Chapter 7
Chemical Formulas and Chemical
Compounds
Table of Contents
Section 4 Determining Chemical Formulas
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Chapter 7
Section 4 Determining Chemical
Formulas
Objectives
• Define empirical formula, and explain how the term
applies to ionic and molecular compounds.
• Determine an empirical formula from either a
percentage or a mass composition.
• Explain the relationship between the empirical
formula and the molecular formula of a given
compound.
• Determine a molecular formula from an empirical
formula.
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Chapter 7
Section 4 Determining Chemical
Formulas
Empirical and Actual Formulas
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7.4
Percent Composition and
Chemical Formulas
>
Empirical Formulas
Empirical Formulas
What does the empirical formula of a
compound show?
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7.4
Percent Composition and
Chemical Formulas
>
Empirical Formulas
The empirical formula gives the lowest wholenumber ratio of the atoms of the elements in a
compound.
The empirical formula of a compound
shows the smallest whole-number ratio
of the atoms in the compound.
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7.4
Percent Composition and
Chemical Formulas
>
Empirical Formulas
Ethyne (C2H2) is a gas used in
welder’s torches. Styrene
(C8H8) is used in making
polystyrene.
These two compounds of
carbon have the same
empirical formula (CH) but
different molecular formulas.
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Chapter 7
Section 4 Determining Chemical
Formulas
• An empirical formula consists of the symbols for the
elements combined in a compound, with subscripts
showing the smallest whole-number mole ratio of the
different atoms in the compound.
• For an ionic compound, the formula unit is usually the
compound’s empirical formula.
• For a molecular compound, however, the empirical
formula does not necessarily indicate the actual
numbers of atoms present in each molecule.
• example: the empirical formula of the gas diborane is BH3,
but the molecular formula is B2H6.
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Empirical Formulas
• To determine a compound’s empirical formula from its
percentage composition, begin by converting
percentage composition to a mass composition.
• Assume that you have a 100.0 g sample of the
compound.
• Then calculate the amount of each element in the
sample.
• example: diborane
• The percentage composition is 78.1% B and 21.9% H.
• Therefore, 100.0 g of diborane contains 78.1 g of B
and 21.9 g of H.
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Empirical Formulas, continued
• Next, the mass composition of each element is
converted to a composition in moles by dividing by
the appropriate molar mass.
• These values give a mole ratio of 7.22 mol B to
21.7 mol H.
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Empirical Formulas, continued
• To find the smallest whole number ratio, divide each
number of moles by the smallest number in the
existing ratio.
• Because of rounding or experimental error, a
compound’s mole ratio sometimes consists of numbers
close to whole numbers instead of exact whole
numbers.
• In this case, the differences from whole numbers may be ignored
and the nearest whole number taken.
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Empirical Formulas, continued
Sample Problem L
Quantitative analysis shows that a compound contains
32.38% sodium, 22.65% sulfur, and 44.99% oxygen.
Find the empirical formula of this compound.
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Empirical Formulas, continued
Sample Problem L Solution
Given: percentage composition: 32.38% Na, 22.65% S,
and 44.99% O
Unknown: empirical formula
Solution:
percentage composition
composition in moles
mass composition
smallest whole-number
mole ratio of atoms
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Empirical Formulas, continued
Sample Problem L Solution, continued
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Empirical Formulas, continued
Sample Problem L Solution, continued
Smallest whole-number mole ratio of atoms: The
compound contains atoms in the ratio 1.408 mol
Na:0.7063 mol S:2.812 mol O.
Rounding yields a mole ratio of 2 mol Na:1 mol S:4 mol O.
The empirical formula of the compound is Na2SO4.
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Molecular Formulas
• The empirical formula contains the smallest possible
whole numbers that describe the atomic ratio.
• The molecular formula is the actual formula of a
molecular compound.
• An empirical formula may or may not be a correct
molecular formula.
• The relationship between a compound’s empirical
formula and its molecular formula can be written as
follows.
x(empirical formula) = molecular formula
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Molecular Formulas, continued
• The formula masses have a similar relationship.
x(empirical formula mass) = molecular formula mass
• To determine the molecular formula of a compound,
you must know the compound’s formula mass.
