Download 2. CHEMICAL ACTIVITY of the METALS 3. PATTERNS of the

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Preliminary Chemistry Topic 2
METALS
What is this topic about?
To keep it as simple as possible, (K.I.S.S.) this topic involves the study of:
1. OUR USE of METALS
2. CHEMICAL ACTIVITY of the METALS
3. PATTERNS of the PERIODIC TABLE
4. QUANTITY CALCULATIONS... the MOLE
5. METALS from their ORES
...all in the context of how Chemistry contributes to cultural development
but first, an introduction...
Technology Needs Metals
Chemistry of the Metals
The great sweep of human cultural development has many
aspects... Language, Religion, Art & Music, and, of course,
Technology.
In the previous topic you learnt about the Elements of the
Periodic Table. In this topic you will concentrate on the
chemistry of the metals, and some of the chemical patterns
that they show.
... and Speaking of Patterns,
in this topic you will find that
The history of technology is closely linked with our use of
metals; in fact historians have named some parts of history
after the metals that changed the way people lived.
The Periodic Table
Sword from the “Bronze Age”
is full of patterns
s
l
a
t
Me
This topic starts with a quick look at the history of metal
use, and ends with a study of how we get metals from the
Earth, and the chemistry of the extraction process.
No
nMe
tal
s
Measuring Chemical Quantities
In this topic you will also be introduced to the concept of
the “Mole”...
not a burrowing mammal!
not a traitor within the group!
not a gangster’s girlfriend!
certainly not a skin blemish!
A Chemical Mole is a clever way to measure quantities;
...essential for analysis & chemical manufacture.
Electrically powered smelter plant
for extracting
Aluminium from its ore
If you know the mass, you can
figure out how many atoms
there are...
thanks to the mole.
Photo courtesy of Comalco Aluminium Ltd
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CONCEPT DIAGRAM (“Mind Map”) OF TOPIC
Some students find that memorizing the OUTLINE of a topic helps them learn and remember the concepts and
important facts. As you proceed through the topic, come back to this page regularly to see how each bit fits the
whole. At the end of the notes you will find a blank version of this “Mind Map” to practise on.
Metals
We Use
Today
The Activity Series
of the Metals
Metal Reaction with
• Oxygen
• Water
• Acids
Metal
Extraction
Needs Energy
History of
Metal Use
Electron Transfer
REDOX
1st Ionization Energy
Chemical Activity
of the
Metals
Our Use of
Metals
Activity & Usage
of Metals
History of the
Periodic Table
Patterns
of the
Periodic Table
METALS
Patterns in
Extracting
Metals
from
Ores
Quantity
Calculations
the Mole
Minerals
Ores
&
Resources
Definition
of the Mole
Avogadro’s
Number
The Case for Recycling
Metals
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Conductivity
Melting Points
Chemical Bonding
Valency
Reactivity
Atomic Radius
1st Ionization
Energy
• Electronegativity
Molar Ratios in
Reactions
Empirical
Formulas
Case Study:
Extracting
Copper
from its Ore
•
•
•
•
•
•
•
Gay-L
Lussac’s Law
&
Avogadro’s Hypothesis
2
Mole Quantity
Calculations
• Masses
• Gas Volumes
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1. OUR USE OF METALS
For most of human existence, people used tools of stone,
wood and bone. Primitive tribes were familiar with gold
which occurs uncombined in nature, but it is too soft to be
useful for anything but jewellery and decoration.
The Iron Age (approx. 2,500 to 1,500 years ago)
About 1,000 B.C. the extraction of iron from its ores was
discovered. This requires much higher temperatures, and
the breakthrough was probably the invention of the
bellows, a device to pump air into a furnace so the wood or
charcoal burns hotter.
About 5,000 years ago, in the Middle East, some people
accidentally discovered that if certain rocks were roasted by
fire, small amounts of copper would be found later in the
ashes. Copper is too soft to be really useful, but there was
a brief “Copper Age” around the eastern end of the
Mediterranean Sea. Copper was used for decoration,
jewellery, small utensils, and occasionally for knives and
spear points.
Iron is stronger and harder than bronze. A warrior armed
with iron weapons will usually beat a bronze-armed man.
Iron tools and even the humble nail allowed new
developments in buildings, ships, wagons... remember that
towns, trade and commerce give wealth and power. An iron
plough allows more land to be cultivated to grow more
food, to feed a bigger army... and so on.
The First Uses of Metals
It is no accident that the dominant world power of this
time was ancient Rome, because their technology was
based on iron.
The big breakthrough was the discovery by these copperusing people that if they roasted copper-bearing rocks
(ores) with tin ores, the resulting “alloy” (mixture) of
copper and tin produced a much harder metal, “bronze”,
which could be cast in moulds, and hammered to shape
many useful tools and weapons... this was the start of
From the Medieval to the Modern
After the collapse of the Roman Empire the various
cultures that dominated the “Dark Ages” still had ironbased technologies.
The Bronze Age (approx 4,500 to 2,500 years ago)
It is no accident that the rise of the great ancient
civilizations occurred about this time. The stone blocks of
the pyramids and temples of ancient Egypt were cut and
shaped with bronze chisels. Egyptians, and later Greeks,
dominated their world because their soldiers were armed
with bronze swords, spears and arrowheads.
The next great technological change was the “Industrial
Revolution” which began about 1750 in England. This
had many aspects, but the big change in technology was the
use of coal (instead of wood) for fuel. As well as steam
engines, coal allowed for large scale smelting of iron and
the invention of steel (an alloy of iron with carbon).
With bronze tools they built better ships and wagons for
transport and trade, which brought wealth and power.
Photo by Jop Quirindongo
Photo: Keith Syvinski
The engines, tools and machinery of the great factories
were based on steel. Transport was revolutionized by steel
locomotives running on steel rails. Steel ships replaced
wooden ones, and steel weapons (machine guns, tanks and
artillery) achieved new heights (depths?) in warfare and
mass destruction.
Photo:
Arian
Kulp
In the 20th century, new metals and alloys became available...
aluminium, titanium, chromium, and many more.
Sad as it might be, the facts of human history are that
progress has been marked by conflict, war and conquest,
and metals have been a vital part of that development.
This was made possible by electricity, which is needed in
large amounts to extract some metals from their ores, or to
purify and process them once extracted.
Metal has many advantages over stone, wood, or bone:
• metal is harder, stronger, and flexible, not brittle.
• metal can be cast, hammered or drawn into shapes not
possible in stone, such as saw blades, swords and armour.
• when tools become blunt, metal can be re-sharpened.
Human Progress has always been linked
to our use of Metals.
Progress in metal usage has always been linked
to the availability of energy to extract the metals.
Basically, a warrior with a bronze sword always beats a
bloke with a stone axe... we call that progress!
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The Metals We Use Today
Solder
is an alloy of 30-50% tin with lead.
In one sense, we are still in the “Iron Age”. Iron is still the
metal we use the most, but nearly always it is mixed with
other elements in a variety of alloys, notably steel.
Its most notable property is a very low melting point,
around 150-200oC.
Its major use is in plumbing for sealing the joints between
pipes, and in electronics for connecting small components
on a “circuit board”.
Metals That Are Used in Their Pure State
Although we use a wide range of alloys, there are some
important metals we use in their pure, elemental state.
Photo by Diana
Aluminium
is very lightweight, yet strong and corrosion resistant
Steel is used for bridges, tools and machinery, bolts, screws
and nails, reinforcing inside concrete structures, engines,
vehicle bodies, trains and their rails, ships, and “tin” cans.
Its lightweight strength is perfect
for aircraft construction.
Why is steel so widely used?
Lightweight and a good conductor,
it is used for electricity power lines.
• Iron ore occurs in huge deposits, so iron is common and
economical to produce.
• Steel (in its various forms) is very hard and strong.
• It can be cast, milled, rolled, worked, bent, cut and
machined into just about any shape or size imaginable.
Malleable and corrosion resistant, it
is ideal for window frames and drink cans.
As always, our usage of the different steel alloys is linked to
their particular properties:
Copper
is used for electrical wiring in buildings and appliances,
because of its great electrical conductivity and its ductility
for ease of wire-making.
Steel
Iron,
Alloy
with...
Mild steel 0.2%
carbon
strong, but
malleable
Metal Extraction Needs Energy
Tool steel
very hard
drills, knives,
hammers
resists
corrosion,
hygenic
food utensils,
medical tools
Stainless
Steel
1-1.5%
carbon
20% nickel
& chromium
Properties
Uses
car bodies,
pipes, roofing
As mentioned previously, our use of different metals
through history can be linked to the availability of energy.
In topic 1, you learned about the process of chemical
decomposition; where a compound breaks down into
simpler substances.
Decomposition is generally an endothermic process;
energy is absorbed by the reactants during the reaction.
Generally, you must supply energy to make the process
happen.
Brass
is a common “non-ferrous” (no iron) alloy.
Metal ores are mineral compounds. To obtain the elemental
metal involves decomposition, which is endothermic and
requires energy. Some compounds require more energy
than others for decomposition.
Copper and tin ores require little energy. A decent wood
fire can “smelt” the metal from its ore. This why copper
and bronze were used in ancient times.
Iron ore requires more energy for decomposition. That’s
why the “Iron Age” came later.
Brass is an alloy of copper and zinc (about 50% each)
Brass is very hard, but easily machined for screw threads, etc.
It is more expensive than steel, but is corrosion resistant, so
it is ideal for taps and fittings for water and gas pipes.
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Aluminium and other “modern” metals require even more
energy, and electricity works better than heat, so these only
became available in quite recent times.
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Worksheet 1
Today, the metal we use most is still t)....................,
in the form of the alloy u)................................. Its
widespread use is because:
• it is common and v)..................................... to
produce.
• it is very w)........................ and ................................
Steel comes in a variety of alloys, including
x)....................... steel (car bodies, pipes, roofing)
and y)................................. steel used for food
utensils and medical tools.
Fill in the blanks. Check answers at the back.
Before metals, people used tools mainly made
from a)............................. or .................................
The first metal used was probably
b)................................., because it occurs in the
elemental state in nature. However, it is too soft
to be used for tools, so was just used for
c)................................
Other alloys used widely include:
• brass, a mixture of z).................... and ...................
• aa)...................................., with a very low melting
point, is an alloy of ab).................................. and
.................................... and is used in
ac)..................................... and ....................................
Metallurgy (the technology of metals) began with
the extraction of d)................................. from ores
that were simply e)............................................
.............................................
A big improvement was the mixing of ores of
f)............................... and ...........................................
This produced the alloy g)......................................,
which made tools and weapons with many
advantages over stone:
• metal is h)............................ and ............................
and is not i)........................................ like stone
• metal can made into intricate shapes, such as
j)......................................., not possible in stone.
As well as many alloys, there are some metals
commonly used in their pure, elemental form:
• Aluminium, which has the advantages of being
ad)................................. and resistant to
ae).......................................
Uses include af).....................................................
and ..................................................
• ag)....................................... is used for electrical
wiring
because
of
its
good
ah)............................................ and because it is
ai)................................ so it is easy to draw out into
wires.
Later, bronze was replaced by k)..............................
which is l)............................... and.............................,
but requires more m).......................................... for
its extraction.
During the “Industrial Revolution”, the use of
n)................. for energy led to the production of
o)............................... which is iron with a small
amount of p)................................ in it. This
allowed the development of machinery, trains
and the modern industrial world.
Chemically, the extraction of metals from ores
involves
aj)..................................................
reactions, which are ak)................-thermic. Some
metals, such as al).................................... require
very
little
energy,
others
such
as
am)...................................................... require much
more. In many cases an)......................................
works better than heat in the extraction and
purification processes. The changes in
ao)............................ usage through history can be
directly linked to society’s changing sources and
uses of ap)......................................
