Evaporation and Intermolecular Attraction Lab

Because it is impossible to make a punny title out of that.

Materials:

• Computer
• Serial Box Interface ULI
• Data Logger
• Two Probes
• 6 Pieces of Filter Paper
• 2 Small Rubber Bands
• Methanol
• Ethanol
• 1-Propanol
• 1-Butanol
• n-Hexane
• n-Heptane

Procedure:

1. Open "Experiment 9" from the Chemistry with Computers from the experiment files of Logger Pro.
2. Wrap filter paper around the tips of both probes, and secure it with the rubber bands. 
3. Place the both probes into a container holding a sample of liquid.
4. After the probes have been in the liquids for around 30 seconds, begin data collection. Monitor the temperature for around 15 seconds to establish the initial temperature of each liquid.
5. Simultaneously remove the probes from the liquids and tape them with the tips of the probes extended of the edge of a table.
6. When both temperatures have reached minimums, stop data collection.
7. Find the maximum and minimum temperatures, then subtract them to find the change of temperature during evaporation.
8. Remove the rubber bands and discard the used filter paper.
9. Predict, using the data just collected, how the next set of liquids will react.
10. REPEAT!

Heptane (red) and Hexane (green) Test

Ethanol (red) and Methanol (green) Test

Propanol (red) and Butanol (green) Test

Data Analysis:

1. n-Heptane and 1-Butanol had nearly the same molecular weights, but had significantly different changes in temperature due to hydrogen bonds. 1-Butanol contains hydrogen bonds, but n-Heptane does not. This made it easier for n-Heptane to evaporate compared to 1-Butanol.
2. Methanol had the weakest intermolecular forces, while 1-Butanol had the strongest. We can see that methanol had the weakest bonds because of the change of temperature. Methanol changed 18.4 degrees. This shows that it had weak bonds, allowing it to evaporate more and quicker than the other liquids. 1-Butanol had the strongest bonds because its temperature did not change very much at all. It's temperature only went down by about 3.07 degrees. This means that it had stronger intermolecular forces that did not allow it to evaporate as much or as quickly as methanol. 
3. In regards of alkanes, n-Hexane had the weakest intermolecular forces, and n-Heptane had the strongest molecular forces. n-Hexane changed a total of 17 degrees, and n-Heptane changes only 9.41 degrees.
4. 
   

So a Limiting and Excess Reactant Walk Into a Bar. . .

. . . and they ask for a 100% yield of H20! Hahahaha! I'm so funny right?!

Ok, for real now! For the last couple of days we have been working on finding the limiting and execs reactants in a reaction, also known as stoichiometry. We were also introduced to percent yield today, and all of this will be shown now.

Ok so let's do problem 21 of chapter 12. This problem plays out as such:

Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. Wite the balanced chemical equation for the reaction. If a plant has 88g carbon dioxide and 64g water available for photosynthesis, determine:

1. The limiting reactant
2. The excess reactant ant the mass in excess
3. The mass of glucose produced.

Alright, so this is what the equations will look like to get the answer:

First the equation needs to be balanced, it will look like 6CO2 + 6H2O = C6H12O6 + 6O2. Now we can start with the actual calculation part.

Let's start with finding what is the limiting and execs reactants. We'll start with carbon dioxide because it comes first. The problem gives you that there is 88grams of CO2. We'll then convert the grams of carbon dioxide to mols. So, there are 44 grams for one mol of CO2 (yes it's labeled wrong in my notes). Now we can find the amount of glucose through the mol ratio between glucose and carbon dioxide. So, for every 6 CO2s there is 1 C6H12O6. The next part is to find how many grams are in one mol of C6H12O6. Putting it all together, there are 180grams of glucose per one mol. Multiply everything together, and then do proper division, and you'll get an answer of 60 grams of Glucose.

