CuSO₄ (aq) + Fe (s) → FeSO₄ (aq) + Cu (s)

Purpose of Lab: By using the equation above, the purpose of the lab was to find the percentage yield by calculating the potential mass of copper that could be created in relation to the actual mass that was created. The lab therefore had two parts - one in the lab, finding the actual yield of copper and one using our previous knowledge on Reacting Masses and limiting reagents to discover the theoretical yield.

Finding the Actual Yield of Copper:

1. An empty beaker was weighed and recorded
2. 6.0g of copper sulphate pentahydrate was placed in the beaker
3. 50ml of distilled water was added to this beaker
4. The beaker was heated to dissolve the crystals using a Bunsen burner
5. When the beaker was no longer being heated, 1.0g of iron filings were added while stirring
6. The beaker needed to cool, which gave time for calculations to determine the theoretical yield
7. The beaker was then drained from its liquid, keeping the newly made copper at the bottom 
8. It was then put in the oven to remove as much water as possible 
9. The beaker containing the copper was weighed

Finding the Theoretical Yield of Copper:

1.              1.    The number of moles of copper sulphate and iron were determined

     2.       From the equation and this calculation, we were able to establish the limiting reagent.   

[Example of Problem using Limiting Reagents]

       3. Working out the mass of copper (see steps 1-3 for reacting masses help)

Finding the Percentage Yield: 

Actual Yield/Theoretical Yield x 100 = Percentage Yield

Uncertainties and Errors:

Many people, including myself, ended up with a percentage yield higher than 100%. This is most likely because of the remaining water that did not have time to fully evaporate; this error is mostly caused by the limited time we have in class to do these labs as accurately as we could. 

Homework: Finish calculations and extra problems on limiting reagents on the back of the lab. 

18/09/14

September The 18 - Flipped Class

The lesson was another flipped class in which the students had time to settle down and do which ever chemistry work they needed. For example, some people:

1. Prepared for Quiz 1.2
2. Had independent questions for the teacher
3. Some people started work on Kerboodle  (http://www.kerboodle.com/users/login)

Mr. Carmichael also wanted the students to hand in their Hydrate lab.

With the free time to work in class, students either worked on the next lab that would be due (MgO Empirical Formula). We also received back our old chapter 1.1 quizzes and had any questions answered.

The Magnesium Oxide Empirical Formula experiment was an experiment in which you react magnesium with oxygen to form magnesium oxide. You did so by burning a piece of pure Mg in a crucible and weighing before and after the experiment. This way you could find out the ratio of O:Mg and hence write your Empirical Formula (which is the simplest whole number ratio).

September 16: The mole concept

Before continuing with the topic 1.2 – The mole concept we:

Took a quiz which included:

v  Balancing equations

v  Writing Formulae

v  Phase Changes

All of these topics were covered in 1.1 - Stoichiometry: Particle Nature of Matter & Chemical Change (see previous entries to review these topics).

In class, we continued to work on moles and applied what we have learned so far in our labs/experiments.

First, we finished the lab on finding the formula of a hydrate:

The purpose of the lab was to find the hydrate’s empirical formula.

Hydrates are inorganic salts which contain a specific number of water molecules loosely attached. Examples include:

Magnesium sulfate heptahydrate (epsom salts) & Sodium carbonate decahydrate (washing soda)

MgSO₄ x 7H₂O                                                          Na₂CO₃ x 10H₂O

However, *when finding the formula of a hydrate you should not assume that it is one with a simple formula*

Hydrates can usually be decomposed into the anhydrous (without water) salt and water by gentle heating à this was done in the experiment using the salt CuSO₄ x XH₂O

The data collected during the experiment was used to find the number of moles of anhydrous solid and the number of moles of water removed (expressed as a ratio). Then, the value of X was identified.

Equation used to find the number of moles = mass/molar mass (mass of one mole)

The actual value was CuSO₄ x 5H₂O. Data may not show the actual/literal value due to possible sources of error. This experiment could have been improved by:

v  Re-heating the hydrate (CuSO₄) more than once to ensure a constant mass is attained (all the water has evaporated)

v  Having more than one trial for the experiment, etc.

We also started another lab on working out the empirical formula (of MgO).

