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Oxford
Resources
for
IB
Diploma Programme
2 0 2 3
E D I T I O N
C H E M I S T RY
CO U R S E
CO M PA N I O N
Sergey Bylikin
Gary Horner
Elisa Jimenez Grant
D avid Tarcy
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Oxford
Resources
for
IB
Diploma Programme
2 0 2 3
E D I T I O N
C H E M I ST RY
CO U R S E
CO M PA N I O N
Sergey Bylikin
Gary Horner
Elisa Jimenez Grant
D avid Tarcy
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Contents
Structure 1. Models of the particulate nature of matter
2
Structure 1.1
Structure 1.2
Structure 1.3
Structure 1.4
Structure 1.5
Structure 2. Models of bonding and structure
94
Structure 2.1
Structure 2.2
Structure 2.3
Structure 2.4
Structure 3. Classic ation of matter
228
Structure 3.1
Structure 3.2
Tools for chemistry
308
Tool 1:
Tool 2:
Tool 3:
Reactivity 1. What drives chemic al reactions?
386
Reactivity 1.1
Reactivity 1.2
Reactivity 1.3
Reactivity 1.4
Reactivity 2. How much, how fast and how far?
460
Reactivity 2.1
Reactivity 2.2
Reactivity 2.3
Reactivity 3. What are the mechanisms of chemic al change?
536
Reactivity 3.1
Reactivity 3.2
Reactivity 3.3
Reactivity 3.4
Cross-topic exam-style questions
652
The inquiry process
655
(authored by Maria Muñiz Valcárcel)
The internal assessment (IA)
(authored by Maria Muñiz Valcárcel)
668
Index
686
Periodic Table
708
Answers:
www.oxfordsecondary.com/ib-science-support
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Introduction
The diploma programme (DP) chemistry course is aimed at students in the 16
to 19 age group.
The curriculum
of the nature of science,
chemistry and
contexts.
seeks to develop
a conceptual understanding
working knowledge of fundamental principles of
practic al skills that
c an be applied in familiar and unfamiliar
As with all the components of
the DP,
this course fosters the IB learner
prole attributes (see page viii) in the members of
the school community.
Nature of science
Nature of science (NOS)
purposes and
is concerned with methods,
outcomes that
•
F alsication
Hypotheses c an be proved false using other
are specic to science.
NOS is a central theme that is present
across the
evidence,
entire course.
NOS features
denitely true.
You will nd
throughout the book and
suggested
but
they c annot be proved to be
This has led
science throughout
are encouraged to come up
to paradigm shis in
history.
with further examples of your own as you work through
•
Models
the programme.
Scientists construct
NOS c an be organized
explanations of
into the following eleven
contain assumptions or unrealistic simplic ations,
aspects:
but
•
models as simplied
their observations. Models oen
the aim of science is to increase the complexity
Observations and experiments
of
the model,
and
to reduce its limitations.
Sometimes the observations in experiments are
unexpected
and lead
to serendipitous results.
•
Theories
A theory is a broad
•
explanation that
takes observed
Measurements
patterns and hypotheses and uses them to generate
Measurements c an be qualitative or quantitative,
predictions.
These predictions may conrm a
but all data are prone to error. It is important to
theory (within observable limitations) or may falsify
know the limitations of your data.
it.
•
Evidence
•
Science as a shared activity
Scientists learn to be sceptic al about their
Scientic activities are oen c arried out in
observations and they require their knowledge to
collaboration,
such as peer review of work before
be fully supported by evidence.
public ation or agreement on a convention for clear
•
Patterns and trends
Recognition of a pattern or trend
communic ation.
forms an
•
important part of
Global impact of science
the scientist’s work whatever the
Scientists are responsible to society for the
science.
consequences of
•
Hypotheses
Patterns lead
environmental,
to a possible explanation. The
hypothesis is this provisional view and
it
requires
knowledge must
and
their work,
whether ethic al,
economic or social. Scientic
be shared
with the public clearly
fairly.
further veric ation.
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Syllabus structure
Topics are organized
below.
into two main concepts:
structure and reactivity.
This is shown in the syllabus roadmap
The skills in the study of chemistry are overarching experimental,
skills that
are integrated
experimental work,
into the course.
inquiries and
Chemistry is a practic al subject,
technologic al,
mathematic al and inquiry
so these skills will be developed
through
investigations.
Skills in the study of chemistry
Structure
Reactivity
Structure refers to the nature of matter
Reactivity refers to how and why
from simple to more complex forms
chemic al reactions occur
Structure determines reactivity, which
in turn transforms structure
Structure 1.
Structure 1.1 — Introduction to
Reactivity 1.
Reactivity 1.1 — Measuring
Models of the
the particulate nature of matter
What
enthalpy changes
particulate nature
drives
chemic al
Structure 1.2 — The nuclear atom
of matter
Reactivity 1.2 — Energy cycles in
reactions?
reactions
Structure 1.3 — Electron
Reactivity 1.3 — Energy from fuels
congurations
Structure 1.4 — Counting
Reactivity 1.4 — Entropy and
particles by mass: The mole
spontaneity (Additional higher
level)
Structure 1.5 — Ideal gases
Structure 2.
Structure 2.1 — The ionic model
Models of
bonding and
Reactivity 2.
Reactivity 2.1 — How much? The
How much,
amount
of chemic al change
how fast and
Structure 2.2 — The covalent
Reactivity 2.2 — How fast? The
structure
how far?
model
rate of chemic al change
Structure 2.3 — The metallic
Reactivity 2.3 — How far? The
model
extent
of chemic al change
Structure 2.4 — From models to
materials
Structure 3.
Structure 3.1 — The periodic
Reactivity 3.
Reactivity 3.1 — Proton transfer
Classic ation of
table:
What
reactions
Classic ation of elements
matter
are the
mechanisms
Reactivity 3.2 — Electron transfer
of chemic al
reactions
change?
Structure 3.2 — Functional
Reactivity 3.3 — Electron sharing
groups:
reactions
Classic ation of organic
compounds
Reactivity 3.4 — Electron-pair
sharing reactions
Chemistry concepts are thoroughly interlinked.
roadmap
above,
You are therefore encouraged
new and
help
For example,
“Structure determines reactivity,
as shown in the
which in turn transforms structure”.
to continuously reect on the connections between
prior knowledge as you progress through the course.
you explore those connections.
identify and
In assessment tasks,
apply the links between dierent
topics.
Linking questions will
you will be expected to
On page 652,
there are three
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examples of DP-style exam
questions that link several dierent
topics in the course.
v
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How to use this book
The
aim
of
this
development
book
and
understanding
Feature
aims,
boxes
by
to
through
and
for
develop
conceptual
opportunities
to
understanding, aid in skills
cement
knowledge and
practice.
sections
signposting
opportunities
is
provide
throughout
content
practice.
relating
This
is
an
the
to
book
are
particular
overview
of
designed to support these
ideas and concepts, as well as
these
features:
Developing conceptual understanding
These
boxes
in
the
Guiding questions
margin
to
Each topic begins with a guiding question to get you thinking.
other
studying a topic,
you might
not
but
by studying that
topic,
Hence,
where
explored
you should
you
the
a
concept
further
dierent
or
context.
in
They
consider these as you work
may
through the topic and
of
you will be able to answer
a
them with increasing depth.
parts
be able to answer these questions
is
condently or fully,
direct
When you
book
start
will
also
direct
you
to
come back to them when you revise your
prior
knowledge
or
a
skill
understanding.
you
a
Linking questions
discussed
there and
other parts of
need,
dierent
about
Linking questions within each topic highlight
content
will
or
way
give
to
think
something.
the connections between
the course.
Nature of science
These illustrate NOS using issues from both modern science and science
history, and show how the ways of doing science have evolved over the
centuries. There is a detailed description of what is meant by NOS and the
dierent aspects of NOS on the previous page. The headings of NOS feature
boxes show which of the eleven aspects they highlight.
Theory of knowledge
This
is
an
thinking
The
TOK
important
and
part
of
the
understanding
features
in
this
IB
how
book
Diploma
we
pose
course.
arrive
It
focuses
on
at
our
knowledge
questions
for
you
that
of
critic al
the
highlight
world.
these
LHA
issues.
Parts of
the book have a coloured
question.
This indic ates that
Chemistry Higher Level.
bar on the edge of the page or next to a
the material is for students studying at DP
AHL means “additional higher level”.
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Developing skills
ATL
Chemistry skills
Approaches to learning
These
ATL
scientists
features
have
communic ation,
thinking
about
and
how
To o l s
These
to
your
as
the
of
and
your
tools
guidance
of
the
on
how
Assessment.
information.
Links
in
for
the
experiments and
research,
you
the
are
DP
use
Flick
experimental or inquiry skills,
of
to
c an be used
think
inquiry
of
reference
Chemistry,
the
to
full
inquiry
this
margin
your mathematic al,
especially through
practic al work. Some of these
as springboards for your Internal
Assessment.
strategies.
book
to
These contain ways to develop
famous
skills
prompt
own
required
how
ATL
c h e m i s t r y,
section
Internal
more
skills,
develop
for
three
examples
self-management,
social
experimental
well
give
demonstrated
material
details
process
section
p r o c e ss
as
in
your
throughout
the
on
for
data
the
all
the
analysis
study
working
book
and
of
the
essential
and
direct
the
you
and
rest
a ss e ss m e n t
mathematic al
modelling
subject
through
will
internal
of
to
work
the
towards
it
and
chemistry,
as
through
book
for
too.
Practicing
Practice questions
Worked examples
These are step-by-step
examples of how to answer
These are designed
questions or how to complete c alculations.
to give you further practice at
You
using your chemistry knowledge and
should
review these examples c arefully,
to allow you to
preferably
check your own understanding and
progress.
aer attempting the question yourself.
Activity
Data-based questions
Part
on
of
the
are
your
also
and
nal
assessment
interpretation
designed
analysis
for
to
of
requires
data.
make
day-to-day
Use
you
you
these
aware
of
experiments
to
answer
questions
the
questions
to
prepare
possibilities
and
for
your
for
that
for
data
are
this.
based
They
acquisition
IA.
These
give
you
opportunity
your
to
an
apply
chemistry
knowledge
oen
in
a
and
skills,
practic al
way.
End-of-topic questions
Use these questions at the end
of
each topic to draw together concepts from
that topic and
to practise
answering exam-style questions.
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Course book denition
The IB Learner Prole
The IB Diploma Programme course books are resource
The aim
of
materials designed
minded
people who work to create a better and
to support
students throughout
all IB programmes to develop internationally
their two-year Diploma Programme course of study
more peaceful world.
in a particular subject. They will help students gain an
develop
understanding of what is expected
described below.
an IB Diploma Programme subject
from the study of
of
the programme is to
while presenting
Inquirers:
content
The aim
this person through ten learner attributes, as
They develop
their natural curiosity. They
in a way that illustrates the purpose and aims
acquire the skills necessary to conduct inquiry and
of the IB.
They reect
the philosophy and
approach of
research and
the IB and encourage a deep
understanding of
snow independence in learning. They
each
actively enjoy learning and
subject
this love of
learning will be
by making connections to wider issues and
sustained throughout
their lives.
providing opportunities for critic al thinking.
Knowledgeable:
They explore concepts,
ideas and
The books mirror the IB philosophy of viewing the
issues that
have loc al and
global signic ance. In so
curriculum in terms of a whole-course approach;
doing,
the use of
they acquire in-depth knowledge and
develop
a wide range of resources, international
understanding across a broad
mindedness,
the IB learner prole and
and
balanced
range of
the IB Diploma
disciplines.
Programme core requirements,
the extended
essay,
and
theory of knowledge,
creativity,
activity, service
Thinkers:
They exercise initiative in applying thinking
(CAS).
skills critic ally and
creatively to recognize and
complex problems,
and
to make reasoned,
approach
ethic al
E ach book c an be used in conjunction with other
decisions.
materials and,
and
indeed,
encouraged
students of
the IB are required
to draw conclusions from a variety
Communic ators:
of resources.
They understand
and
express
Suggestions for additional and further
ideas and
reading are given in each book and
information condently and
suggestions for
more than one language and
how to extend
creatively in
in a variety of modes of
research are provided.
communic ation.
They work eectively and willingly in
collaboration with others.
In addition,
and
the course companions provide advice
guidance on the specic course assessment
Principled:
requirements and
They act
with integrity and
honesty, with
on ac ademic honesty protocol.
a strong sense of
They are distinctive and
fairness,
justice and respect
for the
authoritative without being
dignity of the individual,
groups and communities.
prescriptive.
They take responsibility for their own action and the
consequences that
Open-minded:
accompany them.
They understand
and
appreciate their
IB mission statement
own cultures and
The International Bacc alaureate aims to develop
inquiring,
knowledgeable and c aring young people
who help
to create a better and
more peaceful world
through intercultural understanding and
To this end,
the organization works with schools,
governments and
develop
respect.
international organizations to
challenging programmes of international
educ ation and rigorous assessment.
personal histories,
to the perspectives,
individuals and
seeking and
values and
communities.
are open
traditions of other
They are accustomed to
evaluating a range of points of view, and
are willing to grow from
C aring:
and
the experience.
They show empathy,
towards the needs and
compassion and
feelings of others.
respect
They have
a personal commitment to service, and to act to make
a positive dierence to the lives of
others and to the
environment.
These programmes encourage students across the
world
to become active,
learners who understand
dierences,
compassionate and
that
lifelong
other people, with their
c an also be right.
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Risk-takers:
They approach unfamiliar situations and
uncertainty with courage and
the independence of spirit
ideas and
strategies.
forethought,
and
have
to explore new roles,
They are brave and articulate in
‘Formal’
c ategories (e.g.
and
Balanced:
forms of
you should
use one of
presentation.
separating the resources that
articles,
defending their beliefs.
means that
accepted
books,
internet-based
the several
This usually involves
you use into dierent
magazines,
newspaper
resources, and works of art)
providing full information as to how a reader or
They understand the importance of
viewer of
intellectual,
physic al and
your work c an nd
the same information. A
emotional ballance to achieve
bibliography is compulsory in the Extended
personal wellbeing for themselves and
Essay.
others.
What constitutes malpractice?
Reective:
They give thoughtful consideration to their
M alpractice is behaviour that
own learning and
experience.
you or any student
and
understand
their strengths and
or may result in,
gaining an unfair advantage in one
limitations in order
or more assessment
to support their learning and
results in,
They are able to assess
component.
M alpractice includes
personal development.
plagiarism and collusion.
Plagiarism is dened as the representation of
or work of another person as your own.
the ideas
The following
A note on ac ademic
are some of the ways to avoid plagiarism:
integrity
●
It
is of
appropriately credit
the owners of
information is used
ideas of another person to support one’s
information when
●
that
words and
arguments must be acknowledged
vital importance to acknowledge and
in your work.
Aer all,
passages that
are quoted
verbatim must
owners
be enclosed within quotation marks and
of
ideas (intellectual property) have property rights.
acknowledged
To have an authentic piece of work,
on your individual and
it
must
be based
original ideas with the work of
others fully acknowledged.
Therefore,
●
email messages,
and
any other electronic media
all assignments,
must be treated
in the same way as books and
written or oral, completed for assessment must use your
journals
own language and expression.
Where sources are used
●
or referred
to,
whether in the form of
or paraphrase,
such sources must
direct quotation
the sources of all photographs,
computer programs,
be appropriately
data,
maps,
illustrations,
graphs, audio-visual and
similar material must be acknowledged
acknowledged.
if they are
not your own work
●
How do I acknowledge the work of
dance,
others?
theatre arts or visual arts
creative use of a part
The way that you acknowledge that
ideas of
when referring to works of art, whether music, lm
original artist
you have used the
and
where the
a work takes place, the
must be acknowledged.
other people is through the use of footnotes
Collusion is dened
and
of
as supporting malpractice by
bibliographies.
another student. This includes:
Footnotes (placed
at the bottom of
(placed
of a document) are to be provided
at
the end
a page) or endnotes
●
allowing your work to be copied
assessment
or submitted
for
by another student
when you quote or paraphrase from another document
●
or closely summarize the information provided in
another document.
You do not
need to provide a
duplic ating work for dierent
assessment
components and/or diploma requirements.
footnote for information that is part of a ‘body of
Other forms of
knowledge’.
That
is,
denitions do not
malpractice include any action that
gives
need to be
you an unfair advantage or aects the results of another
footnoted as they are part of
the assumed knowledge.
student.
Examples include,
taking unauthorized
material into an examination room, misconduct during
Bibliographies should
include a formal list of the
an examination and
resources that
you used
falsifying a CAS record.
in your work.
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Experience
the
technology
with
oÏer
for
DP
future
of
education
Oxford’s
digital
Science
You’re already using our print resources,
but
have you tried
our digital course on
Kerboodle?
Developed
in cooperation with the IB and
students and
teachers,
designed
for the next generation of
Oxford’s DP Science oer brings together the IB curriculum
and
future-facing functionality,
and
digital components for the best
Learn
enabling success in DP and
blended
anywhere
optimized
student
access
with
onscreen
resources
to
the
learning experience.
mobile-
access
and
digital
beyond. Use both print
teaching and
to
oΎine
Course
Book
Encourage
of
motivation
engaging
interactive
exercises,
content
activities,
with
a
variety
including
vocabulary
animations,
and
videos
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Embrace
independent
progression
provides
can
a
with
personalized
self-assign
real-time
learning
adaptive
journey
auto-marked
results
and
and
are
Deepen
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students
assessments,
oΊered
and
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understanding
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repetition,
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on
data
For
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to
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Structure
1
Models
the
nature
of
of
particulate
m a tt e r
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Introduction to the
Structure 1.1
particulate nature
of matter
How c an we model the particulate nature of matter?
The universally accepted idea that all matter is composed of
atoms came from experimental evidence that could only be
explained if matter were made of particles.
Early classical theory suggested that all matter was composed
of earth, air
, re, and water
. However
, this theory lacked
predictive power and could not account for the great variety
of chemical compounds, so it was eventually abandoned.
The systematic study of chemical changes led to the
discovery of many chemical elements that could not be
broken down into simpler substances. The fact that these
elements could only combine with one another in xed
proportions suggested the existence of atoms. It was this

way of processing knowledge through observation and
Figure
most
experimentation which led to the modern atomic theory
1
In 2021,
detailed
scientists at
Cornell University c aptured the
picture of atoms to date.
What
do models show us that
microscope images c annot?
Understandings
Structure 1.1.1 —
of
matter,
which
Elements
c annot
be
are the primary constituents
Structure 1.1.2 —
chemic ally
to
broken
down into
and
simpler substances.
Compounds
chemic ally
consist
bonded
explain
of
atoms
together
of
in
dierent elements
a
xed
ratio.
The
physic al
gases)
and
kinetic
properties
changes
of
Structure 1.1.3
—
average
energy (E
kinetic
molecular
of
matter
theory is a model
(solids, liquids,
state.
Temperature
(in
K)
is
a
measure of
) of particles.
k
Mixtures
in
no
so
contain
xed
c an
be
ratio,
more
than
which
separated
are
by
one
not
element
or
chemic ally
physic al
compound
bonded
and
methods.
The composition of matter (Structure 1.1.1)
M atter
and energy
C he mi s t r y
We
and
a re
touch
tho u g h
th e
ex pa n d
of
matter
In
contrast,
up
m a ny
we
to
M atter
is
made
are
considered
of
matt e r
see
we
in
energy
is
energy
are
a
and
its
c o mp o s i t i o n .
c o n s u me
of
ma tt e r.
i t.
The
u n de rs t a n di n g
shown
as
of
ma tte r,
fo r m s
c annot
our
and
study
Air
is
a
u n i ve rs e
of
i t,
of
made
and
M a tte r
s u r ro u n ds
fo r m
is
ma tte r
it
its
ma tte r
of
is
e ve r y w he re.
us,
and
th a t
we
m a tt e r
and
pro pe r t i e s .
we
c an
kn ow
is
c he mi s tr y
The
see
t he re,
seeks
characteristics
gure 2.
anything
closely
property
of
that
exists
but
associated
matter,
such
does
with
as
not
each
the
have
other,
ability
to
these
and
properties.
energy is oen
perform work or
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produce
heat.
3
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Structure
1
Models
of
the
particulate
nature
of
matter
Although
Chemic al
reactions
are
nuclear
in
mass
and
energy
c an
be
converted
into
one
another
studies
only
those
(for
example, in
introduced
reactors
or
inside
stars),
chemistry
transformations of
Reactivity 1.1.
matter
where
both
products
have
the
from
form
to
one
mass
same
and
energy
mass
another
as
rather
are
conserved. In
starting
than
materials,
created
or
chemic al reactions, the
and
the
energy
is
transformed
destroyed.
made up of
particles –
atoms,
molecules,
or ions
u
Figure 2
The characteristics of matter
particles are
occupies a
in constant
volume in
MATTER
motion
space
has a
mass
Thinking skills
ATL
2
The
(E)
famous
are
Einstein
equation,
interconvertible.
chemic al
reactions
8
is
relatively
m s
changes
negligible.
the
is
example
eect
without
What
=
mc
the
small
,
shows
energy
while
that
mass (m)
released
the
or
and
energy
absorbed in
speed of light (c)
is
very
large
–1
(3.00 × 10
This
E
However,
of
a
).
As
a
result,
demonstrates
the
certain
is
factor
compromising
other
the
examples
the
of
loss
gain
importance
minor,
nal
or
it
c an
in
of
mass
c aused
by
chemic al
approximation in science: if
oen
be
ignored
in
c alculations
result.
negligible
eects
have
you
encountered
inchemistry?
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4
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Structure
1.1
Introduction
to
the
particulate
nature
of
matter
The atomic theory
The
law
always
of
of
conservation
combine
elements.
but
could
form
not
water,
consumed
1.0 g
of
It
be
of
mass
denite
was
and
broken
the
monoxide,
the
chemic ally.
of
that
water
with
with
observation that certain substances
led
to
elements
showed
mass
react
and
that
down
experiments
would
and
proportions
theorized
equalled
c arbon
c arbon
in
of
of
idea
Hydrogen
the
mass
formed.
1.33 g
2.66 g
the
that
combined
Other
oxygen
oxygen
of
to
to
and
was
oxygen
hydrogen
c an
and
experiments
through
form
matter
composed
form other substances
react to
oxygen
showed that
combustion
c arbon
to
form
dioxide.
It was proposed that elements, such as hydrogen, oxygen or carbon, are the
The
internal
structure and
primary constituents of matter, and they cannot be chemically broken down into
characteristics of atoms will be
simpler substances. The idea of denite proportions suggested that particles of one
discussed in
Structure 1.2
element, called atoms, would combine with atoms of another element in a xed,
simple ratio, and that atoms of one element have a dierent mass than atoms of a
dierent element. This, and other experimental evidence, led to the atomic theory.
The
atomic
c annot
theory
states
be
created
or
reactions.
Physic al
and
and
that
all
matter
destroyed,
but
chemic al
is
composed of atoms. These atoms
they
are
properties
rearranged
of
during
chemic al
matter depend on the bonding
arrangement of these atoms.
Evidence
Ancie nt
atomists,
Uddāl ak a
Ā runi
Democ ritus
w as
made
postu late d
8th
small
objects
be
of
Similarly,
into
BCE,
5th
that
increasingly
splittable”,
that
next
2000
from
mass
atoms
atomic
in
the
is
wor ld
until
units,
be
on
broken
into
must
What
to
is
is
snap
as
are
due
to
a
atomic
to
to
have
seashell
powder
further.
D alton.
dierent
“kana”.
said
“atomos”, “not
any
of
John
experiments
their
particles
producing
known
development
evidence
theories?
the
Democritus
parts
classied
based
What
natu ra l
matte r
They
parti cles.
c alled
credited
knowledge
evidence.
that
pa rticl es.
successively
not
conservation
“elements”,
s age
proposed that “particles too
He
could
could
be
In dian
re a sone d
the
BCE,
smaller
later,
could
Scientic
Āruni
indivisible
stage
years
the
philos ophers
mass together into the substances and
one
of
in
these
century
composed
The
chan ges
seen
them
Gre ek
indi visible
experience”.
in
observed
tiny,
betwe en
century
to
the
Leuci ppus,
of
that
interactions
In
and
up
among
and
theory,
D alton
over
drew
propose that
types
known as
masses.
be
supported
was
used
evidence?
to
Is
by
veriable
develop these
evidence
shaped

by
our
Figure 3
Top:
Āruni lived
in what
is now modern day
perspective?
Northern India,
depicted
by the Ganges river.
Bottom:
Democritus is
in a Renaissance-era painting
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Structure
1
Models
of
the
particulate
nature
of
matter
Chemic al symbols
In
modern
which
example,
the
Symbol
chemistry,
consist
the
chemic al
of
one
atoms
or
chemic al
symbol
and
two
symbol
for
elements
letters
iron
is
for
Fe
and
are
are
hydrogen
(the
rst
represented
derived
is
two
H
from
(the
letters
rst
of
by
the
the
letter
the
same
element
of
L atin
symbols,
names.
For
hydrogen), and
ferrum
“iron”).
Name
Common
chemic al
elements
and
their
symbols
are
listed in table1; the full list is
given in the data booklet and in the periodic table at the end of this book.
H
hydrogen
C
c arbon
O
oxygen
Na
sodium
Atoms
are
the
properties.
and
form
smallest
While
magnesium
S
sulfur
Cl
chlorine
Fe
iron
bound
is
Table 1
Common chemic al elements

Figure 4
M agnesium
by
chemic al
substance,
another
magnesium

matter
exist
as
elementary
sulde
bound
formula
magnesium
it
atomic
is
forces.
contains
a
still
possess
certain
Elementary substances
substance
(MgS)
chemic ally
of
that
chemic al
individually, they tend to combine together
chemic al compounds
together
elementary
(S)
of
c an
chemic al substances.
element, while
Mg
units
atoms
contain
For
one
type
composed
Mg
of
contain atoms of a
two
or
single
more elements
example, magnesium metal is an
only
chemic al
species,
atoms
of
of
compound,
and
S
atom,
sulfur
as
Mg.
atoms
it
Similarly, sulfur
only.
consists
of
(gure4). MgS is the
In
contrast,
two
dierent,
chemic al
sulde.
(le), sulfur (middle) and magnesium sulde (right)
Pure substances and mixtures
M atter
of
c an
particle
be
classied
arrangement
matter
as
a
pure
substance
or
a
mixture, depending on the type
(gure 5).
– any substance that
occupies space and has mass
pure substance
– has a
uniform chemical
element
–
composed of one
kind
of
atoms,
e.g.,
magnesium (Mg),
sulfur (S)
definite and
mixture
composition
compound
of
two
or
atoms
–
composed
more kinds of
in
a
fixed
ratio,
e.g., magnesium sulfide
(MgS),
– a
combination of two or more pure
substances that retain their indiidual properties
water (H
)

homogeneous
– has
uniform composition
and
properties
throughout,
water,
e.g.,
metal
sea
heterogeneous
– has
nonuniform composition
and
e.g.,
arying
paint,
properties,
salad
dressing
alloy
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
6
Figure 5
How matter is classied
according to the arrangement of particles
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Structure
1.1
Introduction
to
the
particulate
nature
of
matter
Pure substances cannot be separated into individual constituents without a chemical
reaction, which alters their physical properties. In contrast, mixtures can be separated
into individual components that retain their respective physical properties.
Data-based questions
A
student
had
quantitative