If you repeat the same process, but with water instead of carbon dioxide, you'll get 106.6 grams of glucose. This means that the carbon dioxide is the limiting reactant, and water is the execs reactant.
_________________________________________________________________________________
Now to get the percent yield. To get the percent yield we need to use a different problem, in which we are given the actual percent yield. We'll try problem 27 out of the same chapter. It says:

If 14 grams of aluminum hydroxide is present in an  antacid tablet, determine the theoretical yield of aluminum chloride produced when the tablet reacts with stomach acid. If the actual yield of aluminum chloride from this tablet is 22 grams, what is the percent yield?
Al(OH)2 + 3HCl -> AlCl3 + 3H2O

If you do all the math and such you will get a theoretical yield of 23.9 grams of AlCl3. Our actual percent yield is 22 grams of AlCl3. To get the percent yield we need to take the actual yield divided by the theoretical yield.

That would give us: 22/23.9 = .882.
Multiply this by 100 and you get a total of 88.2% yield of Ag.   DONE!

I'm Awesome

So to show you that I know what I'm doing in this class and I'm awesome, here is my test that I aced. It is known as a Chemistry Homework Quiz, and I completed it 14 out of 14. It was over percent composition, empirical and molecular formulas, and hydrations. So yeah, I know my concepts.

And this is what happens when I get distracted while doing my blogs:

An Apple product, within an Apple product, within an Apple product; with the picture being taken with another iPad. It's like a string quartet for Apple.

Silver/Copper Replacement Lab

Which I was gone for, but here it is regardless.

Procedure (Day 1):

1. Obtain 30cm of bare copper wire. Clean the wire, then coil it around a pencil to form a loose spring. Make sure that the other end of the wire reaches the top of your test tube and is uncoiled.
2. Weight the coil with a balance, then place the copper wire in the test tube to check that is is the correct length. Remove the coil and set it aside for later.
3. Weight the weighing dish of silver nitrate and record it's number. 
4. Transfer the contents of silver nitrate to your test tube.
5. Pour distilled water into your test tube until the water is about 2 cm from the top.
6. Cover the top of the test tube with parafilm.
7. Place your thumb on top of the test tube and invert it until all the silver nitrate is disolved. 
8. Weigh the empty weighing dish so that you can determine the mas of silver nitrate that was added to the test tube.
9. Add the copper coil to the test tube.
10. Set the tube aside until the next day or class period.
11. Make sure all data is recorded.

Procedure (Day 2):

1. Weigh a piece of fiter paper for use in separating the silver.
2. Shake the test tube and copper wire to dislodge the silver.
3. Set up a funnel with your filter paper in it.
4. With a waste beaker beneath the funnel, lift the copper wire out of the test tube and hold it over the filter system.
5. Using a water bottle, let D.I. water fun down the wire so that any silver will wash into the filter.
6. Lay the copper wire on a labeled piece of paper to allow it to dry.
7. Pour off the solution in the test tube through the filter paper into the filter, trying to keep the silver precipitate in the test tube.
8. Rinse with distilled water and decant several times into the filter to wash the silver.
9. Finally wash the silver from the test tube onto the filter paper.
10. Allow to dry overnight. 

Procedure (Day 3):

1. Weigh the copper coil and record it's mass.
2. Weigh the silver and filter paper - record the mass.

_________________________________________________________________________________

Mass of silver nitrate = 1.028 grams

Mass of copper coil before reaction = 3.408 grams

Mass of copper coil after reaction = 3.193 grams

Mass of copper reacted = .215 grams

Mass of filter paper & silver = 1.726 grams

Mass of filter paper = 1.420 grams

Mass of silver produced in reaction =.360 grams

Moles of silver produced = .0028 mols

Moles of copper consumed during reaction = .0033 mols

_________________________________________________________________________________

The balanced equation fo this reaction is:  2AgNO3 + Cu  ->  Cu(No3)2 + 2Ag

.65 grams of Ag were formed

.77 grams of Cu became Cu(No3)2

The Baking Soda Lab of Dreams

Materials:

• 20 ml vinegar in a large pipette
• Balance
• 100 ml beaker
• 1 gram of baking soda

Procedure:

1. Get a large plastic pipette filled with vinegar. Measure the mass of the pipette and record the mass.
2. Measure the mass of an empty, clean 00 ml beaker.
3. Transfer about 1 gram of baking soda to the beaker. Record the exact mass of the beaker with the powder.
4. Add vinegar, from the pipette, to the beaker. Swirl the contents and observe the reaction. Continue to add vinegar until no more bubbles are produced. This will take a while so be patient and pay close attention to the reaction.
5. Find the mass of the left-over vinegar in the pipette and record. Subtract the original mass of this pipette to find the mass of the vinegar used in the reaction.