In this experiment we reacted magnesium (Mg) with oxygen (O₂) in the air to form magnesium oxide. This was done by strongly heating magnesium ribbon in a crucible until the magnesium ignited and a white powder was left. Our task was to work out the empirical formula (simplest whole number ratio of atoms) for magnesium oxide (similar to the previous experiment on finding the formula of a hydrate).

                      

Homework: 

Finish the data processing & conclusion on the finding the formula of a hydrate lab 

The molar concept.

We worked more on moles to better our understanding on this very fascinating subject.
A mole is 6.02 x 1023. If you're having trouble remembering this then there is great song that can help you with it, the link will be at the bottom. 

          Ar and Mr        

Ar is the relative atomic mass, and this simply means the average mass of one type of an element to one twelfth of the mass of an atom of carbon-12 and this has no units. For e.g, the relative atomic mass for oxygen is 16. 

Mr is the relative molecular mass, is the weight of all the atoms of a molecule added together. This can be calculated. 

Find the molecular mass of HCL.

H- 1

Cl- 35.5

35.5 + 1 = 36.5. therefore the molecular mass of HCL is 36.5

Percentage compositions. 

The percentage composition indicates how much of an element there is in a compound. This can be done by dividing the relative atomic mass of the element and dividing it by relative molecular mass. 

Which fertilizer, urea(COCNH2)2) or ammonium nitrate (NH4NO3) contains the most nitrogen? 

(14x2)/ (12+16+32) x 100 = 46% or (14x2)/(28+4+48) x 100 = 35% 

therefore the answer is Urea. 

Empirical and molecular formulas. 

The empirical formula of a compound is the simplest ratio of the atoms it contains. It must still be stated in whole numbers though. For e.g

  what is the empirical formula of C3H6?

 the empirical formula would be CH2

The molecular formula on the other hand is the same as or a multiple of the empirical formula, and is based on the actual number of each type in the compound. To find the molecular formula, divide the mass of the element by the atomic mass. 

"2.3g of sodium burn't in air to form 3.1g of an oxide, what is the formula of the oxide?"

2.3/23 = 0.1

(3.1-2.3 = 0.8)

0.8/16 = 0.05

2:1 therefore Na2O

To put what we learnt into application, we did a lab on finding the formula of a hydrate.

Today, we learned about the empirical formula and the molecular formula and how to find the molecular formula. We also learnt how to calculate percentage compositions and the difference between relative atomic formula and relative molecular formula. 

Mole song: http://youtu.be/PvT51M0ek5c

pictures: http://www.revisionworld.com/sites/revisionworld.com/files/rw_files/atomicmass.jpg

http://antoine.frostburg.edu/chem/senese/101/moles/slides/img009.GIF

http://img.docstoccdn.com/thumb/orig/78182635.png

10/09/14- Introduction to moles

In today’s class, we worked on the introduction to moles. A mole is defined as 6.02 x 10²³ (Avogadro's number) particles [Avogadro's number corresponds to the number of atoms as there are in 12g of 12C]. These particles are the smallest units in the sample and could be made of atoms (sodium), ions (fluoride), molecules (water), or 'formula units' (potassium chloride).

               

                                                                                                                           

1 mole of different substances all have different masses. For example, 1 mole of water does not weigh the same as 1 mole of sodium. Due to the size of their smallest particles, the mass of one mole varies.

The equation that can be used to calculate moles is as follows:

Moles = Mass/Molar mass 

To calculate the number of particles:

# of particles = # of moles/Avogadro's number

How many moles are present in 18g of H2O?

1. Add up the molar mass of H2O

H=1; O=16 à 16+2(1) =18    The total molar mass is 18

2. Divide the actual mass by the molar mass

18/18= 1 mole

There is 1 mole of H2O in 18g.

How many water molecules is this?

Water molecules are the particles à use the number of particles equation

# of particles= 1 x 6.02 x 10²³ = 6.02 x 10²³ water molecules

How many hydrogen atoms?

In each H2O molecule, there are two hydrogen atoms.

Multiply two by Avogadro's number to get the number of hydrogen atoms

2 x 6.02 x 10²³ = 1.204 x 10²⁴ hydrogen atoms

How many moles are there in 200g of CaCO₃?

