Chapter 12
Reacting masses
12.1 The mole, Avogadro constant and molar
mass
12.2 Percentage by mass of an element in a
compound
12.3 Chemical formulae of compounds
12.4 Empirical formulae and molecular
formulae derived from experimental data
12.5 Reacting masses from chemical equations
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Key terms
Progress check
Summary
Concept map
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12.1 The mole, Avogadro constant and
molar mass
Mole and Avogadro constant
In daily life, some special units are used to
describe the quantity of items.
Figure 12.1 Socks are in
pairs, eggs are often packed
in dozens, and papers are
often packed in reams.
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Learning tip
1 ream of paper refers to 500 identical sheets of paper.
Unlike socks, eggs and papers, particles (i.e.
atoms, ions or molecules) are too small to be
seen.
A special unit called mole (abbreviation: mol)
is used to describe the quantity of particles in a
substance.
12.1 The mole, Avogadro constant and molar mass
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The number of atoms in exactly 12.0 g of carbon-12
the reference unit for the mole
The number of atoms in exactly 12.0 g of carbon-12
is 6.02 × 1023.
Avogadro constant (symbol: L; unit: mol–1)
One mole of any substance contains 6.02 × 1023
formula units.
12.1 The mole, Avogadro constant and molar mass
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Learning tip
A pure substance has a formula. The simplest unit of a
substance is its formula unit. For example,
Substance
Water
Copper
Carbon
Sodium chloride
Formula unit
H2O
Cu
C
NaCl
12.1 The mole, Avogadro constant and molar mass
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1 mole of water
contains
6.02 × 1023
H2O molecules
1 mole of
copper metal
contains
6.02 × 1023
Cu atoms
1 mole of
carbon contains
6.02 × 1023
C atoms
1 mole of sodium
chloride contains
6.02 × 1023 NaCl formula
units (or contains 1 mole
of Na+ ions and 1 mole
of Cl– ions)
Figure 12.2 One mole of each of the four substances. They all contain
6.02 × 1023 formula units.
12.1 The mole, Avogadro constant and molar mass
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Key point
amount of a substance
One mole is the ________
that contains the same number of formula
units as the number of atoms in exactly 12.0 g
of carbon-12.
Class practice 12.1
12.1 The mole, Avogadro constant and molar mass
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Mole and molar mass
One mole of a substance has a mass equal to its
formula mass expressed in gram unit.
The mass of one mole of a substance is called its
molar mass.
Unit of molar mass: gram per mol (g mol–1)
12.1 The mole, Avogadro constant and molar mass
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Substances consisting of atoms
One mole of carbon and one mole of copper both
contain 6.02 × 1023 atoms.
One mole of carbon weighs 12.0 g while one mole
of copper weighs 63.5 g.
Relative atomic mass of carbon = 12.0
Relative atomic mass of copper = 63.5
Molar mass of carbon = 12.0 g mol–1
Molar mass of copper = 63.5 g mol–1
12.1 The mole, Avogadro constant and molar mass
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Substances consisting of molecules
One mole of water weighs 18.0 g.
Relative molecular mass of water
= 1.0 × 2 + 16.0 = 18.0
Molar mass of water = 18.0 g mol–1
Think about
12.1 The mole, Avogadro constant and molar mass
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Substances consisting of ions
One mole of sodium chloride weighs 58.5 g.
Formula mass of sodium chloride
= 23.0 + 35.5 = 58.5
Molar mass of sodium chloride = 58.5 g mol–1
Key point
mass of one mole of a substance is called
The _____
its molar mass. (Unit of molar mass: g mol–1)
Class practice 12.2
12.1 The mole, Avogadro constant and molar mass
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Relationship between number of formula units,
number of moles and mass
÷ Avogadro constant
number of
formula units
× molar mass
number of
moles
× Avogadro constant
mass
÷ molar mass
12.1 The mole, Avogadro constant and molar mass
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Key point
Example 12.1
Class practice 12.3
Example 12.2
12.1 The mole, Avogadro constant and molar mass
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12.2 Percentage by mass of an element in a
compound
Calculating percentage by mass of an element in the
compound
From the formula of a compound
percentage
by mass of each element in the compound.