two
pure
substances,
observations
were
A
made
and
and
B.
They
were
heated
in

Substance A

M ass
M ass
Observations
of
and
heating
substance
of
Red
Green

Table 2
1.
C alculate
2.
State
3.
Melting
the
4.
A
colour
Results from
a
the
and
colour
ice
is
change
B
a
to
were
qualitative and
Change
in
Observations
aer
aer
mass / g
heating
/ g
26.12
±
0.02
26.62
±
0.02
Black colour
27.05
±
0.02
25.76
±
0.02
Black colour
heating substances A and B
qualitative
changes
some
crucible
contents
/ g
heating
B
and
Substance B
crucible
and
A
crucibles
Appearance aer heating each of the two substances
Substance
before
separate
recorded in table 2.
in
mass
for substances A and B.
observation
physic al
change
substances
both
pure
from
A
the
while
and
B
experiment
rusting
iron
represented
substances,
not
a
mixtures.
is
performed on A and B.
a
chemic al
change.
physic al
change
or
Discuss
whether
a
the
Explain,
chemic al
using
the
observations, whether
change.
experiment
shows
that
A
and
B
are
elements.
5.
Both
A
same
and
B
turned
black
on
heating.
C an
it
be
concluded
that
the
heating
of
these
two
substances
produced the
substance?
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Structure
1
Models
of
the
particulate
nature
of
matter
Melting point determination
Melting
purity
point
of
melting
a
points,
t e m p e ra t u r e
v a l u e.
l ow e rs
over
a
data
The
its
c an
s u b s t a n c e.
which
that
presence
melting
used
of
to
they
matches
impurities
point
a ss e ss
substances
me ans
closely
t e m p e ra t u r e
Relevant
be
Pure
and
melt
Method
the
h av e
at
a
sharp
the
the oretic al
in
substance
c auses
a
(Your teacher will provide specic instructions, depending
on
specific
melting
to
occur
2.
Prepare
3.
•
Inquiry
observations
and
the
solids
of
being
two
analysed.)
organic
solids
(A
and
B)
for
analysis.
skills
and
of
samples
samples
of
each
solid
in
two
separate
c apillary tubes.
Tool 1: Melting point determination
Identify
identity
Obtain
ra n g e.
•
2:
the
1.
record
sucient
relevant
relevant
Following
4.
qualitative
your
amounts
of
the
Prepare,
in
a
teacher ’s instructions, mix small
two
third
solids
together.
c apillary
tube,
a
small
sample of the
mixture of the two solids.
quantitative data
5.
Determine
the
melting
point
of
your
three
samples
Materials
(A,
•
Melting
•
C apillary tubes
•
S amples
point
B
and
the
mixture).
apparatus
Questions
1.
of
two
known
organic
solids,
for
Record
relevant
appropriate
aspirin
and
salol
(phenyl
2.
Comment
points
Wear
eye
protection.
•
Note
that
the
•
You
teacher
melting
will
give
3.
point
you
apparatus
further
gets
safety
and
(for
example,
environmentally
salol
and
aspirin
very hot.
this
on
the
pure
the
results, comparing the melting
substances
structural
information
to
with
impure substances.
formulas of A and B and use
explain
the
dierence in their
4.
To
what
analyse
extent
the
could
success
melting
of
an
point
organic
data
be
used to
synthesis?
hazardous).
contain
more
than
one
element
or
compound
in
no
xed
ratio, which
for determining the
are
melting
quantitative data in an
melting points.
are irritants
Mixtures
Methods
of
Research
prec autions,
depending on the identity of the solids being
analysed
and
format.
2-hydroxybenzoate)
S afety
•
qualitative
example,
point
of
a
substance
not
chemic ally
Mixtures
discussed in the
bonded
and
so
c an
be
separated
by
physic al methods.
are
c an be
homogeneous,
in
which
the
particles
are
evenly
distributed.
Tools for chemistry
Air
is
a
mixture
of
nitrogen,
oxygen, and small amounts of other gases. Air is a
chapter.
homogeneous
oxygen
If
the
the
to
is
particles
mixture
the
E ach
mixture,
consistent
top,
is
are
and
not
evenly
referred to as
which
component
reveals
of
its
composition
regardless
a
of
where
air
distributed,
of
is
such
heterogeneous.
that
milk
mixture
is
a
roughly
as
in
a
its
nitrogen and 20%
mixture of two solids, then
Natural
milk
heterogeneous
maintains
80%
sampled.
physic al
will
have
the
cream rise
mixture.
and
chemic al
properties.
For
The most common homogeneous
example,
hydrogen, H
,
is
explosive,
and
oxygen, O
2
, supports combustion.
2
mixtures, aqueous solutions, will
When
these
substances
are
present
in
a
mixture,
their
properties
stay
the
same.
be discussed in Reactivity 3.1, and
In
contrast,
water, H
O,
is
not
a
mixture
of
hydrogen
and
oxygen
but
a
chemic al
2
the properties of metal alloys in
compound
formed
by
bonding
two
hydrogen
atoms
with
one
oxygen atom.
Structure 2.4.
The
new
gas,
is
with
its
water
substance
not
has
explosive,
own
and
properties
without
a
none
it
and
chemic al
of
the
does
the
properties
not
support
hydrogen
reaction,
which
of
hydrogen
combustion.
and
oxygen
creates
or
It
oxygen. It is not a
is
c annot
a
pure substance
be
separated
from
new substances.
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Structure
1.1
Introduction
to
the
particulate
nature
of
matter
Separating mixtures
Mixtures
mixture
c an
has
separated
in
of
the
Two
the
is
a
c an
be
solid
separate
have
as
is
it
usually
be
a
a
of
A
sand
them.
The
sulfurous
if
we
s u ga r
It
sulfur
is
powders
not.
iron(II)
maintains
pure
This
sulde,
none
of
c an be
dierence
FeS, is not
the
properties
substance.
understand
sugar
and
placed
each component of the
sulfur
compound
sugar
is
and
while
bec ause
between
bec ause
iron
smell.
individual
sugar
and
of
magnetic
separated
from
means
mixture
is
new,
attractions
m i x tu re
physic al
Iron
not
separated
intermolecular
The
to
does
components
c an
by
properties.
magnet.
used
and
solids
S and
separated
using
property
magnetic
be
unique
their
will
intermolecular
dissolve
in
forces.
water, due to
water.
in
water
and
t he
s u ga r
Intermolecular
di ss o l ve s .
The
solution
c an
th e n
be
po u re d
t h ro u g h
fi l t e r
paper
placed
funnel,
not
in
p a ss
th e
w a te r
by
a
p ro c e ss
th ro u gh
w a te r
will
and
p a ss
e v a p o ra te s
e v a p o ra t i n g
th ro u gh
(fi gu re
t he
c alled
re m a i n
on
t h ro u gh
l e av i n g
th e
f i l te r
fil t ra t i on
w a te r
pa p e r.
th e
t he
behind
f ro m
S u ga r
(fi gu re
fi l t e r
f i l te r
th e
the
6 ).
p a p e r,
p a p e r.
pu re
fil tra t e
c r ys ta l s
T he
—
the
fo r m
sand
w h e re a s
Th e
sand.
will
l a rg e
wet
Th e
s u ga r
solution
in
the
sand
t hi s
p a r ti c l e s
sugar
is
which
be
are
discussed
Structure 2.2
will
d i ss o l ve d
dried,
c an
forces
i n s i de
in
a
and
th e
o bt a i n e d
pa ss e d
c r yst al li za t i on
pro c e ss
7 ).
filter paper
filter funnel
residue
(We define
a
residue
as a substance
that
remains
aer
evaporation,
distillation,
filtration
similar
or
any
process)
filtrate

evaporating
Figure 6
Filtration apparatus
sugar solution
basin
solution from
evaporating basin
cold tile
leave for a
few days
t
Figure 7
The crystallization
heat
for
sugar to crystallize
process
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Structure
1
Models
of
the
particulate
nature
of
matter
Distillation can be used to separate miscible liquids with dierent boiling points,
such as ethanol and water. Ethanol has a lower boiling point and will evaporate rst.
Once the vapours rise up a cooling column, they can be condensed to a liquid. As
shown in gure 8, cold water surrounds the condenser and allows the vapours to
condense to liquid ethanol. The water remains mostly in the distillation ask.
thermometer
distillation
u
Figure 8
water out
Distillation apparatus
flask
condenser
ethanol
and
water in
water
distillate
heat
(mostly ethanol)
Paper chromatography can be used to separate substances such as components
Paper
chromatography will
in inks. A piece of chromatography paper is spotted with the mixture. The bottom
be
discussed
in
more detail in
of the paper, below the spot, is placed in a suitable solvent as in gure 9(a).
Structure 2.2
The
substances
in
the
mixture
phase) and the paper (the
u
Figure 9
have
dierent
anities
intermolecular
forces of attraction
between
and
the
or
9(c)
ve
pure substances.
solvent
the
paper.
The stages in 2D paper
Figure
the
shows
a
the
pure
solvent (the
substances
mixture
(b)
(a)
for
mobile
stationary phase). The anity depends on the
that
was
in
the
mixture
composed of
(c)
chromatography
paper
some
some
hours
hours
later
later
solvent
drop of
mixture
turn
and
paper
use
a
90°
clockwise
different
solvent
Data-based questions
Look
at
gure
9.
1.
Which colour dot had the strongest anity for both solvent 1 and solvent 2?
2.
Which
colour
3.
Which
had
a
dots
had
stronger
a
stronger
anity
for
anity
solvent
for
2
solvent
than
1
than
solvent 2?
solvent 1?
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Structure
Table
3
shows
a
summary
of
the
separation
techniques
1.1
Introduction
to
the
particulate
nature
of
matter
discussed.
Components
Technique
Description
removed
mixture
ltration
is
le
poured
through a paper lter or
liquid(s)
solid(s)
soluble
insoluble
substance(s)
substance(s)
solvent, the solution
more soluble
less soluble
cools
substance(s)
substance(s)
other
porous material
dissolution
mixture
(solvation)
or
an
is
added
organic
mixture
is
to
water
solvent
dissolved in
Activity
hot
water
or
an
organic
Suggest
a
suitable
method
for
crystallization
down, and the
crystals
formed
isolated
mixture
by
is
separating
are
each
of
the
following
mixtures:
ltration
heated up
solid(s) and/
evaporation or
a.
salt and pepper
b.
several
c.
sugar
water-soluble
dyes
volatile
until
one
or
more of its
or
distillation
non-volatile
and
water
liquid(s)
components
mixture
is
vaporize(s)
liquid(s)
d.
For
placed on
iron and copper lings
each
mixture,
describe
the
less soluble
a
paper
side of the paper is
chromatography
submerged
a
separation
piece of paper; one
in
more soluble
component(s)
component(s)
move(s)
move(s) faster
or
how
each
technique
and
component
is
outline
isolated.
slower
water or
stay(s) in
solvent; components
place
move along the paper

Ta b l e
3
Summary of separation techniques

Figure 10 An advanced
water for millions of
provided
ltration technique c alled
people.
However,
by fossil fuels. Why might
it
reverse osmosis extracts salt
this process requires vast
be important
from
amounts of energy,
seawater,
most
providing fresh
of which is currently
to consider alternative energy sources?
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Author Name.
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Structure
1
Models
of
the
particulate
nature
of
matter
Planning experiments and risk assessments
Relevant skills
•
Tool
1:
Separation
•
Tool
1:
Addressing
of
mixtures
safety
of
self,
others
and
the
environment
Instructions
1.
Using
to
the
c alcium
Once
a
risk
in
you
have
Identify the
Assess
•
Determine
•
Identify
you
your
doing
of
each
decided
•
the
chapter,
devise
containing
In
properties
•
If
this
mixture
assessment
and
3.
a
c arbonate.
chemic al
2.
ideas
separate
on
protocol
so,
of
a
in
a
sand,
you
must
these
iron
that
would
lings
consider
and
the
allow
you
powdered
physic al and
four substances.
method,
which
method
salt,
identify
the
hazards and complete
you:
hazards
level of
risk
relevant
suitable
control measures
disposal
methods
aligned
with
your
school’
s
health
safety policies.
have
time,
try
methodology
it
out!
and
Remember
risk
that
assessment
your
teacher
should
validate
beforehand.
Extension
You
of
could
each
aer
the
them
evaluate
the
component
separation.
together.
components
the
masses
are
before
salt,
Measure
Then
the
eectiveness
(sand,
mix
all
and
the
them
dry,
iron
to
your
mass
of
together,
and
aer
of
lings,
by
comparing
the
c alcium
c arbonate)
before and
mass
each component prior to mixing
c arry
measure
c alculate
method
and
the
the
out
your
mass
of
separation,
each
percentage
again.
recovery
make
sure
Compare
of
each
component.
Linking questions
What
factors
are
considered
components
of
How
c an
products
How
do
are
contain
mixture?
of
intermolecular
between
Why
the
a
two
a
alloys
generally
choosing
reaction
forces
substances?
metallic
in
be
method
to
separate the
purified?
influence
the
type
(Tool 1)
of
mixture
that
forms
(Structure 2.2)
considered
bonding?
a
(Tool 1)
(Structure
to
be
2.3
mixtures,
and
even though they often
Structure 2.4)
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Structure
1.1
Introduction
to
the
particulate
nature
of
matter
States of matter (Structure 1.1.2)
Solids, liquids and gases
Matter is composed of particles. The types of interactions between these particles
determine the state of matter of a substance: solid, liquid or gas. All substances
can exist in these three states, depending on the temperature and pressure.
The
states
of
matter
of
formula: (s)
for
solid,
•
Water
is
a
solid
•
Water
is
a
liquid
•
Water
is
a
gas
substances
(l)
for
liquid
below
are
and
0 °C: H
shown
(g)
for
by
gas.
letters
For
in
brackets aer the
example:
O(s)
2
between
0
and
O(l)
100 °C: H
2
above
100 °C: H
O(g).
2
A special symbol, (aq), is used for molecules or other species in aqueous solutions.
For example, the expression “NaCl(aq)” tells us that sodium chloride is dissolved
in water while “NaCl(s)” refers to the pure compound (solid sodium chloride). The
properties of the three states of matter are summarized in gure 11.
solid
liquid
•
fixed
•
fixed
•
fixed shape
•
no
c annot be
•
•
volume
compressed
•
attractive
are
forces
•

not
move
Figure 11
no
fixed
•
no
fixed shape
c annot be
attractive
are
vibrate in
•
•
around
Steam,
forces
liquid
c an be
•
attractive
forces
between particles
weaker than
particles
volume
compressed
are
those in solids
fixed positions but
do
•
between particles
strong
particles
volume
fixed shape
compressed
between particles
•
gas
•
vibrate,
negligible
particles
rotate,
vibrate,
and
move
rotate, and
around faster than
move
in a
water and
around
liquid
ice are the three states of water
Changes of state
Substances
ice
will
positions,
but
is
reached.
A
further
reverses
At
c arbon
this
the
certain
melting.
more
This
as
states
it
the
of
until
ice
vaporizes
of
conditions,
change
CO
of
(s),
matter
heated.
a
as
The
they
(changes
accelerates
and
absorb
particles
temperature
melts
temperature
changes
dioxide,
is
violently,
point,
in
water
these
their
energy
increase
eventually
Under
change
absorb
becomes
a
release
to
energy. Solid
vibrate
in
xed
known as the melting point
its
the
or
continue
state
from solid to liquid).
movement of particles, and
gas.
The
turn
into
decrease
in
temperature
state.
solid
state,
substances
known as
gure12),
c an
sublimation,
which
is
is
commonly
gases
directly, without
typic al
used
for
for
dry
ice
(solid
refrigerating ice
2
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cream
and
biologic al
samples.

Figure 12
Sublimation of dry ice
13
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Structure
1
Models
of
the
particulate
nature
of
matter
The
process
water
opposite
vapour
in
the
to
air
sublimation
solidies
and
is
c alled
forms
deposition.
snowakes
of
At
low
temperatures,
various shapes and sizes
(gure13).
When
a
energy
when
a
are
state,
Figure 13
A snowake,
a
becomes
particles
the
becomes
a
liquid
releasing
energy
a
from
by
the
liquid
a
more
condensed
particles
or
a
gas,
from
and
the
when
state
to
a
less
condensed
surroundings. This happens
a
liquid
becomes a gas. These
processes.
substance
the
changes
absorbed
substance,
changes
lose
from
energy
intermolecular
or
to
a
solid,
the
a
to
less
the
forces
and
condensed
state
to
surroundings
and,
for
become
when
a
more
a
condensed
molecular
stronger. This happens when a gas
liquid
surroundings is an
a
becomes
exothermic
a
solid.
The
process of
process.
the product of
The
deposition of
solid
is
endothermic
When

substance
state,
changes
of
state
occurring
in
these
transformations
are
shown
in
gure 14.
water
Non-Newtonian uids
Some
do
substances,
not
behave
Newtonian
to
them.
known
uids
You
as
like
will
maize
known as
non-Newtonian uids,
typic al liquids. The
varies
make
starch
depending
a
on
viscosity
the
2.
of non-
force
or
“oobleck”,
and
add
Continue
applied
a
non-Newtonian uid commonly
slime
Slowly
thick
starch
explore its
3.
properties.
to
or
more
some
smoothly
if
It
as
starch and mix.
the
by
mixture
adding
achieves
more maize
needed.
exploring
should
stirred
maize
until
Adjust
water,
time
mixture.
the
water
consistency.
Spend
your
water
adding
the
harden
if
properties of
tapped,
and
ow
slowly.
Relevant skills
•
Questions
Inquiry 1: Identify dependent and independent
1.
variables
•
Inquiry
2:
Identify
and
record
relevant
qualitative
observations
Describe
the
matter
each
of
properties and identify the state of
•
powdered
•
water
•
the
of
the
following:
maize
starch
S afety
Wear
eye
maize
starch–water
mixture.
protection.
2.
Suppose
you
were
question
relating
asked
to
a
to
maize
develop
a
research
starch–water
mixture.
Materials
Consider
•
Spoon
•
250
•
Powdered
•
Water
or
variables.
3
cm
possible independent and dependent
large spatula
beaker
3.
maize
Research
non-Newtonian uids and identify other
starch
examples of these substances.
4.
How
has
this
experience
changed
the
way
you think
Method
about
1.
Ad d
t h re e
s ta rc h
to
or
t he
fo u r
h e a pe d
b e a ke r.
N o te
spoons
i ts
of
states
of
matter
and
their
properties?
Reect on
ma i z e
a pp e a ra n c e
this,
completing
the
•
I
used to think...
•
Now, I think...
following
sentence
starters:
and
c o n s i s t e n c y.
Linking questions
Why
are
some
conditions?
Why
are
some
(Structure
substances
solid
while
others
are
fluid
under
standard
(Structure 2.4)
2,
changes
of
state
endothermic
and
some
exothermic?
Reactivity 1.2)
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Structure
1.1
Introduction
to
the
particulate
nature
of
matter
gas (g)
freezing
melting
solid (s)
liquid (l)


Figure 14
Endothermic and
Figure 15
Orange growers spray their fruit
with water on cold nights.
exothermic
Freezing of
water is an exothermic process that
releases energy (in the form of
changes of state
heat) to the fruit,
protecting it against cold
Kelvin temperature sc ale (Structure 1.1.3)
As
temperature
measure
of
the
particles
of
a
move
faster,
When
water
a
solid
solid
while
is
the
in
a
energy
to
is
between
a
energies
kinetic
vibrate
in
gas
heated,
changes
added
forces
rises,
average
used
the
they
there
liquid
to
is
and
of
particles
energy
lattice
move
no
increase.
particles.
more,
As
particles
Temperature
substances
in
a
liquid
is a
absorb
vibrate
energy,
more and
faster.
temperature change during the periods when
when
disrupt
of
the
a
liquid
solid
changes
lattice
and
to
a
gas
(gure 16). The
overcome
the
intermolecular
molecules in the liquid.
vaporization
steam
100
kg
+
m
c
/
water
d
C°
condensation
steam
erutarepmet
water
+
s
om
ice
water
melting
K
A
0
freezing

ice
Figure 17
are kilogram
The seven base SI units
(kg) for mass,
length, second
energy input
(s) for time,
electric current,

Figure 16
meter (m) for
ampere (A) for
kelvin (K) for temperature,
Graph of the heating curve for water
mole (mol) for amount
of substance, and
c andela (cd) for luminous intensity. All units
There
were
many
attempts
to
measure
relative
temperature,
but
the
rst widely
of
accepted
temperature
sc ale
was
introduced
by
the
Polish-born
Dutch
measurement
c an be derived
from these
physicist
seven base units
D aniel
Gabriel
F ahrenheit.
You
The
kelvin
is
the
base
unit
of
temperature
will
learn
more about the mole
measurement in the International
in
Structure 1.4.
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System
of
Units
measurements
(SI).
c an
There
be
are
seven base units, and all other units of
derived
from
these
(gure 17).
15
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Structure
1
Models
of
the
particulate
nature
of
matter
Measurement
3
M aking,
recording,
and
communic ating
volume (m
measurements
2
greatly
benets
International
from
the
Bureau
French
mesures),
from
agreed
of
upon
Weights
sc ales. The
and
Measures
=
(BIPM,
in
the
international
organisation
continuously
rene
late
19th
which
The
century, is an
base
)
density
and
units
including
seeks to set up and
measurement
–3
),
(kg m
),
energy
( joule,
J,
where
1 J
–2
s
so
on,
are
derived
from
the
seven base
units.
Bureau international des poids et
established
1 kg m
are
several
dened
that
Boltzmann constant,
standards.
constant,
N
;
and
you
k;
the
according
will
to
seven constants,
recognize, such as the
speed of light,
Plank constant,
c;
the
Avogadro
h
A
The
International
System
of
Units
(SI,
from
the
French
The use of universal and precisely dened units is very
Système international d’unités) is the most commonly
used
system
seven
time
base
of
units:
(second,
(kelvin,
K),
intensity
measurement.
s),
Its
building
length
(metre,
m),
electric
current
(ampere,
amount
of
substance
mass
blocks
important, as it allows scientists from dierent countries
are the
to understand one another and share the results of their
(kilogram, kg),
A),
studies. What other advantages are there to internationally
temperature
shared and continuously updated measurement systems in
(mole, mol) and luminous
the natural sciences? You might want to look up the Mars
(c andela, cd). All other units, such as those of
Climate Orbiter.
Thinking skills
ATL
Throughout
history,
developed,
each
several
with
universal
dierent
temperature
reference
points.
sc ales
Some
have
of
been
these
are
summarized in table 4.
Sc ale
D ate
Reference
Newton
1700s
H
O
points
freezing point = 0°
2
Human
F ahrenheit
1700s
H
O
body
temperature = 12°
freezing point = 32°
2
H
O boiling point = 212°
2
Delisle
1700s
H
O
freezing point = 150°
2
H
O boiling point = 0°
2
Celsius
1700s
O
H
freezing point = 0°
2
H
O boiling point = 100°
2
Kelvin
1800s
Absolute
CGPM
1950s
Triple
BIPM
2018
Kelvin
Table 4
Examples
Temperature
in
the
units

Figure 18
A platinum–iridium cylinder
in the US was used
mass.
to dene a kilogram of
This standard
and
for
m
energy,
and
and
practic al
of
related
s.
It
v arious
to
has
practic al
reasons
( J),
been
could
energy
which
are
decided
reasons”.
=
273.16 K
k.
t e m p e ra t u r e
thermal
joules
water
to
What
sc ales
and
in
as
turn
keep
do
you
such
it
could
be
expressed
dened in terms of the base
kelvin
think
as
an
some
SI
of
base
these
unit
“for
historic al
be?
bec ame obsolete in
c arefully
at
table
4
above.
Identify
one
thing
you
see, one thing it makes
when the kilogram and all other SI
you
units were redened
think
about,
and
one
thing
it
makes
you
wonder.
Share
your
ideas with
as exact quantities
your
based
kg,
historic al
Look
2019,
unit
is
of
dened in terms of the Boltzmann
constant,

zero = 0
point
class.
on physic al constants
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Structure
Kelvin
is
temperature
considered
Absolute
any
to
zero
kinetic
hence
an
an
Under
water
(0 K)
energy
c annot
increase
normal
get
in
is
proportional
absolute
any
that
at
collisions.
colder.
temperature
pressure,
373.15 K.
zero
this
average
kinetic
temperature
M atter
An
of
water
Absolute
the
Introduction
to
the
particulate
nature
of
matter
energy of particles and
sc ale.
implies
on
to
1.1
at
increase
1
degree
boils
on
at
the
absolute
in
the
particles
zero
c annot
temperature
Celsius.
0 °C
is
of
1
c annot
lose
kelvin
equal
to
transfer
heat and
is
equivalent
273.15 K.
100 °C, so that makes the boiling point of
Celsius
sc ale
is
–273.15 °C.
t
Figure 19
The Celsius and
Kelvin
sc ales for temperature (all values are
rounded
to whole numbers)
400 K
water
100 °C
373 K
boils
350 K
50 °C
300 K
water
0 °C
273 K
freezes
40 °C
250 K
50 °C
You
will
learn
more about the
200 K
dry ice
kinetic
78 °C
energy of particles in
195 K
solid CO
2
Reactivity 2.2.
100 °C
150 K
150 °C
100 K
liquid
191 °C
82 K
air
200 °C
50 K
250 °C
absolute
273 °C
0 K
zero
Kelvin
Celsius
Linking questions
What
is
sample
the
graphic al
at
fixed
a
distribution
temperature?
of
kinetic
energy
values of particles in a
(Reactivity2.2)
What must happen to particles for a chemical reaction to occur? (Reactivity 2.2)
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Structure
1
Models
of
the
particulate
nature
of
matter
End-of-topic questions
5.
Which
changes
of
state
are
opposite
to
each other?
Topic review
A.
1.
Using
your
answer
the
knowledge
guiding
from the
question
as
Structure 1.1
fully
as
melting
and
condensation
topic,
B.
vaporization and deposition
C.
deposition and sublimation
D.
sublimation
possible:
How can we model the particulate nature of matter?
6.
Which
of
the
and
freezing
following
statements
is
incorrect?
Exam-style questions
A.
solids
and
liquids
are
almost
incompressible
Multiple-choice questions
2.
Which
of
the
following
are
examples
of
B.
particles
C.
liquids
D.
particles
in
both
solids
and
liquids
are mobile
homogeneous
and
gases
have
no
xed shape
mixtures?
I.
in
solids,
liquids
and
gases
c an
vibrate
Air
7.
II.
Which
elements
c an
be
separated
from
each
other
by
Steel
physic al
III.
Aqueous
KMnO
methods?
potassium manganate(VII),
A.
oxygen
B.
hydrogen
and
nitrogen in air
C.
c arbon
D.
magnesium and sulfur in magnesium sulde
(aq).
4
3.
A.
II only
B.
III only
C.
I and II only
D.
I, II and III
What
correctly
(c arbon
8.
Which
and
change
equivalent
to
decrease
B.
increase
C.
decrease
D.
increase
in
water
temperature
by
increase
in
on
the
Celsius
temperature
by
sc ale is
20 K?
by
20 °C
20 °C
dioxide)?
endothermic?
process
A
exothermic
CO
B
exothermic
CO
C
endothermic
CO
(s)
→
CO
(s)
→
C(g) + O
(s)
→
CO
Extended-response questions
(g)
(g)
9.
Explain
why
the
proportional
Celsius
of
the
obtain
solid
following
(s)
→
C(g) + O
methods
chloride
could
be
temperature
temperature
in
temperature
kinetic
is
directly
energy but the
is
not,
increment
is
even
the
though
same
in
a
1-degree
each
sc ale?
[2]
used to
Ionic
salts
c an
be
broken
down
in
electrolysis. The
from a solution of sodium
unbalanced
chloride
Kelvin
average
(g)
2
10.
sodium
to
(g)
2
2
Which
293.15 °C
293.15 °C
2
2
CO
by
2
2
endothermic
by
Equation describing the
2
4.
in
the
A.
oxygen
oxygen in dry ice
describes the sublimation of dry ice
Exothermic or
D
and
ionic
equation
for
the
electrolysis of molten
water?
lead(II)
I.
evaporation
II.
ltration
III.
distillation
bromide is:
2+
Pb
a.
+
Br
One
→
of
the
formula
A.
Pb + X
of
is
lead,
Pb. State the
X.
[1]
I only
B.
I and II only
C.
I and III only
b.
Balance
c.
The
the
I, II and III
equation.
electrolysis
c arriedout
D.
products
product
point
and
matter
of
at
of
molten
380 °C.
boiling
each
of
Write
equation
gave
data,
(b).
bromide is
reference to melting
species
state
in
lead(II)
With
point
the
temperature.
you
[1]
deduce the state of
in
the
symbols
equation at this
in
the
balanced
[2]
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18
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Structure
11.
The
kinetic
mass
×
the
energy
of
particles
square
of
the
is
c.
equal to half of their
Once
to
2
mv
=
.
Determine
how
much
the
the
obtain
Introduction
excess
removed,
velocity of the particles:
1
E
1.1
the
the
copper(II)
student
pure
to
particulate
oxide
needed
crystals
of
had
to
nature
matter
been
gure
copper(II)
of
out
sulfate
how
from
speed of
k
2
the
molecules
in
a
pure
gaseous
substance
will
solution.
couldfollow
when
the
Kelvin
temperature
is
Describe a method the student
increase
doubled.
to
obtain
pure, dry copper(II)
[2]
sulfatecrystals.
12.
Pure
c aeine
is
a
white
[3]
powder with melting point
14.
Study
the
gure
below.
235 °C.
vaporization
State
A
the
melting
chemist
is
point
of
c aeine
investigating
the
in
kelvin.
ec acy
of
[1]
C°
a.
b.
three
100
condensation
/
yield
in
all
method
yield
extraction
three
once
and
methods.
c ases
and
melting
is
0.960 g.
collects
point
The
of
the
the
theoretic al
She
uses
following
each
data
for the
product:
erutarepmet
c aeine
water
+
steam
water
ice
+
water
melting
0
Method 1
Method 2
Method 3
0.229
0.094
0.380
freezing
ice
Mass of c aeine
energy input
obtained / g
Melting point of
190–220
229–233
188–201
range
mass
caeine product / °C
a.
Explain
why,
input,the
i.
C alculate
the
mean
ii.
C alculate
the
Give
answer
and
of
the
constant
of
c aeine
obtained.
of
signic ant
to
an
gave
c.
Suggest
error
in
the
one
this
gures.
giving
purest
way
solution
to
a
A
student
a
0 °C
for
5.00 g
(at
of
dilute
period
sample
of
energy
remains
time.
[2]
c aeine
product.
minimize
acid
chloride
atmospheric
in
100.0 g of
temperature and
[2]
has
the
following
properties:
the

melting

boiling
point:
–3 °C
[1]
point:
101 °C
random
b.
[1]
copper(II)
sulfuric
sodium
standard
Sketch
sulfate
solution
a
show
graph
the
similar
heating
with
excess
to
the
curve
one
for
a
in
gure 16
sample of this
by
sodiumchloride
reacting
a
increasing
the
reason, which method
experiment.
prepares
of
water
to
13.
at
the
of
appropriate number
pressure)
Determine,
of
[2]
pure
iii.
spite
percentage yield of Method 1.
A
your
in
temperature
copper(II)
solution.
[2]
oxide.
15.
Elemental
iodine
exists
as
diatomic
molecules, I
.
At
2
Copper(II)
The
word
oxide
is
insoluble
equation
for
this
in
water.
reaction
is
as
follows:
room
temperature
black
solid
gently.
sulfuric acid + copper(II) oxide → copper(II) sulfate + water
cold
When
surfaces
pressure,
a.
Write
a
balanced
chemic al
that
and
readily
cooled,
without
solid
iodine
pressure,
forms
it
is
violet
a
lustrous purple-
fumes
when
heated
gaseous iodine deposits on
condensing.
melts
at
Under
114 °C
to
increased
form
a
deep-
equation, including
violet liquid.
state
symbols,
for
this
reaction.
[2]
a.
b.
The
acid
powder
could
was
was
be
heated,
added
observed
then
until
it
copper(II)
was
in
a
use
remove
to
mentioned
that
represent all changes of
b.
State
the
c.
Suggest
how
gaseous
iodine.
above.
[3]
excess and
reason, a method the student could
the
equations
state
melting
point
of
iodine
in
kelvin.
[1]
suspended in the solution,
quickly sinking to the bottom of the beaker. Suggest,
giving
Formulate
oxide
excess
copper(II)
oxide.
liquid
iodine
c an
be
obtained
from
[1]
[2]
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The nuclear atom
Structure 1.2
Understandings
How do nuclei of atoms dier?
Structure 1.2.1
The
answer
to
this
question
was
obtained
by
nucleus
100
years
of
brilliant
In
the
late
is
the
idea
of
atoms
rearranged
in
chemic al
that
electrons
that
Structure 1.2.2
matter
dierent
was
positively
neutrons
charged, dense
(nucleons).
Negatively
was
occupy the space outside the nucleus
Isotopes
of
are
atoms
of
the
same element with
neutrons.
were indivisible and
reactions
gaining
—
M ass
spectra
are
used to determine
(known as
the
theory)
—
numbers
Structure 1.2.3
atomic
a
and
LHA
composed
the
contain
protons
more fascinating than
what is known
1800s,
of
research. Sometimes, the
how we know
the question of
Atoms
composed
charged
question of
—
over
relative
atomic
masses
of
elements
from their isotopic
popularity. The
composition.
discovery
scientists
of
to
electricity
study
the
and
radioactivity
structure
of
the
allowed
atom
itself.
The structure of the atom (Structure 1.2.1)
An
atom
and
contains
neutrons
which
occupy
electrons
are
key
factors
It
is
very
2.
It
is
a
3.
It
has
an
vast
the
in
charged
known
region
as
nucleus,
nucleons).
outside
nucleus
of
the
which
Atoms
nucleus.
itself
also
The
contains
contain
protons,
experiment
structure
neutrons and
particles
made
the
atom
containing
itself.
virtually
all
the
mass of the atom.
charge.
designed
alpha
to
are
by
Ernest
were
given
in
red
Rutherford
toward
a
in
1911,
sheet
of
positively
gold
charged
foil. The main
gure 1.
movable
Rutherford’
s explanation
detector
+
alpha
protons
electrons,
are:
comparison
dense
positive
observations
beam of
of
small
highly
radioactive
alpha particles
the
1.
In
positively
known as subatomic particles.
The
a
a
(collectively
+
source
Most alpha
particles
are
undeflected
atom
+
gold
foil
Some alpha
vacuum
+
particles
are
deflected
slightly
A
few alpha
undeflected
+
large
slight
deflection
deflection
particles
bounce
+
off nucleus
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
20
Figure 1
Rutherford’s gold
foil experiment
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Structure
1.2
The
nuclear atom
F alsic ation
experiment
preceded it, namely the “plum-pudding model”.
vulnerable
The
plum-pudding
claim
that
never
be
amorphous
present
atoms
model
positively
throughout.
particles
red
at
the
undeected.
existing
falsied
model,
If
suggested
charged
this
gold
were
foil
the
way
that
blob
the
would
Rutherford’
s
paving
the atomic model
claims
gold
that
an
foil
Scientic
The
atom
was
electrons
stands
knowledge
have
degree
the
gone
through its
of
new model of the atom.
is
falsiable.
up
to
true
means
directions.
C an
a
means
testing
absolute
always
The
that
This
that
they
are
contradicts them. A scientic
severe
therefore
new
single
that
with
uncertainty.
knowledge
contradicted the
development of a
are
evidence
proven
c ase, all alpha
results
for
the
with
to
is
accompanied
provisional
further
counterexample
strong
c an
a
by a
nature of scientic
evidence
falsify
but
certainty. Scientic
c an
steer it in
claim?
Activity
The
lists
below
properties
which
of
show
the
the
observations
nucleus.
Determine
in
the
which
gold
foil
experiment and the
observation
is
explained
by
property.
Observation
Property
Nearly all the alpha particles went
The
nucleus
has
a
positive
charge.
Occ asionally, some of the alpha
The
nucleus
is
particles
comparison to the size of the atom.
straight through the gold foil.
The
bounced
alpha
when
straight back.
particles
closely
are
repelled
The
approaching the
nucleus
virtually
all
is
the
very small in
very
dense, containing
mass of the atom.
nucleus.
In
1911,
Rutherford
summarized
planetary model of the atom,
In
this
model,
nucleus
of
the
the
in
the
solar
entire
nucleus
negatively
charged
same
as
system’s
atom.
by
the
also
way
mass,
However,
electrostatic
of
electrons
planets
the
results
orbit
atomic
instead
his
experiments
known as the
of
the
orbit
the
Sun.
Just
nucleus
by
positively
as
contains
gravity,
by
proposing the
Rutherford model
the
the
over
Sun
contains
99.9%
electrons
(gure2).
charged atomic
are
of
held
the
99.8%
mass of
around the
attraction.
–
+
–
electron
–
proton
–
neutron
+
+
– nucleus

