Acids and Bases

The last day of actual class at La Junta High School, and we did probably one of the most educational labs I have ever done.  In order to effectively understand Acids and Bases before the year ended, five of us, with the help of Mr. Ludwig, completed the Acid-Base Titration Lab.

Each solution is either determined as an acid or a base depending on it's pH level. If a solution has a pH level of 0-7 it's an acid. If it has a pH level of 7-14 it's a base. If the pH level remains at 7, it's neutral. An example of this would be water. There is a turning point in which the acid turns immediately to a base, and visa versa.
Looking at the difference between acids and bases:

Acids                                                              Neutral                                           Base
low pH                                                             water                                         high pH
bad?                                                                                                                   good?
H+                                                                                                                     OH-

The determining factor between acids and bases is not only dependent on the pH level but also on the amount of hydrogen and hydroxide in the solution. If there is more hydrogen in the solution the solution is an acid, whereas if there is more hydroxide it is a base.

The Experiment:

Our goal for this experiment was to get our solution to the turning point(equivalency point) where it turns from an acid to a base. We were attempting to graph this point on LoggerPro.

Steps:

1. We placed 10 mL of the HCL solution into the beaker along with 50 mL of water.

2. We placed a stirrer in the beaker and placed it on the magnetic stirrer so that it could stir the entire duration of the experiment.
3. We suspended a pH sensor that was connected to LoggerPro into the water in order to track our results. Also, in order to track the pH levels, Mr. Ludwig dropped in a few drops of pH indicator. We did this so that when the acid turned into a base, we would be able to tell immediately because the indicator would scream pink.

4. Using a 50mL buret filled with 0.1 M NaOH, we careful placed the NaOH solution into the distilled water and HCL solution, tracking the amount and pH level every 2 mL.

(because NaOH is a base and HCL is an acid, the solution in the beaker began as an acid. As we slowly poured NaOH into the solution, it was changing into a base.)

Once the solution turned pink, we knew it was fully transformed into a base.

Graph:

As determined by the graph, our experiment went very well as we did accomplish the "S graph" look. As the points on the graph climb, the acid slowly begins turning into a base as the pH level rises. The few dots on the graph that rise so abruptly are the equivalency points where the acid turns into a base, then slowly levels out as a base.

Through the graph and data we collected we were able to determine the concentration of NaOH in the beginning, an the volume of NaOH moles of NaOH, Moles of HCl, and the Concentration of HCL all at the equivalency point.
Concentration of NaOH at beginning: 0.1 M
Volume of NaOH at equivalency point: 15.51 mL or .01551 L
Moles of NaOH at equivalency point: .001551 mol
Moles of HCl at equivalency point: .001551 mol   (the moles of NaOH and the moles of HCl must be equal at the equivalency point.)
Concentration of HCl at equivalency point: .1551 mol/L (to calculate this, we divided the moles of HCl by .01)



In this lab I learned how easy it is for an acid to turn into a base, especially in this case. It only took a drop for the acid to turn pink(into a base). It was a lot of fun working as a class for this last lab. Each of us had a job to do and in this way we accomplished the lab to Mr. Ludwig's expectations. Success!! (:

Crystal Lab

With the extremely busy schedule everyone has had lately, it was no surprise that I missed almost all of the formation of my crystal. I was however involved in all of the process, but lacked many pictures, except the final result.

Here are the steps for the lab:
1. obtain distilled water and put it in a beaker.
2. we put the aluminum potassium sulfate in the distilled water and mixed until it was supersaturated.
3. When there were little crystals still in the bottom of the beaker after it is completely saturated, we put the beaker on the hotplate and repeated the process. The hotter the water is the more it is able to absorb so the aluminum potassium sulfate.
4. we kept adding aluminum potassium sulfate until the heated mixture was supersaturated as well.
5. we left the mixture to cool overnight.

We returned the next day to find that there were crystal seeds all over the beaker.

