Transcript 10.3 Percent Composition and Chemical Formulas

10.3 Percent Composition
and Chemical Formulas >
Chapter 10
Chemical Quantities
10.1 The Mole: A Measurement of
Matter
10.2 Mole-Mass and Mole-Volume
Relationships
10.3 Percent Composition and
Chemical Formulas
1
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10.3 Percent Composition
and Chemical Formulas >
CHEMISTRY
& YOU
What does the percent composition of a
compound tell you?
A tag sewn into the
seam of a shirt usually
tells you what fibers
were used to make
the cloth and the
percent of each.
2
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Percent Composition of a
10.3 Percent Composition
and Chemical Formulas > Compound
Percent Composition of a Compound
How do you calculate the percent
composition of a compound?
3
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Percent Composition of a
10.3 Percent Composition
and Chemical Formulas > Compound
The relative amounts of the elements in a
compound are expressed as the percent
composition or the percent by mass of
each element in the compound.
4
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Percent Composition of a
10.3 Percent Composition
and Chemical Formulas > Compound
The percent composition of potassium
chromate, K2CrO4, is:
K = 40.3%
Cr = 26.8%
O = 32.9%
5
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Percent Composition of a
10.3 Percent Composition
and Chemical Formulas > Compound
These percents must total 100%.
K = 40.3%
Cr = 26.8%
+ O = 32.9%
100%
6
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Percent Composition of a
10.3 Percent Composition
and Chemical Formulas > Compound
These percents must total 100%.
• The percent composition of a compound is
always the same.
7
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Percent Composition of a
10.3 Percent Composition
and Chemical Formulas > Compound
Percent Composition from Mass Data
If you know the relative masses of each
element in a compound, you can calculate
the percent composition of the compound.
8
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Percent Composition of a
10.3 Percent Composition
and Chemical Formulas > Compound
Percent Composition from Mass Data
The percent by mass of an element
in a compound is the number of
grams of the element divided by the
mass in grams of the compound,
multiplied by 100%.
mass of element
% by mass of element = mass of compound ×
100%
9
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.9
Calculating Percent
Composition from Mass Data
When a 13.60-g sample of a
compound containing only
magnesium and oxygen is
decomposed, 5.40 g of oxygen is
obtained. What is the percent
composition of this compound?
10
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.9
1 Analyze List the knowns and the unknowns.
The percent by mass of an element in a
compound is the mass of that element divided by
the mass of the compound multiplied by 100%.
KNOWNS
mass of compound = 13.60 g
mass of oxygen = 5.40 g O
mass of magnesium = 13.60 g – 5.40 g O = 8.20 g Mg
UNKNOWNS
percent by mass of Mg = ?% Mg
percent by mass of O = ?% O
11
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.9
2 Calculate Solve for the unknowns.
Determine the percent by mass of
Mg in the compound.
% Mg =
=
mass of Mg
×
mass of compound 100%
8.20 g
×
13.60 g 100%
= 60.3% Mg
12
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.9
2 Calculate Solve for the unknowns.
Determine the percent by mass of
O in the compound.
%O=
=
mass of O
×
mass of compound 100%
5.40 g
×
13.60 g 100%
= 39.7% O
13
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.9
3 Evaluate Does the result make sense?
The percents of the elements add up
to 100%.
60.3% + 39.7% = 100%
14
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Percent Composition of a
10.3 Percent Composition
and Chemical Formulas > Compound
Percent Composition from the
Chemical Formula
You can also calculate the percent
composition of a compound using its
chemical formula.
• The subscripts in the formula are used to
calculate the mass of each element in a mole of
that compound.
• Using the individual masses of the elements and
the molar mass, you can calculate the percent
by mass of each element.
15
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Percent Composition of a
10.3 Percent Composition
and Chemical Formulas > Compound
Percent Composition from the
Chemical Formula
You can also calculate the percent
composition of a compound using its
chemical formula.