From the picture above, you can see the data we collected throughout the lab. When all was said and done, we calculated that the moles of sodium hydrogen carbonate used was .012 moles. We did this by taking the net mass of baking soda (1.01grams) and dividing it by the mass of NaHCO3 (84grams). 

Next, we calculated the grams of acetic acid added to the beaker for the reaction. First, we were given that there are 5grams of acetic acid per 100grams of vinegar. We then decided to set up a ratio, knowing that we had used 13.34grams of vinegar. If there were 5grams of acetic per 100grams of vinegar, then that was equal to xgrams of vinegar per 13.34grams of vinegar. We cross multiplied to get 100x=66.7, thus resulting in x=.667grams of acetic.

We then calculated the moles of acetic acid used in the reaction. We took the .667grams of acetic acid divided by the actual mass of acetic acid (HC2H3O2). This gave us .011g/mol of acetic acid.

We then found the ratio between the moles of sodium hydrogen carbonate and the moles of acetic acid used in the experiment. Since the .011moles of acetic acid was the smallest number, it served as the 1 in our ratio. We then took the .012 moles of sodium hydrogen carbonate and divided it by the .011 moles of acetic acid, which gave us a ration of 1:1. This, our experimental mole ratio, corresponds to the mole ratio of the balanced equation: NaHCO3 + HC2H3O2 -> CO2 + NaC2H3O2 + H2O, which is a 1:1 ratio.

Mole Classwork

Here is just a lovely picture of some of the work I've been doing in class. I've been working in our Glenco Chemistry book.  Page 221, Chapter Eleven, Problems 42-57 to be exact and if it helps. These have all been over subjects the include: empirical formula, moles, and molecular formula. For example, problem 50 asks:

"What is the empirical formula for a compound that contains 10.89% magnesium, 31.77% chlorine, and 57.34% oxygen?"


To solve this problem I first wrote down everything I knew which looked a lot like this:

Magnesium: 10.89%

Chlorine: 31.77%

Oxygen: 57.43%

Then I found the relative atomic mass in grams of each element (24.3 for Mg, 35.5 for Cl, and 16 for O). I then took the percent of the element and divided it by its weight, getting numbers such as: .448, .895, and 3.584 respectively. 

Now for an empirical formula, you have to find the smallest working ratios between all the elements. Because Mg came up with the smallest number, .448, it will serve as 1 in the ratio. Then take the remaining numbers, .895 and 3.584, and divide them by .448. You will then get about 2 and 8 respectively. This completes your ratio, which is 1:2:8. This means that more every one Magnesium there are two chlorines, and eight oxygens. The complete empirical formula woud then be MCl2O8.

And that's what I've been doing in class. Lots of math. 

  

Hydrates Lab

Procedure:
1. Figure out the molar mass of CuSO4*5H2O
2. Figure out the mass of 5H2O,  then make a prediction about what percent of the entire thing is water.
3. After making this prediction, weigh out 5grams of Copper Sulfate Pentahydrate.
4. Take the Copper Sulfate and place it in a test tube. Use a bunsen burner to heat the test tube until the Copper Sulfate is not longer blue.
5. Wait until the tube is cooled, then weight the Copper Sulfate again.
6. Record the differences of the Copper Sulfate before and after it was heated to find the weight lost. Then calculate exactly how much water was lost.
7. For fun, take remains of the heated Copper Sulfate and pour water on it.

My Group:  
1. My group figured out that the molar mass of CuSo4*5H2O was 250 grams.
2. The mass of 5H2O was 90 grams. Therefore our prediction was that 36% of the Copper Sulfate was water.
3. We completed our procedure and weighted our 5g sample after begin heated. After a date with the bunsen burner, the Copper Sulfate weighed 3.28 grams, being that there was a total weight loss of 1.72 grams.
4. Therefore, we concluded that 34% of our sample of Copper Sulfate was actually water.
5. *SPOILERS* After the Copper Sulfate is heated it turns white. By adding water back to it, it returns to its original blue color!