5.5 Molecular Formula
Percent Composition
Review of
Empirical Formula
A compound is found to contain 77.9% I, 22.1 % O . Calculate the empirical formula.
77.9 g I x
22.1 g O
1 mole
g
=
mol
x 1 mole
g
=
mol
=
=
Molecular Formula
• Actual number of atoms of each element in
a molecule.
• Compare it to empirical formula, which is
the simplest whole number ratio of atoms
in a molecule.
Empirical Formula
Molar Mass
Molecular Formula
Molecular Mass
C4H8NO
86.08 g/mol
C12H24N3O3
258.24 g/mol
C7H12
96.12 g/mol
C14H24
192.24 g/mol
C 4 H8 O
72.08 g/mol
C20H40O5
360.4 g/mol
CH
13.01 g/mol
C6H6
78.06 g/mol
192.24
96.12
= 2
360.4
72.08
= 5
78.06
= 6
13.01
A compound is found to contain 30.8 % C, 4.27 % H, 23.9 % N, and 41.0 % O. The
molecular mass is 468.0 g/mole. Determine the empirical and molecular formulas.
30.8 g C x
1 mol
12.0 g
=
2.567 mol
1.707 mol
= 1.504
4.27 g H
1 mol
1.01 g
=
4.227 mol
1.707 mol
= 2.476
x
=
3
=
5
x2=
23.9 g N x
1 mol
14.0 g
=
1.707 mol
1.707 mol
= 1
=
2
41.0 g O x
1 mol
16.0 g
=
2.563 mol
1.707 mol
= 1.501
=
3
C3H5N2O3
117.0 g/mole
x4=
468.0 g/mole
C12H20N8O12
How to find the molecular formula
1- Find the empirical formula as you always do.
2- Calculate the molar mass of the empirical formula you got.
3- Divide the molecular mass by the molar mass:
Molecular mass
Molar mass
4- Multiply the subscripts in the EF by this number in step(3)
This new formula will be your molecular
formula!
TRY:A compound is found to contain 66.35 g C, 17.54 g H, 33.17 g N, and 63.22 g O.
The molecular mass is 456.44 g/mole. Determine the empirical and molecular formulas.
Hint: use 1.01 g/mol for H.
66.35 g C x 1 mol
=
5.529 mol
= 2.334
=
7
2.369
mol
12.0 g
17.54 g H x
1 mol
1.01 g
=
17.37 mol
2.369 mol
= 7.332
=
22
x3=
33.17 g N x
1 mol
14.0 g
=
2.369 mol
2.369 mol
= 1
=
3
63.22 g O x
1 mol
16.0 g
=
3.9513 mol
2.369 mol
= 1.668
=
5
C7H22N3O5
228.22 g/mole
x2=
456.44 g/mole
C14H44N6O10
Another type of questions
• You are not given the molecular mass.
• What to do?!
• Usually these questions are only about
gases at STP.
A gas has the empirical formula CH2. If 0.85 L of the gas at
STP has a mass of 1.59 g, what is the molecular formula?
What did we just learn?
• We use 22.4 L/mol to help us find the
molecular mass
– For gases only
– At STP
• We use the density (g/L) to find molecular
mass (g/mole)
Homework
Try in this order!
Page 95
#54, 50, 52.