We then took the seed crystal and tied it to a string, then repeated the supersaturated process.

The next day that I actually returned to this class, my crystal was pretty big and my beaker was full of tiny crystals just hanging out at the bottom.

I learned a lot about how crystals can be formed by supersaturating the distilled water. I also learned that creating crystals takes a lot of time and patience as they have to sit and cool. It's interesting to see how quickly the crystals grew overnight though, as they started out with a seed crystal about this size:

They turned into multiple crystals about this size:

This is probably one of the coolest labs I have done in chemistry. Even though it was easy and time consuming, it had a cool result that all of us got to take home with us. I definitely feel like I learned some valuable information through this lab.

Different Types of Chemical Reactions

1. A+B = AB synthesis reaction
    N2+H2 = NH3

2. AB = A+B decomposition reaction

3. A+BC = B+AC  single replacement (bad prom)
    Zn +AgNO3 = Ag+Zn(NO3)2

4. AB+CD = CB+AD (double replacement)

5. C3H8+O2 = CO2 +H2O (combustion) - hardest to balance

ChemTech Project

For my ChemTech project I am going to research the topic of 3D printers. This concept is fairly new and and interesting to the American public.



My Question is: How are 3D objects created from this wonderful, magnificent printer?

Prezi:

http://prezi.com/n4uy8koscrzu/3-d-printing/

Article:http://www.sciencedaily.com/releases/2012/03/120312101918.htm


My resources:
http://en.wikipedia.org/wiki/3D_printing
http://www.objet.com/
http://www.chemistrydaily.com/chemistry/3D_printer
http://hackaday.com/2010/07/15/3d-printing-with-chemicals/
http://www.3dfuture.com.au/articles/11-reasons-why-schools-need-3d-printers/
http://eater.com/archives/2010/12/29/3d-food-printer-futuristically-prints-chocolate-cookies.php#d-food-printer-6

Evaporation and Intermolecular Attractions

Procedure: 

1. We wrapped probe 1 and probe 2 with filter paper and attached them with rubber bands. 

2. For the first round we put probe 1 in the methanol and probe 2 in the ethanol for over 15 seconds so the liquid could soak through the filter paper.

3. We then took the probes out of the containers and attached them to the end of the table with masking tape, making sure they didn't touch. 

4. We allowed the liquids to sit on the edge of the table, while monitoring the temperature. We took apart the probes after the temperature became consistent. 

5. We took the higher temperature and subtracted the lowest temperature to get the change in temp. 

6. We then predicted the change in temperature for the next few chemicals and repeated the steps for heptane and hexane, and propanol and butanol.  

Materials:

Mac computer           rubber bands    masking tape

ULI                           methanol

Data Logger              ethanol

Temperature probes   propanol

filter paper                 hexane

Here are our results:

Heptane and Hexane Lab

As shown in the graph above, during the heptane and hexane section of this lab, it was determined that although at the beginning the temperatures were very similar, as the time progressed, it was obvious that the heptane had a much higher temperature than the hexane did. This was a difference of almost 10 degrees celsius at 174 seconds.

Methanol and Ethanol 

In the methanol and ethanol section of this lab, the results looked a lot lke the heptane and hexane lab with few differences. The temperatures started out very similar to each other but as the chemicals evaporated, it was obvious that the ethanol had a much higher temperature than the methanol. This difference was almost 7 degrees celsius.

 Propanol and Butanol

The propanol and butanol section was the most diverse of the rest of the lab. There might have been a mistake in the lab procedure as the temperature  of the propanol spiked and then immediately decreased. But, as time went on, the temperatures steadied and it was obvious that the butanol had a slightly higher temperature than the propanol. This difference was meerly 5 degrees at 254 seconds.

Popcorn Lab

Popcorn Lab on Prezi

Unknown/Mole Lab

Mole Lab!