% by mass = mass of element in 1 mol compound
×
of element
molar mass of compound
100%
16
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.10
Calculating Percent Composition
from a Formula
Propane (C3H8), the fuel
commonly used in gas grills, is
one of the compounds obtained
from petroleum. Calculate the
percent composition of
propane.
17
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.10
1 Analyze List the knowns and the unknowns.
Calculate the percent by mass of each element
by dividing the mass of that element in one mole
of the compound by the molar mass of the
compound and multiplying by 100%.
KNOWNS
mass of C in 1 mol C3H8 = 3 mol × 12.0 g/mol = 36.0 g
mass of H in 1 mol C3H8 = 8 mol × 1.0 g/mol = 8.0 g
molar mass of C3H8 = 36.0 g/mol + 8.0 g/mol = 44.0 g/mol
UNKNOWNS
percent by mass of C = ?% C
percent by mass of H = ?% H
18
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.10
2 Calculate Solve for the unknowns.
Determine the percent by mass of C
in C3H8.
%C=
=
mass of C in 1 mol C3H8
molar mass of C3H8
×
100%
36.0 g
×
44.0 g 100%
= 81.8% C
19
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.10
2 Calculate Solve for the unknowns.
Determine the percent by mass of H
in C3H8.
%H=
=
mass of H in 1 mol C3H8
molar mass of C3H8
×
100%
8.0 g
×
44.0 g 100%
= 18% H
20
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.10
3 Evaluate Does the result make sense?
The percents of the elements add up to
100% when the answers are expressed to
two significant figures (82% + 18% =
100%).
21
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Percent Composition of a
10.3 Percent Composition
and Chemical Formulas > Compound
Percent Composition as a Conversion
Factor
You can use percent composition to
calculate the number of grams of any
element in a specific mass of a compound.
• To do this, multiply the mass of the
compound by a conversion factor based on
the percent composition of the element in
the compound.
22
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Percent Composition of a
10.3 Percent Composition
and Chemical Formulas > Compound
Percent Composition as a Conversion
Factor
Propane is 81.8% carbon and 18%
hydrogen.
• You can use the following conversion factors
to solve for the mass of carbon or hydrogen
contained in a specific amount of propane.
81.8 g C
18 g H
and
100 g C3H8
100 g C3H8
23
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10.3 Percent Composition
and Chemical Formulas >
CHEMISTRY
& YOU
What information can you get from the
percent composition of a compound?
24
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10.3 Percent Composition
and Chemical Formulas >
CHEMISTRY
& YOU
What information can you get from the
percent composition of a compound?
You can use percent
composition to determine
the mass of an element in
a sample of a compound
of a given size. You can
also determine the
empirical formula of the
compound.
25
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.11
Calculating the Mass of an
Element in a Compound Using
Percent Composition
Calculate the mass of carbon and
the mass of hydrogen in 82.0 g of
propane (C3H8).
26
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.11
1 Analyze List the known and the unknowns.
Use the conversion factors based on the percent
composition of propane to make the following
conversions: grams C3H8 → grams C and grams
C3H8 → grams H.
KNOWN
mass of C3H8 = 82.0 g
27
UNKNOWNS
mass of carbon = ? g C
mass of hydrogen = ? g H
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.11
2 Calculate Solve for the unknowns.
• To calculate the mass of C, first write the
conversion factor to convert from mass of C3H8
to mass of C.
From Sample
81.8 g C
100 g C3H8
Problem 10.10, the
percent by mass of C
in C3H8 is 81.8%.
• Multiply the mass of C3H8 by the
conversion factor.
28
82.0 g C3H8
×
81.8 g C
= 67.1 g C
100 g C3H8
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.11
2 Calculate Solve for the unknowns.
• To calculate the mass of H, first write the
conversion factor to convert from mass of C3H8
to mass of H.
