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Chemistry 8.3 “Metals” PhotoMaster Copyright © 2005-2014 keep it simple science www.keepitsimplescience.com.au
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KEEP IT SIMPLE SCIENCE PhotoMaster Format
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Metals
Stage 6 Science Subject
Chem 8.3 NSW Syllabus Content reference
Chemistry Preliminary Course Topic 2
Firstly, an introduction... Technology Needs Metals The great sweep of human cultural development has many aspects... Language, Religion, Art & Music, and, of course, Technology. 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.
Chemistry of the Metals 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 Periodic Table is full of patterns
Dagger from the “Bronze Age”
s l a t e M
INSPECTION COPY for schools only 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.
Electrically powered smelter plant for extracting Aluminium from its ore
No
n-M
eta
ls
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. 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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Metals
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Topic Outline 5. Extracting Metals from Ores
1. Our Use of Metals
Mineral Ores
History
Case Study: Copper
Metals we use today
The Case for Recycling
2. Chemical Activity of the Metals
4. Chemical Quantities: the Mole
Activity Series Electron Transfer: REDOX 1st Ionisation Energy
The Mole & Avagadro’s Number Mole Ratios in Reactions Mole Calculations
3. The Periodic Table History Patterns of the Periodic Table
Gay-Lusacc’s Law & Avagadro’s Hypothesis Empirical Formulas
What is this topic about? To keep it as simple as possible, (K.I.S.S. Principle) this topic covers:
1. OUR USE OF METALS History of metal usage. Metals we use today and why.
2. CHEMICAL ACTIVITY OF METALS Activity Series. Electron transfer reactions: “REDOX”. First ionisation energy. More patterns.
3. THE PERIODIC TABLE Some chemical history: how the Periodic Table was invented. How to read some patterns within the Periodic Table.
4. CHEMICAL QUANTITIES How we measure chemical quantities: the mole & Avagadro’s number. Mole quantities in reactions. Mole calculations. Gay-Lussac’s Law & Avagadro’s Hypothesis. Calculating an empirical formula.
5. EXTRACTING METALS FROM ORES Metal extraction from ores. Case study: copper. Why recycle? Chemistry 8.3 “Metals” PhotoMaster Copyright © 2005-2014 keep it simple science www.keepitsimplescience.com.au
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1. OUR USE OF METALS
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The First Uses of Metals
The Iron Age (approx. 2,500 to 1,500 years ago)
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.
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 bronzearmed 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 big breakthrough was the discovery by these copper-using people that if they roasted copperbearing 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.
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. With bronze tools they built better ships and wagons for transport and trade, which brought wealth and power. 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. Metal has many advantages over stone, wood, or bone: • metal is harder, stronger, and flexible, not brittle.
It is no accident that the dominant world power of this time was ancient Rome, because their technology was based on iron.
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 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). The engines, tools and machinery of the great factories were based on steel. Transport was revolutionised 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. In the 20th century, new metals and alloys became available... aluminium, titanium, chromium, and many more. 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 can be cast, hammered or drawn into shapes not possible in stone, such as saw blades, swords and armour.
Human Progress has always been linked to our use of Metals.
• when tools become blunt, metal can be re-sharpened.
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
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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.
INSPECTION COPY for schools only 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.
Why is steel so widely used? • Iron ore occurs in huge deposits, so iron is common and economical to produce. • Steel (in its various forms) is hard and strong. • It can be cast, milled, rolled, worked, bent, cut and machined into any shape or size.
Aluminium is very lightweight, yet strong and corrosion resistant Its lightweight strength is perfect for aircraft construction. Lightweight and a good conductor, it is used for electricity power lines. 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 Alloy
Metal Extraction Needs Energy
Iron, with...
Properties
Uses
Mild steel 0.2% carbon
strong, but malleable
car bodies, pipes, roofing
Tool steel
very hard
drills, knives, hammers
resists corrosion, hygenic
food utensils, medical tools
1-1.5% carbon
Stainless 20% nickel Steel & chromium
Brass
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. 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.
is a common “non-ferrous” (no iron) alloy.
Brass is an alloy of copper and zinc (about 50% each)
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.
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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Our use of different metals through history can be linked to the availability of energy.
Iron ore requires more energy for decomposition. That’s why the “Iron Age” came later. Aluminium and other “modern” metals require even more energy, and electricity works better than heat, so these only became available in quite recent times. 5
Complete Worksheet 1
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2. CHEMICAL ACTIVITY OF THE METALS
Metals React With Oxygen One of the most familiar laboratory reactions is the burning of magnesium: Magnesium + Oxygen 2 Mg + O2
Magnesium oxide 2 MgO
In fact, many metals will burn, some a lot more readily and violently than magnesium: Sodium + Oxygen 4 Na + O2
Sodium oxide 2 Na2O
Metals React With Acids The different activity levels of the metals is most clearly seen when metals are reacted with dilute acids. You may have done experimental work to observe how vigorously different metals react with a dilute acid. Metals like calcium and magnesium react vigorously. 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. 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. Aluminium + Oxygen 4 Al + 3 O2
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.
The point is, that different metals have different chemical activities.
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.
2 Na
Water
+ 2 H2O
Hydrogen + Sodium (gas) hydroxide H2 + 2 NaOH
(In fact this is NOT the explosion reaction. The explosion is the reaction of the hydrogen with oxygen, to form water) Once again, some metals react easily and rapidly and form the metal hydroxide, while others react slowly if heated in steam, and form oxides. Zinc + Water Zn + H2O
Copper does not react at all. When there is a reaction, the gas produced is hydrogen.
Hydrogen + Zinc oxide H2 + ZnO
Metals like copper and gold do not react at all.
Bubbles of gas are produced. A flame test gives a “pop” explosion
The metal is “eaten away” and dissolves into the liquid. This is because it forms a soluble ionic compound. Exactly what the compound is, depends on which acid is used. Examples: Zinc + Hydrochloric acid Zn + 2 HCl Magnesium + Nitric acid Mg + 2 HNO3 Iron
What happens is: Sodium +
Lead is very slow indeed.
