QUIZ DAY!

Okay so cool, we finished our quiz on Molecular and Empirical Forumla, along with Percent Composition!
there isnt alot to say about this because we didnt have an actual lesson before or after the quiz...

I found that the quiz was quite straight forward! there was nothing on that quiz that Ms. Chen didnt
teach us. (im not being a suck up here)

Hopefully its not one of those quizzes where i thought i did well, but when i get it back... i didnt do so great.
If your still not getting it here are some VIDEOS:

Molecular and Empirical Formula: http://www.youtube.com/watch?v=gfBcM3uvWfs
and of course... A SONG, pretty intense: http://www.youtube.com/watch?v=slO9kN1XCnE
.... ok so they arent much good at singing, but if u look at the lyrics its about moles and molecular formula!

Percent Composition: http://www.youtube.com/watch?v=xbEeyT8nK84

ENJOY!

Review for Quiz!

Next class is quiz day!! so today we did a lot of review!

Quiz topics

• percent compositions
• empirical formula
• molecular formula
• calculating empirical formula of organic compounds

Percent compositions

ex. CO2
Total MM = 44          %compositions = mass of elements  x 100%
MM of C = 12                                       mass of comp
MM of O = 32

% of C = 12g/mol x 100% = 27.3%
               44g/mol

% of O = 32g/mol x 100% = 72.7%
               44g/mol

talk through of CO2 composition

Empirical Formula

- lowest term ratio of atoms in formula
* All ionic compounds are empirical formula


We have 10.87 g of Fe and 4.66g of O, whats the empirical formula
#1 convert to moles

Fe  10.87g x 1mol  =  0.195 moles
                     55.8g
O    4.66g x 1 mol  =  0.291 moles
                    16g
#2 divide both by smallest molar amount
Fe = 1
O = 1.5
#3 scale to whole numbers
Fe = 2
O = 3

Molecular Formula

empirical formula of C2H5 molar mass of 58g/mol
MM C2H5 = 29g/molE
n = 58g/mol = 2
      29 g
MF= 2( C2H5) = C4H10

really helpful video


New Stuff!!!

Empirical Formula of an organic compound


Found by

• burning wood (reacting with CO20
• collecting and weighing the products
• from mass of a product the moles of each element in the original organic compound can be calculated

5g of a carbon and hydrogen , 15g of CO2 and 8.18g of H2O

15.0g = 1 mol = 0.341
             44g

8.18g = 1 mol = 0.454
             18g

Molecules of C = 0.341 mol CO2 x 1mol C = 0.341 molecules 0.314 = 1 x 3 =3
                                                        1mol CO2                          0.314

Molecules of H = 0.454 mol H2O x 2mol H = 0.908 molecules 0.908 = 2.66 x 3 = 8
                                                        1mol H2O                          0.314

C3H8

Checking Mass

0.314 x 12g  = 3.77
             1mol                                3.77 +  0.908 = 4.68g
0.908 x 1g     = 0.908
             1 mol


Study, Study, Study, Quiz next day!!

Empirical + Molecular Formula

Today's lesson is on empirical and molecular formula.

Empirical Formula = gives the lowest term ratio of atoms (or moles of course) in the formula.
Ionic compounds are empirical formulas as well!

And here we have our example:

Lets say we have 10.87 g of Fe and 4.66 g of O. What is the empirical formula?
Well, first we convert grams to moles.

we want to convert the 10.87 grams so we will put that first. We want to eliminate grams so be sure to put the molar mass of the element in grams on the bottom.
10.87 g x 1 mole
                ------
                55.8 g = 0.195 moles
4.66 g x 1 mole
              -----
              16 g = 0.291 moles

Now the smaller one of the two is the first one, so be use that one and divide by that amount.
0.195/0.195 = 1
0.291/0.195 = 1.5
Now we have to scale ratios to whole numbers. We need to get rid of that 1.5. We do that by multiplying both by two, because then the 1.5 turns into a 3!
Fe 1 x2 --> 2
O  1.5 x2 ---> 3
So the answer would be Fe2O3

Here is a problem for you to try for practice:
A compound used for sweetening is 57.14% C, 6.16% H, 9.52% N, and 27.18% O.  Calculate the empirical formula of the sweetener and find the molecular formula.  (The molar mass of the compound is 294.30 g/mol)
Remember to follow the steps!
1) convert to moles
2) divide both by the smallest molar amount
3) scale ratios

Now onto molecular formula:

Lets use this formula to help answer the questions:

n = molar mass of compound
      --------------------------
      molar mass of the empirical formula

The empirical formula of a gas is CH2. What is the molecular formula if the molar mass is 42 g/mol?

First, lets calculate the molar mass of the compound, because we need it in the equation.
The mass of carbon is 12, and the mass of hydrogen is 1, but there is 2 o them. So the molar mass of the compound is 14.

Now lets plug that into the equation!

n = 42 g/mol
      --------
      14 g/mol
which equals 3 g/mol.
Now write the answer 3(CH2)
make sure to distribute the 3.
so it is actually C3H6 = MF propane
Make sure you write the MF! It stands for molecular formula. You can remember it by thinking that it is Melanie's initals!

If you want a little extra help, I've got videos galore for you!
Molecular and Empirical Formula
A helpful chemistry tutorial
Extra help!
A super cool 16 minute tutorial! woah!