CHM 300
Review and
Introduction to
Stoichiometry
R.8 Classification of Matter
• Matter: anything occupying
space and having mass
• Matter exists in three states:
solid, liquid and gas
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Mixtures
• Most of the matter around us consists of mixtures of pure
substances
• Wood, gasoline, wine, soil, and air all are mixtures.
• Main characteristic of a mixture is that it has variable
composition
• Mixtures can be classified as homogeneous (having visibly
indistinguishable parts) or heterogeneous (having visibly
distinguishable parts).
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Examples of Mixtures
• A homogeneous mixture is called a solution.
• Air is a solution consisting of a mixture of gases. Wine is a complex
liquid solution. Brass is a solid solution of copper and zinc.
• Heterogeneous mixtures
• Examples: sand in water, salt and sand, and iced tea with ice cubes
• Heterogeneous mixtures usually can be separated into two or
more homogeneous mixtures or pure substances.
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Separating Mixtures
• Mixtures can be separated into pure substances by physical
methods.
• A pure substance is one with constant composition.
Pure water is composed solely of H2O molecules
• water found in nature (groundwater or the water in a lake or
ocean) is a mixture.
• seawater, for example, contains large amounts of dissolved
minerals.
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Separating Mixtures based on Physical Properties
A physical change is a change in the form of a substance, not in its
chemical composition. A physical change can be used to separate a
mixture into pure compounds, but it will not break compounds into
elements.
• Boiling seawater produces steam, which can be condensed to pure water,
leaving the minerals behind as solids.
• The dissolved minerals in seawater also can be separated out by freezing the
mixture, since pure water freezes out.
The processes of boiling and freezing are physical changes. When
water freezes or boils, it changes its state but remains water; it is still
composed of molecules.
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Distillation
One of the most important
methods for separating the
components of a mixture
is distillation.
Depends on differences in
volatility.
Works very well when only one
component of the mixture is
volatile. (sand and water)
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Filtration and Chromatography
• Another method of separation is simple filtration
• which is used when a mixture consists of a solid and a liquid. The mixture is
poured onto a mesh, such as filter paper, which passes the liquid and leaves
the solid behind.
• A third method of separation is chromatography.
• Chromatography is the general name applied to a series of methods that use a
system with two phases (states) of matter: a mobile phase and a stationary
phase.
• The stationary phase is a solid, and the mobile phase is either a liquid or a
gas. The separation process occurs because the components of the mixture
have different affinities for the two phases and thus move through the system
at different rates.
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Compounds and elements
• Pure substances are either compounds (combinations of
elements) or free elements.
• A compound is a substance with constant composition that can be
broken down into elements by chemical processes.
• electrolysis of water, chemical change because the water molecules have
been broken down into the free elements hydrogen and oxygen.
A chemical change is one in which a given substance becomes a new
substance or substances with different properties and different composition.
• Elements, composed of atoms, are substances that cannot be
decomposed into simpler substances by chemical or physical
means.
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The Organization of Matter
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Chapter 5
Stoichiometry
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Table of Contents (1 of 2)
• (5-1)
Counting by weighing
• (5-2)
Atomic masses
• (5-4)
The mole
• (5-5)
Molar mass
• (5-6)
Percent composition of compounds
• (5-7)
Determining the formula of a compound
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Questions to Consider
• Can atoms be counted?
• Does the number of atoms in a reaction affect the result?
• Can the mass of a sample be used to determine the number of
moles it comprises?