+
Sulfuric acid Fe + H2SO4
Hydrogen + Zinc chloride H2 + ZnCl2 Hydrogen + Magnesium nitrate H2 + Mg(NO3)2 Hydrogen H2
+ Iron(II) sulfate + FeSO4
The ionic compounds formed are collectively known as “salts”, so the general pattern of the reactions is Metal + Acid
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
Formula HCl H2SO4 HNO3
There is an “Activity Series” of metals. Chemistry 8.3 “Metals” PhotoMaster Copyright © 2005-2014 keep it simple science www.keepitsimplescience.com.au
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The Activity Series of the Metals
keep it simple science 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. From this, and other reaction studies, the common laboratory metals can be arranged in an “Activity Series”:
Most Active
Li
Al Zn
Zinc + Hydrochloric acid +
+
Zn + 2H
IN SP for EC sc TIO ho N C ols O on PY ly
Mg
To understand this, look again at the reaction between a metal and an acid: Hydrogen + Zinc (gas) chloride
2 HCl
H2
+
ZnCl2
HCl and ZnCl2 are both ionic compounds. Here is the equation re-written to show the individual ion “species”.
Na
Ca
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.
Zn
K
Ba
Electron Transfer in Metal Reactions
-
+ 2Cl
2+
H2 + Zn
-
+ 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. 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.
Fe Sn
Since they aren’t actually involved, we can leave the spectators out. This is called a “net equation”.
Pb Cu
Zn + 2H+
H2 + Zn2+
Ag Least Active
Now we can see what really happened; • a zinc atom became a zinc ion and • 2 hydrogen ions became a (covalent) hydrogen molecule.
Au
If you look for these metals on the Periodic Table you will notice a further pattern.
To do this, the zinc atom has to lose 2 electrons, and the hydrogen ions must gain a pair of electrons to share.
Positions of the first 6 metals of the Activity Series.
3 2 6 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.
Zn2+ + 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 memorising 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. 7
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Oxidation & Reduction
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First Ionisation 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 Ionisation Energy of an element is the energy required to remove an electron from an atom.
The reaction between zinc and acid can be like electrons transferred visualised this: +
For technical reasons, the measurement of this energy is carried out for atoms in the gas state. We know that zinc atoms normally lose 2
+
2 Hydrogen ions
Zinc atom
Zn(g) Hydrogen molecule
The energy required for this to happen is the “1st Ionisation Energy”
+2
2+
electrons to form the Zn ion. However, the formal definition for this process involves just the loss of 1 electron.
Zinc ion Covalent bond (2 electrons being shared)
Every element has its own characteristic value, even those elements which would not normally lose electrons, such as non-metals like chlorine.
For historical reasons, the loss of electrons is called “Oxidation”
Cl(g)
The zinc atom has lost 2 electrons,
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 Ionisation Energy”
and the hydrogen ions have gained electrons. -
2H + 2e
H2
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 Ionisation Energy value.
The gain of electrons is called “Reduction”
Neither process can occur alone... they must occur together
Ionisation Energy Determines the Activity Series 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 Ionisation Energy.
The zinc oxidation allows the hydrogen to be reduced, and the hydrogen reduction allows the zinc to be oxidised.
If the value for 1st Ionisation 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.
The total reaction is an “OxidationReduction” and is commonly abbreviated to “REDOX”. 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. Chemistry 8.3 “Metals” PhotoMaster Copyright © 2005-2014 keep it simple science www.keepitsimplescience.com.au
Cl+(g) + e-
Increasing values for 1st Ionisation Energy
Zn2+ + 2e-
Zn
+
Zn+(g) + e-
K Na Li Ba Ca Mg
If its value for 1st Ionisation energy is higher, the atom cannot react so readily or vigorously... its activity is lower. The ACTIVITY SERIES of the Metals is determined by 1st IONIsATION ENERGY 8
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Al Zn Fe Sn Pb Cu Ag Au
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Choice of Metals Based on Activity
keep it simple science Sometimes which metal is chosen for a particular application is based on its position in the Activity Series.
Another example is the choice of metals for water pipes. 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, copper is chosen, especially for hot water supply. Indoors a rusted-out leaking steel pipe would be a disaster, so it’s worth paying more for copper.
Gold’s extremely low chemical activity (due to a relatively high 1st Ionisation Energy) is part of the reason it has always been used for jewellery.
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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.
Gold’s low activity means it will not tarnish or corrode, so it retains its beautiful colour and lustre.
“Galvanised” steel is coated with a thin layer of zinc to prevent (or slow down) corrosion of steel roofing, fence wires, nails, bolts, etc.
Bronze & Gold have been used throughout history in Art and Religion
Complete Worksheets 2, 3 & 4. Chemistry 8.3 “Metals” PhotoMaster Copyright © 2005-2014 keep it simple science www.keepitsimplescience.com.au
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3. PATTERNS OF THE PERIODIC TABLE
Atomic Structure, Atomic Number and Mass Here is a quick reminder of some basics about atoms you need to know:
In the Nucleus are Protons & Neutrons
The Periodic Table In orbit around the nucleus are the Electrons
is firstly a list of the elements, arranged in order, and showing all the basic details. Atomic Number
18
Ar
Electrons = Protons = “Atomic Number”
Argon
Each element’s atoms have a different, characteristic, number of protons (and electrons). Therefore, each element has a different Atomic Number.
39.95
In the Periodic Table the elements are arranged in order of Atomic Number.
Protons + 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.
Equal to the number of protons in each atom. (Also equals the number of electrons in the neutral atom.)
Chemical Symbol Element Name “Atomic Weight” NOT the “Mass Number”
However, the Periodic Table is far more than a simple list. Why is it such a complicated shape? 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 Ionisation Energies. There are lots more...