From Sample
18 g H
100 g C3H8
Problem 10.10, the
percent by mass of H
in C3H8 is 18%.
• Multiply the mass of C3H8 by the
conversion factor.
29
82.0 g C3H8
×
18 g H
= 15 g H
100 g C3H8
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.11
3 Evaluate Does the result make sense?
The sum of the two masses equals 82 g,
the sample size, to two significant figures
(67 g C + 15 g H = 82 g C3H8).
30
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10.3 Percent Composition
and Chemical Formulas >
What data can you use to calculate
percent composition?
31
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10.3 Percent Composition
and Chemical Formulas >
What data can you use to calculate
percent composition?
You can calculate percent composition if you
know the mass of a compound and the masses
of the elements contained in the compound, or if
you know the chemical formula, the molar mass
of the compound, and the atomic weights of the
elements contained in the compound.
32
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10.3 Percent Composition
Empirical Formulas
>
and Chemical Formulas
Empirical Formulas
How can you calculate the empirical
formula of a compound?
33
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10.3 Percent Composition
Empirical Formulas
>
and Chemical Formulas
The empirical formula of a compound gives
the lowest whole-number ratio of the atoms
or moles of the elements in a compound.
• An empirical formula may or may not be the
same as a molecular formula.
34
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10.3 Percent Composition
Empirical Formulas
>
and Chemical Formulas
The empirical formula of a compound gives
the lowest whole-number ratio of the atoms
or moles of the elements in a compound.
• An empirical formula may or may not be the
same as a molecular formula.
– For example, the lowest ratio of hydrogen to oxygen
in hydrogen peroxide is 1:1. Thus, the empirical
formula of hydrogen peroxide is HO.
– The molecular formula, H2O2, has twice the number of
atoms as the empirical formula.
35
– Notice that the ratio of hydrogen to oxygen is still the
same, 1:1.
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10.3 Percent Composition
Empirical Formulas
>
and Chemical Formulas
For carbon dioxide, the empirical and
molecular
formulas
are the
same—
CO2.
36
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10.3 Percent Composition
Empirical Formulas
>
and Chemical Formulas
The figure below shows two compounds of
carbon and hydrogen having the same
empirical formula (CH) but different
molecular formulas.
Ethyne (C2H2), also
called acetylene, is a gas
used in welders’ torches.
37
Styrene (C8H8) is used
in making polystyrene.
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10.3 Percent Composition
Empirical Formulas
>
and Chemical Formulas
The percent composition of a compound
can be used to calculate the empirical
formula of that compound.
38
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10.3 Percent Composition
Empirical Formulas
>
and Chemical Formulas
The percent composition of a compound
can be used to calculate the empirical
formula of that compound.
• The percent composition tells the ratio of masses of
the elements in a compound.
• The ratio of masses can be changed to ratio of moles
by using conversion factors based on the molar mass
of each element.
• The mole ratio is then reduced to the lowest wholenumber ratio to obtain the empirical formula of the
compound.
39
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.12
Determining the Empirical
Formula of a Compound
A compound is analyzed and
found to contain 25.9% nitrogen
and 74.1% oxygen. What is the
empirical formula of the
compound?
40
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.12
1 Analyze List the knowns and the unknown.
The percent composition gives the ratio of the
mass of nitrogen atoms to the mass of oxygen
atoms in the compound. Change the ratio of
masses to a ratio of moles and reduce this ratio
to the lowest whole-number ratio.
KNOWNS
percent by mass of N = 25.9% N
percent by mass of O = 74.1% O
UNKNOWN
empirical formula = N?O?
41
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.12
2 Calculate Solve for the unknown.
Convert the percent by mass of
each element to moles.
1 mol N
25.9 g N
×
= 1.85 mol N
14.0 g N
74.1 g O
×
1 mol O
= 4.63 mol O
16.0 g O
Percent means
“parts per 100,” so
100.0 g of the
compound
contains 25.9 g N
and 74.1 g O.
