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6 Empirical formula

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					Empirical Formula



        Lots of easy math, with
        lots of rules
Empirical formula

l   The empirical formula is the molecular
    formula reduce to it’s lowest common
    denominator
l   Example: Both ethylene (C2H4) and
    cyclopropane (C4H8) have the same
    empirical formula: CH2



      Ethylene               Cyclobutane
Empirical formula



         Ethylene
                             Cyclobutane

l   Quickly find the % composition of both C2H4
    and C4H8
l   What do we notice?
Now find the % composition of these




Cyclopropane
C 3H 6
                   Cyclopentane
                   C5H10


Hint: You can do this in your head
Molecular vs. Empirical formula

l   Many different molecules can have the same
    empirical formula!
l   The molecular formula is more specific
l   The molecular formula of benzene is C6H6
l   What is its empirical formula?
l   Let’s try C6H12O3
    –   This one’s a little trickier
The trick

                C6H12O3

l What is the greatest common factor
  between 3, 6, and 12
l Simply divide all the subscripts by 3
l C2H4O is the empirical formula
Going from % composition to empirical 
formula

l   Determine the empirical formula of a
    molecule that has is
             92.24 % C; 7.76 % H
1) Assume a 100 gram sample
92.24 g carbon
7.76 g hydrogen
2) Convert both into moles
92.24 g C à 7.7 mole carbon
7.76 g H à 7.7 mole hydrogen
Going from % composition to empirical 
formula

2) Convert both into moles
92.24 g C à 7.7 mole carbon
7.76 g H à 7.7 mole hydrogen
3) Divide all mole values by the lowest
7.7 mol C/ 7.7 = 1 mol C
7.7 mol H/ 7.7 = 1 mol H

The ratio is 1:1 è CH is the empirical formula
Going from % composition to empirical 
formula with a trick

l   Determine the empirical formula of a
    molecule that is
             56.34 % 0; 43.66 % P
1) Assume a 100 gram sample
56.34 g O
43.66 g P
2) Convert both into moles
56.34g / 16.0g/mol O à 3.52 mole oxygen
43.66g / 31g/mol P à 1.41 mole phosphorus
Going from % composition to empirical 
formula

2) Convert both into moles
56.34g / 16.0g/mol O à 3.52 mole oxygen
43.66g / 31g/mol P à 1.41 mole phosphorus

3) Divide all mole values by the lowest
3.52 mol O/ 1.41 = 2.5 mol O
1.41 mol P/ 1.41 = 1 mol P
l Since 2.5 isn’t a whole number you must
   multiply everything by 2 to make it work
Try these in your group

l   36.48 % Na; 25.44 % S; 38.08 % O

l   49.99 % C; 5.61 % H; 44.40 % O

l   38.76 % Ca; 19.97 % P; 41.27 % O

l   The last two have a trick at the end
The next level

l   A compound analyzes as 79.08 % C; 
    5.54 % H and 15.38 % N.  What is the 
    molecular formula if the molar mass 
    is 273.36 g/ mol?

l   Determine the empirical formula and then
    ask me for help.
The density trick

l   The molar mass of any gas may be found by
    taking the product of the density (in g/L) and
    the molar volume (22.4 L/mol at STP).
l   For example:
    –   Find the molar mass of a gas with a density of
        6.39 g/L
The last level for Pre-AP

l   A compound analyzes as 55.80 % C; 
    7.03 % H and 37.17 % O.  What is the 
    molecular formula if this gas has a 
    density of 5.764 g/L?

l   Determine the empirical formula.
l   Find the molar mass
l   Find the molecular formula

				
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posted:6/22/2014
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