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Sunday, December 12, 2010

Lab 4C Day a.k.a time consuming heating stuff up lab

    Today we had another lab which featured finding and calculating the mass of various objects such as anhydrous salt, empty crucible, water given off, etc.  To begin the lab we had to wear the awesome looking safety goggles.  Then ms. chen previewed over what we should do, and everyone i mean everyone ran to get all the supplies.  Each group of two acquired a pipestem triangle, iron ring, stand, and bunsen burner.  We set up the equipment so that the bunsen burner is directly underneath the crucible which is on the  pipestem triangle that is held midair by the iron ring locked onto the stand.  Before starting the lab we had to turn on the bunsen burner by connecting it to gas and lighting it up in order to dry out the crucible that is placed on the triangle.  Once dry we weigh the crucible and record it.  Shortly after we obtain some anhydrous salt equal to 1/3 of the max volume the crucible can hold and weighed and recorded them down again.  Then comes the fun part........heating stuff up


anhydrous salt




    Now we had to heat the anhydrous salt for 5 min.  During the heating you can see that the anhydrous salt which WAS BLUE TURNED GREENISH WHITE.  After the 5min and seeing cool colour changing, we let it cool and recorded its mass.  This process was repeated to see if the mass were the same.  Then we had to do the last step in our procedure, add water to the dehydrated salt.  As soon as we added a few drops of water.......IT TURNED BACK TOO NORMAL OMGWTH!! NO WAI!!!  Then we recorded down this amazing discovery and dumped it in the garbage.  Then we filled in the rest of the information we needed to fill in with the data from the lab.  As soon as we were done we handed our lab filled with all our data and observation to ms.chen.  It was fun

Monday, December 6, 2010

The Empirical Formula of an Organic Compound

What is ORGANIC COMPOUND?
Does that mean that the compound is twice as expensive and has absolutely no difference whatsoever? (distastefully biased plug on organic foods) No, it means that the compound contains carbon. BUT it is noteworthy that certain compounds containing carbon are indeed inorganic, like carbides and carbonates. In this lesson we learn how to calculate the empirical formula of such compounds with chemistry's favorite furry little critter, MOLES.
The smell of vinegar is caused by acetic acid (CH3COOH). It is an organic compound

We know that the empirical formula will contain the elements C and H because they are common in all organic compounds. What we do not know, are the amounts of each. If we let x represent the number of carbon and y represent the number of hydrogen, we can use a method to solve for each. THERE ARE 5 STEPS THAT ONE MUST TAKE TO ACHIEVE THIS GLORY. Actually 6, but the 6th is situaltional. It is also noteworthy that when an organic compound combusts, the compounds carbon dioxide and water are ALWAYS formed.

1st STEP:
Convert the mass of the CO2 and H2O 
2nd STEP:
Find the moles of C and H from the moles of CO2 and H2O
There is 1 carbon in CO2 and 2 hydrogens and H2O so basically multiply the value of H2O in moles obtained from the previous step by 2
3rd STEP:
Divide both your values by the smallest molar amount
4th STEP:
Change the ratios until a whole number for both is reached by multiplying by 1 (2/2, 3/3, 4/4) ETC
The empirical formula will be represented by the fraction obtained
5th STEP:
Convert moles back to grams and add them to Check your work

Confused? EXAMPLES HO!

A 3.79 g sample of an organic compound is burned to yield 6.61 g of CO2 and 3.59 g of H2O. Find the empirical formula.

STEP 1:
(6.61g CO2)(1 mol CO2/44.0g CO2) = 0.150 mol CO2
(3.59g H2O)(1 mol H2O/18.02g H2O) = 0.199 mol H2O
STEP 2:
(0.150 mol CO2)(1 mol C/1 mol CO2) = 0.150 mol C
(0.199 mol H2O)(2 mol H/1 mol H2O) = 0.398 mol H
STEP 3:
0.150 mol C/0.150 mol C = 1
0.398 mol H/0.150 mol C = 2.65
STEP 4:
(2.65/1)(3/3) = 8/3 ----> 8 = y 3=x
C3H8
STEP 5:
(0.150 mol C)(12 g/1 mol)= 1.80g C
(0.398 mol H)(1 g/1 mol) = 0.398g H
1.80 + 0.398 = 2.198

BUT WAIT! 2.198 != 3.97.
EXPLAINED: Some organic compounds may contain oxygen aswell. THUS, the sample mass - mass of C and H = mass of oxygen

Dude, most likely jewish, teaching this:

Thursday, December 2, 2010

THE TOTALLY AWESOME!!! CLASS(EMPIRICAL+MOLECULAR FORMULA)

As the title suggested this class was more than just cool IT WAS AWESOME!! You won't be surprised because every single class of ours is super awesome because we are the awesome JOHNNY!!! (the Johnny part was a joke). okay back to our class~. Sooo what's did we learn today? Is that avocado lover with his not so cute moles planning for the next destructive weapon again??? The question remains to be answered but the important part is that we learned something in CHEM AGAIN~~~.

Okay the serious stuff is comming
Empirical Formula mean The simplest whole number ratio of atoms of each element present in a compound. (WOW DOESN"T THAT SOUND COOL?)
 Because the Pure awesomeness of Ionic compounds. They are given in Empirical formula already. (please don't go and ask the atoms why, they are just that awesome).  The Covalent compounds are the strange ones they come in the right ratio but NOT SIMPLIFIED in math class that would be 1 mark off... so if a ionic compound went and did a math test with covalent the covalent would just barely pass while ionic gets perfect omg><.

Formula = n = molar mass of the compound/molar mass of the empirical formula
Okay now lets have some examples
A compound have 47.25% copper and 52.75% chlorine.
Find the empirical formula for this compound.
 Cu 47.25g Atomic mass 63.6
47.25/63.6 = 0.74
Cl 52.75g Atomic mass 35.5
52.75/35.5 = 1.49
Cu 0.74/0.74 = 1
Cl 1.49/0.74= 2.01 = 2
So the empirical formula for Cu Cl
is CuCl2

Ex 2~ what is the molar formula? molecular mass = 132.16 , empirical formula = C2H4O

n = 132.16/42 = 3.2
MM = 42g/mol
MF= 3(C2H4O) = C6H12O3

Ex3  empirical formula of a compound is CH and the molar mass is 104 g/mol, calculate the molecular formula.
mass of C  =  12.0 g/mol
mass of H  =   1.01 g/mol
empirical formula mass  =   13.0 g/mol
 CH =  (104 g/mol)(1 mol/13.0 g)  =   8.00

MF  =   8(CH) or C8H8