Chapter 4:
Chemical
Composition
4-1
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Questions for Consideration
1.
2.
3.
4.
How can we describe the mass composition of
elements in a compound?
How can we determine the number of atoms in a
given mass of a material? The number of
molecules? The number of formula units?
How can we use the masses of elements in a
compound to determine its chemical formula?
How can we express the composition of a
solution?
4-2
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Chapter 4 Topics:
1.
2.
3.
4.
Percent Composition
Mole Quantities
Determining Empirical and Molecular
Formulas
Chemical Composition of Solutions
4-3
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Chapter 4 Math Toolbox:
4.1 Mole Quantities

This math toolbox gives a concise overview of
the mole-quantity calculations that are
introduced in more detail throughout this
chapter.
4-4
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4.1 Percent Composition



We have used chemical formulas to express the
composition of compounds, but where do these
formulas come from?
We cannot directly determine a chemical formula.
However, we can measure the mass of each
element in a sample of a compound by using
appropriate analytical techniques.
Because the mass of each element in a compound
will vary from one sample of the compound to
another, we must have some expression of
composition that is the same for all samples.
4-5
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Percent Composition


A convenient method for expressing composition
is percent composition by mass.
For any element, E, in a compound, the percent
composition by mass is given by the following
equation:
mass of E
%E =
 100%
mass of sample
4-6
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Moles, Masses, and Particles


How can we describe the composition of a
compound if we know the mass of the elements
in the compound?
A 3.67-g sample of the mineral chalcopyrite was
determined to contain 1.27 g Cu, 1.12 g Fe, and
1.28 g S.

What is the mass percent of each element in this
compound?
4-7
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Percent Composition

The percent
composition for all
the elements present
in a compound must
add up to 100%.
Figure 4.5
4-8
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Composition of Chalcopyrite

Would this mass %
differ for a different
sample of this
mineral?

If you had a 100-gram
sample, what mass of
copper would it
contain?
Figure 4.5
4-9
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Activity: Percent Composition


A sample of a copper compound weighs 1.63 g
and contains 1.30 g of Cu. The other element in
the compound is oxygen. What is the percent
composition of this compound?
By difference, the sample contains 0.33 g O.
Applying the formula for percent composition
gives:
1.30 g Cu
%Cu =
 100% = 79.8%
1.63 g sample
0.33 g O
%O =
 100% = 20.2%
1.63 g sample
4-10
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Activity: Percent Composition

What are the percent iron and the percent
sulfur in an 8.33-g sample of chalcopyrite
that contains 2.54 g Fe and 2.91 g S?
4-11
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Activity Solution: Percent
Composition
1.
What are the percent iron and the percent
sulfur in an 8.33-g sample of chalcopyrite that
contains 2.54 g Fe and 2.91 g S?
% Fe 
%S 
2.54 g Fe
8.33 g sample
2.91 g S
 100%  30.5% Fe in sample
 100%  34.9% S in sample
8.33 g sample
4-12
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Activity: Percent Composition

A 4.55-g sample of limestone (CaCO3) contains
1.82 g of calcium. What is the percent Ca in
limestone?
4-13
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Activity Solutions: Percent
Composition

A 4.55-g sample of limestone (CaCO3) contains 1.82 g
of calcium. What is the percent Ca in limestone?
% Ca 
1.82 g Ca
4.55 g limestone
 100%  40.0% Ca in limestone
4-14
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4.2 Mole Quantities

When working with amounts of a substance on a
macroscopic scale, we cannot simply count atoms
or molecules. There are too many. Instead, we
use the moles of atoms, which are related by
Avogadro’s number:
1 mole = 6.022 × 1023 particles



1 mole C = 6.022 × 1023 carbon atoms
1 mole H2S = 6.022 × 1023 H2S molecules
1 mol Cu2O = 6.022 × 1023 Cu2O formula units
4-15
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The Mole



The mole unit acts as a bridge between the
microscopic world and the macroscopic world.
One mole of substance contains as many basic
particles (atoms, molecules, or formula units) as
there are atoms in exactly 12 g of carbon-12.
One mole of a substance contains 6.022 × 1023
particles (molecules, atoms, ions, formula units,
etc.)
 This number is called Avogadro’s number.
4-16
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Moles of Various Elements and
Compounds
Figure 4.8
4-17
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Activity: Atoms in H2S


How many sulfur atoms are in 1 mol of H2S?
How many hydrogen atoms are in 1 mol of H2S?
Figure 4.6
4-18
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Molar Mass
The Mass of 1 Mole





