Friday, May 4, 2012

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

Monday, March 5, 2012

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.

Monday, February 20, 2012

Popcorn Lab

Monday, February 6, 2012

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



Wednesday, January 25, 2012

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%.

Sunday, January 22, 2012

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)