(How and why this is different will be explained in a later topic)
History of the Periodic Table The modern concept of a chemical element developed almost exactly 200 years ago. 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:
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. Similar elements placed in vertical columns
Inert Gases had NOT been discovered
Lithium, sodium & potassium was one “triad”. Chlorine, bromine and iodine formed another “triad”. Mendeleev’s vertical “families” included Dobereiner’s “triads” and Newland’s “octaves”, but had one big difference...
By 1860, with over 60 known elements, Newlands (English) proposed a “Law of Octaves”. 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.
Mendeleev’s genius was to realise that there were probably missing elements that hadn’t been discovered yet. He cleverly left gaps in his table for these undiscovered elements.
The system worked well for the first 20 elements, but then became confused.
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 eka-silicon. Scientists went looking for such a substance and soon found a new element (which was named “Germanium”) with properties exactly as predicted.
The basis of the modern Periodic Table was developed by the Russian, Dmitri Mendeleev in 1869. Chemistry 8.3 “Metals” PhotoMaster Copyright © 2005-2014 keep it simple science www.keepitsimplescience.com.au
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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
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You learned in topic 1 how melting point is determined by the bonding within a substance. 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.
Electrical Conductivity As you go across any row (“period”) of the table, you will move through a number of metals, then one or two semi-metals, 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
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. Conductivity 2,000
decreasing
Melting Points of Elements
Periods 3
V
Si
Valencies are the same down each group
Peaks are Transition Metals or Semi-Metals
Period 4
1,000
o
Melting Point ( C)
Boiling Points follow a similar pattern to Melting Points
Sketch Graph.
Rb
K
0
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
-2
No chemical reactions, no bonding
-1
Activity of Metals Most active at bottom-left. Activity (generally) decreases upwards and to the right.
Metals (+ve ions)
Semi-Metals (Covalent only)
Bonding
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Non-Metals (Covalent or (-ve) ions)
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Activity of Non-Metals Most active at top-right (Fluorine) 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.
The following diagrams are to scale and show the relative sizes of the first 20 elements
He
H
37
Al
Si
160
Ca
143
N
O
F
Ne
P
S
Cl
Ar
70
70
68
66
110
118
102
94
99
Radius decreasing across a period
197
INSPECTION COPY for schools only The Syllabus requires that you produce a table and a graph of the changes in a property • across a period, and • down a group When you do, you can clearly see how the Periodic Table got its name. “Periodic” means “recurring at regular intervals”.
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.
Spreadsheet Plot of Atomic Radii 300
231
Mg
77
rend sing T Increa a group down Na
200
K
C
88
Rb
K
De acr c r e a s i oss n g a p Tren erio d d
Li
Kr
100
186
B
112
Ar Ne
He
g Trend Increasaingroup down
0
Na
Be
Atomic Radius (picometre)
152
Radius increasing down a group
Li
50
The numbers given are the atomic radii in picometres. 1 picometre = 1x10-12 metre
1
10
20
30
Atomic Number 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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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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Successive Ionisation Energies
Ionisation Energy
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. Once the first electron is removed, the
keep it simple science The meaning of the “1st Ionisation Energy” was explained previously in relation to the Activity Series of Metals.
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.
You should remember that the very active metals are the ones with low 1st ionisation energies. They easily lose their outer electron(s) and so react readily.
decreasing
2nd I.E.
A
+ (g)
A
2+ (g)
+
e
3rd I.E.
A
+2 (g)
A
3+ (g)
+
e
Highest value
-
Patterns in Successive Ionisation Energy Data (values shown are energy units) 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.
Element
Electron Config.
Sodium
2.8.1
0.5
Magnesium 2.8.2
0.7
Aluminium 2.8.3
0.6
1st I.E.
2nd I.E.
3rd I.E. 6.9
9.6
1.4
7.7
10.5
1.8
2.8
11.6
4.5
Highest Value Fluorine (values decrease down)
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1.0 1.5 0.9
4th I.E.
as the next ionisation has to remove an electron from the next lower orbit.
Electronegativity
Atoms with a tendency to gain electrons and form negative ions have high values. 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.
-
Once the entire outer orbit has been stripped away, the next ionisation must remove an electron from an underlying orbit, which requires a huge increase in the next ionisation energy. This results in an interesting pattern: Notice how the values “jump” (underlined data)
increasing
is a value assigned to each element to describe the power of an atom to attract electrons to itself.
+
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. Therefore, each successive ionisation requires more energy.
The trend for the whole Periodic table is:
Lowest
e-
A(g)
...and so on, according to how many electrons the atom has
The Periodic Trend in 1st Ionisation Energy
1st Ionisation Energy
A+(g)
1st I.E.
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
0.8
2.8
0.8
2.5
0.7
2.2
0.7
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4. QUANTITY CALCULATIONS & THE MOLE
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Quantities in Chemical Calculations 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. For example, when hydrogen and oxygen react to form water, the balanced equation is 2H2 + O2
2H2O
This is a true description of what is happening to the molecules:
2 Molecules of H2
+
1 Molecule of O2
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. 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
The Mole
1 mole is a quantity of a chemical substance. 1 mole of any element or compound contains exactly the same number of particles.
Defining the Mole For technical reasons, the “atomic standard” used to compare the masses of all atoms is the carbon atom, which contains 6 protons 6 neutrons 6 electrons Atomic Number = 6 Mass Number = 12
+
6p 6n0
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” 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. This WILL be explained in a later topic. If you cannot wait, go find out about “Isotopes”.
Avogadro’s Number Just how many atoms are in 1 mole? Obviously, it is a very large number. We now know that it is about 6,000 billion trillion.
1 mole of each substance has a different mass, because the atoms and molecules all weigh differently.
Avogadro’s Number 23
6.022 x 10
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
C
18
This number is named in honour of an Italian scientist who you will learn about soon.