Avogadro’s number has been defined so that the
mass of 1 mol of C-12 has a mass of exactly 12
grams.
This means that the average mass of an atom of
any substance in amus is the same numerical value
as the mass of 1 mole of that substance in grams
(molar mass).
The molar mass of carbon is 12.01 g/mol.
We’ll use
The molar mass of oxygen is 16.00 g/mol
4 sig. figs
The molar mass of CO2 is:
12.01 + 2(16.00) = 44.01 g/mol
4-19
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Activity: Molar Mass

Complete the table.
Element or
Compound
Atomic Mass
Molar Mass
H
O
Na
Ca(ClO3)2
H2O
NaCl
4-20
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Activity Solutions: Molar Mass
Element or
Compound
Atomic Mass
Molar Mass
H
1.008 amu
1.008 g/mol
O
16.00 amu
16.00 g/mol
Na
22.99 amu
22.99 g/mol
Ca(ClO3)2
206.98 amu
206.98 g/mol
H2O
18.02 amu
18.02 g/mol
NaCl
58.44 amu
58.44 g/mol
4-21
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Molar Mass

Which contains the greatest number of atoms?
 1 mole of copper or 1 mole of gold?
Answer: Because 1 mole is a fixed quantity of
particles (6.022 × 1023 ), 1 mole of copper contains
the same number of atoms as 1 mole of gold.

1 gram of copper or 1 gram of gold?
Answer: Because each copper atom has less mass
than each gold atom, 1 gram of copper contains
more atoms than 1 gram of gold.
4-22
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Converting Between Grams and Moles
Convert 10.0 grams of CO2 to moles.
1 mol CO 2
moles CO 2 = 10.0 g CO 2 
= 0.277 mol CO 2
44.01 g CO 2
1.
2.
Convert 0.50 mol CO2 to grams.
44.01 g
mass CO 2 = 0.50 mol 
= 22.0 g CO 2
1 mol
Figure from p. 137
4-23
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Activity: Converting Between Grams
and Moles

Convert 10.0 g O2 to moles.
1 mol
moles O2 = 10.0 g O2 
= 0.312 mol
32.00 g
Figure from p. 137
4-24
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Grams  Moles  Particles

Once we know the number of moles of a
substance, we can use Avogadro’s number
(6.022×1023) to determine the number of particles
in that sample of the substance.
1 mole = 6.022 × 1023 particles
Figure from p. 139
4-25
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Activity: Number of Particles


What mass of MgCl2 contains 6.022 × 1023 Cl
ions?
The molar mass of MgCl2 is 95.21 g/mol.

1
mol
Cl
mass MgCl2 = 6.022  1023 Cl 
6.022  1023 Cl
1 mol MgCl2
95.21 g MgCl2


= 47.60 g MgCl2
2 mol Cl
1 mol MgCl2
Figure from p. 139
4-27
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Extra Activity: Conversions with Molar
Mass

How many moles of aspartame (C14H18N2O5) are
found in 40.0 mg of aspartame? How many
molecules of aspartame are found in this mass?
4-28
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Extra Activity Solution: Conversions with
Molar Mass

How many moles of aspartame (C14H18N2O5) are
found in 40.0 mg of aspartame? How many
molecules of aspartame are found in this mass?
We first need to convert from mg to g:
40.0 mg C14 H18 N 2O5 
1 g C14 H18 N 2O5
1000 mg C14 H18 N 2O5
 0.0400 g C14 H18 N 2O5
Next, we need to find the molar mass of C14H18N2O5:
12.01 g C
1.008 g H
+ 18 mol H 
1 mol
1 mol H
14.01 g N
16.00 g O
+ 2 mol N 
+ 5 mol O 
= 294.34 g/mol C14 H18 N 2O5
1 mol N
1 mol O
MM = 14 mol C 
4-29
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Extra Activity Solution: Conversions with
Molar Mass

How many moles of aspartame (C14H18N2O5) are
found in 40. mg of aspartame? How many molecules
of aspartame are found in this mass?
Next, convert from grams to moles:
0.0400 g C14 H18 N 2O5 
1 mol C14 H18 N 2O5
 1.36  10-4 mol C14 H18 N 2O 5
294.34 g C14 H18 N 2O5
Finally, we convert from moles to molecules:
6.022  1023 molecules C14 H18 N 2O5
1.36  10 mol C14 H18 N 2O5 
1 mol C14 H18 N 2O5
-4
 8.19  1019 molecules C14 H18 N 2O5
4-30
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Extra Activity: Conversions with Molar
Mass

If one aspirin tablet contains 0.324 g of
acetylsalicylic acid (C9H8O4), then how
many molecules of acetylsalicylic acid are
in 2 aspirin tablets?
4-31
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Extra Activity Solution: Conversions with
Molar Mass

