December 1, 2010

Wow, I almost can't believe it, it is already the start of December!
Today is officially December 1st so that means it is 24 days until Christmas & 17 more days until Winter Vacation! We are almost there, so keep going everybody!!

Well enough about Christmas (which is my favourite holiday) and lets talk Science.
Today we learned a new topic..... are you ready for this?..........
Empirical + Molecular Formaula!!!!!!
Now this topic is very detailed so stay with me here;

Lets first start with Empirical Formula; (EF)
- EF is the lowest term ratio of atoms in a formula. Otherwise known as reducing.

ex. C6H12 = molecular formula
C3H6 = CH2 = empirical formala

ex. consider that we have 10.87g of Fe and 4.66g of O. What is the EF?
1st Step. Convert grams to moles
Fe 10.87g x 1 mol / 55.8g = 0.195 mols of Fe
O    4.66g x 1 mol / 16.0g = 0.291 mols of O
2nd Step. Divide both by the smaller molar amount
Fe 0.195 / 0.195 = 1  
O   0.291 / 0.195 = 1.5
3rd Step. Scale ratios to whole numbers
Fe 1 x 2 = 2 
O   1.5 x 2 = 3                   Therefore the answer is: Fe2O3

ex. A compund contains 31.9 % K, 28.9 % Cl, 39.2 % O. What is the EF?
* Assume you have 100g
1st Step. Convert grams to moles
K 31.9g x 1 mol / 39.1g = 0.816 mol  
Cl 28.9g x 1 mol / 35.5g = 0.814 mol
O 39.2g x 1 mol / 16g = 2.45 mol
2nd Step. Divide each by the smaller molar amount
K 0.816 / 0.814 = 1
Cl 0.814 / 0.814 = 1
O 2.45 / 0.814 = 3
3rd Step. Scale ratios to whole numbers
Dont need to do because all ratios are whole numbers already (1, 1, 3) 
Therefore the answer is: KClO3

Now I will explain Molecular Formula; (MF)
- MF is the multiple of empirical forms. It shows the number of atoms that combine to form a molecule.
- Formula: n = mm of the compund / mm of the empirical formula. (mm = molar mass)

ex. A molecule has an empirical formula of C2H5 and a molar mass of 58 g/mol.
What is the MF?
MM of C2H5 = 29.0 g/mol
n = 58 g/mol
      -----------     = 2
      29.0 g/mol
MF = 2(C2H5) = C4H10

ex. The empirical formula of a gas is CH2. What is the MF if MM is 42 g/mol?
MM of CH2 = 14 g/mol
n = 42 g/mol
      -----------     = 3
      14 g/mol
MF = 3(CH2) = C3H6

ex. A compound contains 7.44g of C, 1.24g of H, 9.92g of O. The MM is 180 g/mol. What is the MF?
1st Step. Find the empirical formula since it is not shown in the question.
C 7.44 x 1 mol / 12g = 0.62 mols
H 1.24 x 1 mol / 1g = 1.24 mols
O 9.92 x 1 mol / 16g = 0.62 mols
2nd Step. Then divide each by the smallest molar amount which is 0.62
C = 1    H = 2    O = 1      therefore EF = CH2O
3rd Step. Now you can find the MF.
MM of CH2O = 30 g/mol
n = 180 g/mol
      -----------     = 6
      30 g/mol
MF = 6(CH2O) = C6H12O6

That concludes of all what we did today, now I see that in the last blog Percent Composition was not explained so I will explain that right now. Percent Compostion is what we did last class.
Percent Composition: (PC)
- PC is percentage by mass of a "species" in a chemical formula.

ex. What is the PC of CO2?
*Assume you have one mole
Total MM = 44.0 g/mole
% of C = 12.0 divided by 44.0 x 100% = 27.3 %
% of O = 32 divided by 44 x 100% = 72.7 %
Notice that they both add up to a hundred percent, in these equations the two or three answers should always add up to a hundred percent or at least very close to.

ex. Compound contains 5.1g of Cl, 22.0g of C, and has a total mass of 44.1g and contains some amount of O. Calculate the PC.
Mass of Oxygen = 44.1 - 5.1 - 22.0 = 17.0 g
% of Cl = 5.1 divided by 44.1 x 100% = 11.6 = 12%
% of C = 22.0 divided by 44.1 x 100% = 49.9%
% of O = 17.0 divided by 44.1 x 100% = 38.5%
All add up to 100.4 percent. This is a way to check your work too by the way.

ex. Calculate the PC of the underlined components of (NH4)2SO3
Total MM = 116.1
MM of all the NH4 = 36
% of NH4 = 36 divided by 116.1 x 100% = 31.0%

Whew, alot of numbers as you can see :). Now its time to treat you with some video's & practice questions for your own practice!
- Worksheets:
Empirical + Molecular Formula
1) http://eaglepoint.lexingtonchristian.org/hs/science/dhumphreys/Shared%20Documents/Honors%20Chemistry/Empirical%20Formula%20Worksheet%205.pdf
2http://www.ccboe.net/Teachers/nutial_laura/files/CBDD907F339B420CB25588150BD481A9.doc
3) http://www.ezclasssites.com/data/captainkirk/EmpiricalFormulaWS.pdf
4) http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson59.htm
5) http://lhs2.lps.org/staff/sputnam/practice/UnitV_EmpForm.htm
6) http://westwood.sjsd.net/~dshoesmith/FOV1-0003789A/FOV1-000378A1/Percent%20Composition%20and%20Molecular%20Formula%20Worksheet.doc?FCItemID=S01C8C50E
7) http://www.cbv.ns.ca/rv/campbell/Resources/M%20F%20EF.pdf

Percentage Composition
1) http://misterguch.brinkster.net/PRA023.pdf
2) http://www.fordhamprep.org/gcurran/sho/sho/worksheets/worksht58a.htm
3) http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson58.htm
4) http://misterguch.brinkster.net/001_024.doc
5http://cmsweb1.loudoun.k12.va.us/52820831134912597/lib/52820831134912597/Moles/Homework/masspercomp.pdf
6http://www.ccboe.net/Teachers/epperson_steve/files/36F0A3C2EC0648DBB63EF8981578B7A0.doc
7http://asd1.schoolwires.com/1741206142323657/lib/1741206142323657/Percentage_Composition_Worksheet.pdf

- Videos
1) http://www.youtube.com/watch?v=xbEeyT8nK84
2) http://www.youtube.com/watch?v=_H009sTvYE0
3) http://www.youtube.com/watch?v=gfBcM3uvWfs

Ooops almost forgot...... HW:
1) Empirical and Molecular Formula
2) Reminder: Quiz next week regarding this material and the material will be covered on Frida (next class).

Thank You and Have A Great Weekend!!!!!