In the 20th century new metals such as
q).............................. became available because the
r).................................. needed to extract it from its
s)................... was available.
WHEN COMPLETED, WORKSHEETS
BECOME SECTION SUMMARIES
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2. CHEMICAL ACTIVITY OF THE METALS
Metals React With Acids
Metals React With Oxygen
One of the most familiar laboratory reactions is the
burning of magnesium:
The different activity levels of the metals is most clearly
seen when metals are reacted with dilute acids.
Magnesium + Oxygen
2 Mg
+ O2
You may have done experimental work to observe how
vigorously different metals react with a dilute acid.
Magnesium oxide
2 MgO
In fact, many metals will burn, some a lot more readily and
violently than magnesium:
Sodium + Oxygen
4 Na
+ O2
Metals like calcium and
magnesium react vigorously.
Sodium oxide
2 Na2O
Zinc and iron are slower.
In these cases there is a violent exothermic reaction, with
light and heat energy produced. The product is often a
powdery, crumbly solid.
Lead is very slow indeed.
Other metals, such as aluminium and zinc, react on the
surface and the oxide compound formed is airtight and
prevents further reaction. That’s why these metals are often
dull-looking... the surface coat of oxide is dull.
When there is a reaction, the
gas produced is hydrogen.
Aluminium + Oxygen
2 Al
+ 3 O2
Copper does not react at all.
A flame test
The metal is “eaten away” and gives a “pop”
dissolves into the liquid. This is explosion
because it forms a soluble ionic
compound. Exactly what the
compound is, depends on which acid is used.
Aluminium oxide
2 Al2O3
Other metals, such as copper, react with oxygen very slowly
and only if heated strongly. Some, like gold, will not react
at all.
Examples:
Zinc + Hydrochloric
acid
Zn + 2 HCl
The point is, that
metals have different chemical activities.
Magnesium + Nitric
acid
Mg + 2 HNO3
Metals React With Water
Another favourite school reaction is when sodium reacts
with water. This is often done outdoors, because it results
in an exciting little explosion.
Iron
What happens is:
Sodium +
2 Na
Water
+ 2 H2O
Hydrogen + Sodium
(gas)
hydroxide
H2
+ 2 NaOH
Sulfuric
acid
Fe + H2SO4
Metal + Acid
Hydrogen + Magnesium
nitrate
H2 + Mg(NO3)2
Hydrogen
H2
+ Iron(II)
sulfate
+ FeSO4
Hydrogen + a Salt
It will help you greatly to learn
the common laboratory acids
Common Name
Chem Name
Hydrochloric
= Hydrogen chloride
Sulfuric
= Hydrogen Sulfate
Nitric
= Hydrogen nitrate
Once again, some metals react easily and rapidly and form
the metal hydroxide, while others react slowly if heated in
steam, and form oxides.
Hydrogen + Zinc oxide
H2 + ZnO
Metals like copper and gold do not react at all.
There is an “Activity Series” among the metals.
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+
Hydrogen + Zinc
chloride
H2 + ZnCl2
The ionic compounds formed are collectively known as
“salts”, so the general pattern of the reactions is
(In fact this is NOT the explosion reaction. The explosion
is the reaction of the hydrogen with oxygen, to form water)
Zinc + Water
Zn + H2O
Bubbles of
gas are
produced.
Formula
HCl
H2SO4
HNO3
Try the WORKSHEET at end of section
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Electron Transfer in Metal Reactions
The Activity Series of the Metals
The chemical reactions that allow us to see the pattern of
the Activity Series are just part of an even greater pattern
in Chemistry... the process of electron transfer.
From these 3 patterns of reaction, it seems there is a
further, underlying pattern. Certain metals, like sodium,
always seem to react readily and vigorously. Others, like
copper, always react slowly or not at all.
To understand this, look again at the reaction between a
metal and an acid:
From this, and other reaction studies, the common
laboratory metals can be arranged in an “Activity Series”:
Most
Active
Zinc + Hydrochloric
acid
K
Zn
+
Hydrogen + Zinc
(gas)
chloride
2 HCl
H2
Na
Li
Zn + 2H+ + 2Cl-
Ca
Mg
Notice that the chloride ions (Cl ) occur on both sides of
the equation unchanged. Nothing has happened to them at
all. We say they are “spectator ions”. Like by-standers at a
car crash they are not involved, while other atoms and ions
undergo serious changes.
Zn
Fe
Sn
Pb
Since they aren’t actually involved, we can leave the
spectators out. This is called a “net equation”.
Cu
Zn + 2H+
Ag
Au
H2 + Zn+2
Now we can see what really happened;
• a zinc atom became a zinc ion
and
• 2 hydrogen ions became a (covalent) hydrogen
molecule.
If you look for these metals on the Periodic Table you will
notice a further pattern.
2 6
H2 + Zn+2 + 2Cl-
Study this carefully and make sure you understand why
there have to be 2 of some ions to agree with the original
balanced equation.
Al
3
ZnCl2
HCl and ZnCl2 are both ionic compounds. Here is the
equation re-written to show the individual ion “species”.
Ba
Least
Active
+
Positions of the first 6 metals
of the Activity
Series.
To do this, the zinc atom has to lose 2 electrons, and the
hydrogen ions must gain a pair of electrons to share.
1 5
4
The highly active metals all lie to the extreme left of the
table, AND the higher their activity, the lower down the
table they are within each column.
Zn+2 + 2e-
2H+ + 2e-
H2
Now it should be clear what really happened: the zinc atom
gave a pair of electrons to some hydrogen ions. Electrons
were transferred from one “species” to another.
This is one of many patterns that allows you to use the
Periodic Table instead of learning many small facts. For
example, instead of memorizing the Activity Series fully,
you can remember the pattern above and always be able to
figure out the order of the most active metals.
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Zn
The equations above are “Half-Equations” and are often
used to describe what is really happening in a reaction.
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Oxidation and Reduction
First Ionization Energy
Although you’re not yet required to know about
Oxidation and Reduction, this bit you have to learn.
The transfer of electrons from one species to another is
one of the most fundamental and important general
reactions of Chemistry.
Definition
The Ionization Energy of an element is the energy
required to remove an electron from an atom.
The reaction between zinc and acid can be visualized like
this:
electrons transferred
+
Zinc atom
For technical reasons, the measurement of this energy
is carried out for atoms in the gas state.
+
Zn(g)
2 Hydrogen ions
Zn+(g) + e-
The energy required for this to happen is the
“1st Ionization Energy”
Hydrogen molecule
We know that zinc atoms normally lose 2 electrons to
form the Zn+2 ion. However, the formal definition for
this process involves just the loss of 1 electron.
+2
Every element has its own characteristic value, even
those elements which would not normally lose
electrons, such as non-metals like chlorine.
Covalent bond
(2 electrons being shared)
The zinc atom has lost 2 electrons,
Cl(g)
Zn
Zn+2 + 2e-
Cl+(g) + e-
Normally a chlorine atom forms a negative ion
by gaining an electron.
Technically though, it is possible for it to lose
an electron if energy is added.
This energy is the “1st Ionization Energy”
For historical reasons,
the loss of electrons is called “Oxidation”
and the hydrogen ions have gained electrons.
2H+ + 2e-
Even the inert gases, which normally do not form ions
at all, can be forced to lose an electron if energy is
added. They too have a 1st Ionization Energy value.
H2
The gain of electrons is called “Reduction”
Ionization Energy Determines the Activity Series
Now back to the metals and the Activity Series.
Neither process can occur alone... they must occur together
In order for a metal to begin reacting with an acid, (or
with water or oxygen) it must lose an electron. This will
require the input of its 1st Ionization Energy.
The zinc oxidation allows the hydrogen to be reduced, and
the hydrogen reduction allows the zinc to be oxidized.
The total reaction is an “Oxidation-Reduction” and is
commonly abbreviated to “REDOX”.
If the value for 1st Ionization energy is very low, the
metal will gain this energy easily and quickly from its
surroundings. It will readily enter the reaction, and the
reaction will proceed vigorously.
Note that the syllabus does NOT require you to know
these definitions yet, but it is worth knowing about Redox
for future topics. You ARE required to know about
electron transfer and its involvement in metal reactions.
If its value for 1st Ionization energy is higher, the atom
cannot react so readily or vigorously... its activity is
lower.
Try the WORKSHEET at end of section
The ACTIVITY SERIES of the Metals
is determined by
1st IONIZATION ENERGY
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Increasing values for 1st Ionization Energy
Zinc ion
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K
Na
Li
Ba
Ca
Mg
Al
Zn
Fe
Sn
Pb
Cu
Ag
Au
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Choice of Metals Based on Activity
Another example is the choice of metals for water pipes.
Sometimes which metal is chosen for a particular
application is based on its position in the Activity Series.
Steel is cheap, but since iron
is about the middle of the
Activity Series it will
corrode (rust) by contact
with water. Is it better to
choose a lower activity metal
such as copper, which will
not corrode as quickly, but is
more expensive?
Example
In critical electronic connections, such as computer
network plugs, it is essential that the electric signals get
through without loss or distortion.
Normally we use copper for electrical wiring, but in a
critical connection plug it is worth the extra expense of
using gold.
The decision is usually to
use cheap steel pipes for
longer, outdoor uses like
your garden taps.
Copper is a low activity metal, but can slowly react with
oxygen to form a non-conducting oxide layer in the
connection. Gold is lower down the activity series and will
not react at all, so the plug connection cannot corrode.
Brass fittings
Copper pipe
Indoors, where distances are
shorter, and a rusted-out
leaking pipe would be a
disaster inside a wall or
ceiling, copper is chosen,
especially for hot water
supply.
Gold’s extremely low chemical activity (due to a relatively
high 1st Ionization Energy) is part of the reason it has
always been used for jewellery.
Interestingly, sometimes the higher activity metals corrode
less. Aluminium and zinc are higher up the Activity Series
than iron. They react rapidly when exposed to oxygen, but
the surface layer of oxide is airtight and waterproof, and
prevents oxygen or water getting to the metal underneath.
Therefore, these metals can be used in situations where
corrosion needs to be prevented.
“Galvanized” steel is coated with a thin layer of zinc to
prevent (or slow down) corrosion of steel roofing, fence
wires, nails, bolts, etc.
Gold’s low activity means it will not tarnish or corrode, so
it retains its beautiful colour and lustre.
Metals in Art and Religion
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Worksheet 2
Part B Practice Problems
Part A Fill in the blanks. Check answers at the back.
1. Write a balanced, symbol equation for the reaction of
each of the following metals with oxygen.
When a metal reacts with oxygen it forms an a)......................
compound.
METAL + OXYGEN
a) Lead
(assume lead(IV) ion forms)
b)......................................
b) Iron
(Assume iron(III) ion)
Some metals will also react with water, forming
c)..................................... gas and a d)......................................
compound.
METAL + WATER
c) Lithium
c)..................... + d).................
2. a) Arrange the metals in Q1 in order of decreasing
chemical activity.
b) Which one(s), if any, might ignite easily and burn in
air with a visible flame?
Most metals will react with acids, forming e)..........................
gas and an ionic compound called a “f).................................”
METAL + ACID
e)....................... + f).................
3. Write a word equation AND a balanced, symbol
equation to describe the reaction of:
In all these reactions the various metals react at
g)................................... rates, showing an order of chemical
h)...................................... From these reactions and others,
the “Activity Series” has been determined.
a) calcium metal with water (reacts spontaneously at room
temperature)
Metals such as i)...................................... and .............................
are the most active. These are the elements located in the
j)........................... columns of the Periodic Table.
b) Tin metal with water (heated in steam) (Assume tin(II))
Some metals such as k)............................. and .........................
have very low activity, and often do not react at all. Other
common metals like l)............................................. and
.................................... are in the middle of the series. They
will react, but generally do so m)............................................