Figure 2
The Rutherford model of the atom
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Structure
1
Models
of
the
particulate
nature
of
matter
Models
Scientists
use
models
to
represent
natural phenomena. All
Atoms
themselves
are
extremely small. The diameter of
–10
models
have
understood.
2.
The
useful
size
limitations,
Consider
of
model
the
of
which
the
nucleus
the
should
depiction
is
be
of
identied and
the
exaggerated
atom
but
it
in
gure
serves as a
most
unit
atoms
used
to
is
in
the
range
1 × 10
–10
to
5 × 10
m. The
describe the dimensions of atoms is the
picometre, pm:
–12
nuclear atom.
1 pm = 10
The vast space in the atom compared to the tiny size of
the nucleus is hard to fully appreciate. Rutherford’
s native
In
X-ray
m
crystallography
dimensions
New Zealand is a great rugby-playing nation. Imagine
is
the
a
commonly
angstrom,
used
unit
for atomic
symbol Å:
–10
1 Å = 10
m
being at Eden Park stadium (gure 3) and looking down at
For
example,
the
atomic
radius of the uorine atom is
the centre of the pitch from the top row of seats. If a golf
–12
60 × 10
m
(60 pm).
To
convert
this
toÅ
we
c an use
ball were placed at the centre of the eld, the distance
dimensional analysis,
using
the
conversion
factors
between you and the golf ball would represent the
given
above:
distance between the electron and the nucleus.
–12
10
m
1 Å
–1
The
relative
volume
of
open
space
in
the
atom
is
vast, and
60 pm ×
×
=
0.60 Å
=
6.0 × 10
Å
–10
1 pm
our
simple
representation
of
gure
2
is
obviously
spite
tiny
of
its
limitations,
Rutherford’
s
work
has
formed the
unrealistic. The nucleus occupies
basis
a
m
Rutherford’s atomic model
In
in
10
of
much
of
our
thinking
on
the
structure of the atom.
volume of the atom and the diameter of an atom is
Rutherford
approximately
is
rumoured
to
have
said to his students:
100 000 times the diameter of the nucleus.
All science is either physics or stamp collecting!

Figure 3
Eden Park, Auckland,
New Zealand.
If the atom were the size of the stadium,
the nucleus would
look like a golf ball in the centre
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of the eld
22
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Structure
1.2
The
nuclear atom
TOK
All
the
could
two
models
be
we
argued
physicists,
have
that
Gerd
Binnig
Switzerland
invented
microscope
that
level.
This
Nobel
gave
Prize
in
discussed
objects
the
scientists
in
and
assume
only
that
“real”
Heinrich
sc anning
generates
Physics
are
atoms
when
Rohrer,
tunnelling
are
they
real.
c an
working
microscope
However, it
be
at
seen. In 1981
IBM in Zurich,
(STM),
an
electron
three-dimensional images of surfaces at the atomic
the
ability
1986
was
to
observe
awarded
individual
to
Binnig
atoms
and
directly. The
Rohrer
for their
groundbreaking work.
You
c an
nd
an
atomic
sc ale
lm
created
by
IBM
c alled
A Boy and his Atom
on
the internet.

Figure 4
Has
A still from A Boy and his Atom
technology
extended
human’
s
c apacity
to
make
observations of the
natural world?
How
important
are
material
tools
in
the
production or acquisition of
knowledge?
Other
experiments
particle,
masses
the
and
have
neutron,
charges
shown
with
of
that
nearly
the
the
the
nucleus
same
subatomic
mass
particles
also
as
are
contains
the
a
proton.
neutral subatomic
The
relative
shown in table 1.
t
Particle
Relative
mass
Relative
charge
Table 1
the proton,
proton
1
+1
neutron
1
0
electron
negligible
–1
Relative masses and
charges for
Loc ation
neutron and
electron
nucleus
The
electric
charge
c arried
by
a
single
electron
is
outside nucleus
known as the
elementary
The
–19
charge
(e)
and
it
has
a
value
of
approximately
1.602 × 10
C.
The
actual
particles
are
commonly
expressed
in
elementary
charge
units.
of
a
the
charge
proton as +e.
of
The
an
electron
symbol
e
is
c an
oen
be
represented as –e,
omitted,
so
it
is
and
customary
the
to
particles
and
c an
be
charges of
found in the
For
data
example,
masses
charges of
these
subatomic
booklet.
charge
say that
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electrons
and
protons
have
charges
of
–1
and
+1,
respectively.
23
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Structure
1
Models
of
the
particulate
nature
of
matter
How small is small?
Relevant
skills
•
Tool
3:
Apply
•
Tool
3:
Use
and
and
use
SI
prexes and units
interpret scientic notation
Instructions
1.
A
variety
lengths,
objects
of
small
but
in
lengths
rather
order
of
are
based
size,
shown in table 2. Without looking at their
on
what
from
you
smallest
Item
proton,
charge
sheet
of
onion
cell,
paper,
0.10 mm
bond,
250 µm
length
267 pm
printed
full
c arbon
atom,
diameter
150 pm
fullerene,
diameter
0.71 nm
C
each item, list these
0.84 fm
thickness
stop,
about
largest.
Length
radius
diameter
iodine-iodine
know
to
diameter
0.30 mm
60

2.
Table 2
Lengths of
various small items
Convert the length values into metres and state them in standard form to
two signicant gures. Refer to the following conversion factors:
–3
•
milli, m: 10
•
micro, µ: 10
•
nano, n: 10
•
pico, p: 10
•
femto, f: 10
List
the
–6
–9
–12
–15
3.
gave
4.
for
a
web
than
Provide
values
question
Conduct
smaller
5.
length
the
in
table
2
in
order
of
increasing
size.
Was
the
list
you
correct?
search
to
values
given
the
full
1
reference
nd
for
three
in
more
table
your
2,
values to add to the list: one
one
larger,
information
and
one
intermediate.
sources in question 4,
ATL
following
your
school’s
Atomic number
citing
and
referencing
and the nuclear
system.
symbol
As of 2023, there are 118 known elements, given atomic numbers 1 to 118. The atomic
number of an element is also the number of protons in the nucleus of that atom. Gold,
atomic number 79, has 79 protons, while carbon, atomic number 6, has 6 protons. As
all the relative mass is in the nucleus, the dierence between the atomic number and
mass number is the number of neutrons in the element. Gold has atomic number 79 and
mass number 197
. Therefore, it has 197 – 79 = 118 neutrons. Each element is neutral,
with no charge, so the number of electrons in a neutral atom must equal the number
ofprotons.
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24
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Structure
1.2
The
nuclear atom
Activity
Determine
Atomic
the
missing
symbol
values
Atomic
from
the
table.
number
M ass
number
Protons
Neutrons
O
Electrons
8
13
27
85
37
80
35
27
32
120
Pb
80
207
69
100
A
Chemists
frequently
neutrons,
protons
use
nuclear
symbol notation,
X, to denote the number of
Z
isotope,
for
Z
is the
example,
and
electrons in an atom.
atomic
with
mass
number,
number
and
197
X
is
and
A
the
represents
chemic al
atomic
the
mass number of the
symbol
number
79,
(gure 5). Gold,
would
have
a
nuclear
197
symbol notation of
Au.
79
mass number
A
N
=
=
+
Z
number
of
N
chemic al
where
symbol
for the element
neutrons
X
Z
atomic number

=
number
of
Figure 5
The nuclear symbol notation
Atoms
atoms
form
protons
compounds
sometimes
are
protons.
For
the
compound
ionic
example,
a magnesium
in
the
no
ion
nucleus
is
sharing
magnesium
a
2+
two
or
transferring
neutral,
magnesium
with
(12)
by
longer
atoms
oxide.
charge,
greater
having
as
than
react
electrons.
more
with
the
number
number
loses
of
of
As
fewer
oxygen
M agnesium
the
or
a
result, these
electrons than
atoms
two
to
produce
electrons
to
form
positively
charged
protons
negatively
charged
electrons
remaining (10).
The
resulting
charge
mass
is
also
number:
displayed
in
the
nuclear
symbol
24
24
(12
protons
+
12
notation
2
below:
+
charge: 2+
electrons)
(12
protons
–
10
electrons)
Mg
12
atomic number: 12
(12
The
oxygen
atom
chemic al element: Mg
protons)
gains
the
(magnesium)
two
electrons
lost
by
magnesium
to
produce
an
16
ion
with
a
2–
negative
charge.
The
nuclear
symbol
for
the
oxide ion is
oxide
2–
O
.
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8
25
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Structure
1
Models
of
the
particulate
nature
of
matter
The overall chemical equation for the reaction between magnesium and oxygen is
1
2+
O
Mg +
Ionic
in
bonding
is
→
Mg
2–
+ O
2
discussed further
2
Structure 2.1
2+
2–
Mg
+ O
two
ions
Ionic
is
more
result
bonds
in
a
hold
commonly
force
the
of
ions
written
attraction
together
as
MgO,
between
to
form
as
the
them
solid
opposite
charges on the
known as an
magnesium
ionic bond.
oxide.
Activity
Linking questions
Deduce
the
nuclear
notation
for
an
protons,
21
electrons,
ion
symbol
with
What determines the different chemical properties of atoms? (Structure 1.3)
24
and
28
How
does
the
atomic
number
relate to the position of an element in the
neutrons.
periodic
table?
(Structure
3.1)
Isotopes (Structure 1.2.2)
Isotopes are dierent atoms of the same element with a dierent number of neutrons.
As a result, they have dierent mass numbers, A, but the same atomic number
, Z.
35
Chlorine, for example, has two isotopes: one with mass number 35,
Cl, and one
17
37
Cl. They have similar chemical properties, as they are both
with mass number 37
,
17
chlorine atoms with the same number of electrons, but dierent physical properties,
such as density, because atoms of one isotope are heavier than atoms of the other
.
Naturally
occurring
(protium)
and
(gure6),
is
hydrogen
hydrogen-2
radioactive,
consists
of
(deuterium).
so
it
does
not
two
The
stable
third
occur
in
isotopes,
isotope
nature
in
of
hydrogen-1
hydrogen, tritium
signic ant quantities.
Activity
Copy
the
table
below
and
complete
it
by
deducing
the
nuclear
symbols and/
or composition of these isotopes.
Isotope
Nuclear
Z
symbol
N
A
1
hydrogen-1
(protium)
H
1
hydrogen-2
(deuterium)
hydrogen-3
(tritium)
Atomic
numbers
of
1
3
isotopes
are
oen
omitted
in
nuclear
symbol
notation.
For
37
example,
‘Cl’
17
,
tells
so
you
isotope
the
including
written
A ,
the
with
listed
a
for
of
chlorine
isotope
the
chlorine
atomic
hyphen,
each
is
number
such
element
with
as
on
mass
and
is
therefore
not
37
must
necessary.
chlorine-37,
the
number
periodic
or
c an
is
not
written as
Cl.
have an atomic number of
These
Cl-37. The
table
be
a
isotopes
c an also be
relative atomic mass,
whole
number
bec ause it
r
is
the
weighted
average of all isotopes of that element.
Natural abundance (NA)

Figure 6
A portable tritium
The radioactive dec ay of tritium
light
of
an
isotope
is
the
percentage of its atoms among
source.
produces
high-energy electrons (beta particles).
all
atoms
of
the
abundances
given
for
all
element
isotopes
of
found
an
on
our
element,
planet.
we
c an
If
we
know
c alculate
the
the
natural
average
A
of that
r
These electrons hit
a uorescent material
element.
The
opposite
task
(c alculation
of
natural
abundances
from
A )
is
possible
r
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and
26
make it
glow in the dark
only
if
the
element
is
composed
of
two
known isotopes.
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Structure
1.2
The
nuclear atom
Worked example 1
C alculate
the
A
r
for
iron
Isotope
using
N atural
the
values
in
the
following
table.
abundance (NA)/ %
54
Fe
5.845
Fe
91.754
56
57
Fe
2.119
58
Fe
0.282
Solution
We
know
The
natural
A
r
=
average
abundance
of
the
natural
values
add
abundance
up
to
57
×
100%
of
so
each
we
isotope
divide
by
multiplied
100
to
by
obtain
their
the
mass
numbers.
average.
Therefore:
54
A
r
× 5.845
+
56
×
91.754
+
2.119
+
58
×
0.282
=
= 55.91
100
Worked example 2
There
are
two
abundance
stable
(NA)
of
isotopes
each
of
chlorine:
isotope
given
Cl-35
that
A
and
for
Cl-37.
chlorine
C alculate
is
the
natural
35.45.
r
Solution
(A
A
of isotope 1
×
NA
of isotope 1)
+
(A
of isotope 2
×
NA
of isotope 2)
=
r
100
Therefore:
(35 × NA
of
Cl-35) + (37 × NA
of
Cl-37)
=
35.45
100
Let
x
=
NA
of
Substituting
Cl-35, then 100
in
35x + 37(100
the
above
x
=
equation
NA
of
Cl-37.
gives:
x)
=
35.45
100
Expanding
3700
the
brackets
and
resolving the
x
terms
gives:
2x
=
35.45
100
Then
rearrange in terms of
3700
x
x:
3545
=
2
x =
77
.5 and 100
x =
22.5.
Therefore,
35
The
actual
natural abundances of
the
natural
abundance
of
Cl-35
is
77.5%
and
Cl-37 22.5%.
37
Cl and
Cl
are
75.8
and
24.2%,
respectively.
Average
A
values
for all elements
r
The results of our calculations are slightly dierent because we used mass numbers,
are
35
which
are
rounded
values
for
the
actual
masses of the
given in the data booklet and in
37
Cl and
Cl
atoms.
the periodic table at the end of this
book.
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Structure
1
Models
of
the
particulate
nature
of
matter
Density
at
Melting
Boiling
point / °C
point / °C
Compound
–3
4 °C / g cm
1
H
O
1.000
0.00
100.0
O
1.106
3.82
101.4
2
2
H
2

Table 3
Physic al properties of normal and
heavy water
235
Naturally

Figure 7
A pellet
of enriched
occurring
dierences
in
(increase
the
properties
of
these
are
used
for the
U. The
enrichment
uranium
235
used
isotopes
238
U and
uranium consists of two main isotopes,
physic al
as fuel in nuclear reactors
in
proportion of
238
U
over
U)
of
nuclear
fuel
(gure7), as most
235
nuclear
reactors
contains
only
Enriching
to
track
0.72%
one
the
require
type
of
of
uranium
this
at
least 3% of
U,
while
natural
uranium
isotope.
isotope
mechanisms
with
and
in
a
particular
progress
of
substance
reactions.
c an
This
is
also
make
oen
it
possible
referred to as
isotope labelling.
Global impact of science
Developments
in
environmental,
politic al,
Nuclear
ssion,
colossal
amounts
of
of
meitnerium
109,
in
Her
published in
bomb
stating
“I
you
will
to
work
with
is
a
technology
is
as
such
well
Frisch
In
later
being
up
the
have
nuclei
as
the
aer
to
years,
the
in
large
It
from
atoms
releasing
led tothe
bomb.
the
discovery
the
has
Meitner,
Meitner
developed
ethic al,
consequences.
of
atomic
Lise
doctorate
led
may
economic
development.
named
physics
Otto
1939.
and
splitting
one
(Mt),
applic ations
cultural
energy,
receive
Nature in
atomic
C an
nuclear
history
their
involves
energy,
Element
(gure8).
Figure 8
which
and
social,
development
woman

science
was
US.
the
second
University of Vienna
of
nuclear
ssion,
invited to work on
She
declined, famously
have nothing to do with a bomb!”
think
of
other
scientic
developments
that
have had important
Austrian-Swedish physicist
ethic alimplic ations?
Lise Meitner in 1906
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Structure
1.2
The
nuclear atom
LHA
Practice questions
Linking question
1.
State
the
nuclear
symbols
numbers
of
2.
Naturally
occurring
protons
and
for
potassium-39
neutrons
in
the
and
copper-65.
nucleus
of
each
Deduce the
How
isotope.
c an
provide
sulfur
has
32
abundances:
C alculate
isotopes
33
S(95.02%),
the
four
average
A
with
the
following
34
S(0.75%),
value
for
mechanism?
36
S(4.21%)
and
isotope
evidence
tracers
for
a
reaction
natural
(Reactivity 3.4)
S(0.02%).
sulfur.
r
3.
The actual
A
value
of
sulfur
is
32.07.
Suggest
why
your
answer to the
r
previous
question
diers
from
this
value.
M ass spectrometry (Structure 1.2.3)
The
mass spectrometer
abundance
of
isotopes
(gure
in
a
9)
is
an
instrument
used
to
detect
the
relative
sample.
detector
lightest particles
positive
ions
(stage 5)
are
(deflected most)
accelerated
field
heating
filament
sample
(stage 1)
to
in
the
electric
(stage 3)
vaporize
magnet
inlet
to
(stage 4)
inject
heaviest particles
sample
(deflected
least)
N
electron
ionize
beam to
sample
(stage 2)
S

Figure 9
The
sample
within
a
the
result,
is
injected
sample
the
known as
Schematic diagram
are
atoms
c ations.
into
then
lose
For
of
the
a mass spectrometer
instrument
bombarded
some
of
example,
their
and
with
vaporized
high-energy
electrons
copper
atoms
to
form
c an
be
(stage 1). The atoms
electrons
positively
ionized
as
(stage
2).
As
charged ions,
follows:
+
Cu(g) + e
→
Cu
(g) + 2e
The resulting ions are then accelerated by an electric eld (stage 3) and deected by
a magnetic eld (stage 4). The degree of deection depends on the mass to charge
ratio (m/z ratio). Particles with no charge are not aected by the magnetic eld and
therefore never reach the detector
. The species with the lowest m and highest z will
be deected the most. When ions hit the detector (stage 5), their m/z values are
determined and passed to a computer
. The computer generates the mass spectrum
of the sample, in which relative abundances of all detected ions are plotted against
their m/z ratios(gure 10).
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Structure
LHA
u
1
Models
Figure 10
of
the
particulate
nature
of
matter
M ass spectrum of a
sample of copper
100
80
ytisnetni
60
evitaler
40
20
0
0
60
62
64
66
68
m/z
The
operational
examination
details
of
the
mass
spectrometer
will
not
be
assessed in
papers.
Worked example 3
Figure
A
,
of
11
shows
boron
a
from
mass
this
spectrum
mass
from
a
sample
of
boron.
C alculate
the
relative
atomic
mass,
spectrum.
r
100
80.1
ytisnetni
evitaler
50
19.9
0
0
2
4
6
8
10
12
m/z

Figure 11
M ass spectrum of boron
Solution
First,
we
number
mass
We
need
of
10,
number
c an
then
to
derive
which
of
11,
the
has
a
which
c alculate
A
information
relative
has
by
a
from
graph.
abundance
relative
finding
the
of
abundance
sum
of
The
19.9%.
the
of
peak at
The
m/z
peak at
=
10
m/z
represents
=
11
an
isotope
with
a
mass
represents an isotope with a
80.1%.
relative
abundance
of
each
isotope
multiplied
by
its
mass
r
number. The relative abundance values add up to 100%, so we divide the result by 100 to obtain the average.
11
×
80.1
+
10
×
19.9
=
10.8
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100
30
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Structure
1.2
The
nuclear atom
LHA
Data-based questions
1.
Estimate
atomic
the
relative
mass,
A
,
for
abundance
this
of
element
each
and
isotope
identify
the
from
gure
12.
Use
your
estimates
to
c alculate
the
relative
element.
r
t
Figure 12
M ass spectrum
of unknown element
6
5
ytisnetni
4
evitaler
3
2
1
0
204
203
205
206
207
208
209
m/z
2.
M ass
spectrometry
including
those
of
is
used
cosmic
for
discovering
origin.
For
the
presence
example,
cobalt
of
and
specic
nickel
elements
are
in
common
geologic al
samples,
components
of
iron
meteorites
(gure14).
Cobalt
and
nickel
compositions
of
have
these
similar
two
properties
metals
are
and
very
nearly
dierent,
identic al
so
they
relative
c an
atomic
easily
be
masses.
However,
distinguished
by
the
mass
isotopic
spectrometry
(gure13).
100
100
80
ytisnetni evitaler
ytisnetni evitaler
80
nickel
60
40
cobalt
60
40
20
20
0
0
0
58
60
0
62
58
60

Figure 13
M ass spectra of cobalt (le) and nickel (right)
Estimate
relative
the
relative
atomic
abundance
mass,
A
and
of
hence
each
isotope
deduce
for
nickel.
Use
whether
cobalt
or
your
nickel
estimates
has
the
The
actual
A
value
for
nickel
to
c alculate
larger
r
3.
62
m/z
m/z
its
A
r
is
58.69.
Suggest
why
your
result
in
question
2
is
dierent.
r
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31
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1
Models
of
the
particulate
nature
of
matter
LHA
Structure

Figure 14
Tamentit
iron meteorite,
found
in 1864 in the S ahara Desert
Mass spectra
M ass
a
spectra
chance
to
c an
be
practice
found
in
various
c alculating
databases
average
atomic
on
the
mass
internet,
values
giving
you
from authentic
data.
Relevant skills
•
Tool
2:
Identify
•
Tool
3:
Percentages
and
extract
data
from databases
Instructions
1.
Using
a
database
of
your
choice,
search
for
the
mass
spectra
of
three
dierent elements.
2.
From
the
mass
spectra,
c alculate
the
relative
atomic
mass
of
each
element.
3.
Compare
booklet.
your
c alculated
Comment
on
relative
any
atomic
dierences
mass
you
to
that
stated in the data
observe.
Linking question
How does the fragmentation pattern of a compound in the mass spectrometer
help in the determination of its structure? (Structure 3.2)
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Structure
1.2
The
nuclear atom
End-of-topic questions
5.
Which
of
the
following
statements
are
correct?
Topic review
I.
1.
Using
your
knowledge
from the
Structure 1.2
Nearly
all
mass
of
the
atom
is
contained within
topic,
its nucleus.
answer
the
guiding
question
as
fully
as
possible:
II.
The
How do nuclei of atoms dier?
mass
number
shows the number of
protons in an atomic nucleus
III.
Isotopes
of
the
numbers
of
protons.
same
element
have
equal
Exam-style questions
Multiple-choice questions
63
2.
What
is
correct
for
2+
Cu
A.
I and II only
B.
I and III only
C.
II and III only
D.
I, II and III
?
29
Protons
Neutrons
Electrons
A
29
34
27
B
29
34
31
6.
Which
of
C
34
63
31
D
34
29
27
1
3.
Which
values
are
the
same
for both
2
H
and
H
2
I.
the
following
species
contain
equal
numbers
A.
cobalt-58 and nickel-58
B.
cobalt-58 and nickel-59
C.
cobalt-59 and nickel-58
D.
cobalt-58 and cobalt-59
?
2
boiling point
II.
∆H
III.
number
Extended-response questions
of combustion
of
protons
7.
IV.
of
neutrons in their nuclei?
density
The
at
gold
gold
page
A.
I and III only
B.
I and IV only
C.
II and III only
D.
I, II and III
a.
foil
foil.
experiment
This
involved ring alpha particles
experiment
is
depicted
in
gure 1 on
20.
An alpha particle is a helium nucleus. State the
nuclear
b.
symbol
Suggest
that
the
would
for
an
results
have
of
been
alpha
the
particle.
gold
foil
observed
in
[1]
experiment
each of the
6
4.
The
naturally
occurring
isotopes
of
lithium
are
following
Li and
alternative scenarios:
7
Li.
Which
shows
the
correct
approximate
percentage
i.
abundances
for
Atoms
are
instead
hard,
dense, solid balls
lithium?
of
Percentage
Percentage
6
abundance of
ii.
positive
Atomic
charge.
nuclei
are
[1]
instead
negatively
7
Li
abundance of
75
25
B
50
50
Li
charged.
[1]
39
8.
There
are
two
stable
isotopes
of
potassium:
K and
41
K. The
A
of
potassium
is
39.10.
Use
this
information
LHA
A
r
C
35
65
to
D
10
determine
isotopes
90
and
the
relative abundances of the two
sketch
the
mass
spectrum
of
potassium
metal.
9.
“Dutch
14%
[3]
metal”
zinc.
oen
used
Dutch
mass
is
This
for
metal
an
alloy
alloy
making
c an
be
spectrometry.
composed of 86% copper and
closely
resembles gold, so it is
costume
jewellery.
distinguished
Explain
how
from gold using
[2]
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33
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Electron congurations
Structure 1.3
How c an we model the energy states of electrons in atoms?
This question is complex with many layers. What are electrons? How do we know they exist in energy states? What various
models about these energy states are there?
According
to
behaviour
has
sizes
of
these
modern
no
views,
analogues
clouds
electrons
in
depend
our
on
are
quantum
everyday
the
life,
energies
of
objects
we
c an
that
behave
visualize
electrons,
which
as
both
electrons
in
c an
only
have
particles
atoms
as
and
fuzzy
certain,
waves.
clouds.
predened
Although
The
such
shapes
and
v alues.
Understandings
Structure 1.3.1
—
atoms
emitting
photons
return
to
energy
lower
Emission
spectra
when
are
produced
electrons
in
by
excited states
Structure 1.3.5
—
state
electron
—
The
line
emission
provides
evidence
given
for
the
and
Sublevels
E ach
c an
orbital
has
a
dened
conguration
hold
two
contain
a
xed
there
a
high
and
energy
chemic al
electrons of opposite
number
of
orbitals,
regions
spectrum of
of
hydrogen
a
environment,
levels.
spin.
Structure 1.3.2
for
space
where
is
probability of nding an
existence of
electron.
electrons
in
discrete
energy
levels,
which
converge at
energies.
Structure 1.3.3
—
The
main
energy
level
is
given an
Structure 1.3.6
—
In
of
at
higher
—
Successive
convergence
an
emission
spectrum, the limit
frequency
corresponds to
LHA
higher
2
integer
number,
n,
and
c an hold a maximum of 2n
ionization.
electrons.
Structure 1.3.7
ionization
energy data
Structure 1.3.4 — A more detailed model of the atom
for
an
element
give
information
about
its
electron
describes the division of the main energy level into s, p, d
conguration.
and f sublevels of successively higher energies.
Emission spectra (Structure 1.3.1)
Much
of
studies
that
our
understanding
involving
sunlight
prism.
This
c an
be
which
each
example
of
a
A
gaseous
pure
glow —
prism,
as an
it
to
the
within
spectrum
a
prism
the
is
subjected
series
and
a
continuous
(gure
into
and
spectrum
words,
a
into
congurations
In
the
1600s,
dierent
next,
the
as
and
atoms
Isaac
has
come
Newton
from
showed
coloured components using a
(gure
appears
in
Sir
a
no
rainbow.
1a).
This
type
continuous
gaps
The
are
of
series
visible.
spectrum
of
The
colours,
classic
wavelength of visible light
700 nm.
element
other
produces
down
merges
emission spectrum
between
lines
400 nm
in
light.
continuous spectrum
continuous
from
electron
with
wavelengths,
colour
ranges
will
broken
generates a
contains light of all
in
of
interaction
it
of
will
lines
(gure
source
to
emit
a
high
light.
against
1b)
of
spectrum
In
dark
contrast,
visible
will
a
voltage
When
light
appear.
under
this
light
reduced
passes
background.
when
of
all
This
a
cold
This
gas
pressure
through a
is
is
known
placed
wavelengths, a series of dark
is
known as an
absorption
1c)
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Structure
1.3
Electron
congurations
a
continuous
emission
spectrum
spectrum
b
hot gas
c
cold gas

