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keep it simple science TM Emmaus Catholic College SL#802440 Preliminary Chemistry Topic 2 METALS What is this topic about? To keep it as simple as possible, (K.I.S.S.) this topic involves the study of: 1. OUR USE of METALS 2. CHEMICAL ACTIVITY of the METALS 3. PATTERNS of the PERIODIC TABLE 4. QUANTITY CALCULATIONS... the MOLE 5. METALS from their ORES ...all in the context of how Chemistry contributes to cultural development but first, an introduction... Technology Needs Metals Chemistry of the Metals The great sweep of human cultural development has many aspects... Language, Religion, Art & Music, and, of course, Technology. In the previous topic you learnt about the Elements of the Periodic Table. In this topic you will concentrate on the chemistry of the metals, and some of the chemical patterns that they show. ... and Speaking of Patterns, in this topic you will find that The history of technology is closely linked with our use of metals; in fact historians have named some parts of history after the metals that changed the way people lived. The Periodic Table Sword from the “Bronze Age” is full of patterns s l a t Me This topic starts with a quick look at the history of metal use, and ends with a study of how we get metals from the Earth, and the chemistry of the extraction process. No nMe tal s Measuring Chemical Quantities In this topic you will also be introduced to the concept of the “Mole”... not a burrowing mammal! not a traitor within the group! not a gangster’s girlfriend! certainly not a skin blemish! A Chemical Mole is a clever way to measure quantities; ...essential for analysis & chemical manufacture. Electrically powered smelter plant for extracting Aluminium from its ore If you know the mass, you can figure out how many atoms there are... thanks to the mole. Photo courtesy of Comalco Aluminium Ltd Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 1 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 CONCEPT DIAGRAM (“Mind Map”) OF TOPIC Some students find that memorizing the OUTLINE of a topic helps them learn and remember the concepts and important facts. As you proceed through the topic, come back to this page regularly to see how each bit fits the whole. At the end of the notes you will find a blank version of this “Mind Map” to practise on. Metals We Use Today The Activity Series of the Metals Metal Reaction with • Oxygen • Water • Acids Metal Extraction Needs Energy History of Metal Use Electron Transfer REDOX 1st Ionization Energy Chemical Activity of the Metals Our Use of Metals Activity & Usage of Metals History of the Periodic Table Patterns of the Periodic Table METALS Patterns in Extracting Metals from Ores Quantity Calculations the Mole Minerals Ores & Resources Definition of the Mole Avogadro’s Number The Case for Recycling Metals Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science Conductivity Melting Points Chemical Bonding Valency Reactivity Atomic Radius 1st Ionization Energy • Electronegativity Molar Ratios in Reactions Empirical Formulas Case Study: Extracting Copper from its Ore • • • • • • • Gay-L Lussac’s Law & Avogadro’s Hypothesis 2 Mole Quantity Calculations • Masses • Gas Volumes www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 1. OUR USE OF METALS For most of human existence, people used tools of stone, wood and bone. Primitive tribes were familiar with gold which occurs uncombined in nature, but it is too soft to be useful for anything but jewellery and decoration. The Iron Age (approx. 2,500 to 1,500 years ago) About 1,000 B.C. the extraction of iron from its ores was discovered. This requires much higher temperatures, and the breakthrough was probably the invention of the bellows, a device to pump air into a furnace so the wood or charcoal burns hotter. About 5,000 years ago, in the Middle East, some people accidentally discovered that if certain rocks were roasted by fire, small amounts of copper would be found later in the ashes. Copper is too soft to be really useful, but there was a brief “Copper Age” around the eastern end of the Mediterranean Sea. Copper was used for decoration, jewellery, small utensils, and occasionally for knives and spear points. Iron is stronger and harder than bronze. A warrior armed with iron weapons will usually beat a bronze-armed man. Iron tools and even the humble nail allowed new developments in buildings, ships, wagons... remember that towns, trade and commerce give wealth and power. An iron plough allows more land to be cultivated to grow more food, to feed a bigger army... and so on. The First Uses of Metals It is no accident that the dominant world power of this time was ancient Rome, because their technology was based on iron. The big breakthrough was the discovery by these copperusing people that if they roasted copper-bearing rocks (ores) with tin ores, the resulting “alloy” (mixture) of copper and tin produced a much harder metal, “bronze”, which could be cast in moulds, and hammered to shape many useful tools and weapons... this was the start of From the Medieval to the Modern After the collapse of the Roman Empire the various cultures that dominated the “Dark Ages” still had ironbased technologies. The Bronze Age (approx 4,500 to 2,500 years ago) It is no accident that the rise of the great ancient civilizations occurred about this time. The stone blocks of the pyramids and temples of ancient Egypt were cut and shaped with bronze chisels. Egyptians, and later Greeks, dominated their world because