1. Calculate molar mass

Ca=40; C=12; O=16 --> 40+12+16(3) =100 The total molar mass is 100

2. Divide actual mass by molar mass

200/100= 2 moles of CaCO₃

Then, we performed a counting and weighing atoms lab. We were to weigh certain substances in a weighing boat (trying to get approximately the same volume for each compound) and then work out the number of moles in that amount of the substance. Below is a section of the data table.*

Name of substance	Molar mass	Mass of substance (±.01)	Number of moles (calculated)	Type of particles in sample	Number of particles (calculated)
Carbon	12	.20	.017	Atoms	1.0234 x 1022
Potassium chloride (KCl)	74.5	.31	.004	Formula units	2.408 x 1021
Sugar (C6H12O6)	180	.36	.002	Molecules	1.204 x 1021

*Results may vary depending on how much of the substance was weighed

In conclusion, we learned what a mole is and how to calculate the moles in a substance given its chemical formula and weight. We learned about Avogadro’s number and its role in working out the number of particles in a substance. Lastly, we learned to identify which type of particle is the smallest.

Images from: education-portal.com; www.bbc.co.uk; faculty.clintoncc.suny.edu; www.webelements.com;

1.1 Stoichiometry: Particle Nature of Matter & Chemical Change

08 /09 /14

*Atomic Theory Test was done today*
*Notes from Youtube Video*

Particle Theory

• All matter is made up of particles 
• Particles have kinetic energy (above 0 K)

Phase Changes

• Solid - particles are packed together forming rigid shape
• Liquid - particles can move a bit more freely creating a substance that flows (takes shape of container)
• Gas - particles have free movement & fill entire volume of container
• Melting = solid  →liquid
• Freezing = liquid → solid
• Vaporization = liquid → gas
• Condensation = gas  → liquid
• Sublimation = Solid → gas
• Deposition = gas  → solid     

Combining Powers: using charts below

• Simple binary compound (ex. Calcium fluoride)
• Transition Metal Compound (ex. Titanium IV oxide)
• Compound ions (ex. Lithium sulfate)

Balancing Equations

1. Write symbols for chemicals being used
2. Balance - have same number of atoms on each side of equation
3. State symbols (s, g, l, aq)

11.1 Uncertainties and Errors in Measurements and Results

In this class we started a new topic on the uncertainties that can occur during measurements and results.

We had to understand the difference between Random Errors and Systematic Errors. 

Random Errors: errors that can from from the equipment used e.g. stopwatch. this can be reduced by doing an experiment more than once.

Systematic Errors: this can occur from a poor experimental design and cannot be fixed by repeated trials.

Also we learned how to find uncertainties in both analogue and digital devices.

For analogue devices = ± half of the smallest increment (reading)

For digital devices = ±smallest reading.

If u are adding or subtracting values with uncertainties, you add both the values together and then both the uncertainties. However, if you are multiplying or dividing you multiply/divide the values BUT you need to change the uncertainties into a percentage and add the two percentages.

e.g.   (a ± Δa)(b ± Δb) = (ab)±[ (Δa/a x 100)% + (Δb/b x 100)%]

Class of August 29th - Electron Configuration and Oribitals

Picked up on the "2.2 Electron Configuration" spreadsheet. Things covered in this lesson included; Emission Spectra, Electromagnetic Spectrum, Electron Arrangements (2n^2) and Orbitals.

Emission Spectra

• The spectrum of light emitted when electricity is passed through an element
• The glow from objects comes from when electricity goes through an element and the electrons go up an energy level. When they come back down they release protons.

Electromagnetic Spectrum

• When the wavelength increases in the electromagnetic spectrum the frequency decreases, and visa versa. 
• the wavelength from longest to shortest goes from: Radio - Microwave - Sub mm - Infrared - Visible - Ultraviolet - Xray - Gamma Ray 

Electron Arrangements

• Pre-IB understanding taught that electron shells would be filled 2,8,8,8... etc. This actually a flithy lie
• Each energy level is made up of different types of orbitals
• The four types are s, p, d, f
• Each orbitals are made up from 2 electrons.
• s has only one shape of orbital, p has 3, d has 5, and f has 7.
• To work out which shell has which amount of orbitals you can use this diagram and follow the arrows to allocate the electrons:

• The only 2 exceptions form this diagram are when you are dealing with Cr and Cu or dealing with an first row transition element that is ionised.