The mole ratio of N to O is N1.85O4.63.
42
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.12
2 Calculate Solve for the unknown.
Divide each molar quantity by the smaller
number of moles to get 1 mol for the
element with the smaller number of moles.
1.85 mol N = 1 mol N
1.85
4.63 mol O
= 2.50 mol O
1.85
The mole ratio of N to O is N1O2.5.
43
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.12
2 Calculate Solve for the unknown.
Multiply each part of the ratio by the
smallest whole number that will convert
both subscripts to whole numbers.
1 mol N × 2 = 2 mol N
2.5 mol O × 2 = 5 mol O
The empirical formula is N2O5.
44
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.12
3 Evaluate Does the result make sense?
The subscripts are whole numbers, and
the percent composition of this empirical
formula equals the percents given in the
original problem.
45
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10.3 Percent Composition
and Chemical Formulas >
You are doing an experiment to try to find
the molecular formula of a compound. You
discover the percent composition. Can
you determine the molecular formula?
46
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10.3 Percent Composition
and Chemical Formulas >
You are doing an experiment to try to find
the molecular formula of a compound. You
discover the percent composition. Can
you determine the molecular formula?
You can determine the empirical formula.
This might be the same as the molecular
formula, or it might not. You would need
more data to be sure of the molecular
formula.
47
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10.3 Percent Composition
Molecular Formulas
>
and Chemical Formulas
Molecular Formulas
How does the molecular formula of a
compound compare with the
empirical formula?
48
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10.3 Percent Composition
and Chemical Formulas >
Interpret Data
Ethyne and benzene have the same
empirical formula—CH.
Comparison of Empirical and Molecular Formulas
49
Formula (name)
Classification of
formula
Molar mass
(g/mol)
CH
Empirical
13
C2H2 (ethyne)
Molecular
26 (2 × 13)
C6H6 (benzene)
Molecular
78 (6 × 13)
CH2O (methanol)
Empirical and molecular
30
C2H4O2 (ethanoic acid)
Molecular
60 (2 × 30)
C6H12O6 (glucose)
Molecular
180 (6 × 30)
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10.3 Percent Composition
and Chemical Formulas >
Interpret Data
Methanal, ethanoic acid, and glucose have
the same empirical formula—CH2O.
Comparison of Empirical and Molecular Formulas
50
Formula (name)
Classification of
formula
Molar mass
(g/mol)
CH
Empirical
13
C2H2 (ethyne)
Molecular
26 (2 × 13)
C6H6 (benzene)
Molecular
78 (6 × 13)
CH2O (methanal)
Empirical and molecular
30
C2H4O2 (ethanoic acid)
Molecular
60 (2 × 30)
C6H12O6 (glucose)
Molecular
180 (6 × 30)
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10.3 Percent Composition
and Chemical Formulas >
Interpret Data
Notice that the molar masses of the compounds in these
two groups are simple whole-number multiples of the
molar masses of the empirical formulas, CH and CH2O.
Comparison of Empirical and Molecular Formulas
51
Formula (name)
Classification of
formula
Molar mass
(g/mol)
CH
Empirical
13
C2H2 (ethyne)
Molecular
26 (2 × 13)
C6H6 (benzene)
Molecular
78 (6 × 13)
CH2O (methanal)
Empirical and molecular
30
C2H4O2 (ethanoic acid)
Molecular
60 (2 × 30)
C6H12O6 (glucose)
Molecular
180 (6 × 30)
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10.3 Percent Composition
Molecular Formulas
>
and Chemical Formulas
Methanal
(formaldehyde),
ethanoic acid (acetic
acid), and glucose have
the same empirical
formula—CH2O.
52
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10.3 Percent Composition
Molecular Formulas
>
and Chemical Formulas
The molecular formula of a compound
is either the same as its experimentally
determined empirical formula, or it is a
simple whole-number multiple of its
empirical formula.