Figure 1
The spectra generated
absorption
from
(a) visible light
of
all wavelengths (b) a heated
spectrum
gas (c) visible light
of all wavelengths
passing through a cold gas

Figure 2
The aurora borealis (Northern Lights) in Lapland,
drawn by the E arth’s magnetic eld
to the polar regions,
Sweden.
Charged
high-energy particles from
where they excite atoms and
the Sun are
molecules of atmospheric gases,
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c ausing them to emit light
35
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Structure
1
Models
of
the
particulate
nature
of
matter
Emission spectra
Emission
spectra
handheld
Discharge
ionized
c an
be
spectroscope
lamps
when
a
observed
by
contain
voltage
through a simple
holding
it
up
low-pressure
is
to
a
gases
light
Method
source.
which
1.
Observe natural light through the spectroscope. Note
2.
O bs e r ve
are
down the details of the spectrum you observe.
applied.
L E D.
Relevant
a r ti fi c i a l
No te
d ow n
light
the
f ro m
a
details
c o m pu te r
of
th e
s c re e n
s pe c tr u m
or
yo u
skills
o bs e r ve.
•
Tool
•
Inquiry
3:
Construct
graphs
2:
and
and
draw lines of best t
3.
Identify
record
relevant
Observe
down
observations
and
sucient
relevant
light
the
Inquiry
2:
Identify
and
describe
patterns,
Inquiry
2:
various
of
the
discharge lamps. Note
emission
colours,
lines
you
observe,
wavelengths and number of lines.
trends and
Q uestions
relationships
•
details
quantitative data.
including
•
from
qualitative
Assess
accuracy
1.
Sketch
2.
Describe
the
spectra
each
as
you
a
observed.
continuous,
emission or
S afety
absorption
•
Wear
•
The
eye
protection.
3.
discharge
lamps
will
get
very
hot.
Look
up
the
discharge
lamps
spectra of the elements in
you
observed.
Compare the
c are.
theoretic al
•
emission
Handle them
the
with
spectrum.
Further
safety
prec autions
will
be
given
by
depending
on
the
exact
observed
emission lines, commenting
your
on
teacher,
and
the
number,
colours and positions of the
nature of the
emissionlines.
discharge lamps.
4.
Next,
you
will
wavelengths
compare
of
the
the
theoretic al
emission
lines.
and
observed
Construct
a
graph
M aterials
of
•
Discharge lamps
•
Handheld
theoretic al
Draw
a
line
of
wavelength
best
t
vs
observed
through
wavelength.
your data.
spectroscope
5.
Comment on the relationship shown in your graph.
6.
Comment
on
the
accuracy
of
the
observed
wavelength data.
E ach
to
element
identify
orange
same
the
light
emission
Figure 3
Sodium
streetlights (le) and
its
own
element.
with
Like
example,
colour
barcodes
spectra
characteristic
For
wavelengths
yellow-orange
substance.

has
c an
be
in
of
589.0
appe ars
a
used
shop
to
line
spectrum,
excited
in
and
a
that
test
be
of
used
chemic al
which
atoms
589.6 nm
ame
c an
identify
sodium
c an
emit
(gure3,
any
to
be
used
yellow-
right).
The
sodium-containing
identify
products,
line
elements.
the line emission spectrum of sodium (right)
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Structure
1.3
Electron
congurations
Observations
Chemists
oen
properties
directly
of
through
instruments.
boundaries
the
of
our
light
in
the
helium
emission
human
features
through
from observing the
senses
in
a
lamps
detail.
As
is
also
of
of
is
in
the
dierence
between
observing
a
natural
directly and with the aid of an instrument?
expand the
to
vapour lamps
gure 3, observing
reveals
the
orange
helium
is
phenomenon
(oen sight), or with
Sodium
seen
region
What
revealing otherwise
spectroscope
yellow
spectrum
or
light.
c an be made
technology
observations,
orange-yellow
emission
from
the
data
Observations
Advancements
imperceptible
emit
generate
matter.
a
strong
spectrum. The light
the
more
naked
complex
eye but the

Figure 4
Helium emission spectrum
(gure 4).
Flame tests
Flame testing is an analytical technique that can be used to
Materials
identify the presence of some metals. The principle behind
•
Flame
ame tests is atomic emission. Electrons are promoted
•
Small
to a higher energy level by the heat of the ame. When
•
Bunsen
they fall back to a lower energy level, photons of certain
•
Small
wavelengths are emitted. Some of these photons are in the
KCl,
test
wire
portion
burner
samples
C aCl
(platinum
of
dilute
and
of
, SrCl
2
or
nichrome)
hydrochloric acid
heatproof mat
various
, CuCl
2
metal
salts
(e.g.
LiCl,
NaCl,
)
2
visible region of the spectrum.
Method
1.
Clean the end of the ame test wire by dipping it into
the HCl solution and placing it in a non-luminous Bunsen
burner ame. Repeat until no ame colour is observed.
2.
Dip
the
end
samples,
Bunsen
metal

Figure 5
Flame test
of
and
burner
in
the
the
ame
place
it
in
ame,
salt
and
test
the
noting
the
wire
into
one
of
the
salt
edge of the non-luminous
down the identity of the
colour(s)
you
observe.
colours for dierent elements
Relevant skills
•
Inquiry
2:
Identify
and
record
relevant
qualitative
observations
S afety
•
Wear
•
Take
eye
protection.
suitable
•
Dilute
•
A
variety
of
which
prec autions
3.
Clean the wire again and repeat with other salt samples.
4.
Clear
around open ames.
up
as
instructed
by
your
teacher.
hydrochloric acid is an irritant.
of
dierent
chloride
salts
will
be
used, some
Q uestions:
are
irritants —
avoid contact with the skin.
1.
•
Dispose
•
Further
of
all
substances
Look up the emission spectra of the metals you tested.
appropriately.
Compare these to the colours you observed. Comment
safety
prec autions
will
be
given
by
your
on any similarities and dierences.
teacher,
depending
on
the
identity
of
the
salts being
2.
Explain
why
the
dierent
metals
show
dierent
analysed.
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amecolours.
37
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Structure
1
Models
of
the
particulate
nature
of
matter
TOK
One
of
the
ways
knowledge
Inductive reasoning
up”:
they
take
is
involves
specic
developed
drawing
observations
is
through
conclusions
and
build
reasoning.
from
general
inductive
Reasoning
experimental
principles
reasoning
c an
be
deductive
observations.
or
Inductive
(“bottom-up”
3.
1.
For
example,
you
might
the
following
Lithium
chloride
Lithium
sulfate
gives
a
red ame test.
Lithium
iodide
gives
a
red ame test.
From
these
observations,
Deductive arguments
when
a
asked
to
apply
are
approach):
c an
“top
make
down”:
scientic
the
conclusion
they
infer
knowledge
hypothesis
about
lithium
salts:
in
that
specic
a
new
all
lithium
salts
conclusions
give
from
red ame tests.
general
reasoning
(“top-down”
approach):
pattern
4.
example,
Lithium
From
this,
What
are
C an
suppose
bromide
salts
you
the
give
is
your
a
scientic
lithium
knowledge
includes
the
following
existing
observation
premises:
salt.
red ame tests.
could
propose
advantages
reasoning
You do this all the time
hypothesis
3.
Lithium
premises.
context.
theory
2.
For
theory
pattern
observations
deductive
1.
“bottom
red ame test.
you
your
are
observation
make
gives
arguments
from them.
4.
2.
inductive.
always
be
and
that
lithium
bromide
disadvantages
neatly
classied
of
into
gives
each
these
a
type
two
red ame test.
of
reasoning?
types?
On
what
grounds
might
we
doubt
a
claim
reached
through
inductive
On
what
grounds
might
we
doubt
a
claim
reached
through
deductive
Visible light is one type of
light,
are
microwaves,
all
The
part
of
energy
the
of
reasoning?
reasoning?
electromagnetic (EM) radiation. In addition to visible
infrared
radiation
electromagnetic
the
radiation
is
(IR),
ultraviolet
(UV),
X-rays
and
gamma
rays
spectrum.
inversely
proportional
to
the
wavelength,
λ:
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1
E
∝
λ
38
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Structure
Electromagnetic
waves
all
travel at the
speed of light,
8
of
light
is
approximately
frequency
of
the
equal
radiation,
to
3.00 × 10
f,
by
the
such
as
gamma
c,
in
a
vacuum.
The
1.3
Electron
congurations
speed
–1
m s
following
.
Wavelength
is
related to the
equation:
c = f × λ
High
energy
EM
waves,
rays,
have
short
wavelengths and high
frequencies while low energy waves, such as microwaves, have long wavelengths
and
low
frequencies.
f
λm
1
10
4
10
gamma
rays
14
10

10
(γ
rays)
λnm
1
10
0
10
400
10
1
10
X-rays
10

ultraviolet
10
00
1
elbisiv
10
ygrene
(UV)

10
14
10
infrared
00
(IR)
4
10
1
10

10
Activity
10
microwaves
10
00
Compare
0
the
colours
red and
10

10
green

in
gure 6. Determine which
colour has:
10

10
radio
waves
a.
the
highest
wavelength
b.
the
highest
frequency
c.
the
highest
energy
4
10
4
10

Figure 6
The wavelength (λ) of electromagnetic radiation is inversely
proportional to both frequency and
energy of that
radiation
Data-based questions
Look
at
the
spectra
below.
Explain
how
we
know
that
stars
are
partly
composed
of
hydrogen.
3900
4000

Figure 7
7600
4500
5000
The hydrogen emission spectrum
5500
(top) and
6000
6500
the absorption spectrum generated
from
7000
7500
the Sun (bottom)
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Structure
1
Models
of
the
particulate
nature
of
matter
The line emission spectrum of hydrogen
(Structure 1.3.2 and 1.3.3)
E ach
line
which
the
A
in
idea
that
photon
radiation
E
=
the
emission
corresponds
is
as
to
a
spectrum
specic
electromagnetic
a
quantum
of
of
an
element
amount
radiation
energy,
of
comes
which
is
has
energy.
in
a
specic
This
is
wavelength,
c alled
quantization:
discrete packets, or quanta.
proportional
to
the
frequency of the
follows:
h × f
Where
E =
the
specic
energy
possessed
by
the
photon,
expressed in joules, J
–34
h =
f
=
Planck’
s
constant,
frequency
of
the
6.63 × 10
radiation,
J s
expressed
in
hertz,
Hz,
or
inverse
–1
seconds, s
In
1913,
Niels
emission
1.
The
electron
These
2.
3.
Bohr
spectra.
orbits
When
an
of
the
right
at
that
When
This
Bohr ’s
c an
are
main
exist
the
in
amount
for
a
a
the
his
only
in certain
the
orbit
stationary
orbits
the
lowest
moves
to
to
the
a
lower
energy
orbiting
making
any
electrons
Since
electrons
attempt
not
electrons
dened
would
prolonged
did
to
in
energies,
the
photons
of
specic
spectra.
By
measuring
the
energies
For
a
of
energy
Bohr
their
radiate
existence
radiate
energy
higher
energy
overcome
based on its
around the nucleus.
level absorbs a photon
energy
level
and
remains
of
model
atoms
wavelengths,
of
level, it emits a photon of light.
between
the
main
Classic al
the
two
levels.
problem of the
electrodynamics
predicted
energy and quickly fall into the nucleus,
when
transitions
the
a
dierence
Rutherford model of the atom (Structure 1.2).
that
atom
were:
time.
returns
rst
it
hydrogen
theory
discrete energy levels
with
energy,
represents
was
of
of
the
of
short
electron
photon
model
postulates
associated with
electron
level
theory
proposed
The
impossible.
Bohr
postulated that
staying in stationary orbits.
the
atom
could
have only certain, well-
between stationary orbits could absorb or emit
producing
wavelengths
of
characteristic lines in the atomic
these
lines,
it
was
possible
to
c alculate
electrons in stationary orbits.
hydrogen
atom,
the
electron
energy (E
)
in
joules
could
be
related to the
n
energy
level number (n)
by
a
simple
equation:
1
E
= –R
n
H
2
n
–18
where
R
≈
2.18 × 10
J
is
the
Rydberg
constant.
This
equation
clearly
represents
H
the
quantum
only
nature
discrete,
of
the
quantized
half-integer
parameters,
number (n)
c an
numbers
mean
take
values.
where
These
known as
only
higher
atom,
positive
the
energy
values
are
of
an
electron
characterized
quantum numbers. The
integer
values
(1,
2,
3,
c an
by
have
integer or
principal quantum
…),
where
greater
energy.
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Structure
The
most
electron
stable
has
state
the
ground state
of
lowest
of
the
atom.
c alled
excited states.
return
to
the
the
ground
hydrogen atom is the state at
possible
In
contrast,
Atoms
state
energy.
by
in
the
excited
emitting
This
energy
energy
states
of
=
is
unstable
specic
1,
Electron
congurations
where the
known as the
levels with
are
photons
n
level
1.3
n
=
and
2,
3,
…
are
spontaneously
wavelengths
(gure8).
+energy
e
+
+
p
+
p
e
p
excitation
dec ay
hf

Figure 8
Energy
rungs.
stand
levels
An
in
atoms
Electrons
between
specic,
the
Electrons returning to lower energy levels emit a photon of light, hf
amount
electron
energy
the
discrete
same
c an
level.
spectrum
of
resemble
c annot
exist
ladders
between
rungs
of
a
amount
of
energy,
of
be
ladder.
with
varying
energy
Jumping
and
levels,
up
distances
between the
much
how
each
jumping
rung
down
a
like
or
rung
level
or
you
c annot
requires a
level
releases
energy.
excited
Electrons
hydrogen
to
any
energy
returning to
n
=
2
level,
will
n,
and
return
to
any
lower
produce distinct lines in the visible
(gure 9).
Note that the red line has a longer wavelength and lower frequency than the violet
line. The energy of the photon released is lower when an electron falls from n = 3 to
n = 2, than from n = 6 to n = 2. In both cases, it represents the dierence between
two of the allowable energy states of the electron in the hydrogen atom.
colour
wavelength / nm
transition
from
violet
blue
cyan
red
410
434
486
656
n
= 6
n
= 5
n
= 4
n
n
= 6
n
= 5
n
= 4
n
= 3
n
= 2
n
=
= 3
1
◂
Figure 9
The visible lines in the
emission spectrum
of hydrogen
show electrons returning from higher
energy levels to energy level n = 2
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Structure
1
Models
of
the
particulate
nature
of
n = 7
matter
Electron
transitions
to
the
ground
state,
n =
1,
release
higher
energy, shorter
n = 6
wavelength
ultraviolet
the
infrared
region
It
important
light,
while
electrons
returning to
n =
3
produce lines in
n = 5
of
the
electromagnetic
spectrum
(gure 10).
n = 4
n = 3
is
required
IR
to
note
that
electrons
between
will
allowable
absorb
energy
or
rele ase
states.
Any
only
the
excess
exact
will
not
energy
be
radiation
absorbed,
n
to
move
not
= 2
and
if
an
insucient
amount
of
energy
is
supplied
the
electrons
will
move.
visible light
Energy
levels
closer
to
the
nucleus
hold
fewer
electrons. The maximum number
2
of
electrons
holds
n
n
=
3
to
has
a
any
two
energy
level,
electrons, at
maximum
of
18
n
n, is 2n
=
2
. For
there
example,
could
electrons, and
n
=
be
4
a
has
the
energy
maximum
a
of
maximum
level with
eight
of
32
n
=
1
electrons,
electrons.
1
UV

=
up
in
radiation
Figure 10
Electron transitions for the
Communic ation skills
ATL
hydrogen atom.
Notice how the allowable
energy levels get closer together when
When
explaining
concepts,
we
sometimes
use
diagrams,
graphs or images to
the electron moves further away from the
help
nucleus.
us
convey
our
ideas
more
clearly.
The energy dierence between
Prepare
n = 3 and
a
written
explanation
of
atomic
emission that does not include
n = 2 is much smaller than that
any
between n = 2 and
diagrams.
Exchange
it
with
a
partner.
Give
each
other
feedback,
n = 1
concentrating on:
•
Use
•
Order
•
Whether
When
of
you
feedback
or
scientic
in
any
have
to
diagram
whether
which
or
are
important
shared
make
to
voc abulary
ideas
each
it
other ’
s
improvements
accompany
not
given
concepts
adds
to
your
the
to
are
missing
from
the
explanation.
feedback, spend some time using the
your
work.
explanation.
Finally,
Discuss
choose
why
a
graph, image
you chose it and
explanation.
Linking questions
What
qualitative
such
as
gas
from
gaseous
How
does
period
quantitative
do
an
tubes
elements
different
How
and
discharge
emission
and
element’s
in
the
data
c an
prisms
in
be
the
collected
study
of
from instruments
emission
spectra
from light? (Inquiry 2)
spectra
elements?
number
and
provide
evidence
for
the
existence of
(Structure 1.2)
highest
periodic
occupied
table?
main
(Structure
energy
level
relate to its
3.1)
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Structure
1.3
Electron
congurations
The quantum mechanic al model
of the atom (Structure 1.3.4)
The
Bohr model
atoms.
It
energy
levels.
of
was
narrow
several
1.
It
3.
It
bec ause
model
more
attempt
than
and
could
one
assumed
the
to
explain
quantization:
allowable
problems
The
2.
in
an
on
According
lines
dierences
was
based
to
the
Bohr,
the
incorrect
electron
It
energy
that
emission
levels.
of
states
electrons
spectra
these
However,
lines
this
of
of
electrons in
existed
in
hydrogen
discrete
consisted
corresponded to the
model
was
limited
by
assumptions:
predict
electron.
the
idea
wavelengths
energy
not
the
the
was
was
a
emission
only
spectra of elements containing
successful
subatomic
with
particle
in
the
a
hydrogen atom.
xed orbit about the
nucleus.
could
not
account
for
the
eect
of
electric and magnetic elds on the
spectral lines of atoms and ions.
4.
It
5.
Heisenberg’s
could
not
explain
molecular
bonding
and
geometry.
The
uncertainty
principle
states
that
it
is
impossible
to
principles
bonding
know
the
loc ation
and
momentum
of
an
electron
simultaneously.
and
stated
that
electrons
exhibited
xed
momentum
in
molecular
geometry
are
Bohr ’
s
Structure 2.2
explained in
model
behind
precisely
specic
circularorbits.
Bec ause
the
of
these
modern
limitations,
quantum
the
Bohr
theory
has
been
eventually
superseded
by
mechanic al model of the atom.
TOK
The
modern
quantization
quantum
with
the
mechanics
combines
Heisenberg’s uncertainty principle
impossible
and
the
that
the
the
less
more
we
it
the
of
of
we
a
particle
know
is
possible
not
of
its
an
probability
the
position
momentum,
to
pinpoint
electron
of
nding
in
the
an
an
of
and
This
an
means
electron,
vice
versa.
electron
we
in
the
the
boundaries
of
waveforms),
are
the
limits
of
human
Wave–particle duality
subatomic
species
to
is
the
The
of
around
obstacles),
all
(combination
obstacles) and
of
which
are
Einstein
(1879–1955)
described
by the
in
equation
1926
of
the
by
the
electron
is
equation,
Austrian
quantitatively
which
was
physicist Erwin
(1887–1961). Solutions to the Schrödinger
give
functions,
duality
Schrödinger
a
series
known as
states
and
of
three-dimensional
wave functions,
energies
of
mathematic al
which
describe the
electrons in atoms.
knowledge?
ability
behave
characteristics
released
even whole atoms and
interference
(bending
through
or
nature.
waves.
wave–particle
as
of
both
The
concept
that
objects
these
of
wave–particle
duality
illustrates the fact
electrons and other
particles
and
species,
such
as
of
study
do
not
always
fall
neatly into the
waves.
discrete
Certain
and
of
absorbed
phenomena of light, but together they do.
possible
What
to
be
particulate
separately neither of them fully explains the
Schrödinger
knowledge?
c apable
diraction
to
their
electrons,
are
(passing
formulated
(1908–1974)
tendency
suggest
We have two contradictory pictures of reality;
Albert
implic ations of this uncertainty principle on
the
photons,
characteristic
space.
Bronowski
and
entities
molecules,
tunnelling
each
has been to prove that this aim is unattainable.
are
of
c an
exact picture of the material world. One achievement …
Jacob
small
loc ation or
atom,
discrete
However,
One aim of the physical sciences has been to give an
What
momentum
as
states that it is
simultaneously.
about
about
trajectory
idea of
accurately both the momentum
know
c alculate the
region
determine
position
Although
predict
to
the
following key principles.
c ategories
we
have
developed.
the
construction
What
is
the
role
mass,
of
c ategorisation
in
of
knowledge?
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Structure
1
Models
of
the
particulate
nature
of
matter
Schrödinger ’s
probability
wave
density,
electrons
are
path,
theory
this
functions
using
uncertain. Instead
gives
of
space
at
region
in
space
where
of
are
two
four
several
electrons.
atomic
a
the
region
There
types
orbitals
in
of
is
a
order
of
that
that
a
an
electrons
electron
probability
orbitals,
has
electrons in atoms in terms of their
idea that the momentum and position of
follow
will
from the nucleus. An
high
atomic
orbital
are
the
saying
distance
there
E ach
of
probability
certain
orbitals,
Subsequent
describe
Heisenberg’s
and
of
increasing
theoretic al,
and
energy
these
orbital
shape
are
are
a
dened
found
in
a
travel
specic
atomic orbital
nding
each
characteristic
be
an
is a
electron.
c an hold a maximum
and
energy.
labelled
labelled
The
rst
s, p, d, and f.
alphabetic ally
(g, h, i, k and so on).
The
principal
main

Figure 11
energy
quantum
levels.
number,
These
n,
energy
introduced
levels
are
by
split
the
into
Bohr
model
sublevels
represents the
comprised of
An s orbital is spheric al. The
atomic
orbitals.
For
example,
for
n
=
1,
2
and
3,
the
s
atomic
orbitals
are 1s, 2s
sphere represents the boundary space
and
3s.
As
n
increases,
the
s
orbitals
are
further
distanced
from the nucleus.
where there is a 99% probability of nding
an electron.
The s orbital c an hold two
Figure
12
shows
that,
for
1s,
there
is
a
high
probability
of
nding
electrons close
electrons
to
the
away
nucleus
from
although
nucleus.
The
the
this
is
a
There
is
zero
is
probability
nucleus.
there
same
from
the
and
true
for
nucleus
For
small
the
with
two
the
of
that
of
zero
probability
an
nding
highest
regions
reaches
highest
probability
probability
3s,
and
2s,
never
electron
the
is
at
we
move further
somewhat
could
electron
probability
zero
when
an
be
further
away,
found closer to the
between
even
the
two
peaks.
greater distance
probability.
1s
0
50
pm
2s
0
50
100
pm
average
radius
3s
0
50
100
150
pm
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
44
Figure 12
The plots of
the wavefunctions for the rst
three s orbitals
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Structure
Imagine
that
8.00am.
they
At
could
you
are
8.15am,
be.
possibly
a
student
there
is
Some
students
in
sta
no
from
is
could
•
may
•
might perhaps be at the airport
•
might
be
Although
at
their
even
the
the
of
nding
to
dene
might
be
certain
orbital
the
a
exact
school
region
A
of
the
of
the
the
the
town
region
of
or
suggest
or
so
that
laboratory,
in
the
lesson
you
the
or
school
to
congurations
begin at
wonder
where
teacher:
the
library
c ar park
Pole!
teacher
the
is
is
where
could
a
town
99%
with
be
a
the
it
drawn
your
possible
of
is
a
to
high
draw
this
nding
them.
is
probability
of
cluster
This
loc ated,
Similarly,
an
a
probability
around
school
airport.
high
is
there
chance
where
includes
space
unknown,
areas
surface
there
that
chemistry
teacher,
Electron
centre
showing
where
perimeter,
class
oce
North
dots
DP
your
chemistry
town
boundary
space
around
represents
electron
of
the
to
loc ation
cluster
the
principal’
s
in
gone
teacher.
region
the
house
have
three-dimensional
room,
school
your
of
your
•
in
for
sign
•
be
the
waiting
still
1.3
or
a
atomic
nding
an
(gure13).
t
Figure 13
Representation of a 1s atomic orbital as
y
y
a cluster of dots (le) and
a sphere that encloses 99%
of the dots (right)
x
x
z
z
A
p orbital
is
orientations
dumbbell
shaped
parallel to the
x,
y
There
and
z
are
axes
three
(gure
p
orbitals,
14).
These
each
are
described with
labelled p
, p
x
p
.
These
shapes
all
describe
boundaries
with
the
highest
and
y
probability of nding
z
electrons in these orbitals.
t
Figure 14
The three p atomic orbitals
are dumbbell shaped,
aligned along the
z
z
z
x, y and
x
x
x
z axes.
There is zero probability
of
nding the electron at
of
the axes between the two lobes of the
dumbbell.
E ach of
the p
the intersection
orbitals c an hold
two electrons
y
y
y
p
x
orbital
p
y
orbital
p
z
orbital
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Structure
1
Models
of
the
particulate
nature
of
matter
Theories and models
Current
atomic
before.
Theories
the
natural
theories
are
theory
are
world.
are
evolved
Contrary
substantiated
amassed,
from
previous
comprehensive
to
by
documented
the
vast
and
systems
use
of
the
amounts
models,
of
ideas
word
of
each
that
“theory”
observations
communic ated
by
a
superseding
model
in
and
everyday
and
the
one
explain
tested
an
that
c ame
aspect of
language, scientic
hypotheses, which
large number of scientists.
+
+
+
+
+
+
+
+
+
+
+
+
+
800–400 BCE
Āruņi’
s
kana
Democritus’ atomos
1897
1913
Thomson’
s “plum
ohr model
pudding” model
“billiard ball”
mechanic al
model
1803
D alton’s
1930
uantum
1912
1926
Rutherford’
s model
Heisenberg’s uncertaint
model
and
regions
of
probabilit
model

Figure 15
The atomic theory has seen the idea of
model where electrons have specic energies and
What
other
examples
of
theories
c an
you
atoms evolve from
are found
think
indestructible spheres to the quantum
mechanic al
in regions of high probability
of ?
Linking question
What
is
the
periodic
relationship
table?
between
(Structure
energy
sublevels
and
the
block
nature of the
3.1)
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Structure
1.3
Electron
congurations
Electron congurations (Structure 1.3.5)
E ach
atomic
spheric al
orbital
and
it
has
type
the
oriented
dierently.
higher
energy
in
has
There
than
s
a
characteristic
lowest
or
are
possible
ve
d
shape
energy.
orbitals
and
There
and
energy. The s orbital is
are
seven
f
three
p
orbitals,
orbitals,
and
each
these
are
p.
z
y
x
s
z
z
z
y
y
x
x
p
z
x
x

E ach
of
z
types
of
there
are
x
level
(table
orbital:
four
dened
1).
s
For
and
types:
n
p.
s,
by
=
1,
For
p,
d,
the
d
=
the
3,
and
s
x
are
three
p
n,
c an hold
there
n
=
types:
s,
p,
and
d.
n
of
of
of
3
to be known
types
are two
For
n = 4,
number
(n)
orbitals
M aximum
number
of
orbitals
per
per
energy
electrons
within
type
2
level
s
1
s
1
p
3
2
3
x
f
2
f.
sublevel
1
x
f
orbitals need
2,
For
y
Number
Type
quantum
number
number,
exists.
Total
Principal
x
1
Only the shapes of s and
quantum
z
y
f
0
orbital
there
z
y
f
f orbitals.
principal
only
n
and
2
z
–1
p,
x
d
y
f
The shapes of the s,
x
1
z
x
y
d
y
–2
energy
orbitals
x
0
y
f
Figure 16
y
d
–1
x
z
y
d
z
–3
z
y
–2
f
1
z
d
y
p
0
y
z
x
p
–1
z
y
s
1
p
3
d
5
(n
2
)
energy
level
1
2
4
8
9
18
(2n
)
t
Table 1
c an hold 2n
sublevels,
s
1
p
3
d
5
Each energy level,
dened
by
2
n,
electrons. The number of
or atomic orbital types, is equal
to n. For n = 4 there are four types of orbitals
(s,
4
16
p,
d,
and
f ) with 16 atomic orbitals in total
32
2
occupied
by a maximum of 2(4)
= 32 total
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electrons
f
7
47
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Structure
1
Models
of
the
particulate
nature
of
matter
Activity
State
the
following
for
the
energy
level with
a.
the
sublevel types
b.
the
number
c.
the total number of atomic orbitals
d.
the
of
maximum
atomic
orbitals
number
of
in
each
electrons
at
n
= 5:
sublevel
that
energy
level.
Orbital diagrams
For
to
convention,
represent
arrangement
u
Figure 17
In orbital diagrams,
represents an orbital.
each box
This diagram
s
sublevel
an
how
of
(one
“arrow
in
electrons
electrons
box
box”
are
in
notation
arranged
orbitals
is
in
c alled an
atomic
orbital diagram
orbitals
is
used
(gure 17). The
c alled electron conguration
representing an s orbital)
shows
the number of orbitals for each sublevel.
Arrows are drawn in the boxes to represent
electrons.
A maximum of two electrons
c an occupy each orbital,
maximum
so each box has a
of two “arrows”
p
sublevel
(three
d
sublevel
(five
f
sublevel
Atomic
two
orbitals
electrons
of
are
solves
Hence
downwards
not
this
with
of
space
charged
be
problem
each
orbital
half-arrow,
three p orbitals p
five
the
d
where
to
by
using
spins
box
is
(gure
a
f
±
, and p
y
)
z
orbitals)
there
and
occupy
, p
x
orbitals)
seven
negatively,
able
opposite
the
the
representing
regions
should
electrons
directions.
one
are
representing
representing
boxes
Electrons
mechanics
pair
boxes
(seven
electrons.
boxes
the
is
a
like
high
same
spin
probability of nding
charges
region
notation
for
repel
of
each
space.
each
other, so
Quantum
electron. A
behave like magnets facing in opposite
shown
18).
This
with
is
an
upwards
known as the
half-arrow,
, and
Pauli exclusion
principle:
Only two electrons c an occupy the same atomic orbital and those electrons
must have opposite spins.
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Structure
N
t
S
Figure 18
1.3
Electron
congurations
Electron spin is represented
by an arrow pointing up
(positive spin) or
down (negative spin)
S
N
N
S
S
N
magnet analogy
half-arrows
representing
3d
electrons
of opposite spin
in an orbital
degenerate
3p
Electron
own
spin
axis.
is
oen
However,
interpreted
this
as
the
interpretation
rotation
has
no
of
the
physic al
electron
basis:
ygrene
TOK
around its
degenerate
3s
electrons in
2p
atoms
the
as
behave
spin
they
like
waves,
and
nor
the
wave-like
have
no
analogues
a
wave
behaviour
in
our
c annot
of
rotate.
electrons
everyday
life
Unfortunately, neither
c an
and
be
c an
visualized
be
in
any
way,
2s
expressed only in
degenerate
mathematic al
theory
the
but
power
form. This lack of visualization does not undermine the quantum
rather
of
shows
the
mathematics
limits
as
the
of
human
language
perception
of
and,
at
the
same
1s
time,
science.
1
To
what
extent
does
mathematics
support
knowledge
2
3
development in the
n
natural
sciences?

