# 6 Empirical formula by hedongchenchen

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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

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
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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