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82
Ar Pb
Carbon
Argon
Lead
12.01
39.95
207.2
1 mole
1 mole
1 mole
= 12.01 grams
= 39.95 grams
= 207.2 grams
207.2 grams of Lead contains 6.022 x 1023 Lead atoms
39.95 grams of Argon contains 6.022 x 1023 Argon atoms
EACH OF THESE HAS THE SAME NUMBER OF ATOMS
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12.01 grams of Carbon contains 6.022 x 1023 Carbon atoms
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Calculating Mole Quantities
Moles & Numbers of Particles Since one mole of any substance contains Avogadro’s Number of particles:
You need to be able to calculate mole quantities in terms of both mass and number of particles.
No. of moles = No. of particles you have Avogadro’s Number
Molar Mass The “Molar Mass” of any chemical species is the mass (in grams) of 1 mole of the substance.
n= N NA
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
(for elements like these just use Atomic Weight)
Oxygen Chlorine
O2 Cl2
Solution
n= N NA
(16.00 x 2) = 32.00 (35.45 x 2) = 70.90
= 7.88x1024 23 6.022x10 = 13.1 mol
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)
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
Solution so N = n x NA = 0.0250 x 6.022x1023 a) n = N NA = 1.51 x 1022 atoms
Worksheet at the end of this section
b) m = n x MM = 0.0250 x 207.2 (molar mass of Pb) = 5.18 g
(add up At.weights of all atoms in the formula)
Mole Quantities in Chemical Equations
Number of Moles in a Given Mass
When you consider an equation like
When you weigh a chemical sample you then need to be able to calculate how many moles this contains.
2H2 + O2 you know it means
2H2O
No. of moles = mass of substance you have molar mass
n= m MM Example Calculations 1. How many moles in
Solution
a)
b)
2 Molecules of H2
n = m = 5.23 = 0.215 mol MM 24.31 =
96.7 (2x1.008 + 16.00) = 96.7/18.016 = 5.37 mol
2 million H2 + 1 million O2
2 million H2O
or, 200 zillion H2 + 100 zillion O2
200 zillion H2O
= 2 moles H2 + 1 mole O2
2 moles H2O
The Balancing Coefficients in a Chemical Equation May be Interpreted as Mole Ratios
so m = n x MM = 1.50 x (22.99 + 35.45) = 1.50 x 58.44 = 87.7 g
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2 Molecules of H2O
or, (let’s use Avagadro’s number) (2 x NA) H2 + NA O2 (2 x NA) H2O
2. What mass is needed if you want to have 1.50 moles of salt (sodium chloride)? n= m MM
1 Molecule of O2
However, the number of molecules reacting is really just a ratio. The actual numbers might be
a) 5.23g of magnesium? b) 96.7g of water?
n= m MM
+
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Calculating Mass Quantities in Reactions
Mole Quantities in Chemical Equations (cont.)
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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. So,
2 H2
+
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.
Solution Always start with the balanced equation:
or any other proportional quantities. Example Problem a) If 0.50 mol of sodium reacted completely with hydrochloric acid, how many moles of products would be formed?
4 Al mole ratio
Solution a) The balanced equation is
so,
+ 2 HCl :
2 mol
H2 :
0.50 mol : 0.50 mol :
3
:
2 = 4.29 26.98 = 0.159 mol
a) Mass O2 consumed: mole ratio Al : O2 = 4 : 3 = 0.119 mol \ moles of O = 0.159 x 3 2 4 \ mass of O : m = n x MM = 0.119 x 32.00 2 = 3.81 g
+ 2 NaCl
1 mol
:
2 Al2O3
No. of moles of Aluminium: n = m MM
b) What mass of each product would be formed?
2 Na mole ratio 2 mol
4
+ 3 O2
: 2 mol.
0.25 mol : 0.50 mol
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 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
\
mass of Al2O3: m = n x MM = 0.0795 x 101.96 = 8.11 g
INSPECTION COPY for schools only Practical Work:
Using Mass & Mole Ratios to Determine a Formula
ceramic crucible
A common experiment is to burn a piece of magnesium in a crucible, as suggested by the diagram. Reaction:
Magnesium + Oxygen
Magnesium oxide
Careful measurement of mass allows the empirical formula for magnesium oxide to be determined.
Typical Measurements Mass of empty crucible = 42.74 g Mass of magnesium = 2.05 g Mass of crucible + product after burning = 46.22 g Mass of magnesium oxide formed = 3.48 g \ Mass of oxygen in compound = 1.43 g \
Analysis of Results Remember that to convert any mass to moles: n = m / MM Elements Ratio of masses: Ratio of moles: (divide by Atomic Weight)
Magnesium : 2.05 g : 2.05 / 24.31 :
= 0.0843 Simplified ratio = 0.0843/0.00843 : = 1.0 : Nearest whole number ratio 1 : \ Empirical Formula is MgO
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Oxygen 1.43 g 1.43 / 16.00
mol : 0.0894 mol 0.0894/0.0843 (divide both by the 1.06 smaller) 1
There is often a large error due to incomplete burning
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A Little History... How the Mole was Invented
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Avogadro’s Hypothesis
Gay-Lussac’s Law
The Italian, Amadeo Avogadro (1776-1856) was trained in Law, but became very interested in Science.
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.
In 1811, he noticed the similarity between GayLussac’s Law (an empirical “law” based on experiment) and the concept that atoms must combine in simple, whole number ratios to form compounds.
After a series of clever experiments, in which the volumes of reacting gases were measured, in 1808 he proposed the “Law of Combining Volumes”:
This led him to make an hypothesis:
Equal Volumes of all Gases Contain Equal Numbers of Molecules
When measured at constant temperature and pressure, the volumes of gases in a chemical reaction show simple, wholenumber ratios to each other.
(when measured at the same conditions of temperature and pressure) This was a vital breakthrough in the history of Chemistry.
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.
