and Empirical Formula

Determine the mass percentage of each
element in the compound.
mass _ of _ element
100
mass _ of _ compound

2 Fe  2 x 55.8 = 111.6 g

3 O  3 x 16.0 = 48.0 g

Formula mass = 159.6 g

% Fe =
111.6

159.6
70.0 % Fe

%O=
48.0

159.6
30.0% O

Gives the lowest whole # ratio of elements in
a compound.

The empirical formula for C6H12O6 is

The empirical formula for C2H6 is

* most basic ratio of elements in the
compound
CH2O
CH3

X (empirical formula) = molecular formula

So. . . .
molecular _ formula _ mass
X=
empirical _ formula _ mass

And. . .
emp _ formula  X  mol _ formula _ mass

What is the molecular formula of the
molecule that has an empirical formula of
CH2O and a molar mass of 120.12 g/mol?
◦ We need to find the molar mass of the empirical
formula first.

Easy level
◦ C4H8O4  What is the empirical formula?

Medium Difficulty
◦ The previous problem

Hardest Difficulty
◦ Next problem


Find the empirical formula of a compound that
contains 53.7% iron and 46.3% sulfur.
Steps:
◦ % composition mass of 100g sample moles mole
ratio
1
mol
53.7 g 

55.8 g
0.962 mol Fe
46.3 g  1mol 
1.44 mol S
32.1g

Take smaller # and divide everything by it
0.962
Fe 
 1.0
0.962
1.44
S
 1 .5
0.962
1:1.5 2 
2:3
Fe2S3