Chapter 10 : The Mole
Sec. 10.4: Empirical &
Molecular Formulas
Objectives
• Explain what is mean by the percent
composition of a compound.
• Determine the empirical and molecular
formulas for a compound from mass
percent and actual mass data.
• Synthetic chemists
are often involved in
developing new
compounds for
industrial,
pharmaceutical and
home uses.
• It is the analytical
chemist’s job to
identify the elements
that a compound
contains and
determine its
chemical formula.
Percent Composition
• The first step in this chemist’s job is to
determine the percent by mass of the
elements in the compound.
• Recall that the percent by mass of any
element in a compound is found by:
Mass of Element
x 100
Mass of Compound
The percent by mass of each element in a
compound is the percent composition of
a compound.
Percent composition can be determined
from
•experimental data
•the chemical formula
From Experimental Data
A chemist has a 100.0 g sample of XY. The
sample contains 55.0 g X and 45.0 g Y.
Percent of X = 55.0 g/100.0 g x 100 = 55%
Percent of Y = 100 – 55 = 45%
From the Formula
Assume you have 1 mol of the compound.
1 mol of H2O = 18.0 g
In 1 mol of H2O, there are 2 moles of H atoms and 1
mol O atoms
2 mole H atoms x
1.0 g
= 2.0 g H
1 mol H atoms
1 mol O atoms = 16.0 g
% H = 2.0g/18.0g = 11 %
% O = 100 – 11 = 89%
Practice Problems
• Calculate the percent composition of
sodium sulfate (Na2SO4).
• Which has the larger percent by mass of
sulfur, H2SO3 or H2S2O8?
• What is the percent composition of a
compound that contains 2.644 g of gold
and 0.476 g of chlorine?
Empirical Formula
• The empirical formula is the simplest
formula for a compound.
• A molecular formula is the same as the
empirical formula or it is a multiple of
the empirical formula; it is the actual
number of atoms of each element in one
molecule or formula unit of the
compound.
Empirical Formula
Benzene
Molecular
formula C6H6
Empirical Formula
CH
Empirical Formula
Acetylene
Molecular Formula C2H2
Empirical Formula ?
Empirical Formula
Glucose
Molecular
Formula C6H12O6
Empirical
Formula ?
Empirical Formula
Carbon Dioxide
Molecular Formula CO2
Empirical Formula ?
Empirical Formula
• The empirical formula may or may not be
the same as the molecular formula.
• If they are different, the molecular
formula will be a multiple of the
empirical formula.
To Determine Empirical Formula
Molar
Mass
Empirical Formula
• The percent composition of a compound
was found to be 40.05% S and 59.95% O.
What is the empirical formula for the
compound?
Empirical Formula
• Step One: Assume you have 100 g of the
compound**; Step Two: use molar mass to
determine moles of each element.
• 40.05 g S x 1 mol S = 1.248 mol S
32. 1 g
• 59.95 g O x 1 mol O = 3.747 mol O
16 g
** If you are given mass data instead of %
composition, determine moles directly from
the grams.
Empirical Formula
• Step 3: Divide both mole values by the
smaller of the two to get the mole ratio of
the elements.
1.248 mol S = 1 mol S
1.248
3.747 mol O = 3 mol O
1.248
Empirical Formula
• Now use the simplest, whole number mole
ratio as subscripts in the empirical
formula:
1 mol S: 3 mol O
SO3
Practice Problems
• What is the empirical formula for a
compound that contains 10.89% Mg, 31.77%
Cl, and 57.34% O?
• Determine the empirical formula for a
compound that contains 74.19% Na and
25.81% O.
• When an oxide of potassium is
decomposed, 19.55 g of K and 4.00 g of O
are obtained. What is the empirical
formula of the compound?
Honors Practice Problem
• Determine the empirical formula for a
compound that contains 35.98%
aluminum and 64.02% sulfur.
• The chemical analysis of aspirin
indicates that the molecule is 60.00% C,
4.44% H, and 35.56% O. Determine the
empirical formulas for aspirin.
Molecular Formula
• Compounds with the same empirical
formula can have very different properties.
• Remember, the empirical formula does not
always indicate the actual number of moles
in the compound. So, different compounds
can have the same empirical formula.
• Acetylene (C2H2) and benzene (C6H6) are
different compounds with the same
empirical formula, CH.
Molecular Formula
• In order to distinguish between different
compounds with the same empirical
formula, a chemist must go one step further
and determine the compound’s molecular
formula.
• The molar mass of the compound is
determined through experimentation and
compared with the molar mass
represented by the empirical formula.
Molecular Formula
• Suppose a compound has an empirical
formula of ClCH2 and a molar mass of 98.96
g/mol. How can its molecular formula be
determined?
• Step One: Find the molar mass of the
empirical formula.
ClCH2 = 35.5 + 12.0 + 2(1.0) = 49.5 g/mol
Molecular Formula
• Step 2: Divide the molar mass of the
compound by the molar mass of the
empirical formula.
98.96 g/mol = 1.999 = 2
49.5 g/mol
The molecular formula is this multiple of
the empirical formula.
Molecular Formula
•
Step 3: Multiply the subscripts in the
empirical formula by this multiple.
ClCH2 becomes Cl2C2H4. This is the
molecular formula of the compound.
Practice Problems
•
•
The empirical formula of a compound is
found to be C2H3O2. It has a molar mass of
118.1 g/mol. Determine the molecular
formula for the compound.
The molar mass for a compound having
the empirical formula of CH is found to be
78.1 g/mol. What is the molecular formula
for the compound?
Honors Practice Problem
• A colorless liquid composed of 46.68%
nitrogen and 53.32% oxygen has a molar
mass of 60.01 g/mol. What is the molecular
formula?
• A compound was found to contain 49.98 g C
and 10.47 g H. The molar mass of the
compound is 58.12 g/mol. Determine the
molecular formula.