For example, consider the reaction:
Examples of Gay-Lussac’s Law Hydrogen(g) + Chlorine(g) 1 litre 1 litre Hydrogen(g) + Oxygen(g) 2 litres 1 litre Hydrogen(g) + Nitrogen(g) 3 litres 1 litre
Hydrogen(g) + Chlorine(g)
Hydrogen chloride(g) 2 litres
Prior to Avogadro, it was assumed that the the reaction involved single atoms, like this:
Water(g) (vapour) 2 litres
H(g)
Hydrogen(g) + Chlorine(g)
1 volume :
2 H2(g) +
Cl2(g)
2 HCl(g)
O2(g)
2 H2O(g)
3 H2(g) + N2(g)
2 NH3(g)
HCl(g)
1 volume
:
Hydrogen chloride(g)
2 volumes
Now, reasoned Avogadro, gases react in simple, whole-number 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!
Now consider the balanced equations for these three example reactions: +
+ Cl(g)
but the combining volumes (discovered by experiment) were
Ammonia(g) 2 litres
Notice that in every case, that the volumes are always in a simple, whole number ratio to each other.
H2(g)
Hydrogen chloride(g)
i.e. Hydrogen is H2(g) and Chlorine is Cl2(g), and the correct equation is H2(g)
The mole ratios are the same as the volume ratios discovered by Gay-Lussac!
+
Cl2(g)
2 HCl(g)
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.
Why should this be?
... enter Avogadro!
Although he did not invent the concept of the mole, we name the number of particles in 1 mole in Avogadro’s honour... Avogadro’s Number. Chemistry 8.3 “Metals” PhotoMaster Copyright © 2005-2014 keep it simple science www.keepitsimplescience.com.au
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Comparing Mass Changes When Metals Burn
Molar Volume of a Gas
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.
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,
That’s why we need the mole... we measure quantities by their mass, but this makes no sense until moles are calculated.
if measured at the same temperature and pressure.
The syllabus requires that you should consider the mass changes involved when various metals combine with oxygen to form their oxide compound.
This volume is the “Molar Volume” and is the same for every gas. Usually, the volume is measured at 25oC and a pressure of 100 kPa. (kPa = kilopascals, the normal unit for measuring gas pressures in Chemistry.)
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)
1 mole of any gas = 24.79 litres o
Mass of Oxide formed
at 25 C and 100 kPa
115
215
Mole Calculations with Gases
Fe2O3
43
143
This additional knowledge opens up the opportunity to carry out quantity calculations which involve mass and volumes of gases.
Zinc
ZnO
49
149
Lead
PbO2
15
115
Lithium
Li2O
Iron
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 25C, 100kPa)?
Empirical & Molecular Formulas You are reminded that a molecular formula really does describe the atoms present in a molecule.
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
The molecular compound methane, has formula CH4, because that’s exactly what each molecule contains... 1 carbon atom and 4 hydrogen atoms. Lattice structures, either ionic or covalent are NOT molecular. Example: sodium chloride, NaCl
\
Volume of CO2 = 0.1564 x 24.79 = 3.877 L (at 25C, 100kPa)
Molar Vol. of all gases at 25C, 100kPa
The formula does NOT describe a molecule, but only gives the simplest ratio between the bonded atoms... this is an empirical formula.
2. What volume of hydrogen gas (at 25C, 100kPa) would be produced if 10.00g of lithium metal was reacted with sulfuric acid?
Earlier was an example of how formulas are determined by analysing the mass composition of a compound.
Solution 2 Li(s) + H2SO4(aq) 2 : 1
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.)
= 10.00 = 1.441 mol Moles of lithium: n = m MM 6.941 Mole ratio is 2:1, so moles of H2 = 1/2 x 1.441=0.7204 \
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:
H2(g) + Li2SO4(aq) 1 : 1
Volume of H2 = 0.7204 x 24.79 = 17.86 L (at 25C, 100kPa)
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5. METALS FROM THEIR ORES
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The Importance of Predicting Yield from an Ore
Ores and Minerals ... and now back to the metals. 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: • 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.
The whole situation of economic feasibility makes the science of Analytical Chemistry vital in the mining and metals industry. 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.
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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.
Photo courtesy of Comalco Aluminium Ltd
Ores are Non-Renewable Resources
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!
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Minerals and ores have been formed over millions and billions of years of geological processes on Earth. Because of that time-frame, the ores are nonrenewable 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. 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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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. Compressed air blown in to create a froth of bubbles
These compounds usually occur as thin “veins” of blue-green minerals embedded in masses of worthless silicate rock.
Froth
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”.
Crushed Ore in a slurry of water and “wetting oil”
Waste Mineral Slurry
Chemistry of Smelting The concentrated copper minerals now undergo Decomposition Reactions.
Compressed air creates a froth of bubbles in a detergent solution.
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
Froth overflows for collection
Copper minerals, sprayed with a special oil, cling to the bubbles and are carried upwards to overflow with the froth.
Copper + Sulfur dioxide
2Cu + SO2
Silicate minerals are wetted by the water and, being denser, sink to the bottom.
The copper collected is about 98% pure. The collected froth is then treated to separate the oil and detergent for re-use.
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.
The ore concentrate is now about 30% copper.
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
Cu2+ + 2e-
The copper dissolves into the solution, but impurities do not.
Impure Copper dissolves into solution
Cu+2
ions migrate through CuSO4 solution
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Pure Copper deposits on electrode
After migrating through the solution, the ions are redeposited as pure copper metal on the other electrode: 2+
Cu
+ 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 nonrenewable 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. 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.
Recycling aluminium requires about 7kJ of energy, a saving of about 96% in energy and environmental impact!
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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.
Scrap Metal awaiting recycling. Photo by Pawel Grabowski
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Worksheet 1
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Our Use of Metals
Fill in the blank spaces.
