
shows the smallest whole-number ratio of
the atoms in the compound

CH only means it is a 1:1 ratio between
carbon and hydrogen, NOT the actual
number of atoms in the compound
 could be C2H2 (ecetylene), or C6H6
(benzene), or C8H8 (styrene)

What is the empirical formula of a compound that’s
mass is 25.9% N and 74.1% O?
 % is not really a unit of measurement we can use
in chemistry at times
 if there was 100g of this compound then 25.9g
would be nitrogen and 74.1g oxygen
 change to moles in order to convert the % to a #
25.9g N x 1 mol N
14.0 g N
 74.1g O x 1 mol O
16.0 g O

= 1.85 mol N
= 4.63 mol O
What is the whole
number ratio?


a formula is not just the ratio of atoms, it is also
the ratio of moles
this leaves a ratio of N1.85O4.63 which is not a a
possible formula/whole number
 therefore, divide each by the lowest quantity
(in this case, 1.85)
 now is N1O2.5

again, still not a whole number, but if you
were to double it, it would be N2O5
 the empirical formula is N2O5


Calculate the empirical formula of a compound
composed of 38.67 % C, 16.22 % H, and 45.11 %N.
Assume 100 g so…



38.67 g C x
1mol C
= 3.220 mole C
12.01 g C
16.22 g H x
1mol H = 16.09 mole H
1.01 g H
45.11 g N x 1mol N
= 3.219 mole N
14.01 g N



3.220 mole C
16.09 mole H
3.219 mole N
• C3.22H16.09N3.219
If we divide all of these by the smallest
number (3.22), it will give us whole numbers and
therefore the empirical formula

C1H5N1 is the empirical formula

A compound is found to contain 16g of C and 4g
of H. What is the empirical formula?


16 g C x
1mol C
= 1.33 mole C
12.01 g C
4gHx
1mol H = 4 mole H
1.01 g H
 1.33 is too far off from 1
 divide both by 1.33



1.33/1.33 = 1
4/1.33 = 3
empirical formula is CH3
Compound
Water
Hydrogen Peroxide
Glucose
Methane
Ethane
Octane
Molecular Formula
Empirical Formula
H 2O
H2O2
C6H12O6
CH4
C2H 6
H2O
HO
CH2O
CH4
CH3
C4 H 9
C8H18

a molecular formula is the same as, or a multiple
of, the empirical formula
 it is based on the actual number of atoms of each
type in the compound, NOT just the ratio
 the following all have same empirical formula
(CH2O at a 1-2-1 ratio) but different molecular
formulas



C2H4O2 is the molecular formula of ethanoic acid
CH2O is the molecular formula of methanol
C6H12O6 is the molecular formula of glucose

divide the experimental molar mass by the
mass of one mole of the empirical formula


this results in the multiplier to convert to the
molecular formula (this sounds confusing but it really isn’t)
ex: calculate the molecular formula of a
compound whose molar mass is found to be
60.0g and has an empirical formula of CH4N


experimental molar mass of ? --- 60.0 g
=2
empirical molar mass of CH4N ---- 30.0 g
2 (CH4N) = C2H8N2



A compound is known to be composed of 71.65 %
Cl, 24.27% C and 4.07% H. Its molar mass is
known to be 98.96 g. What is its molecular
formula?
need to calculate the empirical formula first
assume 100 g so…

71.65 g Cl x

24.27 g C x

4.07 g H x
1mol Cl = 2.02 mole Cl
35.5 g Cl
1mol C = 2.02 mole C
12.0 g C
1mol H
= 4.07 mole H
1.0 g H

Cl2.02C2.02H4.07


divide by lowest (2.02 mol )
ClCH2 is the empirical formula



if ClCH2 was the molecular formula, this would give
a mass of 48.5 g
however, recall the problem asked for the
molecular formula of a compound with a
molar mass of 98.96 g which is twice that of
48.5 g
therefore, Cl2C2H4 is the molecular formula