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