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)................................ 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. 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. In the 20th century new metals such as q).............................. became available because the r).................................. needed to extract it from its s)................... was available.
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Student Name...........................................
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. 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 .................................... 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. 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)......................................
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Worksheet 2
Chemical Activity of the Metals
Fill in the blank spaces. When a metal reacts with oxygen it forms an a)......................... compound. METAL + OXYGEN
b) .............................
Some metals will also react with water, forming c)..................................... gas and a d)...................................... compound. METAL + WATER
c).................. + d).................
Most metals will react with acids, forming e).......................... gas and an ionic compound called a “f)...........................” METAL + ACID
e)....................... + f).................
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. Metals such as i).............................. and ............................. are the most active. These are the elements located in the j)........................... columns of the Periodic Table. 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).......................................
Student Name........................................... 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)........... “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).................................................. 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)........................................... Sometimes the choice of which metal to use is determined by the activity level. An example is ad)............................... .... ....................................... .............................................
Worksheet 3 Practice Problems
Student Name...........................................
a) Lead
(assume lead(IV) ion forms)
4. All the following equations are Metal + Acid reactions. Fill in all blank spaces, then re-write in symbols and balance.
b) Iron
(Assume iron(III) ion)
a) Zinc + Sulfuric acid
1. Write a balanced, symbol equation for the reaction of each of the following metals with oxygen.
................. +.....................
c) Lithium b) Calcium + Hydrochloric acid 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? 3. Write a word equation AND a balanced, symbol equation to describe the reaction of: a) calcium metal with water (reacts spontaneously at room temperature)
................. +...................
c).................. +......................
Hydrogen + Barium nitrate
d).................... + ....................
Hydrogen + iron(II) chloride
5. For each of the reactions in Q4, which chemical species a) lost electrons? b) gained electrons?
b) Tin metal with water (heated in steam) (Assume tin(II))
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c) was a “spectator”?
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Worksheet 4
Test Qestions
keep it simple science sections 1 & 2
Multiple Choice 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 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 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. 4. A metal which reacts readily and vigorously with oxygen, water and dilute acids would probably: A. have a high value for 1st ionisation energy. B. be from the “Transition” block of the Periodic Table. C. have a very low 1st ionisation energy. D. be located at extreme right of the Periodic Table. 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.
Longer Response Questions Mark values shown are suggestions only, and are to give you an idea of how detailed an answer is appropriate. Answer on reverse if insufficient space. 7. (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).
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Student Name........................................... 8. (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. 9. (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.
10. (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.
11. (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. 12. (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.
13. (4 marks) a) Write an equation (including states) for the first ionisation of i) magnesium ii) oxygen b) Describe how the Activity Series of Metals is related to the values of 1st Ionisation Energy.
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Worksheet 5
Patterns of the Periodic Table
Fill in the blank spaces.
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. It was the Russian d)................................. who invented the e)...................... ......................, in more or less its modern form. He realised 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. Sure enough when discovered, the missing elements were found to have properties i)............................................ The patterns in the Periodic Table include: which generally Conductivity, j)............................ to the right, as you go from metals to k)................................. and ............................... 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) 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 Nonmetals can bond w)................................ or can x).................. electrons to form y)................. ions. Chemistry 8.3 “Metals” PhotoMaster Copyright © 2005-2014 keep it simple science www.keepitsimplescience.com.au
Student Name...........................................
Chemical Reactivity is different for metals and non-metals. 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 ........................... 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. 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. Successive Ionisation 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. 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. 25
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Worksheet 6 Test Questions
Patterns of the Periodic Table Student Name...........................................
Multiple Choice 1. The scientist most responsible for development of the Periodic Table was: A. Avogadro B. Newlands C. Gay-Lussac D. Mendeleev
the
6. (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.
2. Element “X” is in Group 2 and element “Y” in Group 7.
If X & Y formed a compound, you would expect it to be A. B. C. D.
ionic, with formula X2Y covalent, with formula X2Y ionic, with formula XY2 covalent, with formula Y2X
3. If the elements “X” & “Y” in Q2 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 ionisation energy than Y.
D. Y to have a higher melting point than X. 4. 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.
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b) Briefly explain the general trend shown in your graph. 7. 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. a) Atomic Radius
C. decreasing attraction of electrons to the nucleus.
ii)
D. increasing mass of the atoms.
Longer Response Questions Mark values shown are suggestions only, and are to give you an idea of how detailed an answer is appropriate. Answer on reverse if insufficient space.
5. (5 marks) a) Write equations to represent the 1st, 2nd, 3rd & 4th ionisations for a calcium atom.
i)
b) Electronegativity Also indicate (“H”&“L”) the position of elements with highest & lowest values.
i)
ii)
c) 1st Ionisation Energy
i)
Show“H”&”L” ii)
b) Between which two of these successive ionisations would you expect a huge increase in the required energy?
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Worksheet 7 Masses, Moles & Particles Student Name........................................... Practice Problems
1. Molar Masses
3. Moles and Number of Particles
Calculate the molar mass of:
a) How many particles (atoms/molecules) in:
a) potassium c) tin e) nitrogen gas g) sodium iodide i) ammonia k) aluminium oxide
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
b) krypton d) bromine (Br2) f) magnesium oxide h) iron(III) sulfide j) copper(II) sulfate l) glucose (C6H12O6)
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?
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? ii) You have a sample containing 2.575x1024 molecules of water. How many moles is this, and what is its mass? iii) If you weigh out 400.0g of water, how many moles is this, and how many molecules are present? iv) If you have 2.569g of pure nickel, how many atoms are there? v) What mass of sulfur would contain 2.500x1023 atoms?
Worksheet 8 Mole Ratios & Mass in Reactions Student Name ......................................... Practice Problems 1. Mole Ratios in Equations
2. Mass Quantities in Reactions
Sodium reacts with water as follows: 2Na + 2H2O H2 + 2NaOH
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?