If one aspirin tablet contains 0.324 g of acetylsalicylic
acid (C9H8O4), then how many molecules of
acetylsalicylic acid are in 2 aspirin tablets?
0.324 g C9 H8O4
1 mol C9 H8O4
6.022  1023 molecules C9 H8O4


1 aspirin tablet
180.17 g C9 H8O4
1 mol C9 H8O4
= 1.08  10
21
molecules C9 H8O 4
1 aspirin tablet
1.08 x 1021 molecules C9 H8O4
2 aspirin tablets 
1 aspirin tablet
= 2.16  1021 molecules C9 H8O4
4-32
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4.3 Determining Empirical and
Molecular Formulas

The formula for a substance also tells us
about the composition of a compound:


A formula unit for an ionic compound tells us
the ratio of ions of different elements in the
compound. (MgCl2 has a 1:2 ratio of Mg2+ to
Cl.)
A molecular formula tells the number of atoms
of each element in a molecule and the atom
ratio. (CO2 has a 1:2 ratio of C to O atoms.)
4-33
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Empirical and Molecular Formulas

Empirical formula



Expresses the simplest ratios of atoms in a compound
Written with the smallest whole-number subscripts
Molecular formula


Expresses the actual number of atoms in a compound
Can have the same subscripts as the empirical formula
or some multiple of them
4-34
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Empirical Formulas

What is the same
about these two
compounds?
Figure 4.13
4-35
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Empirical Formulas

The empirical formula of a
substance is the ratio of atoms
of different elements, in terms
of the smallest whole numbers.

What are the empirical
formulas of these two
compounds?

What are the empirical
formulas of H2O2 and H2O?
Are they different compounds?
Figure 4.13
4-36
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What is the empirical formula for
copper(II) oxide?
Figure 4.12
4-37
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Empirical or Molecular Formulas
4-38
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Activity: Empirical Formula
Determine the empirical formulas of these substances.
For which substances is the empirical formula the same
as the molecular formula?
Figure 4.14
4-39
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Activity Solutions: Empirical Formula
Determine the empirical formulas of these substances.
For which substances is the empirical formula the same
as the molecular formula?
Figure 4.14
4-40
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Determining Empirical Formulas from
% Composition


If we know the masses of the elements in a
compound, or its percent composition, we can
determine its mole ratio, and therefore the
compound’s empirical formula.
Consider the compound commonly called
chalcopyrite. Any sample will have the following
% composition:
Figure 4.5
4-41
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Determining Empirical Formulas
1.
2.
3.
Since the percent composition does not change
from sample to sample, assume any size sample.
The most convenient is 100 grams so % value =
mass value.
Convert grams to moles for each element using its
molar mass as a conversion factor.
Without changing the relative amounts, change
moles to whole numbers. Do this by dividing all
by the same smallest value. If all do not convert
to whole numbers, multiply to get whole
numbers.
4-42
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Determining Empirical Formulas
1. 100 grams chalcopyrite contains:
 30.5 g Fe
 34.6 g Cu
 34.9 g S
2. mol Fe = (30.5g Fe)(1mol/55.85 g) = 0.5461 mol Fe
mol Cu = (34.6g Cu)(1mol/63.55 g) = 0.5444 mol Cu
mol S = (34.9g S)(1mol/32.07 g) = 1.088 mol S
3. (0.5461 mol Fe)/(0. 5444) = 1.003 mol Fe
(0.5444 mol Cu)/(0. 5444) = 1.000 mol Cu
(1.088 mol S)/(0. 5444) = 1.999 mol S
Figure 4.5
FeCuS2
4-43
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Steps for Determining Empirical Formula
Figure from p. 143
4-44
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Activity: Empirical Formula from
Percent Composition

A compound was determined to have the
following percent composition:



50.0% sulfur
50.0% oxygen
What is the empirical formula for the
compound?
SO2
4-45
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Activity: Determining Empirical
Formulas

Determine the empirical formula for the mineral
covellite, which has the percent composition
66.5% Cu and 33.5% S.
4-46
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Activity Solution: Determining Empirical
Formulas
Determine the empirical formula for the mineral covellite,
which has the percent composition 66.5% Cu and 33.5% S.
First, reassign the percentages to units of grams: 66.5 g Cu and
33.5 g S.
Then, convert to moles and divide both numbers by the lowest
number.