4. All the following equations are Metal + Acid reactions.
Fill in all blank spaces, then re-write in symbols and
balance.
All these reactions involve the transfer of n).........................
In the case of the Metal + Acid reaction, the metal atoms
always o)........................... electron(s) while a pair of
p)................................ ions gain 2 electrons (which they
share in a q)........................................ bond) and form a
r)........................................ molecule with formula s)...............
a) Zinc + Sulfuric acid
....................... +.........................
b) Calcium + Hydrochloric
acid
“Oxidation” is the technical term for t)...................................
................................. The opposite is “u)...................................
In the Metal + Acid reaction, the metal is always
v).............................................. while w).............................. ions
are always x)..................................................
........................ +......................
c)....................... +...........................
d).................... + .........................
Hydrogen + Barium
nitrate
Hydrogen + iron(II)
chloride
The “1st y).................................... Energy” of an element is
defined as the energy required to z).........................................
............................... from atoms in the aa)....................... state.
The very active metals are like that because they have very
ab)................................... (high/low) values for this. Metals
further down the series do not react as vigorously because
their values are ac)...................................................
5. For each of the reactions in Q4, which chemical species
a) lost electrons?
b) gained electrons?
c) was a “spectator”?
Sometimes the choice of which metal to use is determined
by the activity level. An example is ad)....................................
.......................................................................................................
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3. PATTERNS OF THE PERIODIC TABLE
History of the Periodic Table
Atomic Structure, Number and Mass
The modern concept of a chemical element as a substance
containing identical atoms was first accepted almost exactly
200 years ago.
Here is a quick reminder of some basics about atoms you
need to know:
In the Nucleus are
Protons &
Neutrons
By 1830 there were about 40 known elements. Even with
such a small sample, people began to notice patterns:
Dobereiner (German) pointed out that there were several
groups of 3 elements with remarkably similar properties:
Lithium, sodium & potassium was one “triad”.
Chlorine, bromine and iodine formed another “triad”.
In orbit around
the nucleus are
the Electrons
By 1860, with over 60 known elements, Newlands
(English) proposed a “Law of Octaves”.
No.Electrons = No.Protons = “Atomic Number”
If the elements were arranged in order of relative weights,
Newlands found that every 8th element (an “octave”) was
similar in properties. These similar elements included
Dobereiner’s triads.
Each element’s atoms have a different, characteristic,
number of protons and electrons. Therefore, each element
has a different Atomic Number.
In the Periodic Table the elements are arranged in order of
Atomic Number.
The system worked well for the first 20 elements, but then
became confused.
No.Protons + No.Neutrons = “Mass Number”
(Electron mass is insignificant)
The Mass Number is always a whole number, but in the
Periodic Table the “Atomic Weight” is shown instead.
(How and why this is different will be explained in a later
topic)
The basis of the modern Periodic Table was developed by
the Russian, Dmitri Mendeleev in 1869.
Mendeleev used many physical and chemical properties:
• atomic weight
• density
• melting point
• formula of oxide compound
• density of oxide
and many more,
and arranged the elements in order of weight, but with
elements with similar properties under each other.
The Periodic Table
is firstly a list of the elements, arranged in order, and
showing all the basic details.
Atomic Number
Equal to the number of electrons
and the number of protons in
each atom
18
Ar
Argon
39.95
Similar elements placed in
vertical columns
Inert Gases had NOT
been discovered
Chemical Symbol
Mendeleev’s vertical “families”
included Dobereiner’s “triads”
and Newland’s “octaves”, but
had one big difference...
Element Name
“Atomic Weight”
NOT the “Mass Number”
Mendeleev’s genius was to realize that there were probably
missing elements that hadn’t been discovered yet. He
cleverly left gaps in his table for these undiscovered
elements.
However, the Periodic Table is far more than a simple list.
Why is it such a complicated shape?
The most famous case was that of the “missing” element
Mendeleev called “eka-silicon”. He used the patterns in his
table to predict, very precisely, the properties for ekasilicon. Scientists went looking for such a substance and
soon found a new element (which was named
“Germanium”) with properties almost exactly as predicted.
The shape and arrangement of the Periodic Table is a very
clever device to allow many patterns and groupings to be
accommodated. You have already learnt one pattern in the
position of the most active metals, and their 1st Ionization
Energies.
There are lots more...
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Patterns of the Periodic Table
In Mendeleev’s day no-one could explain why these patterns existed.
However, when scientists see patterns in nature like this, they know there must be underlying “rules” or
“laws of nature” causing and controlling the patterns.
Perhaps Mendeleev’s great contribution was not just the Periodic Table itself,
but the stimulus it gave other scientists to investigate the reasons behind the patterns.
Within 40 years Science had unravelled the secrets of atomic structure, the electron energy levels, and more.
At this stage, your task is to learn some of the patterns.
Melting Point
You learned in topic 1 how melting point is determined by the
bonding within a substance.
Electrical Conductivity
As you go across any row (“period”) of the table, you will
move through a number of metals, then one or two semimetals, then into the non-metals.
Therefore, the conductivity will start out high, but rapidly
decrease as you encounter a semi-metal, and become
extremely low at the non-metals.
Semi-Metals
NonMetals
Metals
Conductivity
Moving to the right across a period you enter the “Transition
Block” containing typical hard, high melting point metals, held
strongly together by “metallic bonding”.
Further right you hit the Semi-Metals. These often have very high
melting points because of their covalent lattice structure.
Then you enter the Non-Metals which have covalent molecular
structures and quite low mp’s. At the far right column, each period
ends with an Inert Gas which are all single-atom molecules, and
have the lowest mp of each period.
This pattern repeats itself along each period.
2,000
decreasing
At the left side of the table are the very active metals of the
Activity Series. They are also usually soft, and have relatively low
(for metals) melting points.
Melting Points of Elements
Periods 3
V
(oC)
Boiling Points
follow a similar pattern to
Melting Points
Sketch Graph.
0
Melting Point
1,000
Si
Valencies are the same
down each group
Peaks are Transition Metals
or Semi-Metals
Period 4
Rb
K
Na
Inert Gases
Ar
Kr
Atomic Number
Chemical Bonding, Valency & Reactivity
What you’ve already learnt about the Activity Series, Ionic and Covalent Bonding and Valency
will help you make sense of the following:
Group 8 Inert Gases
+1
0
+2
+3
4
-3
3
1
-2
2 -1
Activity of Metals
Most active at
bottom-left.
Activity (generally)
decreases upwards
and to the right.
Metals
(+ve ions)
Bonding
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Activity of Non-M
Metals
Semi-Metals
(Covalent only)
Most active at top-right
(Fluorine)
Non-Metals
(Covalent or (-ve) ions)
12
No chemical reactions,
no bonding
Activity (generally)
decreases downwards
and to the left.
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Atomic Radius
The size of an atom is the distance across its outer electron shell.
You might think that the atoms along each period would be the same size,
because it’s the same orbit being added to.
However, the increasing amount of positive charge in the nucleus pulls that
orbit inwards closer and closer to the centre.
H
37
Na
186
K
231
He
50
The numbers given are the atomic radii in picometres.
1 picometre = 1x10-112 metre
Radius increasing down a group
Li
152
The following diagrams
are to scale and show the
relative sizes of the first
20 elements
Be
112
Mg
160
Ca
C
B
N
77
88
Al
66
P
Si
143
O
70
S
110
118
102
Ne
F
70
68
Ar
Cl
94
99
Radius decreasing across a period
197
Down each group the radius increases.
This is because, as you go down a group, you have added an entire
electron shell to the outside of the previous layer
“Periodic” means “recurring at
regular intervals”.
This graph shows what
a
spreadsheet plot gives for the radii
of the first 37 elements.
Notice how the same graphical
pattern keeps recurring... it is a
periodic pattern.
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Rb
K
De
acr creasi
oss ng
a p Tren
erio d
d
300
200
Na
Li
100
When you do, you can clearly see
how the Periodic Table got its
name.
rend
sing T
Increa a group
down
He
Ne
Ar
Kr
d
ng Tren
Increasai group
n
w
do
0
The Syllabus requires that you
produce a table and a graph of
the changes in a property
• across a period,
and
• down a group
Atomic Radius (picometre)
Spreadsheet Plot of Atomic Radii
1
10
20
30
Atomic Number
There are a number of irregularities and “glitches”
apparent on the graph. It is beyond the scope of
this course (and way beyond the K.I.S.S. Principle)
to attempt an explanation of these.
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Ionization Energy
Successive Ionization Energies
The meaning of the “1st Ionization Energy” was explained
previously in relation to the Activity Series of Metals.
Having added the energy of 1st I.E. and removed an
electron from any atom, it is then possible to add more
energy and remove a 2nd electron, and a 3rd, and so on.
A+(g)
A(g)
+
e-
where “A” stands for any atom
in the gas state
Any atom can lose an electron if enough energy is
supplied... even atoms which do not normally lose
electrons.
decreasing
A+(g)
e-
2nd I.E.
A+(g)
A+2(g)
+
e-
3rd I.E.
A+2(g)
A+3(g)
+
e-
+
Once the first electron is removed, the remaining electrons
are pulled in tighter to the nucleus. Each one experiences
increased force of attraction, so it requires more energy to
remove the next electron.
The trend for the whole Periodic table is:
Lowest
A(g)
...and so on,
according to how many electrons
the atom has
The Periodic Trend in 1st Ionization Energy
You should remember that the very active metals are the
ones with low 1st ionization energies. They easily lose their
outer electron(s) and so react readily.
1st Ionization
Energy
1st I.E.
Highest value
Each successive ionization requires more energy.
Once the entire outer orbit has been stripped away, the next
ionization must remove an electron from an underlying
orbit, which requires a huge increase in the next ionization
energy. This results in an interesting pattern:
increasing
Patterns in Successive Ionization Energy Data
(values shown are energy measurements)
Successive Elements on Period 3
Explanations:
1st I.E. increases to the right because each atom across a
period has more and more (+ve) nuclear charge attracting
and holding electrons in the orbit concerned. Therefore, it
requires more energy to remove an electron.
1st I.E. decreases down each group because, at each step
down, an extra whole layer of electrons has been added to the
outside of the atom. The outer shell is further away from the
nucleus, and is partially “shielded” from nuclear attraction by
the layers of electrons underneath it. Therefore, it becomes
easier and easier to remove an electron.
Atoms with a tendency to lose electrons easily
(low 1st I.E.) and form (+ve) ions have very low
values.
Once again, there is a pattern in these values in
the Periodic Table.
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1st
I.E.
2nd
I.E.
3rd
I.E.
4th
I.E.
Sodium
2.8.1
0.5
4.5
6.9
9.6
Magnesium 2.8.2
0.7
1.4
7.7
10.5
Aluminium 2.8.3
0.6
1.8
2.8
11.6
Highest Value
Fluorine
(values decrease down)
Atoms with a tendency to gain electrons and
form negative ions have high values.
Electron
Config.
Notice how the values “jump” (underlined data) as the next
ionization has to remove an electron from the next lower
orbit.
Electronegativity
is a value assigned to each element to describe the
power of an atom to attract electrons to itself.
Element
1.0 1.5
0.9
0.8
Electronegativity Values
of selected elements
(values decrease to left)
Inert gases
not included
2.0 2.5 3.0 3.5 4.0
3.0
2.8
0.8
2.5
0.7
2.2
0.7
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Worksheet 3
Atomic Radius ae).......................................... across a period
because each successive element has af)...........................
(more/less) positive charge in the ag).................................. to
attract the electron shell and pull it inwards. As you go
down a group the radius ah)................................. as each new
electron shell is added.