Key point
Percentage by mass of element A in a compound
Example 12.3
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Calculating the mass of an element in the compound
The mass of an element in a compound can be
calculated from:
formula of the compound
percentage by mass of that element in the
compound
Example 12.4
12.2 Percentage by mass of an element in a compound
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Calculating the relative atomic mass of an element
The relative atomic mass of an element may be
calculated from:
formula of the compound
percentage by mass of that element in the
compound
Example 12.5
Class practice 12.4
12.2 Percentage by mass of an element in a compound
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12.3 Chemical formulae of compounds
Empirical formula
Empirical formula: formula which shows the
simplest whole number ratio of the atoms or
ions present.
Applicable to all compounds.
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Molecular formula
Molecular formula: formula which shows the
actual number of each kind of atoms in one
molecule of the substance.
Only applicable to molecular compounds and
elements consisting of molecules.
Structural formula
Structural formula: formula which shows how
the constituent atoms are joined up in one
molecule of the substance.
12.3 Chemical formulae of compounds
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Empirical
formula
Molecular
formula
Structural formula
—
N2
N≡N
Carbon dioxide
CO2
CO2
O=C=O
Ethene
CH2
C 2H 4
Propene
CH2
C 3H 6
Ethanol
C 2H 6O
C 2H 6O
Methoxymethane
C 2H 6O
C 2H 6O
SiO2
—
Substance
Nitrogen
Quartz
—
Table 12.1 The different formulae of some substances.
12.3 Chemical formulae of compounds
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Note:
Empirical and molecular formulae of a compound
may be the same or different.
Molecular formula is the empirical formula
multiplied by some whole number.
Different compounds may have the same
empirical formula and the same molecular formula
but they have different structural formulae.
Class practice 12.5
12.3 Chemical formulae of compounds
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12.4 Empirical formulae and molecular
formulae derived from experimental data
Determination of empirical formulae
The empirical formula of a compound can be
calculated from its composition by mass.
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Determining the empirical formula
of an oxide of copper
Pass town gas into a combustion tube.
Heat a known mass of oxide of copper (black) in
the combustion tube.
Hydrogen and carbon monoxide in the town gas
reduce the oxide to reddish brown copper.
Find the mass of copper.
12.4 Empirical formulae and molecular formulae derived from
experimental data
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oxide of copper
hole
excess town
gas burns here
town gas
supply
heat
combustion tube
Figure 12.3 To determine the empirical formula of an oxide of copper by
passing town gas over the heated oxide.
Learning tip
Hydrogen and carbon monoxide reduce the oxide by
removing oxygen from it.
12.4 Empirical formulae and molecular formulae derived from
experimental data
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SBA note
At the beginning of the experiment, town gas is
passed into the combustion tube. This is to expel
the air inside the tube.
The hot copper formed may react with the oxygen
in air again. Therefore, it is necessary to pass the
town gas through the combustion tube, even after
heating has stopped.
12.4 Empirical formulae and molecular formulae derived from
experimental data
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Specimen results
Item
Mass (g)
Combustion tube
18.100
Combustion tube + oxide of copper
18.701
Combustion tube + copper
18.579
Mass of copper in oxide
18.579 – 18.100 = 0.479
Mass of oxygen in oxide
18.701 – 18.579 = 0.122
Table 12.2 The specimen results of the experiment
Empirical formula of the oxide of copper can be
worked out as shown in ‘Problem-solving strategy
12.1’.
Problem-solving strategy 12.1
12.4 Empirical formulae and molecular formulae derived from
experimental data
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Determining the empirical formula
of an oxide of magnesium
Heat a known mass of magnesium strongly in a
crucible (also of known mass) until it catches fire.
Lift the crucible lid slightly from time to time for
letting in air to react with magnesium.
From experimental results
empirical formula
of the oxide of magnesium: MgO
12.4 Empirical formulae and molecular formulae derived from
experimental data
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crucible
pipe-clay
triangle
magnesium
ribbon
rocksil
tripod
heat
very
strongly
Figure 12.4 To find the empirical
formula of an oxide of magnesium
by heating magnesium in air.