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Section 5.1
Counting by Weighing
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Average Mass
• It is used to determine the number of atoms in a set quantity of
a substance
• It is determined using a sample of the substance
Average mass =
total mass of atoms in sample
number of atoms in sample
• Atoms do not need to be identical in order to be counted by
weighing
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Section 5.2
Atomic Masses
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Atomic Masses
• The modern system of determining atomic masses was first
used in 1961
• Based on 12C
• 12 atomic mass units (u)
• The most accurate method currently used to compare the
masses of atoms involves the use of the mass spectrometer
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Mass Spectrometer (1 of 2)
• Atoms are passed into a stream of high-speed electrons that
convert them into positive ions
• Ions are passed through a magnetic field
• Accelerating ions create their own magnetic field, resulting in a
change in the path travelled
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Mass Spectrometer (2 of 2)
• The degree of deviation depends on the mass of the ion
• Ions with higher masses deviate the least
• Deviated ions hit the detector plate
• Comparing the deflected position of each ion gives their mass
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Figure 5.1 - The Mass Spectrometer
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Determining Atomic Masses
• Consider the analysis of 12C and 13C in a mass spectrometer
Mass 13 C
 1.0836129
12
Mass C
• Based on the definition of the atomic mass unit
Mass of 13 C = 1.0836129 12 u   13.003355 u
Exact number by definition
• The average mass for an element is also referred to as the
average atomic mass or atomic mass
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Section 5.4
The Mole
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The Mole and Avogadro’s Number (1 of 2)
• The mole is defined as the number equal to the number of
carbon atoms in exactly 12 grams of pure 12C
• 6.022 × 1023
• Also called Avogadro’s number
• One mole of a substance contains 6.022 × 1023 units of that
substance
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The Mole and Avogadro’s Number (2 of 2)
• A sample of a natural element with a mass equal to the
element’s atomic mass expressed in grams contains 1 mole of
atoms
Element
Number of Atoms Present
Mass of Sample (g)
Aluminum
6.022 × 1023
26.98
Copper
6.022 × 1023
63.55
Iron
6.022 × 1023
55.85
Sulfur
6.022 × 1023
32.07
Iodine
6.022 × 1023
126.9
Mercury
6.022 × 1023
200.6
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Interactive Example 5.4 – Calculating Number of
Atoms
A silicon chip used in an integrated circuit of a microcomputer
has a mass of 5.68 mg. How many silicon (Si) atoms are present
in the chip?
• Information needed
• the atoms of Si in the chip
• Information available
• Chip has 5.68 mg of Si
• Mass of 1 mole ( 6.022 x 1023 atoms) of Si = 28.09 g (from the periodic table)
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Solution to 5.4
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Section 5.5
Molar Mass
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Molar Mass of a Compound
• It is the mass of one mole of the compound measured in grams
• Traditionally called molecular weight
• Formula unit
• Used for compounds that do not contain molecules
• NaCl - Sodium chloride
• CaCO3 - Calcium carbonate
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Interactive Example 5.6 - Calculating Molar Mass I
• Juglone, a dye known for centuries, is produced from the
husks of black walnuts. It is also a natural herbicide (weed
killer) that kills off competitive plants around the black walnut
tree but does not affect grass and other noncompetitive plants.
The formula for juglone is C10H6O3.
• a.) Calculate the molar mass of juglone.
• b.) A sample of 1.56 × 10−2 g of pure juglone was extracted from black
walnut husks. How many moles of juglone does this sample
represent?
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Solution (5 of 34)
• a.) The molar mass is obtained by summing the masses of the
component atoms. In 1 mole of juglone there are 10 moles of
carbon atoms, 6 moles of hydrogen atoms, and 3 moles of
oxygen atoms:
10 C: 10  12.01 g = 120.1 g
6 H: 6  1.008 g =
6.048 g
3 O: 3  16.00 g = 48.00 g
Mass of 1 mol C10 H 6O 3 = 174.1 g
• The molar mass of juglone is 174.1 g
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Solution (6 of 34)
• b.) The mass of 1 mole of this compound is 174.1 g; thus 1.56
× 10−2 g is much less than a mole. The exact fraction of a mole
can be determined as follows:
1.56  10
–2
1 mol juglone
g juglone 
= 8.96  10 –5 mol juglone
174.1 g juglone
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Section 5.6
Percent composition of compounds
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Describing the Composition of a Compound
• In terms of the numbers of its constituent atoms
• In terms of the percentages (mass) of its elements
• Consider ethanol (C2H5OH)
g
= 24.02 g
mol
g
Mass of H = 6 mol ×1.008
= 6.048 g
mol
g
Mass of O = 1 mol ×16.00
= 16.00 g
mol
Mass of 1 mol C 2 H 5OH = 46.07 g
Mass of C = 2 mol × 12.01
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Mass Percent
• It is calculated by comparing the mass percent of the element
in one mole of the compound with the total mass of one mole
of the compound and multiplying the result by 100%.
• Calculating the mass of carbon in ethanol
mass of C in 1 mol of C 2 H 5OH
Mass percent of C =
 100%
mass of 1 mol of C 2 H 5OH
24.02g
=
× 100 = 52.14%
46.07g
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Interactive Example 5.9 - Calculating Mass
Percent
• Carvone is a substance that occurs in two forms having
different arrangements of atoms but the same molecular
formula (C10H14O) and mass. One type of carvone gives
caraway seeds their characteristic smell, and the other type is
responsible for the smell of spearmint oil. Compute the mass
percent of each element in carvone.