Part A Fill in the blanks. Check answers at the back.
As early as 1830, the German a)............................................
noticed patterns in the properties of the elements. In 1860,
the English scientist b)...................................... proposed a
“Law of c).........................................” describing the
repeating pattern of properties.
First ai)........................ Energies aj)............................
across a period, as the increasing amount of nuclear charge
makes it more and more difficult to ak)..................................
an electron. The values al).................................... down a
group because each extra shell of electrons is am).................
(more/less) strongly held than the previous.
It was the Russian d).......................................... who invented
the e)........................................................, in more or less its
modern form. He realized that there were probably many
elements that had not f).............................................................,
so he g)................................................... in his table for later
additions. By studying the details of known elements, he
was able to h)....................................... very precisely the
properties of the missing elements.
Successive Ionization Energies measure the energy
required to an)............................ another, subsequent
electron from an atom. The energy required to remove the
next electron is always ao)..............................................
(higher/lower). When the next electron happens to be in
the next lower shell, the value ap)............................................
by a huge amount.
Sure enough when discovered, the missing elements were
found to have properties i)........................................................
.......................................................................................................
aq)................................................... is a value which describes
the power of an atom to ar)................................... electrons.
The element with the highest value is as)...............................,
and values decrease as you move to the at).........................
and as you move au).................................. the Periodic Table.
The patterns in the Periodic Table include:
Conductivity, which generally j)......................................... to
the right, as you go from metals to k).....................................
and .................................................
Part B Practice Problems
Melting Points: tend to l)........................to about the
middle of each period, then m)............................. The
highest value is usually a n)...................................... metal or
one of the o)................................................. elements. The
lowest value on each period is always the
p)................................ gas member on the extreme
q)................................ (right/left)
1. a) Write equations to represent the 1st, 2nd, 3rd & 4th
ionisations for a calcium atom.
b) Between which two of these successive ionisations
would you expect a huge increase in the required energy?
2. On each of the following Periodic Table diagrams label
the arrows with the word “increasing” or “decreasing” to
correctly describe the trend in the direction shown.
Valencies are r)................................. down each vertical
group. Bonding follows the pattern of the main categories
of elements. s).................................. form t).............................
bonds when they lose electrons and become u)....................
ions. The Semi-metal elements form only v)...........................
bonds. The Non-metals can bond w).......................................
or can x)................................ electrons to form y)....................
ions.
ii) (down)
b) Electronegativity
Also indicate
(“H”&“L”) the
position of
elements with
highest &
lowest values.
Chemical Reactivity is different for metals and nonmetals. The most active metals are located at the left
z)................................... (top/bottom) of the table.
Generally, activity decreases aa)............................. and to the
ab)................................... The Inert Gases show no chemical
activity. Apart from them, the most active non-metals are
located on the right ac).................................. (top/bottom)
of the table. Activity generally decreases as you move
ad)........................................... and ..........................................
i)
ii)
i)
c) 1st
Ionisation
Energy
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i) (right)
a) Atomic
Radius
Show“H”&”L”
ii)
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4. QUANTITY CALCULATIONS & THE MOLE
Defining the Mole
Quantities in Chemical Calculations
For technical reasons, the “atomic standard” used to
compare the masses of all atoms is the carbon atom,
which contains
Atoms, molecules and ions always react with each other in
fixed, whole-number ratios. That’s why balancing an
equation is so important... it actually brings the equation
into line with what is happening at the particle level.
6 protons
6 neutrons
6 electrons
For example, when hydrogen and oxygen react to form
water, the balanced equation is
2H2 + O2
Atomic Number = 6
Mass Number = 12
2H2O
This is a true description of what is happening to the
molecules:
2 Molecules
of H2
+
1 Molecule
of O2
The mass of this atom is defined to be
exactly 12.000000 atomic mass units (a.m.u.)
and all other atoms are given masses relative to this one.
Since this is the standard of comparison, the formal
definition of the mole is:
“the number of atoms contained in
exactly 12 grams of carbon-12”
2 Molecules
of H2O
However, when we carry out chemical reactions in the
laboratory or in Chemical Industry, we cannot see or count
the molecules. Instead, we measure the mass or volume of
substances.
Note: In Topic 1 it was pointed out that the Mass
Number for any atom is a whole number. It has still not
been explained why the “Atomic Weights” in the
Periodic Table are mostly not whole numbers.
To measure out the correct numbers of particles for a
reaction we need a simple way to convert masses and
volumes to numbers of molecules, and vice-versa. That’s
the purpose of
This WILL be explained in a later topic.
If you cannot wait, go find out about “Isotopes”.
The Mole
Avogadro’s Number
Just how many atoms are in 1 mole?
1 mole is a quantity of a chemical substance.
Obviously, it is a very large number. We now know that it
is about 6,000 billion trillion.
1 mole of any element or compound contains exactly the
same number of particles.
Avogadro’s Number
1 mole of each substance has a different mass, because the
atoms and molecules all weigh differently.
6.022 x 1023
particles in 1 mole of anything
The really clever and convenient thing about the mole is its
link to the Periodic Table and the “Atomic Weights” shown.
6
Carbon
C
Argon
Ar
Pb
12.01
39.95
207.2
1 mole
= 12.01 grams
1 mole
= 39.95 grams
This number is named in honour of an Italian scientist
who you will learn about soon.
82
18
6p+
6n0
207.2 grams of
Lead
contains
6.022 x 1023
Lead atoms
Lead
1 mole
= 207.2 grams
39.95 grams of
Argon
contains
6.022 x 1023
Argon atoms
12.01 grams of
Carbon
contains
6.022 x 1023
Carbon atoms
EACH OF THESE HAS THE SAME NUMBER OF ATOMS
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Calculating Mole Quantities
Moles and Numbers of Particles
You need to be able to calculate mole quantities in terms of
both mass and number of particles.
Since one mole of any substance contains Avogadro’s
Number of particles:
Molar Mass
The “Molar Mass” of any chemical species is the mass (in
grams) of 1 mole of the substance.
No. of moles = No. of particles you have
Avogadro’s Number
n=
You need to add up all the Atomic Weights
of all the atoms given in the formula.
Examples:
Name
Argon
Sodium
Formula
Ar
Na
Example Calculations
1. How many moles are present in a sample of lead
containing 7.88 x 1024 atoms?
Molar Mass (g)
39.95
22.99
Solution
(for elements like these just use Atomic Weight)
Oxygen
Chlorine
O2
Cl2
(16.00 x 2)
(35.45 x 2)
32.00
70.90
Water
H2O (1.008x2 + 16.00) 18.016
Carbon Dioxide CO2 (12.01 + (16.00x2) 44.01
Sodium chloride NaCl (22.99 + 35.45) 58.44
b) m = n x MM = 0.0250 x 207.2 (molar mass of Pb)
= 5.18 g
Try the Worksheet at the end of this section
Number of Moles in a Given Mass
Try the Worksheet at the end of this section
When you weigh a chemical sample you then need to be
able to calculate how many moles this contains.
Mole Quantities in Chemical Equations
When you consider an equation like
No. of moles = mass of substance you have
molar mass
Solution
2H2 + O2
m
MM
n = m = 5.23 = 0.215 mol
MM
24.31
b)
n = m = 96.7
MM
(2x1.008 + 16.00)
= 96.7/18.016
= 5.37 mol
2 Molecules
of H2
1 Molecule
of O2
2 Molecules
of H2O
2 million H2 + 1 million O2
2 million H2O
or, 200 zillion H2 + 100 zillion O2
200 zillion H2O
or, (let’s use Avagadro’s number)
(2 x NA) H2 + NA O2
so m = n x MM = 1.50 x (22.99 + 35.45)
= 1.50 x 58.44
= 87.7 g
= 2 moles H2 + 1 mole O2
(2 x NA) H2O
2 moles H2O
The Balancing Coefficients in a Chemical Equation
May be Interpreted as Mole Ratios
Try the Worksheet at the end of this section
Preliminary Chemistry Topic 2
Copyright © 2005-2007 keep it simple science
+
However, the number of molecules reacting is really just a
ratio. The actual numbers might be
2. What mass is needed if you want to have 1.50 moles
of salt (sodium chloride)?
n= m
MM
2H2O
you know it means
a) 5.23g of magnesium?
b) 96.7g of water?
a)
= 7.88x102423
6.022x10
= 13.1 mol
Solution
a) n = N so N = n x NA = 0.0250 x 6.022x1023
NA
= 1.51 x 1022 atoms
(add up At.weights of all atoms in the formula)
Example Calculations
1. How many moles in
n= N
NA
2. a) How many atoms of lead are needed to make
0.0250 mole?
b) What would be the mass of this quantity?
(these elements are diatomic molecules... 2 atoms each)
n=
N
NA
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Mole Quantities
in Chemical Equations (cont.)
Calculating Mass Quantities in Reactions
Mole calculations allow you to calculate the mass of
products and reactants involved in a reaction.
The balancing coefficients of an equation can be
interpreted as the mole ratio of reactants and products.
2 H2
So,
+
O2
Example Problem
Aluminium burns to form aluminium oxide.
If 4.29g of aluminium was burned,
a) what mass of oxygen would be consumed?
b) what mass of aluminium oxide would be formed?
2 H2O
means 2 mol. reacts with 1 mol. to form 2 mol.
or,
4 mol. reacts with 2 mol. to form 4 mol.
or,
100 mol. reacts with 50 mol. to form 100 mol.
or any other proportional quantities.
Solution
Always start with the balanced equation:
Example Problem
a) If 0.50 mol of sodium reacted completely with
hydrochloric acid, how many moles of products would
be formed?
mole
ratio
b) What mass of each product would be formed?
No. of moles of Aluminium: n = m = 4.29
MM
26.98
= 0.159 mol
Solution
a) The balanced equation is
mole
ratio
so,
2 Na
+ 2 HCl
H2
2 mol
:
2 mol
:
0.50 mol
:
0.50 mol :
: 2 mol.
0.25 mol : 0.50 mol
b) Mass of Hydrogen: m = n x MM = 0.25 x 2.016
= 0.50 g
Mass of salt:
m = n x MM = 0.50 x 58.44
= 29 g
3
:
2
∴ mass of Al2O3: m = n x MM = 0.0795 x 101.96
= 8.11 g
Try the Worksheet at the end of this section
Try the Worksheet at the end of this section
Practical Work:
Using Mass & Mole Ratios to Determine a Formula
A common experiment is to burn a piece of magnesium in a crucible, as
suggested by the diagram. The difficulty is to open the lid enough to admit
oxygen for complete combustion, but to limit the loss of powdery product.
Magnesium + Oxygen
:
2 Al2O3
b) Mass Al2O3 produced:
mole ratio Al : Al2O3 = 4: 2 (i.e. 2:1)
∴ moles of Al2O3 = 1/2 x 0.159 = 0.0795 mol
Answer: 0.25 mol of H2 and 0.5 mol of NaCl
The reaction is
4
+ 3 O2
a) Mass O2 consumed:
mole ratio Al : O2 = 4 : 3
∴ moles of O2 = 0.159 x 3 = 0.119 mol
4
∴ mass of O2: m = n x MM = 0.119 x 32.00
= 3.81 g
+ 2 NaCl
1 mol
4 Al
ceramic
crucible
Magnesium oxide
Careful measurement of mass allows the empirical formula for magnesium
oxide to be determined.
