Transcript The Mole-Ch 11-Section 11.4

11.4 – Empirical and Molecular
Formulas
11.4 Objectives:

Explain what is meant by the percent
composition of a compound.

Determine the empirical and molecular
formulas for a compound from mass percent
and actual mass data.
Percent Composition

Every chemical compound has a definite
composition. What law is this referring to?
The composition of a compound is usually
stated as the percent by mass of each
element in the compound.
 The type of chemist whose job it is to identify
the elements & their percent by mass in a
compound is an analytical chemist.

The equation used to determine the percent
composition of an element in a compound is:
 total mass of element in compound 
% by mass of an element = 
 *100
molar mass of compound


Summary of Determining
Percent Composition:

Find the mass of each component element
in the compound

Divide by the total mass of the compound

Multiply that answer by 100 to get a percent
Percent Composition
Example Problems:

1) Determine the percent by mass of each
element in calcium chloride (CaCl2).
K:
UK:
36.11% Ca and 63.89% Cl

2) What is the percent of oxygen in H3PO4?
K:
UK:
65.31% oxygen

3) Calculate the percent composition of a
compound that is 29.0 g of Ag with 4.30 g of S.
K:
UK:
87.1% Ag and 12.9% S

4) Which has the larger percent by mass of
sulfur, H2SO3 or H2S2O8?
K:
UK:
H2SO3  39.06% S
Empirical Formula
The information from the percent
composition can be used to determine the
formula for a compound.
 The empirical formula is the simplest
whole-number ratio of atoms of elements in
the compound. In many cases, the empirical
formula is the actual formula for the
compound.
 EX: H2O or H2SO4

Calculating Empirical Formulas


1.
2.
3.
4.
5.

The data used to determine the formula for a compound may
be in the form of percent composition or it may be the actual
masses of the elements in a given compound.
If percent composition is given:
Assume that the total mass of the compound is 100.00 g
Percentages of each element equals the mass in grams.
Turn grams to moles by using the molar mass of each element
Find the mole ratio by dividing everything by the smallest # of
moles
If those numbers are whole numbers you have just found the
subscripts for the formula
If actual masses are given you can skip to step 3.
Example Problem:

What is the empirical formula for a compound that
contains 10.89% magnesium, 31.77% chlorine and
the rest oxygen?
K:
UK:
MgCl2O8

Determine the empirical formula of a compound
containing 2.644g of gold and 0.476g of chlorine.
K:
UK:
AuCl

BUT…often in determining empirical
formulas, the calculated mole ratios are still
not whole numbers. In such cases all the
mole ratio values must be multiplied by the
smallest factor that will make them whole
numbers.
More Examples:
1. A blue solid is found to contain 36.84% nitrogen
and 63.16% oxygen. What is the empirical formula
for the solid?
K:
UK:
N2O3
2.
Propane is a hydrocarbon. It is composed of
81.82% carbon and 18.18% hydrogen. What is
the empirical formula?
K:
UK:
C3H8
Molecular Formula

For many compounds, the empirical formula is not
the true formula.


The molecular formula identifies the actual number
of elements in a molecule.


Examples: CH2 (empirical formula) vs. C2H4 (molecular
formula)
Sometimes the empirical and molecular formulas are the
same. Ex: Water H2O
To determine the molecular formula the molar
mass of the compound must be determined
through experimentation and compared with the
mass represented by the empirical formula.

Notice that the molecular formula for acetic
acid (C2H4O2) has exactly twice as many
atoms of each element as the empirical
formula (CH2O).

The molecular formula for a compound is
always a whole-number multiple of the
empirical formula.

In order to determine the molecular formula
for an unknown compound, you must know
the molar mass of the compound in addition
to its empirical formula.

Then you can compare the molar mass of
the compound with the molar mass
represented by the empirical formula.

This is done using the following equation:
given molar mass of compound
mass of empirical formula
You get a number to multiply the subscripts
of the empirical formula by to get the
molecular formula.
 Let’s do some practice problems.

Practice Problems:
1) Maleic acid is a compound that is used in the plastics
and textiles industries. The composition of maleic acid
is 41.39% carbon, 3.47% hydrogen, and 55.14%
oxygen. Its molar mass is 116.10 g/mol. Calculate the
molecular formula for maleic acid.
K:
UK:

Start by determining the empirical formula:

What is the mole ratio of the elements? 1C:1H:1O

So the empirical formula is: CHO
Next, calculate the molar mass represented
by the empirical formula. 29.02 g/mol
 As stated in the problem, the molar mass of
maleic acid is known to be 116.10 g/mol.
 To determine the molecular formula for
maleic acid, calculate the whole number
multiple to apply to its empirical formula.

116.10 g / mol
 4.001
29.02 g / mol



This calculation shows that the molar mass of
maleic acid is four times the molar mass of its
empirical formula CHO.
Therefore, the molecular formula must have four
times as many atoms of each element as the
empirical formula.
Thus, the molecular formula is C4H4O4
To Review:
=n
More Practice Problems:
2) Caffeine is 49.48% C, 5.15% H, 28.87% N and
16.49% O. It has a molar mass of 194 g. What is its
molecular formula?
K:
UK:
3) A compound was found to contain 49.98 g carbon and 10.47 g
hydrogen. The molar mass of the compound is 58.12 g/mol.
Determine the molecular formula.
K:
UK:
C4H10