UNIT 7 The Mole
PERCENT COMPOSITION
EMPIRICAL AND MOLECULAR FORMULAS
PERCENT COMPOSITION
CALCULATING PERCENT COMPOSITION

General formula for calculating percentage:
part x 100
%=
whole
The question, then, becomes: What is the part,
and what is the whole?
 For compounds, the percent composition is
calculated by mass.

 What
is the part?
 What is the whole?
UNKNOWN COMPOUND
Recently discovered new compound
 Named cookie
 Known to be comprised of two elements, Cc
and Do, in an unknown ratio

 Chocolate

chip and cookie dough, respectively
Task is two-fold:
1.
2.
Calculate the percent composition by mass
Identify empirical formula of compound
 Empirical
formula indicates the simplest whole number
ratio of atoms present in compound
CALCULATING PERCENT COMPOSITION

General formula for calculating percentage:
part x 100
%=
whole
The question, then, becomes: What is the part,
and what is the whole?
 When calculating the percent composition for
compounds, the formula becomes:

% composition
molar mass of element x 100
of element = molar mass of compound
FOR EXAMPLE
Calculate the percent composition of each
element in sodium bicarbonate (sodium
hydrogen carbonate).
1.
Write the formula for sodium bicarbonate.
Na1+NaHCO3HCO31–
FOR EXAMPLE
Calculate the percent composition of each
element in sodium bicarbonate, NaHCO3.
2. Calculate the molar mass of the compound.
Na 1 mol x 23.0 g/mol
H
1 mol x 1.0 g/mol
1 mol x 12.0 g/mol
C
O
3 mol x 16.0 g/mol

=
=
=
=
P
23.0 g
A
1.0 g
R
T
12.0 g
S
48.0 g
84.0 g/mol NaHCO3
WHOLE
FOR EXAMPLE
Calculate the percent composition of each
element in sodium bicarbonate, NaHCO3.
3. Substitute values in the formula and solve.
% Na =
%H=
%C=
%O=
23.0 g Na
84.0 g NaHCO3
1.0 g H
84.0 g NaHCO3
12.0 g C
84.0 g NaHCO3
48.0 g O
84.0 g NaHCO3
x 100 = 27.4% Na
x 100 =
1.2% H
x 100 = 14.3% C
x 100 = 57.1% O
100.0%
EMPIRICAL AND MOLECULAR FORMULAS
EMPIRICAL FORMULA


Represents the simplest
whole number ratio of
elements in a compound
Examples
MOLECULAR FORMULA


Represents the actual
number of atoms in a
compound
Examples
CH
C2H2
CH2
CH2O
C2H4
C6H12O6
DEFINITIONS
EMPIRICAL FORMULA
DETERMINING EMPIRICAL FORMULA
The percent composition of a sulfur oxide is
40.05% S and 59.95% O. Find the empirical
formula.
 When
percentages are given, use the values
with grams as the unit.
1.
Find the number of moles of each element.
Conversion factor: molar mass
mol S =
40.05 g S
1 mol S
32.1 g S
= 1.25 mol S
mol O =
59.95 g O
1 mol O
16.0 g O
= 3.75 mol O
DETERMINING EMPIRICAL FORMULA
Recall that the subscripts in a chemical formula
indicate the number of moles of each element
 If the compound contains sulfur and oxygen in
the ratio of 1.25 to 3.75, then the formula
could be written as
S1.25O3.75


What’s the problem here?
DETERMINING EMPIRICAL FORMULA
The percent composition of a sulfur oxide is
40.05% S and 59.95% O.
2. Calculate the simplest mole ratio of the
elements in the compound by dividing the
number of moles of each element by the
smallest value in the mole ratio.
The resulting factor of each calculation
becomes the subscript for the element in the
empirical formula.
DETERMINING EMPIRICAL FORMULA
The percent composition of a sulfur oxide is 40.05%
S and 59.95% O.
2. Ratio of S to O is 1.25 : 3.75
1.25
Smallest value: 1.25 or 3.75?
Subscript for S:
1.25 mol
=1
1.25 mol
Subscript for O:
3.75 mol
=3
1.25 mol
If results are not whole numbers, must multiply both by a
factor that results in whole numbers.
DETERMINING EMPIRICAL FORMULA
The percent composition of a sulfur oxide is
40.05% S and 59.95% O.
3. Using the values in Step 2 as subscripts, write
the chemical formula.
Subscript of S = 1 Subscript of O = 3
S O
empirical formula =
SO3
MOLECULAR FORMULA
MOLECULAR FORMULA

Recall:
 The
molecular formula shows the actual number
of atoms of each element in the compound
Molecular formula is always a whole number
multiple of the empirical formula
 Therefore, the molecular mass (MM of
molecular formula) will be the same whole
number multiple of the empirical mass (MM of
empirical formula)

DETERMINING MOLECULAR FORMULA
The empirical formula of propene is CH2. What is
its molecular formula if the molar mass is
determined experimentally to be 42.0 grams?
1. Find the molar mass of the empirical formula.
Molar mass of CH2
C: 1 mol x 12.0 g/mol = 12.0 g C
H: 2 mol x 1.0 g/mol = 2.0 g H
14.0 g/mol CH2
DETERMINING MOLECULAR FORMULA
The empirical formula of propene is CH2. What is
its molecular formula if the molar mass is
determined experimentally to be 42.0 grams?
2. Compare the mass of the molecular formula to
that of the empirical formula (M/E).
molar mass of molecular formula
42.0 g
= 3
=
molar mass of empirical formula
14.0 g
DETERMINING MOLECULAR FORMULA
The empirical formula of propene is CH2. What is
its molecular formula if the molar mass is
determined experimentally to be 42.0 grams?
3. Multiply the subscript of each element in the
empirical formula by the resulting factor (n).
molecular formula = (empirical formula)n
molecular formula = (CH2)3
molecular formula = C3H6