Analysis of Results
Remember that to convert any
Typical Measurements
mass to moles:
n = m / MM
Mass of empty crucible = 42.74 g
Mass of magnesium
= 2.05 g
Mass of crucible
Elements
Magnesium :
Oxygen
+ product after burning = 46.22 g
Ratio of masses:
2.05 g
: 1.07 g
Ratio of moles:
2.05 / 24.31 : 1.07 / 16.00 (divide by Atomic Weight)
∴ Mass of magnesium oxide
=
0.0843 mol : 0.0669 mol
formed = 3.48 g
Simplified
ratio
=
0.0843/0.0669
: 0.0669/0.0669 (divide both by the
∴ Mass of oxygen in
=
1.26
:
1
smaller)
compound = 1.07 g
Nearest whole number ratio 1 : 1
There is often a large error
due to incomplete burning
∴ Empirical Formula is MgO
Preliminary Chemistry Topic 2
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Comparing Mass Changes
When Metals Burn
A Little History...
How the Mole was Invented
Atoms always react in simple whole-number mole ratios,
but atoms have different masses, and compounds have
various formulas, so the result is that chemicals do NOT
react in simple ratios by mass.
The “mole” as a measure of chemical quantities, is a
mathematically convenient device (a “trick”) to help
chemical calculations.
Here’s how it was figured out...
That’s why we need the mole... we measure quantities by
their mass, but this makes no sense until moles are
calculated.
Gay-Lussac’s Law
Joseph Gay-Lussac was a French scientist with an
unfortunate name, by modern standards. He lived 200 years
ago, and was very interested in flight using balloons, so he
investigated the way gases react chemically.
The syllabus requires that you should consider the mass
changes involved when various metals combine with
oxygen to form their oxide compound.
After a series of clever experiments, in which the volumes
of reacting gases were measured, in 1808 he proposed the
“Law of Combining Volumes”:
The following table shows the mass changes for 100g of
the metal in each case:
100g of
Metal
Formula
of oxide
Mass O2
needed(g)
When measured
at constant temperature and pressure,
the volumes of gases in a chemical reaction
show simple, whole-number ratios
to each other.
Mass of
Oxide formed
Lithium
Li2O
115
215
Iron
Fe2O3
43
143
Zinc
ZnO
49
149
Lead
PbO2
15
115
The volume of a gas is easily changed by temperature and
pressure, so it is very important that the volumes are all
measured at the same conditions.
Examples of Gay-Lussac’s Law:
Empirical Formulas v Molecular Formulas
You are reminded that a molecular formula really does
describe the atoms present in a molecule.
Hydrogen(g) + Chlorine(g)
1 litre
1 litre
Hydrogen chloride(g)
2 litres
The molecular compound methane,
has formula CH4, because that’s
exactly what each molecule contains...
1 carbon atom and 4 hydrogen atoms.
Hydrogen(g) + Oxygen(g)
2 litres
1 litre
Water(g) (vapour)
2 litres
Hydrogen(g) + Nitrogen(g)
3 litres
1 litre
Ammonia(g)
2 litres
Lattice structures, either ionic or covalent
are NOT molecular.
Example: sodium chloride, NaCl
Notice that in every case, that the volumes are always in a
simple, whole number ratio to each other.
The formula does NOT
describe a molecule, but only
gives the simplest ratio between
the bonded atoms... this is an empirical formula.
Now consider the balanced equations for these three
example reactions:
H2(g)
+
2 H2(g) +
On the previous page was an example of how formulas are
determined by analysing the mass composition of a
compound.
Cl2(g)
2 HCl(g)
O2(g)
2 H2O(g)
3 H2(g) + N2(g)
2 NH3(g)
The mole ratios are the same as the volume ratios
discovered by Gay-Lussac!
You should note that this method can only produce an
empirical formula. (In fact, the word “empirical” means
something determined by experiment, not by theory.)
Why should this be?
If a molecular compound, with molecular formula X2Y6
was analysed by mass measurements, its empirical formula
would be calculated to be XY3... simplest ratio of atoms.
Preliminary Chemistry Topic 2
Copyright © 2005-2007 keep it simple science
... enter Avogadro!
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Avogadro’s Hypothesis
Molar Volume of a Gas
The Italian, Amadeo Avogadro (1776-1856) was trained in
Law, but became very interested in Science.
If 1 mole of any chemical species contains the same
number of particles (Avogadro’s Number) AND if equal
volumes of gases contain equal number of particles
(Avogadro’s Hypothesis), then it follows that
1 mole of any gas must occupy the same volume,
if measured at the same temperature and pressure.
In 1811, he noticed the similarity between Gay-Lussac’s
Law (an empirical “law” based on experiment) and the
concept that atoms must combine in simple, whole number
ratios to form compounds.
This volume is the “Molar Volume” and is the same for
every gas. It is measured at the standard set of conditions
known as Standard Laboratory Conditions (SLC); 25oC
and 1 standard atmosphere of pressure.
This led him to make an hypothesis:
Equal Volumes of all Gases
Contain Equal Numbers of Molecules
1 mole of any gas = 24.8 litres at SLC
(when measured at the same conditions
of temperature and pressure)
Mole Calculations Involving Gases
This additional knowledge opens up the opportunity to
carry out quantity calculations which involve mass and
volumes of gases.
This was a vital breakthrough in the history of Chemistry.
For example, consider the reaction:
Hydrogen(g) + Chlorine(g)
Example Problems
1.
If 15.65g of calcium carbonate (CaCO3) was completely
decomposed by heat, what volume of carbon dioxide
gas would be produced (if measured at SLC)?
Hydrogen chloride(g)
Prior to Avogadro, it was assumed that the the reaction
involved single atoms, like this:
H(g)
+ Cl(g)
HCl(g)
Solution
Always begin with the balanced equation for the reaction.
CaCO3(s)
CO2(g) + CaO(s)
mole ratio = 1
:
1
:
1
Moles of CaCO3: n = m = 15.65 = 0.1564 mol
MM
100.09
Mole ratio is 1 : 1, so moles of CO2 formed = 0.1564
but the combining volumes (discovered by experiment) were
Hydrogen(g) + Chlorine(g)
1 volume : 1 volume
:
Hydrogen chloride(g)
2 volumes
Now, reasoned Avogadro, gases react in simple, wholenumber volume ratios because each litre of gas has the
same number of molecules in it. Therefore, to get the
volume ratios shown above, each hydrogen molecule, and
each chlorine molecule, must have 2 atoms!
∴ Volume of CO2 = 0.1564 x 24.8
= 3.88 L (at SLC)
2.
What volume of hydrogen gas (at SLC) would be
produced if 10.00g of lithium metal was reacted with
sulfuric acid?
i.e. Hydrogen is H2(g) and Chlorine is Cl2(g), and the correct
equation is
H2(g)
+
Cl2(g)
2 HCl(g)
Solution
2 Li(s) + H2SO4(aq)
2
:
1
Then, for the same reaction, scientists could measure the
masses of these gases as well as volumes. This showed that
chlorine atoms must be about 35 times heavier than
hydrogen... chemists were on the way to figuring out the
relative atomic weights of elements, and being able to
calculate chemical quantities.
:
H2(g) + Li2SO4(aq)
1
:
1
Moles of lithium: n = m
= 10.00 = 1.441 mol
MM
6.941
Mole ratio is 2:1, so moles of H2 = 1/2 x 1.441=0.7204
∴ Volume of H2 = 0.7204 x 24.8
= 17.9 L (at SLC)
Although he did not invent the concept of the mole, we
name the number of particles in 1 mole in Avogadro’s
honour...
23
Avogadro’s Number, NA = 6.022 x 10
Preliminary Chemistry Topic 2
Copyright © 2005-2007 keep it simple science
Molar Vol.
of all gases
at SLC
Try the Worksheet at the end of this section
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Worksheet 4
4. Mole Ratios in Equations
Sodium reacts with water as follows:
H2 +
2Na + 2H2O
Part A Practice Problems
a) If 1 mole of sodium reacted, how many moles of
i) hydrogen formed?
ii) water consumed?
b) If 0.25 mol of NaOH formed, how many moles of
i) sodium consumed?
ii) hydrogen formed?
c) If 0.75 mole of hydrogen formed, how many moles of
i) sodium consumed?
ii) NaOH produced?
1. Molar Masses
Calculate the molar mass of:
a) potassium
b) krypton
c) tin
d) bromine (Br2)
e) nitrogen gas
f) magnesium oxide
g) sodium iodide
h) iron(III) sulfide
i) ammonia
j) copper(II) sulfate
k) aluminium oxide
l) glucose (C6H12O6)
Aluminium reacts with oxygen:
4 Al
+ 3 O2
2 Al2O3
2. No. of Moles in a Given Mass
How many moles in:
a) 100.0g of lead?
b) 100.0g of zinc?
c) 100.0g of water?
d) 100.0g of copper(II) nitrate?
e) 38.55g of magnesium fluoride?
f) 60.00g of carbon dioxide?
g) 1.000g of zinc oxide?
h) 500.0g of glucose (C6H12O6)?
i) 3.258 x 10-3g of salt (sodium chloride)?
j) 128.6g of ammonium carbonate?
d) If 0.5 mole of Al used, how many moles of
i) Alum.oxide formed? ii) oxygen used?
e) If 0.1 mole of alum.oxide formed, how many moles of
i) aluminium used?
ii) oxygen used?
5. Mass Quantities in Reactions
a) Calcium burns in oxygen to form calcium oxide:
2Ca + O2
2CaO
If 8.50g of calcium reacted, what mass of calcium oxide
would be formed?
b) Silver carbonate decomposes when heated:
2Ag2CO3
2CO2 + 4Ag
3. Moles and Number of Particles
a) How many particles (atoms/molecules) in:
i) 3 moles of water?
ii) 2.478 mol of CO2?
iii) 5 mol of salt?
iv) 0.007862 mol of copper
v) 1/1000 mol of helium
+ O2
If 20.0g of silver carbonate was decomposed
i) what mass of silver metal would form?
ii) what mass of CO2 would be produced?
iii) what mass of O2 would be formed?
c) Aluminium reacts with hydrochloric acid:
2Al + 6HCl
3H2 + 2AlCl3
b) Convert between mass, moles and no.of particles.
25
i) If there are 8.800x10 atoms of tin, how many moles is
this, and what would be the mass?
If 6.50g of aluminium reacted
i) what mass of HCl would be consumed?
ii) what mass of hydrogen gas produced?
iii) what mass of aluminium chloride produced?
ii) You have a sample containing 2.575x1024 molecules of
water. How many moles is this, and what is its mass?
d) Tin reacts with steam as follows:
2H2(g) +
Sn(s) + 2H2O(g)
iii) If you weigh out 400.0g of water, how many moles is
this, and how many molecules are present?
SnO2(s)
If 14.8g of tin reacted
i) what mass of tin(IV) oxide would be formed?
ii) What mass of steam would be consumed?
iii) what mass of hydrogen would be produced?
iv) If you have 2.569g of pure nickel, how many atoms are
there?
v) What mass of sulfur would contain 2.500x1023 atoms?
FOR MAXIMUM MARKS SHOW
FORMULAS & WORKING,
APPROPRIATE PRECISION & UNITS
IN ALL CHEMICAL PROBLEMS
Preliminary Chemistry Topic 2
Copyright © 2005-2007 keep it simple science
2NaOH
Worksheet continues next page...
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6. Empirical Formulas from Mass Composition
a) A compound containing only copper and chlorine is
decomposed, and the masses measured to find the mass
composition:
Mass of copper present = 12.84g
Mass of chlorine present = 7.16g
i) Find the empirical formula.
ii) Name the compound.