Figure 19
The three 2p orbitals are
degenerate as they have the same energy.
E ach
of
the
Orbitals
atomic
with
the
orbitals
same
of
the
energy
same
are
type
in
one
referred to as
sublevel
are
degenerate
of
equal
orbitals
energy.
(gure 19).
These three degenerate atomic orbitals
have lower energy than the three 3p orbitals
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Structure
1
Models
of
the
particulate
nature
of
matter
An atom of boron (B) has ve electrons, and its orbital diagram is drawn as follows:
1
2p
2
2s
boron
(B)
2
1s
The
single
equal
the
electron
energies.
together
in
2p
to
The
show
lemost
Hund’
s rule
states
boron
c an
degenerate
their
box,
in
energy
although
that
every
2p
occupy
any
orbitals
are
equivalence.
it
is
a
matter
of
personal
become
electron
doubly
with
the
occupied
orbitals,
by
the
as
boxes
they
have
joined
half-arrow
is
drawn
preference.
degenerate orbital in a sublevel is singly
occupied orbitals have the same spin. This
one
three
Traditionally,
of
occupied before any orbital is doubly occupied
have
the
represented
same
with
spin
an
in
each
electron
and that
means
of
of
that
them
all electrons in singly
the
three p orbitals must
before
opposite
spin
any
orbital
c an
(gure 20).
Practice questions
1.
2.
Look
at
Why
do
State
gure
you
which
of
conguration
State
the
20.
think
the
The
the
and
and
diagrams
based
reason
1s
1s
for
on
the
2s
2s
below
Hund’s
four
orbitals
orbitals
are
are
fully
represents
rule
and
incorrect
occupied
lled
the
a
before
correct
Pauli
diagrams
by
the
orbitals?
electron
exclusion
being
electrons.
2p
principle.
wrong.
A.
1s
2s
2p
1s
2s
2p
1s
2s
2p
1s
2s
2p
1s
2s
2p
B.
C.
1s
2s
2p
D.
1s

distributed
2p
2s
Figure 20
2p
The electrons are evenly
across the three degenerate
E.
orbitals in nitrogen before an orbital is
doubly occupied
The Aufbau principle states that as electrons are added to atoms, the lowest
available energy orbitals ll before higher energy orbitals do. The third and
fourth energy levels contain d and f orbitals (gure 21). These orbitals are typically
lled aer the s orbitals of the following levels because they are higher in energy.
As shown in gure 21, the 3d sublevel has a higher energy than 4s but lower than
4p, so 4s is lled with electrons rst, followed by 3d and nally 4p. For the same
reason, 4d orbitals are lled aer 5s, and 4f orbitals are lled only aer 6s.
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Structure
t
Figure 21
energy and
1.3
Electron
congurations
The 4s sublevel has a lower
will ll before the 3d
sublevel
4f
6s
5p
4d
5s
ygrene
4p
3d
4s
3p
3s
2p
2s
1s
This
is
consistent
c alcium,
Ca
with
have
experimental
electrons
in
the
data
4s
that
show
that
potassium, K, and
sublevel, not in 3d.
t
Figure 22
Potassium
orbital lling diagram showing the outermost
4p
electron in the 4s orbital bec ause 3d
orbitals are higher in energy
3d
4s
3p
3s
ygrene
2p
2s
1s
Generally,
the
following
order
is
observed:
Activity
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p …
Copy
the
orbital
diagram
from
In the IB Diploma Programme, only the electron congurations of atoms and ions
gure21
and
complete
it
for the
up to Z=36 will be assessed. Electrons in these species can ll sublevels up to 4p.
following
elements
in
their
ground
states:
Electron
sharing
reactions,
so
ion.
There
are
it
1.
Full
is
and
important
three
electron
transfer
ways
are
to
to
fundamental
know
show
the
the
to
electron
electron
understanding
chemic al
a.
aluminium, Al
b.
chlorine, Cl
c.
iron,
conguration of an atom or an
conguration:
Fe
conguration
Refer to the periodic table at the
back
2.
Condensed
electron
of
this
number
3.
Orbital
lling
book
to
deduce the
conguration
diagram
(“arrows
in
of
electrons
in
each atom.
boxes” notation)
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The
orbital
lling
diagram
for
potassium
is
given
in
gure 22.
51
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Structure
1
Models
of
the
particulate
nature
of
matter
Full electron congurations
To
write
the
a
full
electron
electrons
Rule
and
the
in
conguration,
successive
Pauli
orbitals
exclusion
we
use
the
according
periodic
to
the
table,
Aufbau
and
“build
principle,
up”
Hund’s
principle.
Worked example 1
Determine
the
full
electron
conguration
for
the
c alcium
atom.
Solution
The Aufbau principle states that as electrons are added to atoms, the lowest
available energy orbitals ll before higher energy orbitals do. From the Pauli
exclusion principle, we know that each orbital will have a maximum of two
electrons.
The atomic number of calcium is 20. Let’
s split the 20 electrons evenly across
each orbital, starting with the lowest energy rst. When writing electron
congurations, the number of electrons within each sublevel is given in
superscript, next to the sublevel:
2
•
The
1s
•
The
2s
orbital
•
The
three
•
The
3s
•
The
three
has
two
electrons: 1s
2
orbital
also
has
two
electrons: 2s
6
2p
orbitals
have
two
electrons
each, six in total: 2p
2
orbital
has
two
electrons: 3s
6
3p
orbitals
up
to
have
three
electrons
each, six in total: 3p
Practice question
This
3.
Determine
the
full
brings
us
18
electrons,
with
two
le
over to go into the orbital with
electron
2
conguration
the
next
So,
for
lowest
energy, 4s: 4s
for the phosphorus
2
c alcium,
the
full
electron
conguration is 1s
2
2s
6
2p
2
3s
6
3p
2
4s
atom.
Condensed electron congurations
As
the
atomic
gets
longer
ions
is
mostly
electrons,
electron
inner
it
c an
of
18
an
be
electrons
as
is
element
by their
inner
to
to
the
the
full
electron
conguration
write. The chemistry of atoms and
valence electrons, that is, the outermost
core electrons.
highlight
having
(known as the
increases,
time-consuming
determined
rather than the
congurations
core
group
number
and
the
same
A
valence
electron
more
convenient
electrons
and
conguration
way of writing
represent the
as
the
previous
noble gases) element in the periodic table:
Condensed electron conguration = [previous noble gas] + valence electrons
Table 2 shows some more examples of full and condensed electron congurations.
u
Table 2
Examples of
full and
condensed
Condensed
electron congurations for selected
Atomic
Element
Full
electron
conguration
electron
elements
number
conguration
2
O
8
1s
Ne
10
1s
Mn
25
1s
2
2s
2
2
35
1s
2
4
2p
[Ne]
6
2p
2
2s
2
[He] 2s
6
2p
2s
2
Br
4
2p
2
2s
2
3s
6
2p
6
3p
2
3s
2
4s
6
3p
5
4s
2
3d
2
[Ar] 4s
10
3d
5
4p
5
3d
2
[Ar] 4s
10
3d
5
4p
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Structure
the
condensed
electron
conguration
for
the
c alcium
Determine
electron
Solution
worked
congurations
atom.
4.
In
Electron
Practice question
Worked example 2
Determine
1.3
the
condensed
conguration
for the
phosphorus atom.
example
1,
2
c alcium to be 1s
we
2
2s
determined
6
2p
2
3s
6
the
full
electron
conguration of
2
3p
4s
.
The previous noble gas in the periodic table is argon, which has an atomic
2
number of 18. Argon has an electron conguration of 1s
2
2s
6
2p
2
3s
6
3p
, and we
2
can therefore write the condensed electron conguration of calcium as [Ar] 4s
Orbital
diagrams
c an
also
sometimes
be
shortened
by
using
a
.
condensed
The
electron
conguration.
The
condensed
orbital
diagrams
for
periodic
according
manganese
are
shown
table
is
structured
oxygen and
to
the
type
of
sublevel
below.
that
valence
appear
in.
electrons of elements
This
is
discussed further
oxygen: [He]
in
2s
Structure 3.1.
2p
manganese: [Ar]
4s
3d
Self-management skills
ATL
The
ideas
in
mechanic al
Write
a
Write
three
List
the
Write
this
chapter
key
ve
chapter.
topic
model
of
span
the
summary,
key
M ake
range
no
takeaways
voc abulary
brief
a
atom,
the
should
concepts
to
than
that
test
an
key,
then
write
a
and
skills:
electron
sheet
of
A4
from the quantum
congurations.
paper.
chapter.
know
questions
answer
of
how
longer
from
you
to
your
try
from
this
chapter.
understanding
them
out
on
of
one
the
of
ideas in this
your
peers.
Exceptions to the Aufbau principle
The
Aufbau
most
in
principle
elements.
the
correctly
However,
predicts
when
atoms
the
lose
order
of
to
of
atomic
form
ions,
sublevel with the highest principal quantum number ( n)
2
So,
lling
electrons
for
Mn,
with
electron
conguration
[Ar] 4s
are
orbitals
the
lost
for
electrons
rst.
5
3d
,
the
4s
electrons will be lost
2+
rst.
This
gives the manganese ion, Mn
2
not [Ar] 4s
All
you
These
look
at
sc andium
are
electron
conguration
[Ar] 3d
,
.
with
are
the
(Sc)
3d
valence
electrons
known as the 3d
periodic
to
oxidation states
There
with
3
3d
elements
ions.
5
,
some
table
copper
(Cu).
at
tend
to
lose
two
4s
transition elements, or
the
back
These
of
this
transition
book,
metals
electrons
to
form 2+
transition metals. If
these
c an
elements
also
have
are
from
variable
in compounds.
exceptions.
With
only
one
electron
in
its
3d
orbital,
sc andium
3+
readily
forms only Sc
ions,
by
losing
this
3d
electron
and
the
two
4s
electrons.
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Structure
1
Models
of
the
particulate
nature
of
matter
The
Ionization
and
oxidation
ground
those
covered in
Structure 2.1
state
congurations
of
copper
and
chromium
are
also
dierent
from
are
predicted
by
Aufbau
principle.
and
2
Structure 3.1
The predicted electron conguration of copper is [Ar]4s
9
3d
, as the Aufbau principle
suggests that the lower-energy 4s orbital should be lled rst. However
, the observed
1
ground-state electron conguration for copper is [Ar]4s
2
chromium, the predicted conguration is [Ar]4s
10
3d
(gure 23). For
4
1
3d
and the observed is [Ar]4s
5
3d
Activity
(gure 23). In each case, promoting a 4s electron to a 3d level leads to a more stable
electron conguration. In the case of copper
, this gives a full d sublevel, and in the
Deduce
the
electron
conguration
case of chromium, there are no paired electrons but rather six half-occupied orbitals,
2+
of the Cu
c ation.
each containing an electron with the same spin.
u
Figure 23
and
The expected
observed
Cu (Z = 29)
Cr ( Z = 24)
electron
expected
[Ar]
congurations of copper and
[Ar]
configuration
chromium
9
2
3d
4
2
3d
4s
4s
observed
[Ar]
[Ar]
configuration
10
1
The
electron
two
exceptions
LHA
sublevels
congurations
are
that
lled
5
4s
3d
you
with
of
need
chromium
to
1
3d
(Cr)
and
copper
4s
(Cu)
are the only
know. In all other elements up to
electrons
according
to
the
general
Z=36, the
order.
Ionization energy (Structure 1.3.6 and 1.3.7)
The
quantum
mechanic al
discontinuities
is
the
in
minimum
molecule
in
the
rst
energy
itsground
model
of
the
atom
helps
to
ionization energies (IE) of
required
to
eject
an
electron
explain
the
elements.
out
of
a
trends and
Ionization
energy
neutral atom or
state.
+
X(g)
Ionization
energy
→
X
(g) + e
energy and periodic
The
table
+
trends
are
columns
in
the
periodic
table
are
known
as
groups,
and
the
rows
are
known
discussed further in
as
periods.
Going
across
the
periodic
table,
the
groups
are
numbered
from 1 to
Structure 3.1
18.
The
periodic
sublevels
s,
electrons
for
First
p,
table
c an
and
(gure
d,
each
ionization
f
be
element
energy (IE
)
shown
24).
are
as
The
also
four
blocks
sublevels
corresponding
holding
the
to
the
outermost
four
valence
shown.
generally
decreases
down
the
groups of the periodic
1
table
and
Going
down
electrons
energy
the
increases
are
the
group,
shielded
sublevels
shielding
outermost
Going
a
and
the
number
from
(so-c alled
therefore
the
“inner
less
of
pull
sublevels
of
the
electrons”).
energy
is
increases. The outermost
nucleus
by
The
required
the
more
to
electrons
sublevels,
remove
in
the
the
electrons
lower
greater
from the
sublevel.
ac ross
a
outermost
nucle ar
across the periods.
period,
the
ele ctrons
charge.
At
constantbe c ause
the
the
number
are
held
s ame
time,
number
of
of
protons
closer
to
the
in
the
nucleus
nucleus
by
the
shielding
ee ct
inner
ele ctrons
does
the
inc re ases,
remains
not
so
inc re ase d
ne arly
change.
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Structure
1.3
Electron
congurations
LHA
s-block
1
18
transition elements
1s
1s
13
2
2s
14
15
16
17
2p
d-block
3s
3p
3
4
5
6
7
8
9
10
11
12
4s
3d
4p
5s
4d
6s
5d
6p
7s
6d
7p
4f
f-block
5f

Figure 24
Therefore,
ionization
The
more
a
energy
energy
general
across
The blocks of the periodic table correspond
trend
period
is
is
re quire d
inc re ases
of
ac ross
decreasing
shown
Period
in
to
remove
the
period.
ionization
to the sublevels s,
outermost
energy
down
a
p, d and f
ele ctrons,
group
and
so
increasing
gure 25:
2
Period
3
Period
4
Period
5
2500
He
1
Ne
lo
2000


gree
Ar
1500
N
Kr
 o i a  i  o i
Xe
O
H
1000
Be
sriF
B
500
Al
Li
Na
Rb
K
0
10
2
18
36
54
Aoi ber

Figure 25
Plot
of
rst
ionization energy against
atomic number for the elements from
hydrogen to xenon
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55
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1
Models
of
the
particulate
nature
of
LHA
Structure
matter
There
are
two
clear
discontinuities
1.
Between the group 2
The
valence
The
paired 2s
across the period:
and group 3 elements
2
electron
conguration
of
2
beryllium is 2s
while
for
boron it is 2s
1
2p
.
2
making
the
electrons
electron
shield
slightly
the
easier
single
to
2p
electron
in
boron
from the nucleus,
remove.
+
Be
Be
0
0
2p
2p
2
1
2s
2s
+
B
B
1
0
2p
2p
2
Scientists
trends
The
—
in
look
the
presence
results
overall
that
of
to
for
patterns
they
and
collect.
discrepancies
do
pattern
conclusions
out
data
—
not
t
the
allows
be
2
2s
Patterns and trends
The
s ame
trend
c an
2s
be
observe d
in
comparing
group
2
to
group
2
in
any
period.
aluminium,
For
example,
the
rst
so
the
3s
ionization
3
elements
1
ele ctrons
energy
of
shield
the
aluminium
is
lone
3p
lower
ele ctron
than
that
in
of
magnesium.
further
Suppose you have a two-story building and you need to remove one oor to meet
drawn.
new height regulations. Which oor would you remove? Obviously, it will be the
What can be inferred from the
top oor, as the building would collapse otherwise! The same reasoning can be
patterns in successive ionization
applied to the ionization of atoms — electrons are removed rst from the highest
energies?
occupied energy level, and from the highest energy sublevel within that level.
2.
Between the group 15 and 16 elements
From
group
15
to
16
there
is
also
2
u
Figure 26
A half-lled
conguration
of
Nitrogen
a
has
lled
p
from
nitrogen
more
sublevel,
stable
and
(gure
the
same
region
of
the
three
electrons
a
2s
in
ionization
energy.
The
3
2p
electron
therefore
more
26).
This
is
space
and
have
in
2p
the
drop
2
nitrogen is 1s
2
while
for
conguration
energy
bec ause
is
the
increased
orbitals
oxygen it is 1s
do
not
than
required
paired
to
remove
electrons
into
4
2p
oxygen as it has a half-
repulsion.
come
electron
2
2s
in
However,
close
an
electron
oxygen
in
occupy
nitrogen,
proximity.
p sublevel is
+
N
N
3
more stable than p sublevels with 2 or 4
2
2p
2p
2
2
2s
electrons
2s
+
O
O
4
3
2p
2p
2
2s
2
2s
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Structure
1.3
Electron
congurations
6
most
stable
p
orbital
conguration is p
,
a
completely
lled
p
LHA
The
sublevel,
3
followed
by p
,
a
half-lled
10
example, d
why
sublevel.
This
is
generally
true
for
other
sublevels.
For
5
and d
chromium
and
electron
copper
congurations
do
not
obey
the
are
also
Aufbau
stable,
which
principle
partly
explains
(gure 23).
C alculating ionization energy from spectral data
As
the
principal
between
lines
the
quantum
levels
converging
level
in
number
converges
the
to
hydrogen
of
a
energy
levels
continuum.
emission
increases, the distance
This
c an
spectrum,
be
observed
shown
in
by
gure
spectral
27.
t
∞
level 5
Figure 27
Ultraviolet and visible light
transitions in hydrogen and
the resulting
level 4
emission spectrum
level 3
level 2
level 1
high energy
low energy
ultraviolet
light
visible light
–8
The spectral lines in the hydrogen emission spectrum converge at 9.12 × 10
m, or
912 Å (gure 28). This represents the wavelength of light at which the hydrogen
atom is ionized.
This
wavelength
c an
be
used
to
c alculate
the
rst
ionization
energy
of
hydrogen.
t
912 Å
Figure 28
Hydrogen is ionized at
the wavelength where the spectral lines
converge in the emission spectrum
900
950
1000
1050
1100
1150
1200
1250
Wavelength / Å
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1
Models
of
the
particulate
nature
of
matter
LHA
Structure
Worked example 3
8
Spectral
lines
converge
at
9.12 × 10
m
in
the
emission
spectrum
of
hydrogen
in
the
1
hydrogen
atom.
C alculate
the
rst
ionization
energy
of
kJ
mol
Solution
First,
c alculate
the
frequency
of
radiation using
to
3.00
8
of
light,
approximately
8
3.00
equal
1
× 10
m s
=
Then,
3.29 × 10
c alculate
f
the
=
6.63 × 10
=
2.18 × 10
=
f
λ, where
×
c
is
the
speed
×
9.12 × 10
m
1
Hz
(s
)
energy
using
Planck’s
34
E
c
1
m s
8
=
15
f
× 10
15
J s
×
3.29
constant
and
the
equation
E
=
h
×
f
1
× 10
s
18
Alternatively,
The
values
of
the
J
these
two
steps
c an
be
merged
into
one
by using the
speed of light,
h × c
equation
Planck’
s
constant
constant
are
booklet.
The
constant
are
and
=
Avogadro’
s
λ
This
given in the data
mole
E
and
represents
absorbed
Avogadro’s
level,
discussed further in
or
in
the
energy of a single photon of light which would be
exciting
removing
the
one
electron
electron
in
a
hydrogen
atom
to
the
convergence
from the atom.
1
Ionization
Structure 1.4.
energies
are
usually
given
in
kJ mol
.
You
c an
convert the ionization
1
energy
value
to
kJ mol
using
Avogadro’s constant (N
, the number of atoms
A
in
1mol)
and
the
following
equation:
1
The
ionization
(energy
energy
in
needed
kJ mol
to
remove
one
electron
from an atom)
× N
A
=
1000
–18
2.18 × 10
23
J × 6.02 × 10
–1
mol
=
1000
3
=
1
kJ mol
1.31 × 10
Worked example 4
1
The
rst
ionization
booklet.
C alculate
spectrum
in
energy
the
of
Na
is
496
wavelength
of
kJ
mol
as
given
convergence
for
by
the
the
IB
data
sodium
atom
Å.
Practice questions
Solution
5.
In
the
the
emission
helium
spectrum of
atom,
the
First,
nd
the
from
kJ
J
energy
of
ionization
for
one
atom
by
converting
the
given
value
spectral
to
and
dividing
it
by
Avogadro’
s constant.
–8
lines
converge
C alculate
the
at
5.05 × 10
m.
1
23
496 000 J mol
/
Then
the
6.02 × 10
1
mol
19
=
8.24 × 10
J
rst ionization
h × c
c alculate
wavelength of light using
E
=
–1
energy,
in
kJ mol
, of helium.
λ
–34
6.63
× 10
8
J s × 3.00
× 10
–1
m s
19
6.
The
rst
ionization
8.24 × 10
energy of
J
=
λ
–1
c alcium
is
590 kJ mol
.
C alculate
7
λ = 2.41 × 10
the
wavelength
of
=
in
Å,
for
the
m
convergence,
2410 Å.
c alcium atomic
This
corresponds
to
the
UV
region
in
the
electromagnetic
spectrum.
spectrum.
Successive ionization energies
It
requires
more
atom
bec ause
while
the
energy
the
to
remove
number
of
electron–electron
the
protons
repulsion
second and
exceeds
the
successive
number
of
electrons
remaining
from an
electrons
decreases.
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Structure
a
the
so
result,
electron
increased
that
only
remove
the
clouds
electrostatic
the
stable
next
are
noble
electron
pulled
attraction.
gas
closer
Once
to
all
conguration
increases
sharply,
as
the
the
nucleus
valence
and
remains,
the
shown
gure
in
held
electrons
energy
tighter
are
Electron
congurations
LHA
As
1.3
by
removed
required to
29.
t
Figure 29
ygrene
from
Removing 10 electrons
magnesium
gives the noble-gas
2
conguration 1s
or [He].
There is a
)
1–
n o it a z i n o I
lom Jk(
considerable increase in energy required to
remove the 11th electron
0
1
2
3
4
5
6
7
Numer
o
8
9
10
eletron
11
12
remoe
Worked example 5
The
rst
ve
successive
ionization
energies
for
an
unknown
element
X
have
1
the
following
the
group
values:
403,
2633,
3860,
5080
and
which
element
6850 kJ mol
.
Deduce
Practice question
of
the
periodic
table
in
X
is
likely
to
be
found.
7
.
The
rst
ve
successive ionization
energies
of
an
have
the
following
unknown element
801,
2427,
Solution
values:
1
The
largest
increase
in
energy
occurs
from
the
rst
ionization
(403 kJ mol
) to
3660,
25 026 and
1
the
second
(2633 kJ mol
).
This
means
that
the
second
electron is likely to be
–1
32 827 kJ mol
removed
from
a
stable
noble
gas
conguration
of
the
atom.
.
Deduce the
Therefore, the
group of the periodic table in
outermost
energy
level
of
the
element
contains
one
electron, so the element
which this element is likely to
belongs
to
group
1
of
the
periodic
table.
befound.
Data-based question
Using
gure
successive
30
and
the
ionization
periodic
energies
table,
for
explain
the
two
large
jumps
in
the
sodium.
6.0
5.5
5.0
EI
4.5
gol
4.0
3.5
3.0
2.5
1
2
3
4
number

Figure 30
5
of
6
7
electrons
8
9
10
11
removed
Successive ionization energies for sodium
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LHA
Structure
1
Models
of
the
particulate
nature
of
matter
Ionization energy data
Relevant
Part 3: Graphing the logarithm of the ionization
skills
•
Tool
2:
Extract
•
Tool
2:
Use
represent
•
data
energies
from databases
6.
spreadsheets to manipulate data and
data
in
graphic al
Title
the
third
(ionization
form
column
energy)”
in
as
your
spreadsheet “log
shown
below:
T
ool 3: Construct and interpret graphs
A
Instructions
Part 1: Data collection
1.
B
C
element name:
1
Identify a database that contains successive
2
ionization energy data for the elements (for example,
ionization
log
energy/
(ionization
WebElements).
2.
Choose
one
of
the
following
elements:
sulfur,
–1
chlorine,
3.
Collect
argon,
potassium
successive
spreadsheet,
or
ionization
labelling
the
energy data in a
columns
as
4
1
follows:
5
2
6
A
energy)
kJ mol
ionization
3
c alcium.
etc
B
7
.
Compute
the
logarithm
of
each
ionization
energy
element name:
1
using
2
8.
ionization
the
spreadsheet
Construct
a
ionization
energies
vs
graph
ionization
LOG
showing
by
(or
LOG10) function.
the
plotting
logs
log
of
successive
(ionization
energy)
number.
energy/
9.
–1
Answer
the
following questions:
kJ mol
ionization
3
a.
4
Identify
the
large
increases
in
ionization
energy
1
that
5
2
6
b.
indic ate
a
change
in
main
energy
level.
Why is it useful to plot the logs of the ionization
etc
energies?
Part 4: Evidence for the existence of sublevels
10.
Part 2: Graphing successive ionization energies
4.
Plot
a
M ake
axis
5.
line
sure
that
labels,
Answer
a.
graph
IE
and
present
sc ales
your
and
a
graph
the
suitably, with
descriptive
values
that
outermost
ionization
11.
electrons.
increases
with
a
graph
the
graph
that
electrons
and
“zoom
will
in
in”
allow
energy
to
you to closely
level
inspect
n
=
the
2.
Enlarge
increases
closely.
title.
correspond to the
energy
Construct
examine
vs ionization.
12.
each
C an
any
of
Experimental
later
the
data
is
oen
transformed
role
of
large
Explain
how
increases in
they
relate to the
sublevels.
into
graphic al
advancement
existence of main energy levels in the atom.
unusually
existence
is
Explain how the graph provides evidence for the
see
energy?
and
electron.
you
ionization
of
representations
organized into tables
graphic al
forms.
What
KOT
why
successive
c.
you
energy
following questions:
Identify the
Explain
ionization
suitable
the
innermost
b.
of
representations in the
scientic
knowledge?
employed
in
other
Are
subject
graphic al
areas?
Linking questions
How
in
does
the
properties
trend in
of
metals
IE
values
and
across
non-metals?
a
period
and
(Structure
down
a
group
explain
the
trends
3.1)
+
Why are log scales useful when discussing [H
LHA
How
the
do
patterns
variable
of
successive
oxidation
states
of
ionization
these
] and ionization energies? (Tool 3, Reactivity 3.1)
energies
elements?
of
transition
(Structure
elements
help
to
explain
3.1)
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Structure
1.3
Electron
congurations
End-of-topic questions
5.
What
is
the
maximum
possible
number
of
electrons in
Topic review
the
1.
Using
your
answer
knowledge
the
guiding
from the
question
as
Structure 1.3
fully
as
third
energy
level?
topic,
A.
3
B.
6
C.
9
D.
18
possible:
How can we model the energy states of electrons
inatoms?
6.
Exam-style questions
What
is
in
ground
the
the
electron
conguration
of
chromium (Z = 24)
state?
Multiple-choice questions
7
A.
2.
Which
row
is
correct
for
the
following
[Ar] 3d
regions of the
2
electromagnetic
B.
[Ar] 4s
C.
[Ar] 4s
D.
[Ar] 4s
4
3d
spectrum?
1
Ultraviolet
(UV)
Infrared
short
low
energy
wavelength
energy
1
(IR)
5
4p
low
7.
A.
Which
of
the
low
energy
frequency
low
energy
A.
IE
>
correct?
IE
B.
Molar
C.
The
4
long
B.
high
is
frequency
3
high
following
LHA
high
5
3d
ionization
energies
are
measured in kJ.
wavelength
short
high
third
ionization
energy
of
the
atom
X
represents
long
C.
the
frequency
wavelength
energy
wavelength
high
long
low
low
frequency
wavelength
frequency
process:
2+
X
3+
(g)
→
X
–
(g) + e
energy
D.
D.
Ionization
energies
decrease
across a period going
from le to right.
3.
Which
of
the
following
sources
of
light
will
produce a
8.
line
spectrum
when
placed
behind
a
Which
statement
nitrogen
I.
a
II.
an
gas
and
oxygen
rst
is
ionization
energies of
correct?
IE
(N)
<
IE
1
(O)
bec ause
oxygen
has
two
paired
two
paired
1
inc andescent lamp
electrons
an
the
atoms
discharge tube
A.
III.
about
prism?
alkali
metal
salt
placed
in
a
in
its
partly
lled
sublevel
Bunsen burner
B.
IE
(N)
>
IE
1
(O)
bec ause
oxygen
has
1
ame
electrons
A.
I and II only
B.
I and III only
C.
IE
in
(N)
its
<
partly
IE
1
C.
II and III only
D.
I, II and III
IE
a
(N)
higher
>
IE
1
4.
An
electron
n = 2
in
an
transition
between
atom
energy
levels
n = 4 and
produces a line in the visible
9.
The
bec ause
rst
sublevel
oxygen
(O)
than
loses
an
electron
bec ause
an
electron
nitrogen
oxygen
loses
1
from
isolated
sublevel
1
from
D.
(O)
lled
a
higher
ve
unknown
sublevel
successive
element
are
than
nitrogen
ionization
578,
1817,
energies
2745,
for an
11 577 and
–1
spectrum.
likely
to
Which
produce
electron
a
line
A.
from
n = 4 to
n = 1
B.
from
n = 4 to
n = 3
C.
from
n = 3 to
n = 2
D.
from
n = 5 to
n = 3
in
transition
the
UV
in
the
same atom is
spectrum?
14 842 kJ mol
this
element
A.
1
B.
2
C.
13
D.
14
is
.
In
which
likely
to
group of the periodic table is
be
found?
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Structure
1
Models
of
the
particulate
nature
of
matter
16.
Extended-response questions
Sketch
an
number
10.
Explain,
in
your
own
words,
why
orbital
of
lling
unpaired
diagram
for
Al
and
deduce the
electrons.
[2]
gaseous atoms
3+
produce
line
spectra
instead
of
continuous
spectra.
[3]
17
.
A
transition element ion, X
,
has
the
electron
5
conguration [Ar] 3d
11.
State
the
full
and
condensed
electron
congurations
for
element
the
following
species
in
their
selenium
c.
silicon
d.
Ti
[1]
atom
Sketch
the
condensed
orbital
lling
diagram
for
[1]
germanium
atom
and
deduce the total number of p orbitals
[1]
containing
atom
one
or
more
electrons.
[2]
[1]
19.
3+
c ation
Describe,
in
energy
an
your
own
words,
how
the
rst ionization
[1]
of
atom
c an
be
determined
from
its
emission
2–
e.
S
anion
[1]
spectrum.
12.
Determine
which
of
the
congurations
20.
impossible.
Explain
why
it
c annot
exist.
Using
the
2
2
2s
7
2p
2
3s
13.
2
2s
Deduce
6
2p
and
2
1s
1s
14.
Sketch
15.
The
these
which
represents
6
2p
2
2
3s
6
6
3p
2
2s
2p
the
shape
period
explore
3
the
elements,
rst ionization
from sodium to
3s
Explain
the
general
trend and discontinuities in
6
3p
explain,
2
2s
2
booklet,
the
5
2
3s
congurations
of
3p
argon.
2
1s
data
[2]
energies
1s
[2]
below is
of
10
3d
2
4s
an
s
the
following
ground
2
4s
6
3p
a
of
state.
5
4p
10
4p
21.
[2]
The
rst
four
unknown
1
[3]
successive
element
X
are
ionization
given
in
energies
table3.
for an
Deduce the
group of the periodic table in which element X is likely
5s
6
3d
electron
energies.
to
1
be
found.
[1]
5s
orbital.
[1]
–1
n
IE
/ kJ mol
n
diagram
electron
below
energy
(not
levels
in
to
sc ale)
the
represents some of the
n
= 7
n
= 6
n
an
energy
arrow
on
transition
spectrum
of
the
in
diagram
the
hydrogen.
738
1451
visible
to
represent
region
of
the
the
3
7733
4
10543
= 5