Tenchita, Timmy, Serena and I had a lot of fun in this lab just because we felt like really scientists. We were told to figure out the atomic mass given only the number of moles. We weighed each cup to find out the grams and then took the number of grams divided by the number of moles to get the atomic mass. Given the atomic mass we find out what the element was. 


Avogrado's number - (6.0221367 x 10^23).



Data
Sample.             Number of moles.        Grams.           Atomic mass.      Element
     A.                            .163                     6.53             40.061                 Ca
     B.                            .910                    24.51            26.934                  Al
    C.                             .160                     9.44             59                       CO
     D.                            .258                    53.90           208.9                     Bi
     E.                            .220                    14.26           64.81                    Zn 
     F.                             .756                    49.26           65.158                 Zn
     G.                            .492                     28.66           58.25                  Ni
     H.                             .430                    89.16           207.349              Pb
     I.                               .381                    21.28            55.85                Fe
     J.                             .259                     30.70           118.53               Sn

Hydrate Lab

Although I wasn't here for the lab Timmy, Serena, and Tenchita are nice enough to let me use their data to do a write up about this experiment. They chose to do the experiment involving epsom salt. They put the epsom salt in a test tube and weighed it before any other experimentation. The weight before heating the salt was 4.07 grams. They then heated the salt (telling me that Kandace almost killed them a couple times with matches and the gas). While heating the tablets turned white which they then let cool. After letting the flaming salt cool they weighed it after, noting that it weighed 3.66 grams. They then added water, learning that the crystals turned back to blue.
By subtracting the weight after heating from the weight before heating, we discovered that the crystal only had about 10% water. This is different from the expected percentage because of crystals that got stuck in the tube and crystals that were still blue.
The formula for the crystals is CuSO4 5H20. After multiplying the total CuSO4 mass by 5 parts water, I learned that the total water mass is 90 grams while the total mass was 249.6 grams. After dividing the 90 grams by the 249.6 grams, we discovered that the expected mass was 36%.

Calculations

Moles!
No not the animals, more like the number of moles in an element how the amount of moles needed for a compound.
One mole equals 6.02x10^23 atoms.
So, how do you find how many grams per mole an element has? For example, the element Bromine has a mass of approximately 79.9 so it is 79.9 grams/mol. Now, looking at a compound such as K2O. Potassium has a mass of 39.098 and Oxygen has a mass of 15.9994. Since there are two potassium atoms, we have to multiply the 39.098 by 2 and then add the 15.9994. The final answer is 94.196 K2O. This means there are two potassium atoms for every one oxygen atom, but I'll get into that a little bit later.
Looking at a more complex compound such as C14H18N2O5, it's easier to find the measurement of moles in the compound. First you must calculate the grams per mole, which turns out to be 294. Then you take the mass of the smallest element, which is 225 grams. Divide the 225 by the 294 to get the answer. C14H18N2O5 is .765 of a mole.

Percent of Composition
Let's find the percent of composition of Carbon in an artificial sweetener. The compound is C5H8NO4. To get the percent of composition one has to divide the mass of the total Carbon by the total weight of the compound. Carbon has a 12.0107 mass, but there are 5 atoms of Carbon, so the total mass is 60.055. The total weight of the sweetener is 146.1212. 60.055/146.1212 is approximately 41%. So, Carbon is 41% of the artificial sweetener compound.

The Empirical Formula
The Empirical Formula helps a scientist (or student) determine the compound that is put in front of them based on the given weights. For example, when 2.34 grams of Nitrogen and 5.34 grams of Oxygen are put in front of you, by dividing by the weight we find out that there are .167 moles of Nitrogen and .333 moles of Oxygen. Take the .167 and divide it by .167(the least amount of moles) and do the same for .333 which rounds out to 1 to 2. So, the ratio is one to two. But, that could mean that the compound is NO2 or N2O4. The fact that the compound weighed 93grams was given. Through multiplication, it is found that the compound is N2O4 (Dinitrogren Tetroxide)