8. Mass & Gas Volume Calculations
a) To “scrub” the air and remove poisonous CO2 on board
the Space Shuttle, the air is continually pumped through
canisters containing 5.0kg of lithium oxide. The reaction is
Li2O(s) + CO2(g)
Li2CO3(s)
b) i) Find the empirical formula of a compound containing
carbon and hydrogen; a sample was found to contain 1.5g
of carbon and 0.5g of hydrogen.
ii) Name the compound, given that its empirical and
molecular formulas are the same.
b) Iron reacts with oxygen:
4Fe(s) + 3O2(g)
i) How many moles of lithium oxide in each canister?
ii) How many moles of CO2 can this amount absorb?
iii) What volume of CO2(g) is this? (measured at SLC)
i) If 10.0L of O2 at SLC reacted, what mass of iron(III)
oxide would be formed?
ii) If 100g of iron reacted, what volume of oxygen (at SLC)
would be needed?
c) A compound was found to contain 30% nitrogen and
70% oxygen by mass.
i) Find the empirical formula.
ii) It is later found that its molecular formula is a 2 times
multiple of the empirical. Write the molecular formula.
iii) Name the compound.
c) The electrolysis of water causes decomposition:
2H2O(l)
2H2(g) + O2(g)
i) If 1.00g of water was decomposed, what volumes of
each gas (measured at SLC) would be formed?
In an electrolysis experiment, 50mL (0.050 L) of oxygen
was produced. (measured at SLC)
ii) What volume of hydrogen (at SLC) was produced?
iii) What mass of water must have been decomposed?
7. Volumes of Reacting Gases
( Assume all are measured at same temperature & pressure)
2 H2(g) + O2(g)
2 H2O(g)
Part B Fill in the blanks. Check answers at the back
a) If you used 5 litres of hydrogen, how many litres
i) of oxygen consumed?
ii) of water vapour formed?
The formal definition of the mole is “the a)..........................
of atoms in 12.00 grams of b).................................................”
One mole of any substance contains the same number
of c).................................. The mass of 1 mole of any
substance is equal to its d)...........................................................
in grams. The actual number of particles in one mole of
anything is known as “e).................................................’s
Number” and has a value of f)..................................................
b) If you used 0.25 litres of oxygen, how many litres of
i) water vapour formed?
ii) hydrogen consumed?
c) If this reaction produced 20 litres of steam, how many
litres of
i) hydrogen consumed?
ii) oxygen consumed?
In a balanced chemical equation, the “balancing numbers”
(coefficients) may be interpreted as being g)...........................
.............................. of reactants and products.
By converting between the h).......................... of substances
and the number of i)..............................., it becomes possible
to calculate all the quantity relationships during a chemical
j).................................... From the mass composition it is also
possible to determine the k).................................. formula of
compounds.
Ammonia gas is formed by reaction of hydrogen with
nitrogen
3 H2(g) + N2(g)
2 NH3(g)
d) In order to make 9 litres of ammonia, what volume
i) of hydrogen needed?
ii) of nitrogen needed?
Historically, the mole concept arose from the work of 2
men: The Frenchman l).................-.............................
discovered that “the m)........................... of gases in chemical
reactions always show simple, n)............................................
ratios to each other”. Soon after, the Italian
o)................................................. suggested that “Equal
p)................................... of all gases contain q).......................
numbers of r)................................... (when measured at the
same conditions of s).............................. and ..........................)
The standard conditions usually used are known as
t)...................... (abbreviation) and is a pressure of
o
u)........................................... and temperature v)............... C.
e) If 0.6 litre of hydrogen reacted, what volume
i) of ammonia formed?
ii) of nitrogen need?
WHEN COMPLETED, WORKSHEETS
BECOME SECTION SUMMARIES
Preliminary Chemistry Topic 2
Copyright © 2005-2007 keep it simple science
2Fe2O3(s)
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5. METALS FROM THEIR ORES
The Importance of Predicting
Yield from an Ore
Ores and Minerals
... and now back to the metals.
The whole situation of economic feasibility makes the
science of Analytical Chemistry vital in the mining and
metals industry.
Minerals are naturally occurring compounds. “Rocks” are
mixtures of various minerals. Most minerals are lattice
structures, both ionic and covalent. Some very common
minerals include:
Mining operations cost millions of dollars to set up. To do
so, the operators need to be sure that the ore contains
enough metal to be profitable. Chemical analysis in the
laboratory is used to measure the mineral content of the
ore body, to predict the final yield of the metal.
• Silica, which is chemically silicon dioxide (SiO2) and is the
most common mineral on Earth. Other compounds are
often included in the silica lattice to make “silicate”
minerals. These occur in virtually all rocks.
• Calcite, which is calcium carbonate (CaCO3) is the main
mineral in limestone and marble.
Photo:
Thanks Ulrik
Some minerals contain significant quantities of metal(s),
chemically combined in the compound.
Ores are rocks and/or minerals from which it is
economically worthwhile to extract a desired metal.
It is the economic part of this definition which is critical.
For example, there are many rocks and minerals that
contain significant amounts of iron and aluminium. These
are not “iron ore” or “aluminium ore” unless it is
economically worthwhile to mine and process them to get
the metal.
Ores are Non-Renewable Resources
Minerals and ores have been formed over millions and
billions of years of geological processes on Earth.
Photo courtesy of
Comalco Aluminium Ltd
Because of that time-frame, the ores are non-renewable in
the sense that once we use them up, they cannot be
replaced.
There is no immediate concern for running out of the
most important ores, but unlimited exploitation of any
non-renewable resource is:
• irresponsible, to future generations.
• unsustainable, because all non-renewable things must
eventually run out.
• economically stupid, because it may be cheaper to re-use
and recycle, than to constantly extract “new” materials.
• environmentally damaging, because mining and metal
smelting have a history of pollution and ecosystem
destruction.
What Makes It Economically Worthwhile?
Basically, economic feasibility is the balance between:
• the Commercial Price for which the metal can be sold
and
• the Production Costs of mining and transporting the
ore, and chemically extracting and purifying of the metal.
Another factor is the abundance of the metal and its ores
on Earth. For example, iron is relatively cheap because it is
very common in huge deposits of iron ores. Platinum is
very rare, so it commands a high price. This makes it
worthwhile to mine even very low-grade ores. A low-grade
iron ore would not be worth mining!
Preliminary Chemistry Topic 2
Copyright © 2005-2007 keep it simple science
In the not-too-distant future it may become economically
worthwhile to begin “mining” the old rubbish dumps
around our cities, to recover the discarded metals in
society’s garbage.
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Case Study:
Extraction of Copper from its Ores
Copper Ores
include a variety of compounds of copper, including:
Froth Flotation to Concentrate the Ore
• copper(I) sulfide, Cu2S
• copper(II) hydroxide mixed with
copper(II) carbonate, Cu(OH)2.CuCO3
The ore is crushed into a powder and the copper
minerals are separated from the silicates by a process of
“Froth Flotation” which relies on differences in
“wettability” and density.
These compounds usually occur as thin “veins” of bluegreen minerals embedded in masses of worthless silicate
rock.
Froth overflows
for collection
The copper content of the entire ore body might be only
3% or less. Therefore, the first step after mining is to
separate the copper minerals from the “rock”.
Compressed air creates froth bubbles
Compressed air creates a froth of bubbles in a detergent
solution.
Copper minerals, sprayed with a special oil, cling to the
bubbles and are carried upwards to overflow with the
froth.
Chemistry of Smelting
The concentrated copper minerals now undergo
Decomposition Reactions.
Silicate minerals are wetted by the water and, being
denser, sink to the bottom.
In Australia, the main copper ores contain copper(I)
sulfide. If this is heated in a furnace supplied with plenty
of air the reaction is:
Copper(I) sulfide + oxygen
Cu2S + O2
The collected froth is then treated to separate the oil and
detergent for re-use.
Copper + Sulfur dioxide
2Cu + SO2
The ore concentrate is now about 30% copper.
The copper collected is about 98% pure.
Sulfur dioxide is a serious pollutant if released
from the smelter.
These days it is collected and used to
manufacture sulfuric acid... a useful by-product.
Final Purification by Electrolysis
The major use of copper is for electrical wiring. For this it needs to be 99.9% pure.
Copper is purified by electrolysis:
The impure copper
is immersed in
CuSO4 solution and
electrified:
Cu
Cu
+2
-
+ 2e
The copper dissolves
into the solution, but
impurities do not.
-
+
Impure
Copper
dissolves
into
solution
Preliminary Chemistry Topic 2
Copyright © 2005-2007 keep it simple science
Cu+2
ions
migrate through
CuSO4 solution
Pure
Copper
deposits
on
electrode
After migrating
through the solution,
the ions are redeposited as pure
copper metal on the
other electrode:
Cu+2 + 2e-
Cu
Impurities
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The Case for Recycling
Producing the electricity usually involves the burning of
coal at a power station. The burning of fossil fuels like coal
is a major contributor to the “Greenhouse Effect” which
many scientists are now convinced is causing massive
climate changes to the entire Earth.
The point that mineral ores are non-renewable has already
been made. Eventually, any non-renewable resource must
run out, so recycling is inevitable.
There is also a strong environmental case for recycling of
metals, especially aluminium.
Recycling aluminium requires about 7kJ of energy, a saving
of about 96% in energy and environmental impact!
Extracting aluminium from its ore requires about 200kJ
(kilojoules) of energy per kg of metal. This energy is
mainly in the form of electricity, which is needed in huge
quantities for the electrolytic smelting process.
Most local councils now operate “Recycling Centres”
which can sort out paper, glass, plastic, etc from our
garbage, as long as we remember to put recyclables in the
correct bin.
Aluminium (mainly drink cans) collected this way is
returned to scrap-metal businesses which clean and re-melt
the metal to return it to manufacturing industry for re-use.
Photo by Griszka
Worksheet 5
Fill in the blanks. Check answers at the back.
“Minerals” are naturally occurring a)........................................
which are mixed together in rocks.
An “ore” is a b).................................. from which it is
c)......................................... worthwhile to extract a desired
d)..............................
Whether it is worthwhile (or not) to mine an ore depends on
the balance between the e)...........................................................
and the f)........................................................... of mining,
transporting and g)..................................... the metal.
h)..................................... analysis of an ore deposit is vital to
predict the i)..................................... from the ore, to
determine if it is worth mining.
Ores are j)............................................... resources because
once used, they cannot k)........................................................
due to the immense time it takes for l).......................................
processes to form them.
WHEN COMPLETED, WORKSHEETS
BECOME SECTION SUMMARIES
Preliminary Chemistry Topic 2
Copyright © 2005-2007 keep it simple science
25
Copper ores contain compounds such as m)...........................
............................ and .....................................................................
After mining, the ore is crushed, then concentrated by
“n)........................... .......................................”. This process
uses a froth of bubbles to separate the
o)................................... density copper compounds from the
worthless rock which is mainly p)........................................
minerals.
The “smelting” process involves q)............................................
reactions. For a sulfide ore, it reacts with r).............................
to form s).............................. metal and t)....................................
gas.
The final step is to u)................................... the copper by a
process of v).......................................................
There are many good reasons to w).........................................
metals, especially x)........................................ which requires
large amounts of y)................................... energy to extract
from its ore. Producing the electricity required is often done
by burning z).......................... fuels such as aa)......................
This contributes to the “ab)....................................................
Effect”, responsible for global climate changes. Recycling
aluminium requires only a fraction of this energy.
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CONCEPT DIAGRAM (“Mind Map”) OF TOPIC
Some students find that memorizing the OUTLINE of a topic
helps them learn and remember the concepts and important facts.
Practise on this blank version.
METALS
Preliminary Chemistry Topic 2
Copyright © 2005-2007 keep it simple science
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Practice Questions
7.
The scientist most responsible for the development of the
Periodic Table was:
A. Avogadro
B. Newlands
C. Gay-Lussac
D. Mendeleev
When you have confidently mastered this level, it is
strongly recommended you work on questions from past
exam papers.
8.
Element “X” is in Group 2 and element “Y” in Group 7.