Draw
1
2
hydrogen atom.
n
= 4
n
= 3
n
= 2
n
=
Table 3
Successive ionization energies for element X
1
lowest
emission
[1]
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62
LHA
titanium
b.
X.
ground states:
18.
a.
. Determine the atomic number of
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Counting particles
Structure 1.4
by mass: the mole
How do we quantify matter on the atomic sc ale?
Atoms
are
contains
more
a
extremely
huge
atoms
of
the
in
in
all
of
substance,
a
small,
so
any
number
of
glass
water
oce ans
the
of
these
than
combined.
mole,
physic al
particles.
The
enables
glasses
unit
of
chemists
object
comfortably
There
the
of
the
to
are
same
water
relative
amount
for
with
time,
large
the
molecular
expressing
numbers
concepts
masses
masses
of
of
of
very
molar,
allow
the
atomic
small
particles.
relative
use
of
atomic
small
At
and
numbers
species.
de al
Understandings
Structure 1.4.1
of
substance.
elementary
— The mole (mol) is the SI unit of amount
One
mole
entities
contains
given
by
the
Structure 1.4.4
exactly the number of
gives
Avogadro constant.
in
the
that
—
M asses
of
atoms
are
The
ratio
compound.
number
Structure 1.4.2
—
simplest
of
atoms
The
of
empiric al
of
atoms
formula of a compound
of
molecular
each
each
element
present
formula
gives the actual
present
in
element
a
molecule.
compared on a
12
sc ale
relative to
mass (A )
and
C
and
relative
are
expressed
formula
as
relative atomic
Structure 1.4.5
mass (M ).
r
by
the
amount
—
of
The
molar
solute
and
concentration
the
is
determined
volume of solution.
r
–1
Structure 1.4.3 — Molar mass (M) has the unitsg mol
Structure 1.4.6
volumes
of
of
all
—
Avogadro’s
gases
temperature
and
law
measured
pressure
states
under
contain
the
that
equal
same conditions
equal
numbers
ofmolecules.
The mole (Structure 1.4.1)
Atoms
and
Even
million
a
molecules
atoms
are
of
so
small
lead,
Pb,
that
the
their
masses
heaviest
c annot
stable
be
element,
measured
would
directly.
have
a
mass
–16
of
only
3.4 × 10
analytic al
g.
balance.
This
At
is
the
too
small
same
to
time,
be
the
weighed
number
of
even
Pb
on
the
atoms
in
most
1 g
of
sensitive
lead is
21
huge,
about
chemists
masses
19th
2.9 × 10
need
and
a
unit
very
century
,
that
large
and
which
is
allows
hard
to
them
imagine,
to
work
let
alone
comfortably
numbers of atoms. This unit, the
quickly
bec ame
one
of
the
count.
most
with
mole,
useful
Therefore,
both
was
concepts
very small
devised in the
in
chemistry.
The mole (with the unit “mol”) is the SI unit of amount of substance that contains
23
6.02214076 × 10
c an
be
an
will
use
atom,
a
elementary
molecule,
entities
an
of
electron
that
or
substance. An elementary entity
any
other
species. In this book, we
23
the
rounded
value
of
the
mole:
1mol
=
6.02 × 10

Figure 1
One mole quantities
of dierent substances (le to right):
aluminium,
water,
copper, sucrose and
sodium chloride
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Structure
1
Models
of
the
particulate
nature
of
matter
Avogadro’s constant
Prex
Symbol
(N
)
is
the
conversion factor linking the number of particles
A
F actor
–1
and
amount of substance
in moles. It has the unit of mol
:
–12
pico
p
10
23
–9
nano
n
10
micro
µ
10
milli
m
10
centi
c
10
N
=
–1
6.02 × 10
mol
A
–6
In
–3
–2
chemic al
other
we
c alculations,
conversion
need
to
factor
multiply
substance (n)
into
the
the
Avogadro’s
(table1).
mass
in
number
For
kg
of
constant
is
example,
by
1,000.
atoms
or
used
to
in
Similarly,
any
the
convert
other
same
way
kilograms
to
as
into
any
grams,
convert the amount of
structural units (N),
we
need
–1
deci
d
10
to
multiply
that
amount
by
N
:
A
3
kilo
k
10
mega
M
10
N =
n×N
6
9
giga
G
10
A
In
chemistry
texts,
the
term
“amount
of
substance”
is
oen
abbreviated to just
“amount”.

Table 1
Decimal prexes
Worked example 1
C alculate
the
amount
of
lead
(Pb),
in
mol
and
mmol,
in
a
sample
containing
21
2.9
×
10
atoms
of
this
element.
Solution
To nd
n,
we
c an
rearrange
the
equation
N
=
n
×
N
as
follows:
A
N
n
=
N
A
21
2.9
×
10
3
Therefore,
n(Pb)
=
≈
4.8 × 10
mol
23
6.02
×
10
3
According
The
use
gures
of
is
chemistry
correct
signic ant
discussed in the
Tools for
to
table1,
1 mmol
=
3
mol, so 4.8
10
×
10
mol
=
4.8 mmol.
3
In this example, both answers (4.8 × 10
mol and 4.8 mmol) have been
rounded to two signicant gures, the same as in the least precise value used in
21
chapter.
the division (2.9 × 10
Research skills
ATL
The
are
).
mole
so
is
a
small.
convey
just
huge
number,
Measuring
how
and
it
amounts
is
of
useful
for
everyday
counting
objects
particles
in
moles
bec ause they
c an help use to
large this number is.
Activity
Choose
one
of
the
following
and
C alculate:
approximate
a.
the number of atoms in
•
How
many
moles
of
grains
2.5 mol of copper metal
•
How
many
moles
of
water
conduct
the
necessary
research
to
reach an
answer.
of
sand
are
molecules
in
are
a
desert
in
a
of
large
your
sea
or
choice?
ocean
of
your
choice?
b.
the
number
molecules in
•
One
•
What
the number of atoms in
•
How
tall
0.25 mol
•
How
many
0.25 mol
c.
of
of
of
mole
of
human
cells
represents
roughly
how
many
people?
water
water
is
the
is
age
a
of
the
universe,
stack
of
one
moles
of
air
mole
are
in
in
of
moles
of
seconds?
sheets of paper?
your school building?
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64
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Structure
Relative molecular
1.4
Counting
particles
by
mass: the mole
mass and molar mass
(Structure 1.4.2 and 1.4.3)
In
Structure 1.2,
is
the
we
introduced the concept of
relative atomic mass,
A , which
r
ratio
atom.
of
the
Similarly,
mass
of
a
certain
atom
relative molecular
to
mass,
one-twelh
M ,
is
the
of
ratio
the
of
mass
the
of
mass
a
c arbon-12
of
a
molecule
r
or
other
A
and
multiatomic
M
r
are
ratios,
species
so
they
to
one-twelh
of
the
mass
of
a
c arbon-12 atom. Both
have no units.
r
To nd the
M
of
a
molecule,
we
need to add together the
A
r
in
that
values
for all atoms
r
molecule.
Worked example 2
C alculate
the
M
for
a
molecule
of
water.
r
Solution
Water, H
O,
is
composed
of
two
hydrogen atoms (A
2
atom (A
You
=
1.01)
and
one
oxygen
r
=
16.00).
Therefore
M (H
r
r
should
always
use
the
O)
=
2
×
1.01
+
16.00
=
18.02.
2
actual
(not
rounded)
values of
A ,
which
are
given
r
in
the
keep
data
all
booklet
signic ant
and
the
gures
periodic
in
table
at
c alculated M
the
end
values
of
and
this
book.
never
Similarly,
round them to the
r
nearest
integer
substance
is
number.
If
a
is
c alculated using the smallest
composed
of
ions
instead
of
molecules, the
M
for that substance
r
formula unit.
For
example,
c alcium chloride
2+
(C aCl
)
is
an
ionic
compound
that
consists
of
many
c alcium
c ations
(C a
) and
2
2+
twice
as
many
and two Cl
bec ause
chloride
ions.
the
The
masses
anions
ions
of
(Cl
have
).
Its
smallest
approximately
electrons
are
formula
the
unit
same
contains
masses
as
one
Ca
neutral atoms
negligible.
The
in
Therefore,
M (C aCl
r
M any
form
ionic
is
A (C a) + 2×A (Cl)
r
compounds
coordination
hydrates
) =
2
copper(II)
40.08
+
(2×35.45)
=
and
compounds
will
structure of
be
discussed
Structure 2.1.
110.98.
r
form
bonds
=
composition
ionic
hydrates:
(Structure
sulfate
compounds
2.2)
with
the
in
ions.
pentahydrate, CuSO
which
•5H
4
water
molecules
One of the most common
O.
Copper(II) sulfate
2
Activity
pentahydrate
coecient
forms
“5”
large,
clear,
before “H
O”
deep-blue
means
that
crystals
one
(gure2). The
stoichiometric
formula unit of copper(II) sulfate is
C alculate the
2
M
values
for the
r
bound
with
ve
molecules
of
water.
Therefore, the
M
value
for
this
be
c alculated
as
hydrate
c an
following
r
species:
follows:
a.
ammonia, NH
b.
sulfuric acid, H
3
M (CuSO
r
•5H
4
O) =
2
A (Cu) +
r
A (S) +
4×A (O) + 5×M (H
r
r
r
O)
2
SO
2
=
63.55
+
=
249.72
32.07
+
(4
×
16.00)
+
(5×18.02)
c.
sodium
sulfate
Na
•10H
SO
2
4
4
dec ahydrate,
O
2
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Structure
1
Models
of
the
particulate
nature
of
matter

Figure 2
Crystals of
copper(II) sulfate pentahydrate, CuSO
•5H
4
Molar
mass,
Molar
mass
M,
is
of
a
chemic al
numeric ally
substance
equal
to
is
relative
the
mass
of
molecular
O
2
1 mol
mass
of
that
substance.
(for substances with
molecular and ionic structures) or relative atomic mass (for substances with atomic
–1
structure).
For
example,
M(Na)
=
22.99 g mol
–1
and
M(H
O)
=
18.02 g mol
2
Science as a shared endeavour
A
shared
understanding of common terminology helps scientists to
communic ate
Hi s to r i c a l l y,
c o n ta i n e d
or
o t he r
eectively.
th e
as
mo l e
ma ny
pa r t i c l e s)
This
was
terminology
defined
e l e me n ta r y
as
t h e re
as
entities
w e re
is
th e
constantly
amount
(a to ms ,
a to ms
in
of
being
mo l e c u l e s ,
0.0 12 k g
updated.
substance
(o r
t ha t
ions,
12 g )
of
e l e c tro n s
c a r b o n -1 2 .
23
How e ve r,
to
be
s c i e n ti s t s
On
t he
re v i s e d
16
to
n u me r i c a l
v alue
f re qu e n tl y,
m e a s u re
November
as
m a ss
2018,
of
th e
th e
mo l e
( a p prox i m a t e l y
i m prove me n ts
with
gre a te r
scientists
from
here
physic al
that
all
constants
SI
base
instead
units,
of
)
had
a l l ow e d
pre c i s i o n .
including
physic al
6 .0 2  ×  10
i n s t r u me n t a t i o n
more than 60 countries met at the
General Conference on Weights and Measures
agreed
in
the
objects.
in
Versailles,
mole,
were
France.
It
was
dened in terms of
Following these changes, one
23
mole
of
entities
The
no
a
substance
of
2018
that
match
now
dened
of
equals
exactly.
two
the
exactly
as
6.02214076 × 10
12 g
exact
SI
numeric al
the
mole
As
quantities,
v alues
of
me ans
a
that
result,
the
their
the
kilogram
the
mass
experimentally
dierences
determined
and
respective
between
these
mass
of
numeric al
a
A
the
or
are
1 mol
of
c arbon-12
v alues
of
M
mole)
M
no
(dened
longer
(dened
through
r
c arbon-12
v alues
of
numeric al
r
the
elementary
substance.
redenition
longer
through
is
so
atom).
small
However,
the
(approximately
–8
4 × 10
%)
that
they
Why
are
constants
How
do
scientists
c an
and
be
ignored
values
achieve
a
for
all
continuously
shared
practic al
being
purposes.
revised
and
updated?
understanding of changes made to
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existing
66
denitions?
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Structure
The amount (n),
mass (m)
and
molar
the
common
mass (M)
of
any
substance
are
related as
it
used in almost
1.4
Counting
particles
by
mass: the mole
follows:
m
n
=
M
This
all
is
probably
stoichiometric
the
masses
of
most
c alculations.
chemic al
expression
Although
substances
are
the
in
chemistry,
base
SI
traditionally
unit
of
as
is
mass
expressed
in
is
the
kilogram,
grams, and
–1
molar
masses
in
g mol
Worked example 3
Table
of
sugar
sucrose.
C
H
12
O
22
.
is
oen
Sucrose
sold
in
is
organic
an
the
form
cubes
that
are
compound
of
with
the
made
almost
molecular
entirely
formula
C alculate:
11
a.
the molar mass of sucrose
b.
the amount
c.
the number of
of
sucrose in one cube (2.80 g) of sugar
oxygen atoms in one cube of sugar
Solution
a.
M (C
r
H
12
O
22
)
=
12
×
12.01
+
22
×
1.01
+
11
×
16.00
=
342.34
11
1
M(C
H
12
O
22
)
=
342.34 g mol
11
Activity
m
b.
n
=
M
C alculate:
2.80 g
H
n(C
12
O
22
)
=
≈
11
0.00818 mol
a.
1
the
molar
mass
of
sulfuric
acid,
342.34 g mol
H
SO
2
c.
One
mole
n(O)
=
of
sucrose
contains
11 mol
of
oxygen atoms, so
b.
11
×
n(C
H
12
=
11
=
n(O)
×
O
22
11
0.00818 mol
N
=
amount
1.00 g
≈
of
of
substance
sulfuric
in
acid
0.0900 mol
c.
×
the
)
23
N(O)
4
0.0900mol
×
6.02 × 10
–1
mol
the
number
of
hydrogen
22
atoms
≈ 5.42 × 10
in
1.00 g
of
sulfuric
acid
A
19

Figure 3
There are more oxygen atoms in one sugar cube than the estimated
total insect
population on Earth (10
) and
total grains of
21
sand
on Earth’s beaches (10
)
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Structure
1
Models
of
the
particulate
nature
of
matter
Empiric al formula, molecular
formula and
chemic al analysis (Structure 1.4.4)
The
composition
represented
of
each
element
formula
present
shows
in
the
substance
empiric al
ratio
of
by a
of
in
the
chemic al
the
c an
be
in
the
the
of
ratio
The
identic al
is
substance
formula,
molecule
simplest
substance.
formula
ions
a
molecular
dierent
as
compound
the
a
molecular
of
the
and
In
contrast, the
dierent
empiric al
(table2).
formula
structure
c an be
shows the actual number of atoms
substance.
atoms
molecular
or
same
that
of
with
which
empiric al
elements
formulas
of
that
the
are
same
For ionic compounds, the
unit,
which
represents the simplest
(gure4).
Substance
Molecular
oxygen
O
ozone
O
water
H
formula
Empiric al
formula
O
2
O
3
O
H
2
hydrogen
peroxide
O
H
2
butane
4
glucose
C
sucrose
C
HO
2
H
C
C
10
H
6
Table 2

Figure 4
is used
The
Molecular and
12
H
6
number
of
atoms
(N
=
of
a
C
11
).
H
12
supplement
certain
n×N
O
2
O
22
element
Therefore,
11
with the empiric al formula NaF. It
to prevent
is
the
O
22
selected substances
Sodium uoride is an ionic compound
element in mol
5
CH
empiric al formulas of
in some countries as a food
H
2
O
12

O
2
tooth dec ay
proportional to the amount of that
empiric al
formula
also
shows the
A
mole ratio
water, H
of
O,
elements
contains
in
two
a
chemic al
atoms
of
compound.
hydrogen
and
For
example,
one
atom
of
one
molecule of
oxygen, so the
2
atomic
ratio
of
hydrogen
to
oxygen
in
water
is
2:1.
Similarly,
one
mole
of
water
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Structure
contains
mole
The
two
ratio
of
moles
elemental
mass,
mole
which
ratio
of
hydrogen
hydrogen
to
composition
is
c an
atoms
oxygen
of
a
in
and
compound
referred to as the
be
c alculate
to
mole
of
Counting
particles
by
mass: the mole
oxygen atoms, so the
water is also 2:1.
commonly
used
one
1.4
the
is
oen
expressed
in
percent
percentage composition,
by
ω. The
percentage composition of a compound.
Worked example 4
C alculate
the
percentage
composition
of
water.
Solution
Let
n(H
O)
=
1 mol, then
n(H)
=
2 mol and
n(O)
=
1 mol. Using
m
=
n
×
M:
2
1
m(H)
=
2 mol
×
1.01 g mol
=
=
1 mol
×
16.00 g mol
2.02 g
1
m(O)
=
16.00 g
1
m(H
O)
=
1 mol
×
18.02 g mol
×
100%
=
18.02 g
2
2.02 g
ω (H) =
≈
11.2%
18.02 g
ω (O) = 100%
In
practice,
11.2%
chemists
=
88.8%
more
oen
face
the
opposite
problem
of
deducing
Practice question
the
empiric al
other
formula
experimental
destruction
for
a
data.
analysis,
in
compound
The
from
percentage
which
the
its
percentage composition or
composition
compound
is
c an
be
determined
by
combusted
or
decomposed, and
C alculate
the
percentage
composition of sulfuric acid, H
SO
2
the
masses
The
mass
a
the
combustion
percentages
analytic al
In
of
techniques,
typic al
products
into
mass
decomposition
elements
such
experiment,
combustion
converted
of
or
the
are
as
fully
sample
trapped
percentages
in
a
sample
automated
is
burned
and
of
products
c an
in
excess
weights
elements
in
by
elemental
oxygen,
These
the
4
measured.
determined
combustion
weighed.
chemic al
be
are
and
various
analysis.
the
volatile
are then
original
sample.
Worked example 5
Iron
and
formula
oxygen
of
an
form
oxide
several
that
compounds
contains
72.36%
(iron
of
oxides).
Deduce
the
empiric al
iron.
Solution
If
ω (Fe) = 72.36%, then
Let
m(Fe
O
x
)
=
ω (O) = 100%
100 g, then
m(Fe)
=
72.36%
72.36 g and
=
27.64%.
m(O)
=
27.64 g
y
m
Use
n
=
to
determine
the
amount
of
each element:
M
72.36 g
n(Fe)
≈
=
1.296 mol
–1
55.85 g mol
27.64 g
n(O)
=
≈
1.728 mol
–1
16.00 g mol
The
mole
ratio
Therefore,
the
x : y
=
1.296 : 1.728
empiric al
formula
of
≈
1 : 1.333
the
oxide
≈
is
3 : 4
Fe
O
3
.
4

Figure 5
Fe
O
3
is the main component
of the mineral
4
magnetite, a common iron ore
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Structure
1
Models
of
the
particulate
nature
of
matter
Worked example 6
Hydroc arbons
unknown
to
are
organic
hydroc arbon
produce
26.41 g
compounds
has
undergone
of
c arbon
dioxide,
empiric al
formula
of
of
c arbon
and
combustion
CO
,
and
in
hydrogen.
excess
13.52 g
of
An
oxygen
water,
H
2
Deduce
the
the
O.
2
hydroc arbon.
Solution
1
M(CO
)
=
12.01
+
2×16.00
=
44.01 g mol
2
26.41 g
n(CO
)
=
≈
0.6001 mol
2
–1
44.01 g mol
n(C)
=
n(CO
)
=
0.6001 mol
2
1
M(H
O)
=
2×1.01
+
16.00
=
18.02 g mol
2
13.52 g
n(H
O)
=
≈
0.7503 mol
2
–1
18.02 g mol
n(H)
×
n(H
c arbon
=
2
and
O)
=
2×0.7503 mol≈1.501 mol
2
All
the
hydrogen
hydroc arbon, C
H
x
The
mole
ratio
Therefore,
the
x:y
atoms
in
the
combustion
products
=
0.6001 : 1.501
empiric al
formula
of
≈
the
1 : 2.5
=
express
known as
comprised
number
ratio.
a
worked
as
whole
example
number
5,
the
ratios.
ratio
we
H
non-integer
values:
1.296 and 1.728.
ratio,
you
divide
term
the
gives
by
a
ratio
which
this
number
you
ratio
ratio
each
of
1
:
1.333.
should
by
of
3,
in
Then,
multiply
and
then
the
ratio
you
Whole
To
numbers
are also
c alculated
was
convert it to a whole
by the smallest number in the
c an
ratio
5
initially
two
Multiplying
whole
formulas
In
of
This
factor
empiric al
integers.
from
2 : 5
hydroc arbon is C
2
We
originate
, so:
y
to
use
trial
obtain
subsequently
and
the
rounding
error to determine
whole
the
number
result,
ratio.
gives a
3 : 4.
The molecular formula of a compound can be deduced from the empirical formula
if we know the molar mass of the compound. For example, you might determine
experimentally that the molar mass of the hydrocarbon in worked example 6 is
–1
58.12 g mol
. The molar mass of the empirical formula can be calculated:
–1
(12.01
The
×
2)
value
have
+
of
twice
(1.01
×
5)
29.07 is
the
=
29.07 g mol
roughly
number
of
half
atoms
of
as
58.12,
the
therefore
empiric al
the
molecular
formula: C
Determining
the
molar
substances
is
10
masses of
Table2
gaseous
formula must
H
4
suggests
that
this
hydroc arbon
could
be
butane, C
H
4
discussed in
c annot
be
sure
about
it
without
further
analysis,
as
there
is
.
However, we
10
another
hydroc arbon,
Structure 1.5.
methylpropane,
c an
be
with
distinguished
comparing
their
the
by
same
molecular
formula.
Butane
and
methylpropane
measuring their boiling points ( Structure 1.1) or
infrared
spectra (Structure 3.2).
Practice questions
1.
Deduce
a.
an
b.
a
the
empiric al
oxide
of
formulas
manganese
hydroc arbon
that
of
that
the
following compounds:
contains
produces
36.81%
5.501 g
of
of
c arbon
oxygen
dioxide
and
2.253 g of
water upon complete combustion
2.
Deduce
the
molecular
formula
of
the
hydroc arbon
from
1b
if
its
molar
mass
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–1
is
70
42.09 g mol
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Structure
1.4
Counting
particles
by
mass: the mole
Experimental determination of empiric al formula
Relevant
skills
•
Tool
1:
•
Tool
3:
Instructions
Measure
C arry
mass
out
c alculations
involving
decimals
1.
Weigh
a
clean,
dry
2.
Obtain
a
piece
of
andratios
and
•
Tool
3:
Use
•
Tool
3:
Construct
and
•
Inquiry
3:
realistic
1.0 g)
an
your
teacher.
ribbon
(between
Measure
its
exact
0.3 g
mass.
approximation and estimation
interpret
3.
Twist the magnesium into a loose coil and place it
4.
Heat
graphs
inside
to
from
crucible.
magnesium
Explain
and
relevant
the
crucible.
improvements
the
crucible,
with
its
lid
on,
over
a
roaring
investigation
Bunsen
air
to
ame.
enter
Periodic ally
the
li
the
crucible
lid
to
allow
crucible.
S afety
•
Wear
•
Take
•
The
eye
5.
protection.
suitable
prec autions
equipment
prec autions
will
get
around open ames.
very
hot.
around it and do not touch it while it
•
M agnesium
burns
with
a
heating until the magnesium no longer
lights
Then,
up.
crucible
Take suitable
ishot.
Continue
6.
When
7
.
Heat
to
the
the
directly at it.
Repeat
remove
for
a
the
heat
source
and
allow the
few minutes.
crucible is cool, weigh it.
crucible
additional
very bright light. Do not look
cool
and
minute.
this
its
contents
Allow
to
cool
strongly
and
for an
re-weigh.
heating-cooling-weighing
cycle until the
mass is constant.
M aterials
•
crucible and lid
Q uestions
•
balance
1.
•
pipeclay triangle
•
tripod
•
heat-proof mat
•
tongs
•
magnesium ribbon
(±0.01 g)
Process
of
2.
Compare
actual
3.
data
a
your
to
determine
the
empiric al
formula
oxide.
experimental
empiric al
formula to the
one.
Obtain
Plot
•
the
magnesium
mass
graph
data
of
from
mass
of
other
members
magnesium
of
oxide
your
vs
class.
mass of
Bunsen burner
magnesium.
lid
4.
Identify
t
5.
crucible
any
on
Explain
of
6.
line
anomalies
the
what
the
magnesium
Explain
why
(if
applic able)
and
draw a best
graph.
graph
shows about the composition
oxide.
you
repeatedly
heated
and
weighed the
coiled magnesium
crucible
until
a
constant
mass
was
achieved.
ribbon
7
.
Identify
and
explain
two
major
sources
of
error in this
procedure.
8.
Suggest
that
realistic
could
improvements to the methodology
minimize
the
sources
of
error
you
have
Bunsen burner
identied.
9.
Reect
on
empiric al