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?
b) Silver carbonate decomposes when heated: 2Ag2CO3 2CO2 + 4Ag + 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?
d) If 0.5 mole of Al used, how many moles of i) Alum.oxide formed? ii) oxygen used?
c) Aluminium reacts with hydrochloric acid: 2Al + 6HCl 3H2 + 2AlCl3 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?
e) If 0.1 mole of alum.oxide formed, how many moles of i) aluminium used? ii) oxygen used?
d) Tin reacts with steam as follows: Sn(s) + 2H2O(g) 2H2(g) +
Aluminium reacts with oxygen: 4 Al + 3 O2 2 Al2O3
FOR MAXIMUM MARKS SHOW FORMULAS & WORKING, APPROPRIATE PRECISION & UNITS IN ALL CHEMICAL PROBLEMS
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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?
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Worksheet 9 Practice Problems
®
keep it simple science
1. 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. 2. 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.
Empirical Formulas Student Name ......................................... 3. 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.
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Worksheet 10 Reactions Involving Gases Student Name ......................................... Practice Problems 1. Volumes of Reacting Gases
( Assume all gases are measured at the same temperature & pressure)
2 H2(g) +
O2(g)
2 H2O(g)
a) If you used 5 litres of hydrogen, how many litres: i) of oxygen consumed?
2. Mass & Gas Volume Calculations o
All volumes measured at 25 C, 100kPa
a) To “scrub” the air and remove poisonous CO2 on board the Space Shuttle, the air is continually pumped through canisters containing 5.00kg of lithium oxide. The reaction is Li2O(s) + CO2(g)
Li2CO3(s)
ii) of water vapour formed?
i) How many moles of lithium oxide in each canister?
b) If you used 0.25 litres of oxygen, how many litres of i) water vapour formed?
ii) How many moles of CO2 can this absorb?
ii) hydrogen consumed? c) If this reaction produced 20 litres of steam, how many litres of i) hydrogen consumed? ii) oxygen consumed? 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? e) If 0.6 litre of hydrogen reacted, what volume i) of ammonia formed? ii) of nitrogen need?
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iii) What volume of CO2(g) is this? b) Iron reacts with oxygen: 4Fe(s) + 3O2(g)
2Fe2O3(s)
i) If 10.0L of O2 reacted, what mass of iron(III) oxide would be formed? ii) If 100g of iron reacted, what volume of oxygen would be needed? c) The electrolysis decomposition: 2H2O(l)
of
water
2H2(g)
+
causes O2(g)
i) If 1.00g of water was decomposed, what volumes of each gas (measured at 25C, 100kPa) would be formed?
In an electrolysis experiment, 50mL (0.050 L) of oxygen was produced. (at 25C, 100kPa) ii) What volume of hydrogen (at 25C, 100kPa) was produced? iii) What mass of water must have been decomposed?
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Worksheet 11
Mole Concept
Fill in the blank spaces.
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)....................... In a balanced chemical equation, the “balancing numbers” (coefficients) may be interpreted as being g)........................... .............................. of reactants and products.
Student Name........................................... 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. 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 a o pressure of t)................ and temp. of u)........ C.
Worksheet 12 Calculations & the Mole Student Name........................................... Test Questions 5. Carbon monoxide gas reacts with oxygen gas to Multiple Choice form carbon dioxide gas as follows: 1. 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. 2. 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 3. Aluminium reacts with oxygen to form aluminium oxide. 4 Al + 3 O2 2 Al2O3 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 4. 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: A. 2 x NA C. 8 x NA
B. 4 x NA D. 10 x NA
Chemistry 8.3 “Metals” PhotoMaster Copyright © 2005-2014 keep it simple science www.keepitsimplescience.com.au
2CO(g) + O2(g)
2CO2(g)
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
Longer Response Questions 6. (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 25C, 100kPa) 7. (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. 8. (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.
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Worksheet 13
Metals from Ores
Fill in the blank spaces.
“Minerals” are naturally occurring a).............................. which are mixed together in rocks.
Student Name...........................................
An “ore” is a b)............................. from which it is c)................................ worthwhile to extract a desired d)..............................
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.
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.
The “smelting” process involves q)...................................... reactions. For a sulfide ore, it reacts with r)....................... to form s)...................... metal and t)............................. gas.
h)........................... analysis of an ore deposit is vital to predict the i)..................................... from the ore, to determine if it is worth mining.
The final step is to u)........................... the copper by a process of v)............................................
Ores are j)........................................... resources because once used, they cannot k)........................................... due to the immense time it takes for l)....................................... processes to form them. Copper ores contain compounds such as m)........................... and ......................................
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.
Worksheet 14 Metals from Ores Test Questions Multiple Choice 1. The “smelting” of a metal ore always involves: A. separating the metal mineral from the rock. B. decomposing a compound of the metal. C. purifying the extracted metal by electrolysis. D. all of the above.
Longer Response Questions Mark values shown are suggestions only, and are to give you an idea of how detailed an answer is appropriate. Answer on reverse if insufficient space.
2. (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.
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Student Name........................................... 3. (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”.
INSPECTION COPY for schools only 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))
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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Answer Section 4. a)
Worksheet 1 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
b) Ca
Worksheet 4 1. B
c) 4Li
+ O2
2. a) Li, Fe, Pb b) Lithium
Ca
+
2H2O
b) Tin + water Sn + H2O
3. A
4. C
5. A
6. B
10. (example answer) 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: Mg > Zn > Fe > Pb > Cu 11. a) Mg + 2HCl
2Fe2O3
b) Ca + 2H2O
2Li2O
c)
hydrogen + calcium hydroxide H2 + Ca(OH)2
4K
+ O2
H2 + MgCl2 H2 +
Ca(OH)2
2K2O
12.
Ba Ba2+ + 2e2H+ + 2eH2 (Barium lost, hydrogen ions gained) 13. a) i) Mg(g) ii) O(g)
hydrogen + tin(II) oxide H2 + SnO
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2. D
9. 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.