66.5 g Cu  1 mol Cu = 1.05 mol Cu
63.55 g Cu
1 mol S
33.5 g S 
= 1.05 mol S
32.07 g S
1.05 mol Cu
=1
1.05 mol
1.05 mol S
=1
1.05 mol
The whole numbers then become our subscripts.
The empirical formula is therefore: CuS
4-47
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Activity: Determining Empirical
Formulas

Shattuckite is a fairly rare copper mineral. It has
the composition 48.43% copper, 17.12% silicon,
34.14% oxygen, and 0.31% hydrogen. Calculate
the empirical formula of shattuckite.
4-48
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Activity Solutions: Determining Empirical
Formulas

Shattuckite is a fairly rare copper mineral. It has the
composition 48.43% copper, 17.12% silicon, 34.14%
oxygen, and 0.31% hydrogen. Calculate the empirical
formula of shattuckite.
1 mol Cu
0.7621 mol Cu
=
= 2.5 mol Cu  2 = 5 mol Cu
63.55 g Cu
0.3069
1 mol Si
0.6095 mol Si
17.12 g Si 
=
= 2 mol Cu  2 = 4 mol Si
28.09 g Si
0.3069
1 mol O
2.134 mol O
34.14 g O 
=
= 7 mol O  2 = 14 mol O
16.00 g O
0.3069
1 mol H
0.3069 mol H
0.31 g H 
=
= 1 mol H  2 = 2 mol H
1.008 g H
0.3069
48.43 g Cu 
Therefore, the empirical formula is: Cu5Si4O14H2
4-49
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Molecular Formulas from
Empirical Formulas



Benzene and acetylene
have the same empirical
formulas but different
molecular formulas.
How much greater in
mass is benzene than
acetylene?
How much greater in
mass is each of these
than the empirical
formula?
Figure 4.13
4-50
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Empirical and Molecular Formulas

What are the ratios of masses for the
molecular and empirical formulas?
4-51
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Activity: Molecular Formulas from
Empirical Formulas

A compound was determined to have an empirical
formula of CH2. Its molar mass was determined
to be 42.12 g/mol. What is the molecular formula
for this compound?
The mass of the empirical formula is
14.03 g/mol, so the ratio of masses is 3.00.
The ratio of the formulas therefore is 3,
giving a molecular formula of (CH2)3 or
C3H6.
4-52
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Molecular Formulas


To determine a molecular
formula, the problem must
give a piece of experimental
data, such as a molar mass,
MM.
To find the molecular
formula:
1. Find the empirical formula
first.
2. Divide the empirical
formula’s molar mass into
the experimental molar
mass (which is given).
Figure 4.15
4-53
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Activity: Molecular Formulas

Potassium persulfate is a strong bleaching agent.
It has a percent composition of 28.93% potassium,
23.72% sulfur, and 47.35% oxygen. The
experimental molar mass of 270.0 g/mol. What
are the empirical and molecular formulas of
potassium persulfate?
4-54
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Activity Solution: Molecular Formulas

Potassium persulfate is a strong bleaching agent. It has a
percent composition of 28.93% potassium, 23.72% sulfur, and
47.35% oxygen. The experimental molar mass of 270.0 g/mol.
What are the empirical and molecular formulas of potassium
persulfate?
First, find the empirical formula:
1 mol K
0.7399 mol K
28.93 g K 
=
= 1 mol K
39.10 g K
0.7396
1 mol S
0.7396 mol S
23.72 g S 
=
= 1 mol S
32.07 g S
0.7396
1 mol O
2.959 mol O
47.35 g O 
=
= 4 mol O
16.00 g O
0.7396
Thus, the empirical formula is KSO4.
4-55
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Activity Solution: Molecular Formulas
Potassium persulfate is a strong bleaching agent. It has a
percent composition of 28.93% potassium, 23.72% sulfur,
and 47.35% oxygen. The experimental molar mass of 270.0
g/mol. What are the empirical and molecular formulas of
potassium persulfate?
To find the molecular formula:
Calculate the MM of KSO4 = (1 mol K × 39.10 g/mol K) + (1 mol
S × 32.07 g/mol S) + (4 mol O × 16.00 g/mol O) = 135.17
g/mol KSO4

270.0 g/mol (experimental MM )
=2
135.17 g/mol (empirical formula MM )
Thus, the molecular formula is K2S2O8.
4-56
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Determining Percent Composition from
Empirical or Molecular Formula


If you know the formula for a compound, you can
determine the mass percent composition.
Assume you have 1 mole of compound, and
convert moles of the element and compound to
grams.
Mass Percent
Composition
Empirical or Molecular
Formula
Mass Ratio
Atom or Mole Ratio
4-57
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Determining Percent Composition