Figure 6
The experimental set-up
round
C an
to
the
role
formula
the
of
approximation
c alculations.
nearest
whole
and
rounding in
When is it suitable to
number?
When is it not?
you come up with a rule of thumb of when to
round
and
when
not
to
round?
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Structure
1
Models
of
the
particulate
nature
of
matter
Measurement
Atoms,
molecules
impossible.
particles
As
with
to
all
Consider
The
and
ions
concept
mass,
which
mass
of
a
so
the
mole
c an
measurements,
the
are
of
be
is
easily
mass
sample
small
has
of
that
counting
powerful
them
bec ause
directly is virtually
it
relates number of
measured.
an
uncertainty
c alcium
associated with it.
c arbonate,
C aCO
,
is
found to be
3
3.500
up
to
g
±
0.001
0.001
moles
does
c alculation
terms
Is
a
of
g.
in
it
represent?
and
either
nd
measurement
doing
means
g
particles,
questions
This
as
you
it
direction.
out.
is
How
You
quite
that
mass
This
many
will
is
measurement
clearly
particles
see
that
in
a
c an
minuscule
does
moles
it
be
inaccurate
mass.
How
by
many
represent? Do a quick
the
uncertainty
is
tiny, but in
large.
uncertainty
proceed
experiments
the
ever
negligible?
through
involve
the
DP
making
If
so,
when? Think about these
chemistry
course, particularly when
measurements.
Solutions and concentration
(Structure 1.4.5)
M any
chemic al
handle
aect
and
the
component
of
the
the
properties
of
solutes.
For
gases.
dissolved
and
the
or
so
a
in
solutions.
Sometimes
substances
more
the
solvent.
example,
out
mixtures
one
solution,
of
c arried
or
of
solution
of
is
participate
or
more
two
solutes.
other
Solutions
solvent
or
properties
The
a
The
of
the
in
in
easier to
bec ause
chemic al
components.
it
c an
reactions.
E ach solution
solvent is usually the major
whole
components
sugar
are
used
water
solution
of
the
is
more
are similar
solution
like
are
water
(clear
heterogeneous
colourless
are
are
solids
homogeneous
solvent
c alled
mixtures
are
consists of a
to
and
than
properties
Solutions
Homogeneous
reactions
mix
liquid)
than
sugar
(white
crystalline
powder),
so
water
is
the
solvent
discussed in
while
sugar
(from
the
is
the
solute. In this topic, we will consider only
aqueous solutions
Structure 1.1.
L atin
aqua
meaning
“water ”),
in
which
the
solvent
is
water.
solute

In
some
c ases,
the
ethanol
and
water
present
it
is
is
not
water ”
the
water,
in
major
rather
Figure 7
identity
of
each
these
the
of
the
mixture,
component.
than
How a solution is formed
“4%
it
solvent
liquids
is
For
solution
of
is
unclear:
c an
be
traditionally
example,
for
c alled
example, if we mix
a
solvent.
regarded
we
as
the
However, if
solvent,
even if
say “96% solution of ethanol in
water in ethanol”.
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Structure
Solutions
solute
solute,
small
the
are
and
oen
and
so
has
proportion
term
classied
solvent. A
a
of
high
ratio
solute,
“concentrated”
solute
per
much
less
100 g
than
of
the
10 g
according
to
the
concentrated solution
of
of
and
refers
solvent,
the
solute
so
to
has
the
per
or
mole
a
ratio
with
term
100 g
of
much
of
the
to
more
Counting
particles
by
mass: the mole
between the
proportion of
dilute solution
solute
“dilute”
ratio
large
solvent, while a
low
solutions
and
solute
to
a
mass
contains
1.4
solvent.
than
has a
Generally,
10 g of the
refers to solutions with
solvent.
TOK
Some
words
do
interpretation
are
not
is
not
precisely
chemists
would
“dilute”,
as
have
context
dened
c all
a
precise denitions and their choice and
dependent.
and
The
should
solution
of
be
terms
used
“concentrated”
with
c are.
5 g of sulfuric acid (H
For
SO
2
used
in
much
higher
laboratories.
At
proportions
the
permanganate (KMnO
)
in
same
100 g
of
time,
of
sulfuric
a
acid
solution
water
would
of
to
5 g
be
)
and
“dilute”
example, most
in
100 g
of
water
4
water
of
are commonly
potassium
considered
very
4
concentrated
by
permanganate
The
To
any
in
antiseptic
concentrations
what
extent
in
does
communic ation
of
the
concentration.
Molar
of
a
the
worker,
solutions
examples
expressing
composition
solute
a
as
typic al
are
less
above
concentrations
than
could
quantity
0.1 g
be
per
of
100 g
expressed
potassium
of
water.
numeric ally.
numeric ally help or hinder the
knowledge?
Quantitatively,
amount
medic al
to
of
solutions
concentration,
the
c,
is
also
expressed in terms of
known as
molarity,
is
the
ratio of the
volume of the solution:
n
solute
c
=
solute
V
solution
–3
The
most
common
units
for
molar
concentration
are
mol dm
–1
as
mol L
also
be
(which
is
the
–3
).
For
very
dilute
solutions,
smaller
units
same
–3
(mmol dm
or
µmol dm
)
c an
used:
–3
1 mmoldm
–3
=
1 × 10
–3
–3
mol dm
–6
1 µmoldm
=
1 × 10
–3
mol dm
–3
The
units
of
molar
concentrations
are
sometimes
abbreviated
as
M
(for
mol dm
)
–3
or
mM
(for
mmol dm
).
For
example,
the
expression
“2.5 MNaOH”
means that
3
each dm
of
Note
the
that
whole
the
solution
term
solution.
contains
“molar
For
2.5 mol
concentration”
example,
it
is
of
sodium
refers
incorrect
to
to
say
a
hydroxide.
specic
that
“the
substance, not the
concentration of a
–3
sodium
chloride
about
the
would
be
solution
concentration
is
of
1.0 mol dm
sodium
”,
as
chloride
it
is
or
not
clear
water.
The
whether
we
are talking
correct statement
–3
Molar
“the
concentration
concentration
is
oen
of
sodium
chloride
represented
by
in
a
square
solution
brackets
is
1.0 mol dm
”.
around the solute
–3
formula.
For
example,
the
expression [NH
]
=
0.5 M
refers
to
a
0.5 mol dm
3
–
solution
of
ammonia.
Similarly,
the
expression
[Cl
]
refers to the molar
concentration of chloride ions in a solution.
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Structure
1
Models
of
the
particulate
nature
of
matter
Worked example 7
–3
C alculate
the
a
prepared
solution
molar
concentration
by
dissolving
of
sodium
3.60 g
of
chloride,
N aCl(s)
in
in
mol dm
water
to
,
in
make
3
25.0 cm
of
the
nal
solution.
Solution
First,
c alculate
the
molar
mass of sodium chloride:
1
M(NaCl)
=
22.99
+
35.45
=
58.44g mol
m
Then use
n
=
to
c alculate the amount of solute:
M
3.60 g
n(NaCl)
≈
=
0.0616 mol
–1
58.44 g mol
3
Convert
the
volume to dm
by
dividing
Activity
by
1,000:
3
V(solution)
=
25.0 cm
3
=0.0250 dm
n
C alculate
the
H
50.0 cm
mass of sulfuric acid,
Use
c
=
to
c alculate
the
concentration:
V
3
SO
2
,
in
of a
solution
0.0616 mol
4
3
c(NaCl)
–3
where [H
SO
2
]
=
≈
=
1.50 mol dm
2.46 mol dm
3
0.0250 dm
4
The
composition
concentration,
of
a
ρ
solution
,
of
the
is
sometimes
solute,
which
expressed as the
is
the
ratio
of
the
mass
mass of the solute to
solute
the
volume of the solution:
m
solute
ρ
=
solute
V
solution
Worked example 8
C alculate
worked
the
mass
example
concentration
of
sodium
chloride
in
the
solution
from
7.
Solution
If
Activity
we
know
c alculate
the
the
mass
mass
of
the
solute
concentration
and
as
the
volume
of
the
solution,
we
c an
follows:
3.60 g
3
ρ(NaCl)
C alculate
the
molar
=
=
concentration
144 g dm
3
0.0250 dm
–3
of
sulfuric
acid,
in
in a solution with
mol dm
ρ(H
SO
2
Alternatively,
) =
the
concentration
mass
and
concentration
molar
mass,
using
of
NaCl
the
c an
be
found
relationship
4
ρ
from its molar
=
solute
c
×
M
solute
:
solute
–3
0.150 g cm
3
ρ(NaCl) = c(NaCl)× M(NaCl) = 2.46 mol dm
1
×58.44 g mol
–3
The
most
common
concentration
mass,
as
and
units
molar
for
mass
concentration
concentration
of
the
are
same
3
≈ 144 g dm
–3
g dm
and
solute
are
g cm
related
.
M ass
by molar
follows:
ρ
solute
c
=
solute
M
solute
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Structure
1.4
Counting
particles
by
mass: the mole
Worked example 9
A
standard
solution
was
prepared
by
dissolving
6.624g
of
sodium
c arbonate,
used
transfer
Na
CO
2
a
,
in
deionized
water
3
3
using
3
250 cm
volumetric
ask.
An
analytic al
pipette
was
to
10.0 cm
sample
of
this
solution
to
a
3
100cm
volumetric
ask,
and
the
ask
was
topped
up
to
the
graduation
mark
with
deionized
water.
C alculate
the
–3
concentration,
in
moldm
,
of
sodium
c arbonate
in
the
new
solution.
Solution
First,
we
need
to
nd
the
concentration
of
sodium
c arbonate
in
the
standard solution:
1
M(Na
CO
2
)
=
2×22.99
+
12.01
+
3×16.00
=
105.99 g mol
3
6.624 g
n(Na
CO
2
)
=
≈
3
0.06250 mol
–1
105.99 g mol
3
V
=
3
250 cm
=
0.250 dm
standard
Note
that
the
accuracy
of
a
typic al
volumetric
ask
is
three
signic ant
gures.
0.06250 mol
3
c
(Na
standard
CO
2
)
=
=
3
0.250 mol dm
3
0.250 dm
Then
we
First,
c alculate
need
to
c alculate
the
concentration
of
sodium
c arbonate
in
the
new solution.
3
the
amount
of
Na
CO
2
in
3
V
=
10.0 cm
the
sample.
Remember
to
convert
all
volumes to dm
3
3
=
0.0100 dm
sample
3
c
(Na
standard
CO
)
2
3
CO
)
=
c
(Na
sample
CO
2
)
=
0.250 mol dm
3
3
n
(Na
sample
When
the
2
sample
=
0.250 mol dm
3
×0.0100 dm
=
0.00250 mol
3
is
diluted
with
deionized
water
to
produce
the
new
solution,
the
amount
of
solute
does
not
change.
Therefore
n
(Na
sample
Now
you
c an
CO
2
)
=
n
3
work
out
(Na
new
the
CO
2
)
=
0.00250 mol
3
concentration
of
Na
CO
2
volume
of
the
3
V
in
the
new
solution
by
dividing
the
amount
of
Na
3
CO
2
by the
3
new solution:
=
3
100 cm
=
0.100 dm
new
0.00250 mol
3
c
(Na
new
CO
2
)
=
=
3
0.0250 mol dm
3
0.100 dm
It
is
a
common
practice
to
store
chemic als
in
the
form
of
concentrated solutions
Practice question
(so-c alled
needed.
stock solutions)
Stock
solutions
and
with
a
dilute
them
known
to
the
required
concentration
of
concentration when
the
solute
are
c alled
3.
standard solutions.
A
standard
by
copper(II)
To
determine
the
concentration
of
the
standard
solution
in
worked
example
did
the
following
two
was
prepared
2.497 g of
sulfate
pentahydrate,
9,
CuSO
we
solution
dissolving
• 5H
O,
4
2
using
a
in
deionized
c alculations:
3
water
100 cm
volumetric
3
ask.
1.
n
=
c
sample
×
sample
A
5.00 cm
sample of this
V
sample
3
solution
was
diluted
to
250.0 cm
.
n
new
2.
c
C alculate
=
new
–3
mol dm
nal
know that
n
=
sample
gives
the
following
c
n
,
so
you
c an
substitute
equation
1
into
, of copper(II) sulfate in
the
solution.
equation 2. This
new
expression:
The
× V
sample
c
concentration, in
V
new
You
the
process
for
preparing
standard
sample
=
solutions
new
is
discussed in the
Tools
V
new
for chemistry
Therefore,
need
to
to
c alculate
know
solution,
and
the
the
the
original
volume
concentration
concentration
of
the
of
of
a
solute
the
in
solute,
a
new
the
solution,
chapter.
you just
volume of the original
new solution. In summary:
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c
×
1
V
=
1
c
×
2
V
2
75
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Structure
1
Models
of
the
particulate
nature
of
matter
C ase study: spectrophotometry and c alibration curves
Spectrophotometry
the
is
intensity
of
commonly
is
visible,
used
for
an
analytic al
ultraviolet
technique
and
determining
based
near-infrared
concentrations
on
the
radiation.
of
measurement of
This
technique
coloured substances in
solutions.
A
spectrophotometer
through
a
intensity
of
is
a
value
standard
curve
is
and
the
in
solute.
and
of
their
used
the
general
absorbance,
the
the
light
and
studied
for
of
of
a
certain
solution.
absorbed
substance
are
determining
wavelength,
The
are
by
These
c alibration curve
the
unknown
passes
measures the
absorbance. Absorbance
the
prepared
measured.
which
photodetector
converts it into the
light
producing a
sample.
Initially,
several
by serial dilution
absorbances
(gure8).
concentration
of
are
plotted
The
c alibration
the
coloured
studied solution.
or
a
c alibration
electric al
unknown
the
light
studied
absorbances
c ase,
pH
The
plotting
the
amount
the
concentrations,
then
of
transmitted
solutions
substance
In
the
produces
sample
describing
(gure9),
against
small
result
of
curve
relates
conductivity)
concentration
the
of
c an
measurement
a
measurable
the
be
on
solution
found
the
by
to
property
the
measuring
c alibration
(such as
concentration of
that
property
curve.
0.40
ecnabrosba
0.30
0.20
0.10
0
0
0.10
0.20
0.30
0.40
0.50
3
concentration/mmol dm

Figure 8
A typic al c alibration curve
Data-based question

Figure 9
A series of standard
solutions of
potassium permanganate
The calibration curve in gure8
Ideally, the calibration curve should be linear, pass through the origin and have a
was obtained using a series of
tilt of approximately 45°. If the curve does not meet any of these requirements, it
standard solutions of potassium
should be constructed again using a slightly dierent wavelength of light and/or
permanganate, KMnO
. A solution
4
dierent set of standard solutions. Sometimes linearity can only be achieved within
with unknown concentration of
a narrow range of concentrations. In this case, the studied solution can be diluted,
KMnO
has an absorbance of
4
so the concentration of the studied substance falls within the range of calibration
0.285. Determine the concentration
curve. In the last case, some additional calculations will be required to relate the
of KMnO
in that solution.
4
concentrations of the studied substance in the diluted and original solutions.
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76
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Structure
Another
technique,
spectrophotometry
colorimetry,
but
“spectrophotometry”
correct
but
is
based
on
the
1.4
Counting
particles
by
mass: the mole
same principles as
limited to visible light. The terms “colorimetry” and
are
oen
used
interchangeably,
which
is
not
entirely
very common.
Concentration uncertainty of a standard solution
A
standard
In
this
solution
of
copper(II)
activity,
you
is
a
will
sulfate,
solution
prepare
each
by
of
known
two
using
concentration.
standard solutions
dierent
equipment.
M aterials
•
Wash
•
Weighing boats (2)
•
100 cm
•
Stirring
•
Funnels (2)
•
Pipettes
•
Spatula
•
Reagent
•
Blank labels
•
Colorimeter
•
Cuvettes
•
C alibration
bottle
containing
distilled
water
3
By
propagating
assess
will
the
determine
using
a
the
measurement
precision
the
of
Relevant
the
This
values.
concentration
of
will
you
then
allow
you will
You
your solutions
to
assess the
beakers (2)
rods (2)
bottles (2)
skills
Tool
1:
Measuring
•
Tool
1:
Standard
Tool
uncertainties,
concentration
concentrations.
•
•
the
actual
colorimeter.
accuracy
of
3:
volume
solution
C alculate
and
and
mass
preparation
interpret
percentage
error and
curve
relating
concentration of copper(II)
percentage uncertainty
sulfate and absorbance
•
Tool
3:
Express quantities and uncertainties to an
•
Copper(II)
sulfate
pentahydrate, CuSO
•5H
4
appropriate
number
of
signic ant
O
2
gures
•
Tool
3:
Record
•
Tool
3:
Propagate uncertainties
•
100 cm
•
Inquiry
2:
•
Milligram
measurement uncertainties
Additional equipment for solution 1:
3
Assess
accuracy
and
precision
S afety
volumetric ask
balance
(three
decimal
places)
Additional equipment for solution 2:
3
•
Wear
eye
•
Solid
copper(II)
the
protection.
sulfate
is
an
irritant
and
toxic to
•
100
•
Centigram
cm
measuring
balance
cylinder
(two
decimal
places)
environment
Instructions
•
Dispose
of
all
solutions
appropriately.
1.
Use
the
equipment
copper(II)
sulfate
provided
to
prepare two
standard solutions, both with
–3
concentration
solution
1,
milligram
0.020 mol dm
you
should
balance.
use
For
.
the
When
preparing
volumetric ask and
solution
2,
use
the
measuring
meniscus of the solution
cylinder
2.
etched
line
indic ating
Record
and
the
centigram
balance.
measurements
you
make
along
the
way,
including their uncertainties.
3
volume,
e.g.
250 cm
3.
Following
the
your
teacher ’s
colorimeter,
instructions
measure
the
on
how to use
absorbance
of
your
solutions.
4.
a
fixed
volume of solution
when the meniscus is on
the
etched
Refer
to
the
c alibration
concentration
volumetric flask contains
of
curve to determine the actual
your solutions.
Q uestions
1.
line,
Determine
the
uncertainty
of
the
concentrations of
solutions 1 and 2.
3
e.g.
250 cm
2.
C alculate
the
percentage
error
of
the
concentrations
of solutions 1 and 2.
3.
Assess
the
precision
and
accuracy of the
concentrations of solutions 1 and 2.
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77
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Structure
1
Models
of
the
particulate
nature
of
matter
5.
4.
Consider
the
way
you
have
presented
The
construction
of
c alibration
curves
involves
your
ATL
preparing
c alculations
for
the
questions
above.
Do
samples
concentrations.
think
they
convey
your
thinking?
Do
reader
would
be
able
to
easily
solutions
that
cover
Instead
of
measuring
a
and
range of
dissolving
you think
a
a
of
you
follow
certain
mass
of
solute
the
way
you
have done
your
here, chemists oen start with a stock solution and
thought
process?
How
could
you
improve
perform a
the
presentation
of
your
c alculations?
want
to
look
serial dilution.
the
advantages
disadvantages of using a serial dilution in the
through some of the
preparation
worked
Discuss
You
and
may
examples
in
this
textbook
for
of
samples
for
a
c alibration
curve.
ideas.
Avogadro’s law (Structure 1.4.6)
In
1811,
at
the
This
Amedeo
same
Avogadro
temperature
hypothesis
has
suggested
and
been
pressure
conrmed
that
equal
contain
in
many
volumes
equal
of
any two gases
numbers
experiments
of
molecules.
is
now
and
known as
Avogadro’s law
Since
to
the
each
volumes
are
amount
other,
of
two
a
substance
amount
reacting
proportional
n
of
the
to
the
of
a
and
gas
gaseous
is
the
number
of
proportional
species
amounts
of
these
reactants
and
particles
to
measured
its
are
volume.
under
the
proportional
Therefore, the
same conditions
species:
V
1
1
=
n
V
2
In
turn,
the
2
amounts
stoichiometric
know
of
the
other
of
coecients in
volume
gaseous
of
any
gas
a
products
balanced
consumed
substances
c an
be
are
chemic al
or
proportional to their
equation.
produced
found
without
in
the
As
a
result, if we
reaction,
the
volumes
c alculating their amounts.
Worked example 10
The
combustion
of
hydrogen
sulde,
H
S,
proceeds
as
follows:
2
2H
S(g)
+
3O
2
(g)
→
2H
2
C alculate
the
O(l)
+
2SO
2
volumes
of
(g)
2
oxygen,
O
(g),
consumed
and
sulfur
dioxide,
2
SO
(g),
produced
if
the
volume
of
hydrogen
sulde
combusted
was
2
3
0.908 dm
.
All
volumes
are
measured
under
the
same
conditions.
Solution
Practice question
The
ratio of the stoichiometric coecients of H
you
c an
S and O
2
4.
is
2 : 3.
Therefore,
2
3
Incomplete combustion of
multiply
the
volume
of
combusted H
S
by
to
nd
the
volume of
2
2
hydrogen
sulde
elemental
sulfur
produces
combusted O
:
2
instead of sulfur
3
V(O
)
=
3
V(H
2
S)
=
2
The
3
×
0.908 dm
3
≈
1.36 dm
2
dioxide:
2
ratio of the stoichiometric coecients of H
S and SO
2
2H
S(g)
+
O
2
(g)
→
2
2H
O(l)
+
is
1 : 1.
Therefore,
2
2S(s)
2
the
volume
of
combusted H
S
is
the
same
as
the
volume
of
produced SO
2
C alculate
the
3
)
V(SO
=
V(H
2
combusted
of
S)
=
0.908 dm
2
hydrogen sulde if the
Note
volume
:
2
volume of
oxygen
that
the
volume
of
liquid
water
c annot
be
found
in
the
same
manner, as
consumed in
Avogadro’s
law
applies
to
gases
only.
3
this
reaction
was
1.25 dm
Linking question
Avogadro’s
law
applies
to
ideal gases. Under what conditions might the
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behaviour
78
of
a
real
gas
deviate
most
from
an
ideal
gas?
(Structure 1.5)
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Structure
1.4
Counting
particles
by
mass: the mole
End-of-topic questions
C alculate:
Topic review
a.
1.
Using
your
knowledge
from the
Structure 1.4
The
molar
KAl(SO
)
4
answer
the
guiding
question
as
fully
as
mass
of
potassium alum,
topic,
•12H
2
O.
[1]
2
possible:
b.
The
c.
The
d.
The
How do we quantify matter on the atomic scale?
amount
potassium
total
of
substance,
in
mol,
in
1.00 g of
alum.
[1]
number
of
atoms
in
1.00 g of
Exam-style questions
potassium
alum.
[1]
Multiple-choice questions
2.
What
is
the
copper(II)
number
sulfate
of
oxygen
atoms
pentahydrate,
in
CuSO
amount
of
0.400 mol of
by
•5H
potassiumalum.
4
O?
complete
water,
in
mol,
that
decomposition
of
c an
be
produced
1.00 g of
[1]
2
24
A.
3.60
C.
2.16
×
B.
9
D.
5.40
e.
10
The
potassium
24
×
A
sample
containing
0.70 g
of
c alcium
composition,
by
mass, of
alum.
[2]
10
8.
3.
percentage
To
visualize
the
mole,
a
chemistry
student
decided to
nitrate,
23
pile
3
C a(NO
)
3
,
is
dissolved
in
water
to
a
volume
of
200 cm
is
6.02 × 10
needed
–
What
up
grains
of
sand. Estimate the time
.
2
the
concentration of NO
ions
in
this
to
complete
this
project
if
an
average
grain
solution?
3
ofsand
–3
A.
3.5 g dm
B.
7.0 g dm
0.021 mol dm
D.
0.043 mol dm
50 kg
–3
For
which
the
molecular
5 mg,
and
the
student
c an
shovel
of
sand
per
minute.
[3]
–3
9.
4.
weighs
–3
C.
molecule
is
the
empiric al
formula
the
Deduce
the
empiric al
formulas
for
the
following
same as
compounds:
A.
CH
CH
3
B.
CH
formula?
CH
2
OH
C.
CH
2
COOCH
3
CH
3
CH
2
D.
CH
3
CH
2
a.
an
b.
an oxygen-containing organic compound, 5.00 g of
sulfur
that
contains
59.95%
of
oxygen
[1]
COOH
which produces 9.55 g of carbon dioxide and 5.87 g
3
Which volume of a 5.0 mol dm
of
3
–3
5.
oxide
CH
2
sulfuric acid (H
SO
2
of water upon complete combustion.
)
[2]
4
3
stock solution is required to prepare 0.50 dm
of a
10.
A
standard
solution
of
potassium
sulfate, K
SO
2
,
was
4
solution whose concentration of hydrogen ions is
3
prepared
from
8.714 g
of
the
solid
salt
using
a
250 cm
–3
0.10 mol dm
?
volumetric
3
A.
0.010
3
cm
C.
5.0
B.
6.
A
0.0050
student
an
D.
obtained
experimental
g dm
3
cm
the
of
the
empiric al
the
mass
concentration, in
–3
,
and
potassium
10 cm
11.
following data during an
determination
C alculate
–3
cm
3
ask.
formula of
molar
sulfate
The
c alibration
ve
standard
potassium
oxide of tin:
the
in
concentration,
the
curve
in
solutions,
nal
mol dm
, of
solution.
gure7
in
in
was
which
the
permanganate, KMnO
,
[2]
constructed using
concentration of
varied
from
0.100
4
–3
to

M ass
of
tin

M ass
of
oxide
before
heating
=
0.500 mmol dm
these
mass=
of
tin
aer
.
Describe
how
you
would
prepare
1.78 g
solutions
using
serial
dilution.
[3]
heating to a constant
12.
2.26 g
C arbon
monoxide,
produces
c arbon
CO,
is
dioxide,
a
toxic gas. Its combustion
CO
(g).
2
According
to
these
data,
what
is
the
correct
formula of
a.
the
oxide
of
Deduce
balanced
equation
SnO
C.
c arbon
monoxide.
[1]
SnO
3
3
b.
B.
for the combustion
tin?
of
A.
the
SnO
D.
C alculate
the
volumes, in dm
,
of
consumed
c arbon
SnO
2
5
monoxide
and
oxygen
if
the
combustion
produced
3
2.00 dm
Extended-response questions
7.
Alums
are
XAl(SO
)
4
salt
hydrates
•12H
2
O,
of
the
where
X
measured
general
is
an
of
c arbon
under
dioxide.
the
same
All
volumes
conditions.
are
[2]
formula
alkali metal or other
2
singly-charged
c ation.
When
heated, most alums
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decompose
XAl(SO
as
)
4
follows:
•12H
2
O(s)
2
→
XAl(SO
)
4
(s)
2
+
12H
O(l)
2
79
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Structure 1.5
Ideal gases
How does the model of ideal gas behaviour help us to predict the behaviour of real gases?
As
is
with
a
any
theoretic al
simplic ation
many
c ases,
precision
it
that
predicts
sucient
model,
has
for
its
the
the
concept
properties
most
of
an
ideal gas
at
advantages and limitations. In
practic al
of
real gases with a
purposes.
low
real
temperatures
gases
ideal
gas
deviates
model
and
high
pressures
signic antly
c annot
be
from
the
the
behaviour of
prediction, so the
used under these conditions.
However,
Understandings
Structure 1.5.1
collisions
—
An
between
ideal
gas
particles
consists
are
gases
of
moving
particles
with
Structure 1.5.2
—
Real
Structure 1.5.3
—
The
molar
Structure 1.5.4
—
The
relationship
deviate
volume
from
of
an
the
ideal
ideal
between
gas
the
gas
model,
is
constant
a
pressure,
gas
equation
pV
=
nRT
and
the
combined
gas
at
V
1
ideal
volume
particularly
volume,
p
the
negligible
and
no
intermolecular
forces. All
considered elastic.
a
low
temperature
temperature
temperature
p
1
law
at
specic
and
and
amount
and
high
pressure.
pressure.
of
an
ideal
gas
is
shown in
V
2
2
=
T
T
1
2
Assumptions of the ideal gas
model (Structure 1.5.1)
The
ideal
gas
model
states
that
an
ideal
gas
conforms
to
the
following
ve
assumptions:
1.
2.
Molecules of a gas are in constant random motion
This
means
until
they
gas
molecules
with
another
inelastic
or
sound.
collisions
However,
elastic
and
of
larger
the
Vaporized
water
(0
100 kPa
°C)
and
no
energy
They
the
move
side
of
a
in
straight lines
container.
is
energy
between
lost
from
c an
be
transferred
molecules
the
in
an
ideal
as
heat
gas
are
system.
gas
conditions.
same
in
occupies
phase
but
is
the
which
the
about
In
both
same
volume
the
1600
times
the
volume
of
liquid
water
at
273.15K
pressure (standard temperature and pressure, STP).
the
space
4.
or
they occupy
occupies
Nitrogen
changed,
are
stationary.
The volume occupied by gas molecules is negligible compared to the
gaseous
forces
not
molecule
objects,
collisions
volume of the container
Intermolecular
are
gas
Collisions between molecules are perfectly elastic
In
perfectly
3.
that
collide
gas
650
and
of
times
c ases,
the
the
There are no intermolecular
size
gas
molecules
the
is
are
the
volume
number
of
to
of
liquid
nitrogen under
molecules in liquid and
individual
>99.9%
free
of
molecules has not
empty
space. This is the
move.
forces between gas particles
studied in
For an ideal gas, the intermolecular forces are negligible compared to the kinetic
Structure 2.2.
energy of the molecules. As such, an ideal gas will not condense into a liquid.
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Structure
5.
1.5
Ideal gases
The kinetic energy of the molecules is directly proportional to Kelvin
temperature
This
relationship
is
studied in
Reactivity 2.2
Pressure–volume relationships
Robert
Boyle
pressure
of
(1627–1691)
a
given
relationship,
now
established
amount
of
known as
a
gas
is
that,
at
constant
inversely
Boyle’s law,
c an
temperature, the
proportional
be
expressed
to
as
its
volume. This
follows:

Figure 1
An ideal gas consists of
1
p
or
∝
pV = k (a
constant)
or
p
V
1
=p
1
particles that
V
2
collide elastic ally,
have no
2
V
intermolecular forces and
In
gure
the
1,
walls
the
of
the
volume
If
of
so
walls,
so
of
container.
pressure.
space,
molecules
the
the
every
is
is
gas
are constantly striking and bouncing o
force
halved,
second
pressure
a
The
there
these
there
are
doubled
of
are
twice
impacts
twice
as
as
produces
many
a
measurable
molecules
in
volume when compared
each unit
many impacts with the container
(gure 2).
pressure
Figure 2
to the volume of
the gas (the container)
volume

occupy negligible
halved
doubled
Halving the volume of a container doubles the pressure
p
p
,erusserp
,erusserp
TOK
Models are simplied
representations of natural
phenomena. The ideal gas model is
volume, V
reciproc al
volume, 1/V
built on certain assumptions related