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Solution (7 of 34)
• Objective
• To find the mass percent of each element in carvone
• Information available
• Molecular formula, C10H14O
• Information needed to find the mass percent
• Mass of each element (1 mole of carvone)
• Molar mass of carvone
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Solution (8 of 34)
• Step 1 - Determine the mass of each element on one mole of
C10H14O
g
= 120.1 g
mol
g
Mass of H in 1 mol = 14 mol ×1.008
= 14.11 g
mol
g
Mass of O in 1 mol = 1 mol ×16.00
= 16.00 g
mol
Mass of C in 1 mol = 10 mol × 12.01
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Solution (9 of 34)
• Step 2 - Determine the molar mass of C10H14O
120.1 g + 14.11 g  16.00 g  150.2 g
C10 +
H14
+ O
 C10H 14O
• Step 3 - Determine the mass of each element
120.1 g C
Mass percent of C =
× 100% = 79.96%
150.2 g C10 H14 O
14.11 g H
Mass percent of H =
× 100% = 9.394%
150.2 g C10 H14 O
16.00 g O
Mass percent of O =
× 100% = 10.65%
150.2 g C10 H14 O
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Section 5.7
Determining the formula of a compound
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Formulas
• Determined by using a weighed sample and one of the
following techniques
• Decomposing it into its component elements
• Introducing oxygen to produce substances such as CO2, H2O, etc.,
which are collected and weighed
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Figure 5.5 - Analyzing for Carbon and Hydrogen
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Empirical and Molecular Formula (1 of 2)
• Empirical formula: The simplest whole-number ratio of the
various types of atoms in a compound
• Can be obtained from the mass percent of elements in a compound
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Empirical and Molecular Formula (2 of 2)
• The molecular formula varies for molecules and ions
• For molecular substances, it is the formula of the constituent
molecules
• Always an integer multiple of the empirical formula
• For ionic substances, it is the same as the empirical formula
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Figure 5.6 - Substances Whose Empirical and
Molecular Formulas are Different
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Empirical Formula Determination (1 of 2)
• Since mass percentage gives the number of grams of a
particular element per 100 grams of compound, base the
calculation on 100 grams of compound
• Each percent will then represent the mass in grams of that element
• Determine the number of moles of each element present in
100 grams of compound using the atomic masses of the
elements present
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Empirical Formula Determination (2 of 2)
• Divide each value of the number of moles by the smallest of
the values
• If each resulting number is a whole number (after appropriate
rounding), these numbers represent the subscripts of the elements in
the empirical formula
• If the numbers obtained in the previous step are not whole
numbers, multiply each number by an integer so that the
results are all whole numbers
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Interactive Example 5.10 - Determining Empirical
and Molecular Formulas I
• Determine the empirical and molecular formulas for a
compound that gives the following percentages on analysis (in
mass percents):
• 71.65% Cl
• 24.27% C
• 4.07% H
• The molar mass is known to be 98.96 g/mol
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Solution (10 of 34)
• Objective
• To find the empirical and molecular formulas for the given compound
• Information available
• Percent of each element
• Molar mass of the compound (98.96 g/mol)
• Information needed to find the empirical formula
• Mass of each element in 100.00 g of compound
• Moles of each element
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Solution (11 of 34)
• Step 1 - Determine the mass of each element in 100.00 g of
compound
• Cl 71.65 g
• C 24.27 g
• H 4.07 g
• Step 2 - Determine the moles of each element in 100.00 g of
compound
71.65 g Cl ×
1 mol Cl
= 2.021 mol Cl
35.45 g Cl
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Solution (12 of 34)
1 mol C
24.27 g C 
= 2.021 mol C
12.01 g C
1 mol H
4.07 g H 
= 4.04 mol H
1.008 g H
• Step 3 - Determine the empirical formula for the compound
• Dividing each mole value by 2.021 (the smallest number of moles
present), we find the empirical formula ClCH2
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Solution (13 of 34)
• Step 4 - Determine the molecular formula for the compound
• Compare the empirical formula mass to the molar mass
• Empirical formula mass = 49.48 g/mol
• Molar mass is given = 98.96 g/mol
Molar mass
98.96 g/mol
=
=2
Empirical formula mass
49.48 g/mol
Molecular Formula = (ClCH 2 ) 2 = Cl 2C 2 H 4
• The substance comprises molecules of Cl2C2H4
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Homework
• Questions at the end of the Review section (1-14)
• Review naming simple compounds Chapter 3.10
• Chapter 5 questions:
• Review questions: 2, 3, 5, 6
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