These are not intended to be "HSC style" questions, but to
challenge your basic knowledge and understanding of the
topic, and remind you of what you NEED to know at the
K.I.S.S. Principle level.
Part A
Multiple Choice
If X & Y
formed a
compound,
you would
expect it to be
1.
Which list shows metals used by humans in the correct
chronological order of their history of usage?
A. bronze, aluminium, iron
B. copper, bronze, iron
C. gold, iron, bronze
D. copper, steel, bronze
A. ionic, with formula X2Y
B. covalent, with formula X2Y
C. ionic, with formula XY2
D. covalent, with formula Y2X
2.
Which list correctly identifies an alloy, and the elements it
contains?
A. Steel; iron and tin
B. Bronze; tin and zinc
C. Solder; copper and lead
D. Brass; copper and zinc
9.
If the elements “X” & “Y” in Q8 lie in the same period of
the table, you would expect:
A. X to have a smaller radius than Y.
B. Y to have a higher electronegativity than X.
C. X to have a higher 1st ionization energy than Y.
D. Y to have a higher melting point than X.
3.
The metals used by humans have changed over the course
of history. The availability of new metals has often been
dependent on the:
A. availablity of energy to extract metals from ores.
B. discovery of new minerals as people explored the world.
C. invention of new alloys.
D. development of new technologies to use the metals.
10.
The reason for the trend in atomic radius as you move
across a period to the right, is:
A. increasing nuclear charge.
B. addition of extra electron shells.
C. decreasing attraction of electrons to the nucleus.
D. increasing mass of the atoms.
4.
A metal which reacts readily and vigorously with oxygen,
water and dilute acids would probably:
A. have a high value for 1st ionization energy.
B. be from the “Transition” block of the Periodic Table.
C. have a very low 1st ionization energy.
D. be located at extreme right of the Periodic Table.
11.
An atom of argon is about twice as heavy as an atom of
neon. You would expect:
A. a mole of argon to contain about half as many atoms as
a mole of neon.
B. equal masses of each element to contain about the same
number of atoms.
C. 2g of argon to contain about the same number of atoms
as 1g of neon.
D. the molar mass of neon to be about twice the molar
mass of argon.
5.
If nickel reacted with sulfuric acid, the products of the
reaction would be:
A. hydrogen gas and nickel sulfate
B. carbon dioxide gas and nickel sulfate.
C. nickel sulfide and hydrogen gas.
D. sulfur dioxide gas and nickel hydroxide.
6.
During the reaction in Q5, the basic underlying change
occurring is:
A. the breaking covalent bonds.
B. the transfer of electron(s) from one species to another.
C. chemical changes in “spectator ions”.
D. physical dissolving of metal in the acid.
Preliminary Chemistry Topic 2
Copyright © 2005-2007 keep it simple science
12.
Which line shows correctly the molar mass (to the nearest
gram) of the named substance?
A. water, 18g
B. carbon dioxide, 28g
C. oxygen gas, 16g
D. helium gas, 8g
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13.
Aluminium reacts with oxygen to form aluminium oxide.
4 Al
+ 3 O2
2 Al2O3
20. (5 marks)
Give an outline of an experiment you have done to
investigate the relative chemical activity of some metals.
Include the observation(s) you made to assess metal
activity, and state the conclusion(s) reached.
If 1 mole of aluminium (about 27g) was to be reacted, you
would need how many moles of oxygen gas?
A. 0.75 mol
B. 3 mol
C. 1 mol
D. 1.3 mol
21. (6 marks)
Write a balanced symbol equation for the reaction of:
a) magnesium with hydrochloric acid.
b) calcium with water (reacts at room temperature).
c) potassium with oxygen.
14.
Avogadro’s number can be described by the abbreviation NA.
If you had 2 moles of methane (CH4), then the number of
hydrogen atoms present is:
B. 4 x NA
A. 2 x NA
C. 8 x NA
D. 10 x NA
22. (4 marks)
When barium metal reacts with an acid there is an exchange
of electrons such that hydrogen gas and barium ions are
formed.
Write 2 “half-equations” to show clearly the species
gaining, and the species losing, electrons.
15.
Carbon monoxide gas reacts with oxygen gas to form
carbon dioxide gas as follows:
2CO(g) + O2(g)
2CO2(g)
23. (4 marks)
a) Write an equation (including states) for the first
ionization of
i) magnesium
ii) oxygen
b) Describe how the Activity Series of Metals is related to
the values of 1st Ionisation Energy.
If 100mL of carbon dioxide was produced, then the total
volume of reactants (all measured at the same temp. &
pressure) before the reaction would have been:
A. 100mL
B. 150mL
C. 50mL
D. 250mL
24. (5 marks)
a) Sketch a graph (values are not required) to show the
general changes in melting points of the elements across
one period of the Periodic Table.
b) Briefly explain the general trend shown in your graph.
16.
The “smelting” of a metal ore always involves:
A. separating the metal-containing mineral from the rock.
B. decomposing a compound of the metal.
C. purifying the extracted metal by electrolysis.
D. all of the above.
25. ( marks)
a) Write equations (including states) to describe the
i) 1st
ii) 2nd
and
iii) 3rd
ionisations of potassium.
b) At which ionization would you expect a huge increase in
the ionization energy? Explain your answer.
Longer Response Questions
Mark values shown are suggestions only, and are to give
you an idea of how detailed an answer is appropriate.
17. (5 marks)
Give an example of
a) a metal used in its elemental state, and
b) a non-ferrous alloy (naming its components)
in common use. For each, relate the properties of the metal
to its particular use(s).
26. (6 marks)
a) Write a balanced equation for the reaction of aluminium
metal with hydrochloric acid.
b) If 6.58g of aluminium reacted fully, calculate:
i) the number of aluminium atoms involved.
ii) the mass of aluminium chloride formed.
iii) the volume of hydrogen gas (at SLC)
18. (3 marks)
Give a reason why
a) metal tools are superior to stone tools.
b) iron replaced bronze in the history of metallurgy.
c) aluminium did not come into common use until the 20th
century.
27. (4 marks)
It was found by experiment that a compound containing
only tin and oxygen, contained 88% tin, by mass.
Showing your working, determine the empirical formula for
this compound, and give its correct chemical name.
19. (6 marks)
The most common metal in use today is steel, which comes
in a variety of forms, with different properties and uses.
Compare 3 different types of steel, stating the composition
of each and relating its properties to a common use.
Preliminary Chemistry Topic 2
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28. (4 marks)
In the reaction of nitrogen and hydrogen gases to form
ammonia gas, it was found by experiment that 300mL of
hydrogen reacted completely with 100mL of nitrogen.
200mL of ammonia gas was produced. All the gas volumes
were measured at a pressure of 10 standard atmospheres
and 150oC.
a) Write a balanced equation for the reaction.
b) Explain how the experimental measurements are in
agreement with Gay-Lussac’s Law.
30. (8 marks)
a) Give the name and formula for a compound commonly
found in copper ores.
b) Name, and briefly describe the process by which a
copper ore is concentrated and separated from the
surrounding “rock”.
c) Write a chemical equation to describe the reaction which
occurs in the smelting of the ore. (Involving the compound
you named in part (a))
29. (5 marks)
a) Differentiate between a “mineral” and an “ore”.
b) Outline the role of Chemical Science in assessing the
economic feasibility of mining a mineral resource.
c) Briefly discuss the sustainability of using the Earth’s
mineral resources, and outline a strategy for conservation.
d) Name the process by which the smelted copper is
purified, and relate the need for purification to a common
use of the metal.
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Preliminary Chemistry Topic 2
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Answer Section
4.
a)
Worksheet 1
b)
a) stone or wood/bone
b) gold
c) decoration/jewellery
d) copper
e) roasted by fire
f) copper and tin
g) bronze
h) hard and strong
i) brittle
j) a saw blade
k) iron
l) harder and stronger
m) temperature/energy
n) coal
o) steel
p) carbon
q) aluminium
r) energy
s) ores
t) iron
u) steel
v) cheap/economical
w) hard and strong
x) mild
y) stainless
z) copper and zinc
aa) solder
ab) tin and lead
ac) plumbing and electronics
ad) lightweight
ae) corrosion
af) drink cans/window frames/aircraft construction
ag) copper
ah) conductivity
ai) ductile
aj) decomposition
ak) endothermic
al) copper
am) aluminium
an) electricity
ao) metal
ap) energy
c) 4Li
+ O2
Part B Practice Problems
1.
Ca+(g)
a) 1st
Ca(g)
+ FeCl2
+
e-
2nd
Ca+(g)
Ca+2(g)
+
e-
3rd
Ca+2(g)
Ca+3(g)
+
e-
4th Ca+3(g)
Ca+4(g) + eb) Between 2nd and 3rd, because 3rd ionisation takes an
electron from an inner orbit.
2.
a) i) decreasing
ii) increasing
b) i) increasing
ii) decreasing
H= fluorine L= francium
c) i) increasing
ii) decreasing
H = helium L = francium
2Fe2O3
2Li2O
2.
a) Li, Fe, Pb
b) Lithium
3.
a) calcium + water
Ca
+ 2H2O
b) Tin + water
Sn + H2O
Ba(NO3)2
a) Dobereiner
b) Newlands
c) Octaves
d) Mendeleev
e) Periodic Table
f) been discovered
g) left gaps
h) predict
i) almost identical to the predictions
j) decreases
k) semi-metals & non-metals
l) increase
m) decrease
n) Transition
o) semi-metal
p) inert
q) right
r) identical
s) metals
t) ionic
u) positive
v) covalent
w) covalently
x) gain
y) negative
z) bottom
aa) upwards
ab) right
ac) top
ad) down and left
ae) decreases
af) more
ag) nucleus
ah) increases
ai) Ionisation
aj) increase
ak) remove
al) decrease
am) less
an) remove
ao) higher
ap) increases
aq) Electronegativity
ar) attract
as) fluorine
at) left
au) down
Part B Practice Problems
1.
a) Pb + O2
PbO2
3O2
+
Worksheet 3
Part A
a) oxide
b) METAL OXIDE
c) hydrogen
d) hydroxide (or oxide)
e) hydrogen
f) salt
g) different
h) activity
i) potassium and sodium j) left
k) copper and gold
l) iron and tin/lead/zinc
m) slowly
n) electrons
o) lose
p) hydrogen
q) covalent
r) hydrogen
s) H2
t) loss of electrons
u) Reduction
v) oxidized
w) hydrogen
x) reduced
y) Ionisation
z) remove one electron
aa) gas
ab) low
ac) higher
ad) gold used in electronics, because it will not corrode.
+
hydrogen + zinc sulfate
H2 + ZnSO4
hydrogen + calcium chloride
H2 + CaCl2
Ca + 2HCl
c) Barium + nitric acid
Ba + 2HNO3
H2
d) Iron + hydrochloric acid
Fe + 2HCl
H2
5.
a) the metals: Zn, Ca, Ba, Fe
b) hydrogen ions (from the acid)
c) sulfate, chloride and nitrate ions.