Empirical formula of a compound can also be
determined if the percentage by mass of each
element in the compound is known.
Experiment 12.1
Experiment 12.1
Example 12.6
Experiment 12.2
Experiment 12.2
Class practice 12.6
12.4 Empirical formulae and molecular formulae derived from
experimental data
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Determination of molecular formulae
A compound with known empirical formula and
relative molecular mass
determine the
molecular formula of the compound
Example 12.7
Example 12.10
Example 12.8
Experiment 12.3
Example 12.9
Class practice 12.7
Experiment 12.3
12.4 Empirical formulae and molecular formulae derived from
experimental data
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12.5 Reacting masses from chemical
equations
Chemical equations and reacting masses
Consider the equation representing the reaction
between magnesium and oxygen:
2Mg(s) + O2(g) → 2MgO(s)
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2Mg(s)
+
O2(g)
2MgO(s)
→
2 magnesium atoms
1 oxygen molecule
2 formula units of
magnesium oxide
2 × 6.02 ×
magnesium atoms
1 × 6.02 ×
oxygen molecules
2 × 6.02 × 1023
formula units of
magnesium oxide
1023
react
with
1023
2 moles of
magnesium atoms
1 moles of oxygen
molecule
2 mol × 24.3 g mol–1
= 48.6 g of
magnesium atoms
1 mol × 32.0 g
= 32.0 g of
oxygen molecules
to
form
mol–1
2 moles of formula units
of magnesium oxide
2 mol × 40.3 g mol–1
= 80.6 g of
formula units of
magnesium oxide
12.5 Reacting masses from chemical equations
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Learning tip
The molar masses of Mg, O2 and MgO are
24.3 g mol–1, (16.0 × 2 = 32.0) g mol–1, and
(24.3 + 16.0 = 40.3) g mol–1 respectively.
A balanced equation shows the quantitative
relationship of the reactant(s) and the product(s)
in a reaction.
Stoichiometric coefficients in the equation
indicate the relative number of moles (i.e. mole
ratio) of reactants and products involved in the
reaction.
12.5 Reacting masses from chemical equations
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The total mass of reactants is equal to the total
mass of products (i.e. conservation of mass).
Learning tip
The quantitative study of reactants and products in a
reaction is called stoichiometry.
12.5 Reacting masses from chemical equations
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Calculations from chemical equations — reacting
masses
The mass of one of the substances in the reaction
is known.
Masses of other substances reacted or formed can
be calculated based on the balanced chemical
equation.
Problem-solving strategy 12.2
12.5 Reacting masses from chemical equations
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Steps for determining the reacting masses from a
chemical equation:
Known
mass of A
divided by
Mass of the substance
asked in the question
molar
mass of
A
Number of
moles of A
multiplied by molar mass of
that
substance
by mole ratio
(shown in
the equation)
Number of moles of the
substance asked in the question
(Note: A represents the chemical formula of a particular substance.)
Example 12.11
12.5 Reacting masses from chemical equations
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Example 12.12
Limiting reactant
Consider the reaction between hydrogen and
oxygen to form water:
2H2(g) + O2(g) → 2H2O(l)
O2 molecules
H2 molecule
H2O molecule
2 H2 molecules + 2 O2 molecules 2 H2O molecules + 1 O2 molecule
Figure 12.5 Two hydrogen molecules require one oxygen molecule
for complete reaction. Therefore, oxygen is in excess.
12.5 Reacting masses from chemical equations
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Only 1 molecule of oxygen is required to react with
2 molecules of hydrogen for complete reaction.
In this case, oxygen is in excess.
Amount of water produced is limited by the
amount of hydrogen used.
hydrogen is called the limiting reactant.
Limiting reactant limits the amount of the product
formed in a reaction.
Example 12.13
12.5 Reacting masses from chemical equations
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Theoretical yield, actual yield and percentage yield
Theoretical yield: amount of product expected if
the reaction proceeds exactly as shown in the
chemical equation.
Actual yield: amount of product actually obtained
from a reaction.