• Dividing the experimental formula mass by the empirical
formula mass gives the value of x.
• A compound’s molecular formula mass is numerically
equal to its molar mass, so a compound’s molecular
formula can also be found given the compound’s
empirical formula and its molar mass.
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Chapter 7
Section 4 Determining Chemical
Formulas
Comparing Empirical and Molecular Formulas
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Chapter 7
Visual Concepts
Comparing Molecular and Empirical Formulas
Click below to watch the Visual Concept.
http://my.hrw.com/sh/hc6_003036809x
Visual Concept
/student/ch07/sec04/vc00/hc607_04_v
00fs.htm
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Chapter 7
Visual Concepts
Empirical Formula Problem Activity
Click below to watch the PROBLEM ACTIVITY.
PROBLEM ACTIVITY
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Chapter 7
Visual Concepts
Molecular Formula Problem Activity
Click below to watch the PROBLEM ACTIVITY.
PROBLEM ACTIVITY
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Molecular Formulas, continued
Sample Problem N
In Sample Problem M in textbook, the empirical formula
of a compound of phosphorus and oxygen was found to
be P2O5. Experimentation shows that the molar mass of
this compound is 283.89 g/mol. What is the
compound’s molecular formula?
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Molecular Formulas, continued
Sample Problem N Solution
Given: empirical formula
Unknown: molecular formula
Solution:
x(empirical formula) = molecular formula
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Molecular Formulas, continued
Sample Problem N Solution, continued
Molecular formula mass is numerically equal to molar mass.
molecular molar mass = 283.89 g/mol
molecular formula mass = 283.89 amu
empirical formula mass
mass of phosphorus atom = 30.97 amu
mass of oxygen atom = 16.00 amu
empirical formula mass of P2O5 =
2  30.97 amu + 5 × 16.00 amu = 141.94 amu
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Chapter 7
Section 4 Determining Chemical
Formulas
Calculation of Molecular Formulas, continued
Sample Problem N Solution, continued
Dividing the experimental formula mass by the
empirical formula mass gives the value of x.
(P2O5) = P4O10
The compound’s molecular formula is therefore P4O10.
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7.4 Section Quiz.
Assess students’ understanding of
the concepts in Section
7.4.
Continue to:
-or-
Launch:
Section Quiz
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7.4 Section Quiz.
1. Calculate the percent by mass of carbon in
cadaverine, C5H14N2, a compound present in
rotting meat.
a. 67.4% C
b. 58.8% C
c. 51.7% C
d. 68.2% C
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7.4 Section Quiz.
2. Which of the following is NOT an empirical
formula?
a. NO2
b. H2N
c. CH
d. C3H6
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7.4 Section Quiz.
3. Determine the molecular formula of a
compound that contains 40.0 percent C, 6.71
percent H, and 53.29 percent O and has a
molar mass of 60.05 g.
a. C2H4O2
b. CH2O
c. C2H3O
d. C2H4O
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Online Self-Check Quiz
Complete the online Quiz and record answers.
Ask if you have any questions about your
answers.
click here for online Quiz 7.4
(7 questions)
You must be in the “Play mode” for the
slideshow for hyperlink to work.
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Chapter 7
Practice Problems pg. 249
• Complete Section Review 7.4 pg. 249 #1-4 as
homework. You may record in your notes and
separate notebook paper as needed. #5 is bonus.
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Answers for HW pg. 249 #1-4 (#5 bonus)
FOR ME ONLY.
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VIDEOS FOR ADDITIONAL INSTRUCTION
Additional Videos for Section 7.4: Determining Chemical
Formulas
Empirical Formula - Molecular Formula
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SCI LINKS FOR CHAPTER
Additional Student SCI LINKS for CHAPTER 7
The NSTA-sponsored SciLinks Web site contains links to accurate and upto-date science
information on the Internet. Just click on the button below to go to the
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research. The following is a list of the SciLinks codes for this chapter.
Chapter 7: Chemical Bonding
Topic: Chemical Formulas
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SciLinks code: HC60271
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