Figure 3
volume of
Graphs showing the inverse relationship between pressure and
–2
The
SI
other
unit
of
pressure is the
units
of
pressure
atmosphere
inch
(psi).
found
in
to the behaviour of ideal gases.
an ideal gas
(atm),
databases
and
pasc al (Pa),
commonly
millimetres
Standard
temperature
are
of
temperature
for
mercury
and
comparative
100.0 kPa
where
used
in
1 Pa
=
1 N m
–3
=
1 J m
.
M any
dierent countries, including the
(mm Hg),
pressure
purposes.
bar,
and
conditions
STP
for
pounds
(STP)
gases
is
are
0 °C
per
square
frequently
or
What
is
in
development of scientic
the
the
role
of
assumptions
models?
What
not
are
the
implic ations of
acknowledging
a
model’s
limitations?
273.15 K
pressure.
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Structure
1
Models
of
the
particulate
nature
of
matter
Worked example 1
3
A
weather
is
released
balloon
at
sea
lled
level.
with
The
32.0 dm
balloon
of
helium
reaches
an
at
a
pressure
altitude
of
of
100.0 kPa
4500 m,
where
3
the
atmospheric
balloon
helium
at
in
that
the
pressure
altitude.
balloon
is
57.7 kPa.
Assume
remain
that
C alculate
the
the
volume,
temperature
and
in
the
dm
,
of
amount
the
of
constant.
Solution
V
× p
1
From
Boyle’s
law,
it
follows that
V
1
=
, so:
2
p
2
3
32.0 dm
× 100 kPa
3
=
V
≈
55.5 dm
2
57.7 kPa
Practice question
1.
At
a
certain
altitude,
a
weather
balloon
has
a
temperature
of
–35.0 °C and
3
a
volume
contains
of
0.250 m
16.0 g
of
.
C alculate
helium,
the
pressure,
in
kPa, inside the balloon if
He(g).
Real gases vs ideal gases (Structure 1.5.2)
When
the
begin
to
move,
of
volume
occupy
of
a
relationship
of
between
doubling
the
gas
the
against
no
of
become
pressure.
pressure
pressure
decreases
proportion
forces
reducing
pressure
real
large
intermolecular
collisions,
graph
a
and
volume
longer
signic antly,
the
container.
signic ant.
This
means
volume
for
a
real
halves
no
and
molecules
so
may
little space to
decreases the number
for
longer
gas
the
This
that,
the
With
a
real
applies.
an
ideal
gas,
the
Figure
gas.
For
4
inverse
shows a
the
real gas,
volume.
real gas
ideal gas
p
erusserp
0.5p
V
2V
0
volume
0

Figure
an
ideal
4
gas
Doubling
but
not
for
the
a
pressure
real
halves
the
volume
for
gas
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Structure
For
a
gas
to
deviate
intermolecular
the
molecules
from
forces
ideal
and/or
themselves.
gas
a
behaviour,
signic ant
This
there
volume
commonly
occurs
must
of
the
at
a
be
Ideal gases
detectable
gas
low
1.5
must
be
occupied
by
temperature and high
pressure.
Low temperature:
is
reduced.
form
and
High
The
As
As
At
they
molecules
pressure:
volume
of
molecules
At
the
low
temperature,
collide
may
high
with
not
pressure,
themselves
the
relationship
between
not
considered
to
an
c annot
keep
for
temperature.
in
the
At
low
container,
negligible.
At
so
high
intermolecular
ideal
space
of
are
of
more
molecules
signic ant
volume
is
the
gas
forces
molecules
of
attraction
elastic ally.
compressed,
and
energy
intermolecular
rebound
a
gas
there
far
are
the
attraction
to
apart
part
only
no
and
behaviour
occupied
temperature,
forces
be
molecules
an
pressure,
the
kinetic
of
in
the
the
a
reduced
space
longer
space.
volume of the gas.
between them,
inverse, so the gas is
ideal gas.
them.
conditions
there
becomes
pressure
Ideal gas conditions
The
the
another,
necessarily
molecules
be
one
very
by
few
the
prevent
are
low
molecules
molecules
molecules
are
interaction
between
pressure and high
per
unit
of
volume
themselves is
moving
too
fast
to
allow
for
form.
Activity
1.
Outline
the
2.
Discuss
what
deviations
3.
Consider
main
assumptions
conditions
from
how
ideal
each
of
behind
pressure
the
and
ideal gasmodel.
temperature
are
likely
to
lead to
behaviour.
of
the
following
might
affect
the
validity
of
the
ideal
gasmodel:
a.
4.
Strong
b.
L arge
For
each
ideal
a.
intermolecular
molecular
of
the
gas
at
low
volume
following
behaviour
and
forces
give
pairs,
a
predict
which
is
more
likely
to
exhibit
reason:
pressure
or
gas
b.
gas
at
at
high
low
pressure
temperature
or
gas
c.
at
high
hydrogen
temperature
fluoride,
HF(g)
or
hydrogen
d.
bromide,
methane, CH
HBr(g)
(g)
4
or
dec ane, C
H
10
e.
(g)
22
propanone, CH
COCH
3
(g)
3
or
butane, C
H
4
(g)
10
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Structure
1
Models
of
the
particulate
nature
of
matter
Real gases
Gases
that
deviate
from
the
ideal
gas
model
are
known as
n
real gases.
(
2
V – nb
) (
p + a
V
)
nRT
=
Relevant skills
measured
•
Tool
2:
•
Inquiry
Use
spreadsheets to manipulate data.
pressure
1:
Select
sufficient
and
relevant
correction
sources of
correction
information.
forces
for
for
of
between
volume
molecules
molecules
•
Inquiry
1:
Demonstrate
creativity in the designing,
measured
implementation
or
presentation
of
the
investigation.
volume
Instructions
The
relationship
temperature
Waals
of
between
real
gases
pressure,
is
volume, amount and
modelled
by
the
van der
Parameter
a corrects
for
intermolecular
parameter
b
for
molecular
and
b
for
corrects
various
gases
are
force
volume.
strength and
Values of
a
shown in table 1.
equation:
–1
Substance
a
/
× 10
ammonia, NH
6
Pa m
–2
mol
–3
b
/
× 10
3
m
4.225
0.0371
1.355
0.0320
13.89
0.1164
–1
mol
3
argon,
Ar
H
butane, C
4
10
H
butan-1-ol, C
4
OH
20.94
0.1326
7.566
0.0648
5.580
0.0651
9
Cl
chloromethane, CH
3
H
ethane, C
2
6
ethanol, C
H
2
helium,
OH
12.56
0.0871
0.0346
0.0238
5
He
hydrogen
bromide,
HBr
4.500
0.0442
hydrogen
chloride,
HCl
3.700
0.0406
hydrogen
uoride,
9.565
0.0739
5.193
0.0106
krypton,
HF
Kr
methane, CH
2.303
0.0431
9.476
0.0659
0.208
0.0167
4
OH
methanol, CH
3
neon,
Ne
H
pentane, C
5
19.09
0.1449
OH
25.88
0.1568
9.39
0.0905
OH
16.26
0.1079
5.537
0.0305
4.192
0.0516
12
H
pentan-1-ol, C
5
11
H
propane, C
3
8
H
propan-1-ol, C
3
7
O
water, H
2
xenon,
Xe

Table 1
Van der Waals parameters, a and
b,
for a selection of gases
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Structure
1.
Use
of
a
the
selection
factors
instance,
of
the
data
affecting
the
in
table
1
values of
to
a
explore some
and
You
b. For
you could look at:
•
intermolecular force strength and the value of a
•
molar
•
the
effect
gas
behaviour.
mass
and
the
value of
will
how
to
need
to
analyse
decide
it.
how
much
Depending
choose
to
and/or
look up additional data.
b
explore,
2.
you
Consider
may
on
how
need
you
data
which
to
to
Ideal gases
select, and
option
perform
could
1.5
you
c alculations
present
your data
ATL
graphic ally.
of
volume
on
the
deviation
from
Prepare a one-page summary of
ideal
your
exploration
to
share
with
your
class.
Linking question
Under
comparable
behaviour
than
conditions,
others?
The molar
why
do
some
gases
deviate
more
from
ideal
(Structure 2.2)
volume of an ideal gas
Avogadro’s
law
is
covered in
Structure 1.4.
(Structure 1.5.3)
Avogadro’
s law states that equal volumes of any two gases at the same
temperature and pressure contain equal numbers of particles. The molar volume
of an ideal gas is a constant at specied temperature and pressure. For example,
3
at STP
, the molar volume of an ideal gas, V
, is equal to 22.7dm
–1
mol
.
m
28.3 cm
H
CH
He
2
O
4
Cl
2
2
3
V = 22.7 dm

Figure 5
1
1

Figure 6
1
4.00 g mol
2.02 g mol
Molar volume of
1
16.05 g mol
any gas is identic al at
Molar volume of an ideal gas
1
32.00 g mol
70.90 g mol
a given temperature and
compared with a soccer ball
pressure
Worked example 2
3
A
2.00 dm
sample
of
an
unknown
gas
at
STP
has
a
mass
of
2.47 g.
1
Determine
the
molar
mass,
in
g mol
,
of
the
gas.
Solution
3
V
n
=
2.00 dm
=
=
3
V
m
m
M
=
2.47 g
=
n
0.0881 mol
1
mol
22.7dm
1
=
28.0 g mol
0.0881 mol
Practice question
–1
2.
Determine
the
molar
mass,
in
g mol
,
of
an
elemental
gaseous substance
–3
that has a density of 3.12 g dm
at STP. Identify the substance if its molecules
are diatomic.
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Structure
1
Models
of
the
particulate
nature
of
matter
Hypotheses
Amedeo
contain
and
Avogadro
equal
pressure.
postulated
numbers
This
of
that
particles
bec ame
equal
under
known
as
volumes
the
of
same
Avogadro’
s
dierent gases would
conditions
of
temperature
hypothesis.
A hypothesis is a tentative and falsiable explanation or description of a
phenomenon,
be
used
What
to
test
from
the
predictions
which
predictions
c an
be
deduced.
Predictions
natural
c an then
hypothesis.
might
be
derived
from
Avogadro’s
hypothesis?
Experimental determination of the molar mass of a gas
The
ideal
gas
molar
mass
of
under
known
equation
a
gas
by
c an
be
used to determine the
collecting
a
known
Instructions
volume of it
1.
conditions
of
temperature
and
Measure
ambient
Alternatively,
this
practic al
you
will
of
butane
you
c an
with
search
a
barometer.
loc al
weather
data
for
experimentally determine the molar
atmospheric
mass
pressure
pressure. In
found
in
disposable
plastic
pressure
in
your
geographic
loc ation on
lighters.
the
day
you
do
the
experiment.
Relevant skills
2.
•
Tool
3:
range
Record
to
an
uncertainties
appropriate
in
measurements as a
in
precision
and
the
Tool
•
Inquiry
3:
Fill
the
measuring
C alculate
and
2:
accuracy
Assess
interpret
and
3:
Identify
systematic
and
percentage
Inquiry
3:
and
discuss
random
Evaluate
Measure the
the
sources and impacts of
implic ations
limitations
it
in
the
cylinder
to
the
brim
with
water
trough so that its mouth is under
If
done
clamp
be
full
correctly,
of
the
measuring
cylinder
water. Hold it in this position with a
(figure 7).
error.
of
Submerge
and
the
lighter
in
water, then take it out again
methodologic al
and
weaknesses,
water.
error.
4.
•
with
water.
precision.
should
Inquiry
trough
the
propagate
water.
•
of
processed data.
andinvert
•
plastic
temperature
3.
uncertainties
Half-fill
dry
it
thoroughly
with
a
paper
towel.
Weigh the
assumptions on
lighter.
conclusions.
5.
•
Inquiry
3:
Explain
realistic
and
relevant
Hold
an
the
lighter
under
water
and
press
the
button
improvements
on
to
the
lighter
to
release the gas so that it bubbles up
investigation.
inside the measuring cylinder (figure 7). Continue until
S afety
3
you
eye
have
exact
•
Wear
protection.
•
Butane gas is flammable. Keep away from open flames
collected
around
100 cm
of
gas.
Record the
volume.
6.
Release
7.
Dry
8.
If
the
gas
in
a
well-ventilated
area.
and sparks.
the
lighter
as
thoroughly
as
possible and
Materials
reweighit.
•
Disposable plastic lighter
•
L arge
container,
for
•
100 cm
•
Balance (±0.01 g)
•
Clamp and stand
•
Thermometer
example
a
large
plastic
you
have
time,
repeat
to
get
three
sets
of
results.
trough
3
measuring
cylinder
3
100 cm
measuring
cylinder
•
Barometer
(if
available)
water

Figure 7
Experiment
apparatus
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Structure
1.5
Ideal gases
Questions
1.
Design
a
suitable
results
table
for
8.
your data.
Suggest
realistic
determining
2.
Process
the
your
molar
data
mass
of
to
obtain
an
experimental
value
9.
The
the
Propagate the uncertainties.
4.
Compare
value
by
5.
Assess
6.
Discuss
your
experimental
c alculating
the
the
accuracy
value
to
percentage
and
precision
the
theoretic al
error.
of
the
relative
impacts
of
pressure
sum
your
of
Comment
on
the
vapour
pressure
adjust
your
of
data
What
additional
need
to
this
method
for
the
measuring
pressure
butane.
processing
data
and
to
of
cylinder is in fact
water and the
How
could
account
information
for
do
you
this?
you
research?
systematic
and
Consider
alternative
methods
for determining the
random
mass of a gas that could be done in a school
results.
laboratory.
7.
inside
the
partial
molar
on
to
mass of a gas.
your data.
10.
errors
improvements
molar
for
butane.
3.
the
at
least
two
major
If
you
have
time,
show
your
ideas
to
your
sources of
teacher and try them out.
experimental
error.
Linking question
Graphs
data
c an
be
points.
presented
What
representation?
are
the
(Tools
2
as
sketches
advantages
and
3,
or
as
and
accurately
limitations
plotted
of
each
Reactivity 2.2)
Pressure, volume, temperature and amount
of an ideal gas (Structure 1.5.4)
There
are
four
variables
of
ideal
1.
The
pressure
2.
The
volume the gas occupies,
3.
The
absolute
4.
The amount of the gas,
The
eect
keeping
up
with
of
the
any
changed.
of
two
He
of
two
law:
a
gas
by the gas,
that
he
gas
these
V
variables
This
performed
observed
each other:
T
n
constant.
were
aect
p
temperature of the gas,
other
Boyle’s
temperature
exerted
an
kept
that
the
an
is
on
each
what
other
Robert
experiment
constant,
but
pressure
and
the
c an
Boyle
be
investigated
did
when
he
by
c ame
where the amount and the
volume
volume
of
of
the
container
the
gas
were
was
inversely
proportional.
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Structure
1
Models
of
the
particulate
nature
of
matter
Graphing the gas laws
Online
simulations
allow
relationships
between
temperature,
and
amount
ideal
of
gas
gas.
In
data
about
and
direct
to
easily
and
and
temperature
this
you
will
task,
which
analysis
inverse
will
collect
allow
skills,
as
3.
explore the
volume,
pressure
simulation,
spreadsheet
you
pressure
you
well
data
to
as
for
a
for
fixed
a
five
from an
the
simulation,
pressure
certain
amount
different
vary
and
the
record
of
gas.
temperatures
temperature at a
the
resulting
Collect
in
a
data
suitable
volume
for
at
table
least
in
your
spreadsheet.
practice
reinforce
Using
constant
volume and
ideas
4.
Compute the temperature values in both °C and K.
5.
Construct
proportionality.
two
graphs of
V
vs
T; one with
T
in °C and
Relevant skills
the other with
•
Tool
2:
Generate
•
Tool
2:
Use
data
in K.
from simulations.
6.
Using
the
simulation,
vary
the
volume at a constant
spreadsheets to manipulate data.
temperature
3:
T
•
Tool
Understand
direct
and
inverse
•
Inquiry 1: Identify dependent, independent and
a
proportionality.
certain
five
and
amount
different
record
the
of
Collect
gas.
volumes
in
a
resulting
data
suitable
pressure
for
table
at
in
for
least
your
spreadsheet.
controlled
variables.
7.
Construct
8.
Use
a
graph of
p
vs
V
Materials
1
•
Simulation
that
allows
you
to
change
your
spreadsheet
to
compute
values
for
.
pressure,
V
volume
and
temperature
for
an
ideal
gas.
It
must
1
have
9.
an
option
to
hold
one
variable
constant
and
Construct
a
graph of
p
vs
V
vary the
Questions
other
•
two.
Spreadsheet
1.
software
What
were
variables
your dependent and independent
in
each
c ase?
Which
variables
were
Instructions
controlled?
1.
Using
the
simulation,
vary
the
temperature at a
2.
constant
volume
and
record
the
resulting
Describe
a
certain
amount
of
gas.
Collect
data
the
relationship
shown
in
each
graph
pressure
as
for
for
at
direct
proportionality,
inverse
proportionality,
least
orother.
five
different
temperatures
in
a
suitable
table
in
your
3.
spreadsheet.
When
studying
temperature
2.
Construct
a
graph of
p
vs
gases,
values
it
into
is
SI
important
units
to
convert all
(kelvin).
Discuss
why
T.
this
is
the
volume
c ase
units
for
c an
temperature,
vary
whereas
depending
on
the
pressure and
source.
The combined gas law
We
have
seen
proportional
that
to
pressure
absolute
is
inversely
proportional
to
volume
and
directly
temperature.
1
p
∝
;
p
∝
T
V
Combining
the
two
relationships
gives:
pV
pV ∝ T
= k (a
or
constant) or
T
p
V
p
1
V
2
1
2
=
T
T
1
This
equation
is
known as the
2
combined gas law
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Structure
1.5
Ideal gases
Experiments
The
gas
laws
variables
arose
were
manipulated.
from
experiments in which certain
controlled,
Inspect
the
while
others
apparatus
were
shown
c arefully
in
gure8.
pressure gauge
What
might
is
independent
the
be
explored
with
variable?
this
set-up?
What
thermometer
What
variables must
250
be
controlled?
What
is
the
purpose
of
mL
round-bottomed flask
each of the
containing air
itemsdepicted?
water bath
u
Figure 8
experiment
Apparatus for conducting an
into the behaviour of a gas
Worked example 3
3
A
of
weather
balloon
100.0 kPa
is
lled
released
altitude
of
35,000 m,
–50 °C.
C alculate
with
at
32.0 dm
sea
where
level.
the
of
The
helium
balloon
pressure
is
at
25 °C
and
eventually
475 Pa
and
the
a
pressure
reaches
an
temperature
is
3
the
volume,
in
m
,
of
the
gas
in
the
balloon
under
these
conditions.
Solution
List
the
conditions
of
the
gas
in
the
p
weather balloon.
=
100.0 kPa
=
32.0 dm
1
3
V
1
Remember
to
convert
temperature to kelvin:
T
=
25
+
273.15
=
298.15 K
1
Then,
list
the
Remember
conditions
to
make
of
sure
the
the
gas
units
in
the
weather
balloon
at
35,000 m.
are consistent with the initial conditions of
the balloon.
p
=
0.475kPa
2
V
=
unknown
2
T
= –50
+
273.15
=
223.15 K
2
Substitute
the
numbers
into
the
combined
p
p
V
1
gas
law:
V
2
1
2
=
T
T
3
100.0 kPa
×
Practice question
2
1
0.475kPa
32.0 dm
×
V
2
3.
=
Rearranging
the
A
sample
of
an
a
volume
of
1.00 dm
ideal gas has
223.15 K
298.15 K
expression in terms of
V
3
at
STP.
3
gives:
C alculate
the
volume, in dm
,
2
3
V
≈
5.04
×
10
3
dm
3
=
of
that
sample
at
50.0 °C and
5.04 m
2
50.0 kPa.
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Structure
1
Models
of
the
particulate
nature
of
matter
TOK
Throughout
were
of
this
chapter
developed
alpha
particles
you
through
with
have
explored
observations
gold
atoms,
to
of
the
models
the
related
natural
world
manipulation
of
to
or
the
particulate
obtained
gases
in
the
nature
through
gas
laws,
of
matter.
M any of these concepts
experimentation:
to
from
the
interaction
explorations of subatomic particles
atCERN.

Figure 9
The set-up used
The ATLAS detector at
by Joseph Louis Gay-Lussac to investigate the thermal expansion of gases (le)
CERN used
to investigate elementary particles (right)
How do scientists investigate the behaviour of particles that are too small to be observed directly? How have advances in
technology inuenced scientic research into what matter is made up of ?
Ideal gas equation
The
the
combined
three
gas
law
parameters,
suggests
p,
V
or
T,
that
for
aects
any
the
given gas, the change in one of
other
two
in
such
a
way that the
pV
expression
remains
constant.
The
exact
value of that constant must be
T
proportional to the amount of the gas,
pV
pV
∝ n
or
=
T
nR
T
where
is
n:
R is the
universal gas constant, or simply
known as the
pV
=
ideal gas equation,
which
is
gas constant.
traditionally
The
written
as
last
expression
follows:
nRT
The value and units of
R depend on the units of
p,
V, T and
n. If all four parameters
3
are
expressed
R
8.31 J K
in
–1
≈
The
same
standard SI units (p
in
Pa,
V
in m
,
T
in K
and
n in mol), then
–1
mol
.
value and units of
R
c an
be
used
if
pressure
3
volume in dm
is
expressed
in
kPa and
3
,
as
the
two
conversion
factors (10
–3
for
kPa
to
Pa and 10
3
for dm
to
3
m
)
c ancel
each other out.
Linking question
How
c an
the
experimental
ideal
gas
data?
law
be
used
to
(Tool 1, Inquiry 2)
c alculate
the
molar
mass
of
a
gas
from
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Structure
1.5
Ideal gases
Worked example 4
A 3.30 g sample of
an unknown organic compound
was vaporized at
T
=150 °C and
p =101.3 kPa to produce
3
1.91 dm
3
of a gas.
The gas was combusted
in excess oxygen to produce 3.96 g of water,
2.49 dm
of
c arbon dioxide
3
and
1.25 dm
of
nitrogen at STP.
Determine the following for the compound:
a.
molar mass
b.
empiric al formula
c.
molecular formula
Solution
a.
To
determine
the
molar
mass,
we
need
to
nd
out
the
amount
of
the
compound
using
the
ideal
gas
equation:
pV
n
=
T
=
n
=
RT
150
+
273.15
=
423.15 K
3
101.3 kPa
×
1.91 dm
≈
–1
0.0550 mol
–1
8.31 J K
mol
× 423.15 K
3.30 g
–1
Therefore,
M
=
=
60.0 g mol
0.0550 mol
b.
All
c arbon,
amounts
hydrogen
of
these
O)
nitrogen
in
atoms
c arbon
in
the
dioxide,
combustion
water
and
products
nitrogen
originate
are
the
same
we
need
from
as
the
those
organic compound, so the
in
the
original
sample:
3.96 g
m
n(H
and
elements
=
=
≈
0.220 mol
2
–1
M
n(H)
=
2
×
18.02 g mol
O)
n(H
=
2
×
0.220 mol
=
0.440 mol
2
3
2.49 dm
V
n(CO
)
=
=
≈
2
3
V
22.7 dm
0.110 mol
–1
mol
M
n(C)
=
)
n(CO
=
0.110 mol
2
3
V
n(N
)
1.25 dm
=
=
≈
0.0551 mol
=
0.110 mol
2
3
V
22.7 dm
M
n(N)
= 2
×
n(N
)
=
2
×
–1
mol
0.0551
2
The
of
original
the
three
compound
elements
could
also
(hydrogen,
contain
c arbon
oxygen.
and
To
check
nitrogen)
with
this,
the
mass
of
to
the
compare
original
the
total
mass
sample:
–1
m(H)
=
0.440 mol
×
1.01 g mol
≈
0.444 g
–1
m(C)
=
0.110 mol
×
12.01 g mol
m(N)
=
0.110 mol
×
14.01 g mol
≈
1.32 g
≈
1.54 g
–1
m(total)
=
0.444 g
Therefore,
the
+
1.32 g
organic
+
1.54 g
compound
≈
did
3.30 g
not
contain
oxygen,
so
its
formula
c an
be
represented as C
H
x
x : y : z
The
=
0.110 : 0.440 : 0.110
empiric al
=
N
y
.
z
1 : 4 : 1
formula of the compound is CH
N.
4
–1
c.
M(CH
N)
=
12.01
+
4
+
1.01
+
14.01
=
30.06 g mol
–1
.
This
value
is
half
the
experimental
value
(60.0 g mol
),
4
so
the
molecular
formula
of
the
compound
will
have
twice
the
number
of
atoms
of
each element: C
H
2
N
8
.
2
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Structure
1
Models
of
the
particulate
nature
of
matter
End-of-topic questions
0.58
Topic review
×
8.31
×
373
B.
3
100 ×
1.
Using
your
knowledge
from the
Structure 1.5
6
×
10
250
×
10
topic,
3
answer
the
guiding
question
as
fully
as
0.58 ×
100
0.58 ×
100
×
10
6
×
250
×
10
×
10
C.
possible:
8.31
×
100
3
How does the model of ideal gas behaviour help us to
×
10
6
×
250
D.
predict the behaviour of real gases?
8.31
Exam-style questions
6
Which
the
Multiple-choice questions
2.
Which
of
the
following
The
volume
are
graph
pressure
×
373
correctly
and
shows
volume
of
the
an
relationship
between
ideal gas, at constant
temperature?
assumptions
of
the
ideal gas
A.
B.
model?
P
P
I.
occupied
by the gas particles is
negligible
II.
There
are
no
intermolecular
forces
V
V
between
gas particles
C.
III.
Zero
particle
D.
movement
P
A.
I and II only
B.
I and III only
C.
II and III only
D.
I, II and III
The
temperature
7.
3.
of
an
ideal
gas
is
the
pressure
27 °C.
P
of
the
gas
aer
Which
is
of
volume
is
A.
162 °C
B.
450 °C
C.
1527 °C
D.
1800 °C
the
of
following
hydrogen
balloons
atoms,
at
contains
constant
the
largest
temperature
doubled and
and
the
1
V
What is the
number
temperature
1
V
pressure?
tripled?
3
4.
A
syringe
contains
What
gas
will
volume
57 °C
at
40
cm
of
an
ideal
gas
at
27 °C.
H
the
of
the
gas
be
aer
it
is
NH
(g)
2
(g)
1.9
3.6
C.
44
cm
D.
84
cm
HF(g)
3
2 dm
3
1 dm
4 dm
pressure?
8.
3
A.
B.
(g)
4
3
3
2 dm
constant
CH
3
warmed to
What
are
the
conditions
for
the
ideal
gas
behaviour of
cm
real
gases?
3
cm
3
3
5.
A
0.58
g
sample
of
an
ideal
gas
at
100
kPa
and
a
volume
expressions
0.58
of
is
×
250
equal
8.31
×
cm
to
.
Which
the
molar
of
the
mass
Low
temperature
and
low
Low
temperature
and
high
pressure.
C.
High
temperature
and
low
D.
High
temperature
and
high
pressure.
pressure.
100 °C
3
has
A.
B.
pressure.
following
of
the
gas?
100
A.
3
100 ×
10
6
×
250
×
10
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Structure
9.
Which
are
of
the
following
statements
about
an
14.
ideal gas
At
forms
constant
temperature,
the
several
gaseous
empiric al
compounds
p
I.
C arbon
Deduce
correct?
using
and
the
compounds
molecular
data
from
the
1.5
with
formulas
table
Ideal gases
uorine.
for these
below.
[3]
= constant
V
3
C arbon /
M ass
of
1.00 dm
p
Compound
II.
At
constant
volume,
= constant
mass
%
at
STP / g
T
X
13.65
Y
24.02
3.88
4.41
Z
17
.40
6.08
V
III.
At
constant
pressure,
= constant
T
A.
I and II only
B.
I and III only
C.
II and III only
D.
I, II and III
15.
An
organic compound
9.1%
of
hydrogen
vaporized
and
sample of
A
contains
36.4%
A
with
a
of
54.5%
oxygen
mass
of
of
by
c arbon,
mass. A
0.230g occupies
3
a
volume
of
at
0.0785 dm
T=95 °C and
p=102 kPa.
Extended-response questions
10.
Explain,
from
in
ideal
your
own
words,
behaviour
at
low
why
real gases deviate
a.
Determine
the
empiric al
b.
Determine
the
relative
formula of
c.
Using your answers to parts a and b, determine the
molecular
A.
[2]
mass of
A.
[1]
temperatures and high
pressures.
[2]
molecular formula of A.
11.
A
c ar
tyre
inated
to
2.50 bar
(250 kPa)
at
[1]
10 °C contains
16.
A
closed
steel
cylinder
contains
0.32 mol
of
hydrogen
3
12.0 dm
of
compressed
air.
Aer
a
long
journey, the
gas
tyre
temperature
increases
to
25 °C
and
the
and
0.16 mol
of
oxygen
gas.
The
volume of the
pressure
3
cylinder
to
261 kPa.
Determine
the
tyre
mixture
conditions.
Assume
that
is
25 dm
there
was
no
air
is
[2]
a.
C alculate
mixture
Ammonium
c arbonate, (NH
)
4
CO
2
when
(s),
the
)
4
initial
temperature of the gas
pressure,
in
the
cylinder.
the
gas
mixture
in
kPa, of the gas
[1]
decomposes
When
is
ignited,
both
reactants
are
heated:
consumed
(NH
initial
3
b.
readily
the
25 °C.
loss during the
journey.
12.
and
volume under these
CO
2
(s)
→
2NH
3
(g)
+
CO
3
(g) + H
2
the
O(l)
completely
cylinder
rises
to
and
800
the
°C.
temperature inside
C alculate
the
pressure
2
inside
the
cylinder
at
that
moment.
[2]
3
Determine
the
volumes, in dm
at
STP, of the individual
–3
gases
produced
ammonium
on
decomposition
of
17.
2.25 g of
c arbonate.
[2]
An
unknown gas
4.00 g
sample of
X
has
X
was
a
density
of
2.82 g dm
combusted
in
excess
at
STP. A
oxygen to
3
produce
13.
The
gases
produced
in
question12
were
2.50 g
of
hydrogen
uoride
and
2.84 dm
of
transferred to a
c arbon
dioxide
at
STP.
3
sealed
vessel
with
a
volume
of
1.50 dm
,
and
the
vessel
Determine
was
heated
vessel
react
at
up
that
with
to
200 °C.
C alculate
temperature.
each
other.
the
the
following
for
X:
pressure in the
Assume that the gases do not
a.
molar
b.
empiric al
mass
c.
molecular
[1]
[2]
formula
formula
[2]
[1]
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