Assume 1 mole of compound, and remember the
mass of 1 mole is the molar mass.
Figure from p. 149
4-58
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Activity: Determining Percent
Composition

What is the percent sodium in Na2O?
45.98
 100% = 74.19%
61.98
4-59
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Activity: Percent Composition

Which of these
Cuprite,
Cu2O
compounds has the
greater percentage of
Cu?
Answer: cuprite
Chalcocite,
Cu2S
Figure 4.16
4-60
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4.4 Chemical Composition of Solutions



A solution is a
homogeneous mixture.
This is a solution being
prepared by adding the
CuSO4 (solute) to water
(solvent).
The composition of the
solution depends on the
relative amounts of the
solute and solvent.
Figure 4.17
4-61
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Concentration

Which aqueous CuSO4 solution has the greatest
concentration (most concentrated)? Which is most
dilute?
Figure 4.18
4-62
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Percent by Mass of Solute

Solution concentration is often expressed as the
mass percent of solute:
mass of solute
Percent Mass Solute =
100
total mass of solution
4-63
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Activity: Percent by Mass of Solute

What is the mass percent of NaCl in a solution that
is prepared by adding 10.0 g NaCl to 50.0 g water?
mass of solute
Percent Mass Solute =
100
total mass of solution
10.0 g NaCl
Percent Mass NaCl =
100 = 16.7 %
60.0 g solution
4-64
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Molarity (M)

Another common way to express the
concentration of a solution is in molarity units:
moles solute
Molarity =
liters of solution
4-65
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Activity: Preparing a CuSO4 Solution



6.25 grams (0.0250 mol) of CuSO45H2O is added
to a 250-mL volumetric flask.
Water is added to the mark so that the total
volume is 250.0 mL.
What is the molarity of this solution?
0.0250 mol
Molarity =
= 0.100 M
0.250 L
Figure 4.19
4-66
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Activity: Molarity

How many moles of NaCl are in 1.85 L of a
0.25 M NaCl solution?
moles NaCl = 1.85 L × 0.25 mol/L = 0.46 mol
4-67
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Activity: Solution Concentration

Bluestone is copper(II) sulfate pentahydrate,
CuSO4•5H2O, with a molar mass of 249.7 g/mol. A
sample of pond water was found to have a
concentration of 6.2  105 M copper(II) sulfate. If
the pond has a volume of 1.8  107 L, then what mass
of bluestone did the farmer add to the pond as an
algicide?
4-68
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Activity Solution: Solution Concentration

Bluestone is copper(II) sulfate pentahydrate, CuSO4•5H2O, with
a molar mass of 249.7 g/mol. A sample of pond water was found
to have a concentration of 6.2  105 M copper(II) sulfate. If the
pond has a volume of 1.8 x 107 L, then what mass of bluestone
did the farmer add to the pond as an algicide?
5
6.2

10
mol
249.7 g
7
1.8  10 L 

= 2.8  1015 g CuSO 4 5H 2O
1L
1 mol
4-69
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Dilution

Suppose you want to dilute a 0.25 M solution to a
concentration of 0.025 M. What are some ways to
do this?
4-70
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Dilution
Figure 4.21
4-71
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Describe this process
Figure from p. 168
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Dilution

Molesinitial= Molesfinal



Moles = Molarity × Volume
Moles = mol/L × L
MinitialVinitial = MfinalVfinal
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Activity: Dilution

MinitialVinitial = MfinalVfinal

What is the concentration of a solution prepared
by adding water to 25.0 mL of 6.00 M NaOH to a
total volume of 500.0 mL?
6.00 M  0.0250 L
M final =
= 0.300 M
0.500 L
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Activity: Dilution

If 42.8 mL of 3.02 M H2SO4 solution is diluted to a
final volume 500.0 mL, what is the molarity of the
diluted solution of H2SO4?
4-75
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Activity Solution: Dilution

If 42.8 mL of 3.02 M H2SO4 solution is diluted to a
final volume 500.0 mL, what is the molarity of the
diluted solution of H2SO4?
MconVcon = MdilVdil
3.02 M × 42.8 mL = Mdil× 500.0 mL
M dil
3.02 M  42.8 mL
=
= 0.259 M
500.0 mL
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Activity: Dilution

What is the concentration of a solution prepared by
diluting 35.0 mL of 0.150 M KBr to 250.0 mL?
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Activity Solution: Dilution

What is the concentration of a solution prepared by
diluting 35.0 mL of 0.150 M KBr to 250.0 mL?
MconVcon = MdilVdil
0.150 M × 35.0 mL = Mdil × 250.0 mL
M dil
0.150 M  35.0 mL
=
= 0.0210 M
250.0 mL
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