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3. a) calcium + water
Ba(NO3)2
8. a) not brittle/ can be re-sharpened/shape possiblities (e.g. saw) b) Iron is stronger and harder... tools are superior. c) Needs electricity for smelting.
PbO2 3O2
+
7. 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.
Worksheet 3 +
H2
d) Iron + hydrochloric acid + FeCl2 Fe + 2HCl H2 5. a) the metals: Zn, Ca, Ba, Fe b) hydrogen ions (from the acid) c) sulfate, chloride and nitrate ions.
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.
b) 2Fe
hydrogen + calcium chloride H2 + CaCl2
+ 2HCl
c) Barium + nitric acid Ba + 2HNO3
Worksheet 2
1. a) Pb + O2
hydrogen + zinc sulfate H2 + ZnSO4
Zn + H2SO4
Mg+(g) + O (g)+ e-
+ e-
b) The lower the ist Ionisation Energy the more active the metal, because the metal readily loses electron(s) to enter a reaction.
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Worksheet 5 a) Dobereiner
2. Moles in a Given Mass
b)
1. D
2. C
3. B
5. a) 1st Ca(g)
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. 23 21 iv) N = 0.007862 x 6.022x10 = 4.734x10 atoms 23 20 v) N = 1/1000 x 6.022x10 = 6.022x10 atoms 25
+
e-
2nd Ca+(g)
Ca2+(g)
+
e-
3rd Ca2+(g)
Ca3+(g)
+
e-
4th Ca3+(g)
Ca4+(g)
+
e-
Worksheet 8 1. 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
b) Between 2nd and 3rd, because 3rd ionisation takes an electron from an inner orbit.
2. 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
Melting Point
6. a) graph b) At the beginning of a period (left) the elements are soft metals with moderate to low mp’s.
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
Moving right, the mp’s rise to a maximum at a transition metal, or semi-metal. Then mp’s fall rapidly at the non-metals. Lowest values are the inert gases on far right.
7. a) i) decreasing b) i) increasing H= fluorine c) i) increasing H = helium
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
ii) increasing ii) decreasing L= francium ii) decreasing L = francium
Worksheet 7 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
23
b) i) n = N/NA = 8.800x10 /6.022x10 = 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
4. A Ca+(g)
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 / 187.6 = 0.5330 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 -3 -5 i) n = 3.258x10 / 58.44 = 5.575 x 10 mol j) n = 128.6 / 96.094 = 1.338 mol
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
Worksheet 6
use
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
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Worksheet 11
Worksheet 9
1. i) Copper masses = 12.84g No moles = 12.84/63.55 = 0.2020 ratio = 1 \ emp. formula is CuCl ii) Copper(I) chloride
: : : : :
Chlorine 7.16g 7.16/35.45 0.2019 1
2. i) Carbon masses = 1.5g No moles = 1.5/12.01 = 0.125 = 0.125/0.125 = 1 ratio @ 1 \ emp. formula is CH 4 ii) methane
: : : : : : :
Hydrogen 0.5g 0.5/1.008 0.496 0.496/0.125 3.96 4
3. i) masses = No moles = = = ratio =
: : : : : : :
Oxygen 70% 70/16 4.375 4.375/2.14 2.04 2
Nitrogen 30% 30/14.01 2.14 2.14/2.14 1 @ 1 \ emp. formula is NO 2 ii) 2 x (NO2) = N2O4 iii) dinitrogen tetra-oxide
a) number b) carbon (carbon-12) c) particles d) formula mass e) Avogadro’s f) 6.022 x 1023 g) ratio of moles h) mass i) moles (particles) j) reactions k) empirical l) Gay-Lussac m) volume n) whole-number o) Avogadro p) volumes q) equal r) molecules/particles s) temperature & pressure t) 100 kPa u) 25
Worksheet 12 1. C
ii) ii) ii) ii) ii)
3. A
4. C
5. B
6. a) 2Al + 6HCl 3H2 + 2AlCl3 b) n(Al) = m / MM = 6.58 / 26.98 = 0.244 mol i) N(Al)= n x NA=0.244x6.022x102= 1.47x1023atoms 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 7. : Oxygen Tin % mass 88 : 12 moles = 88/118.7 : 12/16.00 = 0.74 : 0.75 @ 1 : 1 \ empirical formula is SnO. Tin(II) oxide
Worksheet 10 1. a) i) 2.5L b) i) 0.5 L c) i) 20 L d) i) 13.5 L e) i) 0.4 L
2. A
5L 0.5 L 10 L 4.5 L 0.2 L
8.a)
N2 + 3H2
2NH3
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.
2. a) i) n(Li2O) = m/MM = 5,000/29.882 = 167 mol ii) n(CO2) = 167 mol
Worksheet 13
iii) v(CO2) = 167 x 24.8 = 4.14x103 L (>4,000L !)
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
b) i) n(O2) = 10.0/24.8 = 0.403 mol \ n(Fe O ) = 0.403 x2/3 = 0.269 mol 2 3 m(Fe2O3) = n x MM = 0.269 x 159.7 = 42.9g ii) n(Fe) = m/MM = 100/55.85 = 1.79 mol \ n( O ) = 1.79 x 3/4 = 1.34 mol 2 V(O2) = 1.34 x 24.8 = 33.2 L c) i) n(H2O) = m /MM = 1.00/ 18.016 = 0.0555 mol \ n(H ) = 0.0555, v(H ) = 0.0555 x 24.8 = 1.38 L 2 2 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(H O) = 0.00403 mol 2 m(H2O) = n x MM = 0.00403 x 18.016 = 0.073g
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Worksheet 14
1. B 2. 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.
FOR MAXIMUM MARKS SHOW FORMULAS & WORKING, APPROPRIATE PRECISION & UNITS IN ALL CHEMICAL PROBLEMS
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. 3. 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.
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