Download empirical formula

The Mole & Chemical Formulas
A chemical formula represents the
ratio of atoms that always exists for
that compound
 Example: Water – H2O
 Always 2 H atoms to 1 O atom

The Mole & Chemical Formulas
This also means if you had a mole of
water…
 You would have 2 moles of Hydrogen
atoms
 And 1 mole of Oxygen atoms

Empirical Formula
This is the simplest ratio of atoms in a
chemical formula.
 Example:
 Glucose, a sugar, has the molecular
formula C6H12O6
 The empirical formula is CH2O

Empirical Formula
Example:
 A chemical has 80.0% C atoms and
20.0% H atoms. What is the empirical
formula of the compound?
 Use the %’s as masses to find
moles…

Empirical Formula
80.0 g C 1 mol C
12.0 g C
= 6.67 mol C
20.0 g H 1 mol H
1.0 g H
= 20.0 mol H
Empirical Formula

Use the moles to determine the ratio:
Carbon:
6.67 mol
6.67 mol
Hydrogen:
=1
20.0 mol
6.67 mol
This is the Empirical
Formula!
CH3
=3
Molecular Formula



This gives the actual number of atoms for
each element in a compound.
Example:
If the molar mass of the “CH3” compound
from the previous problem is 30.0 g/mol,
what is the molecular formula?
Molecular Formula
Empirical formula mass for CH3 is
15.0 g/mol
 Then find the ratio of molar to
empirical mass.

30.0 g/mol (molecular)
15.0 g/mol (empirical)
= 2.00
Molecular Formula
Multiply the empirical formula by your
new ratio:
 C(1x2)H(3x2) = C2H6 (molecular formula)

Molecular Formula
Example:
 A chemical is 48.4% C atoms, 8.12%
H atoms and the rest is Oxygen. If
the molecular mass is 222 g/mol, find
the molecular formula.
 First find empirical formula…

Molecular Formula
48.4 g C 1 mol C
12.0 g C
= 4.03 mol C
8.12 g H 1 mol H
1.0 g H
= 8.12 mol H
43.5 g O 1 mol O
16.0 g O
= 2.72 mol O
Molecular Formula
Carbon:
Hydrogen:
Oxygen:
4.03 mol
2.72 mol
8.12 mol
2.72 mol
2.72 mol
2.72 mol
= 1.48 x 2 = 3
= 2.99 x 2 = 6
= 1.00 x 2 = 2
C3H6O2
Molecular Formula
222 g/mol (molecular)
74.0 g/mol (empirical)
C3x3 H6x3 O2x3
= 3.00
C9H18O6
Molecular Formula!