Worksheet 2
Part A
b) 2Fe
Zn + H2SO4
hydrogen + calcium hydroxide
H2 + Ca(OH)2
hydrogen + tin(II) oxide
H2 + SnO
Preliminary Chemistry Topic 2
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Worksheet 4
Part A
1. Molar Masses
a) 39.10g
d) 159.8g
g) NaI= 149.9
j) CuSO4=159.6g
b) 83.80g
e) N2 = 28.02g
h) Fe2S3=207.9g
k) Al2O3=102.0g
d) n(Sn) = m / MM = 14.8 / 118.7 = 0.125 mol
i) ∴ n(SnO2) = 0.125 mol
m(SnO2) = n x MM = 0.125 x 150.7 = 18.8g
ii) ∴ n(H2O) = 0.125 x 2 = 0.250 mol
m(H2O) = n x MM = 0.250 x 18.016 = 4.50g
iii) ∴ n(H2) = 0.125 x 2 = 0.250 mol
m(H2) = n x MM = 0.250 x 2.016 = 0.504g
c) 118.7g
f) MgO = 40.31g
i) NH3=17.03g
l) 180.2g
6. Empirical Formulas
a) i)
Copper
masses =
12.84g
No moles =
12.84/63.55
=
0.2020
ratio
=
1
∴ emp. formula is CuCl
ii) Copper(I) chloride
b) i)
Carbon
masses =
1.5g
No moles =
1.5/12.01
=
0.125
=
0.125/0.125
=
1
ratio
≅
1
∴ emp. formula is CH4
ii) methane
c) i)
Nitrogen
masses =
30%
No moles =
30/14.01
=
2.14
=
2.14/2.14
ratio
=
1
≅
1
∴ emp. formula is NO2
ii) 2 x (NO2) = N2O4
iii) dinitrogen tetra-oxide
2. Moles in a Given Mass
use n = m/MM
a) n= 100.0/207.2 = 0.4826 mol
b) n = 100.0/ 65.39 = 1.529 mol
c) n = 100.0 / 18.016 = 5.551 mol
d) n= 100.0 / 251.12 = 0.3982 mol
e) n = 38.55 / 62.31 = 0.6187 mol
f) n = 60.00 / 44.01 = 1.363 mol
g) n = 1.000/ 81.39 = 0.01229 mol
h) n = 500.0 / 180.2 = 2.775 mol
i) n = 3.258x10-3/ 58.44 = 5.575 x 10-5 mol
j) n = 128.6 / 96.094 = 1.338 mol
3. Moles & Particles
a) use n = N/NA and N= n x NA
i) N = 3 x 6.022x1023 = 1.807x1024 molecules
ii) N = 2.478 x 6.022x1023 = 1.492x1024 molecules
iii) N = 5 x 6.022x1023 = 3.011x1024 “sets” of ions.
iv) N = 0.007862 x 6.022x1023 = 4.734x1021 atoms
v) N = 1/1000 x 6.022x1023 = 6.022x1020 atoms
b) i) n = N/NA = 8.800x1025/6.022x1023 = 146.1 mol
m = n x MM = 146.1x118.7 = 17,340g (=17.34kg)
ii) n = N/NA = 2.575x1024/6.022x1023 = 4.276 mol
m = n x MM = 4.276 x 18.016 = 77.04g
iii) n = m/MM = 400.0/ 18.016 = 22.20 mol
N = n x NA = 22.20 x 6.022x1023 = 1.337x1025 molecules
iv) n = m/MM = 2.569/58.69 = 0.04377 mol
N = n x NA = 0.04377 x 6.022x1023 = 2.636x1022 atoms
v) n = N/NA = 2.500x1023/6.022x1023 = 0.4151 mol
m = n x MM = 0.4151 x 32.07 = 13.31g
: Hydrogen
: 0.5g
: 0.5/1.008
: 0.496
: 0.496/0.125
:
3.96
:
4
: Oxygen
: 70%
: 70/16
: 4.375
: 4.375/2.14
:
2.04
:
2
7. Volumes of Reacting Gases
a) i) 2.5L
ii) 5 L
b) i) 0.5 L
ii) 0.5 L
c) i) 20 L
ii) 10 L
d) i) 13.5 L
ii) 4.5 L
e) i) 0.4 L
ii) 0.2 L
4. Mole ratios in Equations
a) i) 0.5 mol
ii) 1 mol
b) i) 0.25 mol
ii) 0.125 mol
c) i) 1.5 mol
ii) 1.5 mol
d) i) 0.25 mol
ii) 0.375 mol
e) i) 0.2 mol
ii) 0.15 mol
8. Mass & Gas Volume
a) i) n(Li2O) = m/MM = 5,000/29.882 = 167 mol
ii) n(CO2) = 167 mol
iii) v(CO2) = 167 x 24.8 = 4.14x103 L (>4,000L !)
b) i) n(O2) = 10.0/24.8 = 0.403 mol
∴ n(Fe2O3) = 0.403 x2/3 = 0.269 mol
m(Fe2O3) = n x MM = 0.269 x 159.7 = 42.9g
ii) n(Fe) = m/MM = 100/55.85 = 1.79 mol
∴ n( O2) = 1.79 x 3/4 = 1.34 mol
V(O2) = 1.34 x 24.8 = 33.2 L
c) i) n(H2O) = m /MM = 1.00/ 18.016 = 0.0555 mol
∴n(H2) = 0.0555, v(H2) = 0.0555 x 24.8 = 1.38 L
and n(O2) = 0.0555/2, v(O2) = (0.0555/2) x24.8=0.688L
ii) use Gay-Lussac’s Law: v(H2) = 100mL (0.10 L)
iii) n(H2) = 0.10 / 24.8 = 0.00403 mol
∴ n(H2O) = 0.00403 mol
m(H2O) = n x MM = 0.00403 x 18.016 = 0.073g
5. Mass Quantities in Reactions
a) n(Ca) = m/MM = 8.50/40.08 = 0.212 mol
∴ n(CaO) = 0.212 mol
m(CaO) = n x MM = 0.212 x 56.08 = 11.9g
b) n(Ag2CO3) = m/MM = 20.0/275.81 = 0.0725 mol
i) ∴n(Ag) = 0.0725 x 2 = 0.145 mol
m(Ag) = n x MM = 0.145 x 107.9 = 15.6g
ii) ∴ n(CO2) = 0.0725 mol
m(CO2) = n x MM = 0.0725 x 44.01 = 3.19g
iii) ∴ n(O2) = 0.0725 / 2 = 0.03625 mol
m(O2) = n x MM = 0.03625 x 32.00 = 1.16g
c) n(Al) = m / MM = 6.50/26.98 = 0.241 mol
i) ∴ n(HCl) = 0.241 x 3 = 0.723 mol
m(HCl) = n x MM = 0.723 x 36.458 = 26.4g
ii) ∴ n(H2) = 0.241 x 3/2 = 0.3615 mol
m(H2) = n x MM = 0.3615 x 2.016 = 0.729g
iii) ∴ n(AlCl3) = 0.241 mol
m(AlCl3) = n x MM = 0.241 x 133.33 = 32.1g
Preliminary Chemistry Topic 2
Copyright © 2005-2007 keep it simple science
: Chlorine
: 7.16g
: 7.16/35.45
: 0.2019
:
1
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Worksheet 4
Part B
a) number
c) particles
e) Avogadro’s
g) ratio of moles
i) moles (particles)
k) empirical
m) volume
o) Avogadro
q) equal
s) temperature & pressure
u) 1 standard atmosphere
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21.
a) Mg + 2HCl
b) carbon (carbon-12)
d) formula mass
f) 6.022 x 1023
h) mass
j) reactions
l) Gay-Lussac
n) whole-number
p) volumes
r) molecules/particles
t) SLC
v) 25
c) 4K
22.
2K2O
+ O2
Ca(OH)2
Melting Point
Practice Questions
13. A
14. C
15. B
16. B
Part B Longer Response
In some cases there may be more than one
correct answer possible. The following “model”
answers are correct, but not necessarily perfect.
17.
a) Copper. Used for electrical wiring, due to its excellent
conductivity and high ductility.
b) Solder, an alloy of tin & lead. Used for joining pipes in
plumbing, and joining wires in electronics, because of its
very low melting point.
18.
a) not brittle/ can be re-sharpened/shape possiblities (saw)
b) Iron is stronger and harder... tools are superior.
c) Needs electricity for smelting.
19.
Mild Steel (0.2% carbon). Used for car bodies & sheet
metal, because it is strong but very malleable.
Tool Steel (1.5% carbon). Used for hammers, drills, etc
because it is very hard and strong.
Stainless Steel (20% nickel & chromium). Used for food
utensils and medical equipment because it resists corrosion
and is very hygenic.
20.
Small pieces of metal added to dilute acid in test tubes.
(To keep expt. fair, the acid must be same strength, and
metal pieces same size.)
Observe the rate of gas production to assess reactivity.
Conclusion: order of activity is: Mg > Zn > Fe > Pb > Cu
Preliminary Chemistry Topic 2
Copyright © 2005-2007 keep it simple science
H2 +
Ba
Ba+2 + 2e2H + 2e
H2
(Barium lost, hydrogen ions gained)
23.
a) i)
Mg(g)
Mg+(g) - + e+
ii) O(g)
O (g)+ e
b) The lower the ist Ionisation Energy the more active the
metal, because the metal readily loses electron(s) to enter a
reaction.
24
a) graph
b) At the beginning of a
period (left) the elements
are soft metals with
moderate to low mp’s.
Moving right, the mp’s rise
to a maximum at a transition
metal, or semi-metal. Then mp’s fall rapidly at the nonmetals. Lowest values are the inert gases on far right.
25.
K+(g) -+ ea) 1st:
K(g)
+
+2
2nd: K (g) +2
K (g) +
e
3rd:
K (g)
K+3(g) + eb) Between 1st & 2nd, because the 2nd ionization involves
an electron from an inner orbit, which will require a big
increase in energy to remove.
26.
a) 2Al + 6HCl
3H2
+ 2AlCl3
b) n(Al) = m / MM = 6.58 / 26.98 =2 0.244 mol23
i) N(Al)= n x NA=0.244x6.022x10 = 1.47x10 atoms
ii) n(AlCl3) = 0.244 mol
m(AlCl3) = n x MM = 0.244 x 133.33 = 32.5g
iii) n(H2) = 0.244 x 3/2 = 0.366 mol
V(H2) = 0.366 x 24.8 = 9.08 L
27.
Tin
:
Oxygen
% mass
88
:
12
moles =
88/118.7 :
12/16.00
=
0.74
:
0.75
≅
1
:
1
∴ empirical formula is SnO. Tin(II) oxide
28.a)
N2 + 3H2
2NH3
a) compounds
b) mineral
c) economically
d) metal
e) commercial price
f) production cost
g) extracting (smelting)
h) Chemical
i) yield
j) non-renewable
k) be replaced
l) geological
m) copper(I) sulfide & copper carbonate/hydroxide
n) Froth-flotation
o) lower
p) silicate
q) decomposition
r) oxygen
s) copper
t) sulfur dioxide
u) purify
v) electrolysis
w) recycle
x) aluminium
y) electrical
z) fossil
aa) coal
ab) Greenhouse
Multiple Choice
5. A
9. B
6. B
10. A
7. D
11. C
8. C
12. A
b) Ca + 2H2O
+
Worksheet 5
Part A
1. B
2. D
3. A
4. C
H2 + MgCl2
b) volumes =
100mL 300mL
200mL
Vol. ratio =
1
: 3
:
2
The volumes of the gases are in a simple, whole number
ratio to each other. This is Gay-Lussac’s Law.
29. a) A mineral is a naturally-occurring crystalline compound.
An ore is a mineral which is economically worth mining
to extract a metal from. All ores are minerals; not all
minerals are ores.
b) Chemical analysis allows an ore body to be analysed to
predict the yield of metal.
c) Ores are non-renewable resources, and once used
cannot be replaced. Therefore, it is wise to conserve these
resources by recycling metals wherever possible.
30. a) Copper(I) sulfide, Cu2S.
b) Crushed ore is separated by “froth flotation”. Low
density ore is carried in a detergent froth, while silicates
fall to the bottom.
c) Cu2S + O2
2Cu
+ SO2
d) Electrolysis. Copper needs to be very pure for its main
use in electrical wires. If impure, conductivity is lower.
32
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