12.5 Reacting masses from chemical equations
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The actual yield of a reaction is often less than
the theoretical yield because:
the reaction is incomplete.
impurities are present in the reactants.
side reactions occur in which unwanted side
products are produced.
some product is lost during different
experimental processes, such as purification.
12.5 Reacting masses from chemical equations
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The efficiency of a chemical reaction can be
expressed by the percentage yield.
Key point
Example 12.14
Class practice 12.8
12.5 Reacting masses from chemical equations
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Key terms
1.
2.
3.
4.
5.
6.
7.
8.
9.
actual yield 實際產量
Avogadro constant 亞佛加德羅常數
composition by mass 質量組成
empirical formula 實驗式
limiting reactant 限量反應物
molar mass 摩爾質量
mole 摩爾
molecular formula 分子式
percentage by mass 質量百分比
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10.
11.
12.
percentage yield 百分產率
structural formula 結構式
theoretical yield 理論產量
Key terms
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Progress check
1.
2.
3.
4.
What is the meaning of mole?
What is the Avogadro constant?
What is the meaning of molar mass?
How is the mole of a substance related to its mass
and number of formula units?
5. How can we calculate the percentage by mass of
an element in a compound?
6. What are empirical formula, molecular formula and
structural formula?
7. How can we determine the empirical formula of a
compound?
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8. How can we determine the molecular formula of a
compound?
9. What are the interrelationship between masses of
reactants and products in a reaction?
10. How can we calculate masses of reactants and
products in a reaction from the relevant equation?
11. What is the meaning of a limiting reactant?
12. What are the meanings of actual yield and
theoretical yield?
13. How can we calculate the percentage yield of a
chemical reaction?
Progress check
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Summary
12.1 The mole, Avogadro constant and molar
mass
1.
2.
3.
Chemists use mole (abbreviation: mol) to
describe the quantity of particles in a substance.
The Avogadro constant (L) is the number of
atoms in exactly 12.0 g of carbon-12. It is equal
to 6.02 × 1023 mol–1.
The molar mass of a substance is its formula
mass expressed in gram unit. The unit of molar
mass is g mol–1.
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4.
Important relationships involving moles:
12.2 Percentage by mass of an element in a
compound
5.
The percentage by mass of an element in a
compound can be found by the equation:
Percentage by mass of element A in a compound
Summary
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12.3 Chemical formulae of compounds
6.
Chemical formulae are part of the language of
chemistry. Some common chemical formulae
include empirical formula, molecular formula
and structural formula.
12.4 Empirical formulae and molecular formulae
derived from experimental data
7.
Empirical formula of a compound is the formula
which shows the simplest whole number ratio of
the atoms or ions present.
Summary
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8.
9.
The empirical formula of a compound can be
calculated from its composition by mass. The
composition of a compound has to be
determined by experiment.
Molecular formula may be determined from
empirical formula and relative molecular mass.
This is because molecular formula is a whole
number multiple of empirical formula.
12.5 Reacting masses from chemical equations
10.
The theoretical amounts of substances used up
or produced in a reaction can be calculated from
its balanced equation.
Summary
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11.
12.
13.
Limiting reactant is the reactant that is
completely used up in a reaction. It limits the
amount of product(s) formed in the reaction.
The theoretical amounts of product predicted by
calculation from its balanced equation is called
theoretical yield. The actual yield of a reaction is
often less than the theoretical yield.
Percentage yield is the ratio of actual yield and
theoretical yield. It is a measure of the efficiency
of a chemical reaction.
Summary
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Concept map
Mass
Molar
mass
equals
Number
of moles
equals
Number of
formula units
Avogadro
constant
equals
6.02 × 1023
formula units
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Molar mass
without unit, equals
Formula mass
Relative
molecular mass
equals
relative
Sum of ______________
atomic masses
_____________________
of all atoms/ions in a
formula unit of a substance
equals
Sum of relative
atomic masses of
all atoms in a
_____________
molecule
Concept map
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Relative
molecular
mass
Empirical
formula
determines
Molecular
formula
Chemical
formulae
Structural
formula
Concept map
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