53
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10.3 Percent Composition
Molecular Formulas
>
and Chemical Formulas
The molecular formula of a compound
is either the same as its experimentally
determined empirical formula, or it is a
simple whole-number multiple of its
empirical formula.
• Once you have determined the empirical
formula of a compound, you can determine
its molecular formula, if you know the
compound’s molar mass.
54
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10.3 Percent Composition
Molecular Formulas
>
and Chemical Formulas
You can calculate the empirical formula
mass (efm) of a compound from its empirical
formula.
• This is simply the molar mass of the empirical
formula.
55
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10.3 Percent Composition
Molecular Formulas
>
and Chemical Formulas
You can calculate the empirical formula
mass (efm) of a compound from its empirical
formula.
• Then you can divide the experimentally
determined molar mass by the empirical formula
mass.
• This quotient gives the number of empirical
formula units in a molecule of the compound and
is the multiplier to convert the empirical formula
to the molecular formula.
56
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.13
Finding the Molecular Formula of a
Compound
Calculate the molecular formula of a
compound whose molar mass is 60.0 g/mol
and empirical formula is CH4N.
57
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.13
1 Analyze List the knowns and the unknown.
• Divide the molar mass by the empirical
formula mass to obtain a whole number.
• Multiply the empirical formula subscripts by
this value to get the molecular formula.
KNOWNS
empirical formula = CH4N
molar mass = 60.0 g/mol
UNKNOWN
molecular formula = C?H?N?
58
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.13
2 Calculate Solve for the unknown.
First calculate the empirical formula
mass.
efm of CH4N = 12.0 g/mol + 4(1.0 g/mol) + 14.0 g/mol
= 30.0 g/mol
59
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.13
2 Calculate Solve for the unknown.
• Divide the molar mass by the
empirical formula mass.
molar mass
efm
60.0 g/mol = 2
=
30.0 g/mol
• Multiply the formula subscripts by this
value.
(CH4N) × 2 = C2H8N2
60
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10.3 Percent Composition
and Chemical Formulas >
Sample Problem 10.13
3 Evaluate Does the result make sense?
The molecular formula has the molar
mass of the compound.
61
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10.3 Percent Composition
and Chemical Formulas >
What information, in addition to
empirical formula, is necessary to
determine the molecular formula of a
compound?
62
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10.3 Percent Composition
and Chemical Formulas >
What information, in addition to
empirical formula, is necessary to
determine the molecular formula of a
compound?
Molecular formula can be determined if
the empirical formula and the molecular
mass of a compound are known.
63
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10.3 Percent Composition
Key Concepts
>
and Chemical Formulas
The percent by mass of an element in a
compound is the number of grams of
the element divided by the mass in
grams of the compound, multiplied by
100%.
The percent composition of a
compound can be used to calculate the
empirical formula of that compound.
64
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Key Concepts and
10.3 Percent Composition
and Chemical Formulas > Key Equations
The molecular formula of a compound is
either the same as its experimentally
determined formula, or it is a simple
whole-number multiple of its empirical
formula.
% by mass
mass of element
of element = mass of compound ×
100%
% by mass = mass of element in 1 mol compound
×
of element
molar mass of compound
100%
65
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10.3 Percent Composition
Glossary Terms
>
and Chemical Formulas
• percent composition: the percent by
mass of each element in a compound
• empirical formula: a formula with the
lowest whole-number ratio of
elements in a compound; the
empirical formula of hydrogen
peroxide (H2O2) is HO
66
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10.3 Percent Composition
and Chemical Formulas >
BIG IDEA
The Mole and Quantifying Matter
• The molecular formula of a compound
can be determined by first finding the
percent composition of the compound
and determining the empirical formula.
• Using the empirical formula mass and
the molar mass of the compound, the
molecular formula can be determined.
67
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10.3 Percent Composition
and Chemical Formulas >
END OF 10.3
68
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