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!!