their soldiers were armed with bronze swords, spears and arrowheads. The next great technological change was the “Industrial Revolution” which began about 1750 in England. This had many aspects, but the big change in technology was the use of coal (instead of wood) for fuel. As well as steam engines, coal allowed for large scale smelting of iron and the invention of steel (an alloy of iron with carbon). With bronze tools they built better ships and wagons for transport and trade, which brought wealth and power. Photo by Jop Quirindongo Photo: Keith Syvinski The engines, tools and machinery of the great factories were based on steel. Transport was revolutionized by steel locomotives running on steel rails. Steel ships replaced wooden ones, and steel weapons (machine guns, tanks and artillery) achieved new heights (depths?) in warfare and mass destruction. Photo: Arian Kulp In the 20th century, new metals and alloys became available... aluminium, titanium, chromium, and many more. Sad as it might be, the facts of human history are that progress has been marked by conflict, war and conquest, and metals have been a vital part of that development. This was made possible by electricity, which is needed in large amounts to extract some metals from their ores, or to purify and process them once extracted. Metal has many advantages over stone, wood, or bone: • metal is harder, stronger, and flexible, not brittle. • metal can be cast, hammered or drawn into shapes not possible in stone, such as saw blades, swords and armour. • when tools become blunt, metal can be re-sharpened. Human Progress has always been linked to our use of Metals. Progress in metal usage has always been linked to the availability of energy to extract the metals. Basically, a warrior with a bronze sword always beats a bloke with a stone axe... we call that progress! Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 3 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 The Metals We Use Today Solder is an alloy of 30-50% tin with lead. In one sense, we are still in the “Iron Age”. Iron is still the metal we use the most, but nearly always it is mixed with other elements in a variety of alloys, notably steel. Its most notable property is a very low melting point, around 150-200oC. Its major use is in plumbing for sealing the joints between pipes, and in electronics for connecting small components on a “circuit board”. Metals That Are Used in Their Pure State Although we use a wide range of alloys, there are some important metals we use in their pure, elemental state. Photo by Diana Aluminium is very lightweight, yet strong and corrosion resistant Steel is used for bridges, tools and machinery, bolts, screws and nails, reinforcing inside concrete structures, engines, vehicle bodies, trains and their rails, ships, and “tin” cans. Its lightweight strength is perfect for aircraft construction. Why is steel so widely used? Lightweight and a good conductor, it is used for electricity power lines. • Iron ore occurs in huge deposits, so iron is common and economical to produce. • Steel (in its various forms) is very hard and strong. • It can be cast, milled, rolled, worked, bent, cut and machined into just about any shape or size imaginable. Malleable and corrosion resistant, it is ideal for window frames and drink cans. As always, our usage of the different steel alloys is linked to their particular properties: Copper is used for electrical wiring in buildings and appliances, because of its great electrical conductivity and its ductility for ease of wire-making. Steel Iron, Alloy with... Mild steel 0.2% carbon strong, but malleable Metal Extraction Needs Energy Tool steel very hard drills, knives, hammers resists corrosion, hygenic food utensils, medical tools Stainless Steel 1-1.5% carbon 20% nickel & chromium Properties Uses car bodies, pipes, roofing As mentioned previously, our use of different metals through history can be linked to the availability of energy. In topic 1, you learned about the process of chemical decomposition; where a compound breaks down into simpler substances. Decomposition is generally an endothermic process; energy is absorbed by the reactants during the reaction. Generally, you must supply energy to make the process happen. Brass is a common “non-ferrous” (no iron) alloy. Metal ores are mineral compounds. To obtain the elemental metal involves decomposition, which is endothermic and requires energy. Some compounds require more energy than others for decomposition. Copper and tin ores require little energy. A decent wood fire can “smelt” the metal from its ore. This why copper and bronze were used in ancient times. Iron ore requires more energy for decomposition. That’s why the “Iron Age” came later. Brass is an alloy of copper and zinc (about 50% each) Brass is very hard, but easily machined for screw threads, etc. It is more expensive than steel, but is corrosion resistant, so it is ideal for taps and fittings for water and gas pipes. Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science Aluminium and other “modern” metals require even more energy, and electricity works better than heat, so these only became available in quite recent times. 4 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Worksheet 1 Today, the metal we use most is still t)...................., in the form of the alloy u)................................. Its widespread use is because: • it is common and v)..................................... to produce. • it is very w)........................ and ................................ Steel comes in a variety of alloys, including x)....................... steel (car bodies, pipes, roofing) and y)................................. steel used for food utensils and medical tools. Fill in the blanks. Check answers at the back. Before metals, people used tools mainly made from a)............................. or ................................. The first metal used was probably b)................................., because it occurs in the elemental state in nature. However, it is too soft to be used for tools, so was just used for c)................................ Other alloys used widely include: • brass, a mixture of z).................... and ................... • aa)...................................., with a very low melting point, is an alloy of ab).................................. and .................................... and is used in ac)..................................... and .................................... Metallurgy (the technology of metals) began with the extraction of d)................................. from ores that were simply e)............................................ ............................................. A big improvement was the mixing of ores of f)............................... and ........................................... This produced the alloy g)......................................, which made tools and weapons with many advantages over stone: • metal is h)............................ and ............................ and is not i)........................................ like stone • metal can made into intricate shapes, such as j)......................................., not possible in stone. As well as many alloys, there are some metals commonly used in their pure, elemental form: • Aluminium, which has the advantages of being ad)................................. and resistant to ae)....................................... Uses include af)..................................................... and .................................................. • ag)....................................... is used for electrical wiring because of its good ah)............................................ and because it is ai)................................ so it is easy to draw out into wires. Later, bronze was replaced by k).............................. which is l)............................... and............................., but requires more m).......................................... for its extraction. During the “Industrial Revolution”, the use of n)................. for energy led to the production of o)............................... which is iron with a small amount of p)................................ in it. This allowed the development of machinery, trains and the modern industrial world. Chemically, the extraction of metals from ores involves aj).................................................. reactions, which are ak)................-thermic. Some metals, such as al).................................... require very little energy, others such as am)...................................................... require much more. In many cases an)...................................... works better than heat in the extraction and purification processes. The changes in ao)............................ usage through history can be directly linked to society’s changing sources and uses of ap)...................................... In the 20th century new metals such as q).............................. became available because the r).................................. needed to extract it from its s)................... was available. WHEN COMPLETED, WORKSHEETS BECOME SECTION SUMMARIES Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 5 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 2. CHEMICAL ACTIVITY OF THE METALS Metals React With Acids Metals React With Oxygen One of the most familiar laboratory reactions is the burning of magnesium: The different activity levels of the metals is most clearly seen when metals are reacted with dilute acids. Magnesium + Oxygen 2 Mg + O2 You may have done experimental work to observe how vigorously different metals react with a dilute acid. Magnesium oxide 2 MgO In fact, many metals will burn, some a lot more readily and violently than magnesium: Sodium + Oxygen 4 Na + O2 Metals like calcium and magnesium react vigorously. Sodium oxide 2 Na2O Zinc and iron are slower. In these cases there is a violent exothermic reaction, with light and heat energy produced. The product is often a powdery, crumbly solid. Lead is very slow indeed. Other metals, such as aluminium and zinc, react on the surface and the oxide compound formed is airtight and prevents further reaction. That’s why these metals are often dull-looking... the surface coat of oxide is dull. When there is a reaction, the gas produced is hydrogen. Aluminium + Oxygen 2 Al + 3 O2 Copper does not react at all. A flame test The metal is “eaten away” and gives a “pop” dissolves into the liquid. This is explosion because it forms a soluble ionic compound. Exactly what the compound is, depends on which acid is used. Aluminium oxide 2 Al2O3 Other metals, such as copper, react with oxygen very slowly and only if heated strongly. Some, like gold, will not react at all. Examples: Zinc + Hydrochloric acid Zn + 2 HCl The point is, that metals have different chemical activities. Magnesium + Nitric acid Mg + 2 HNO3 Metals React With Water Another favourite school reaction is when sodium reacts with water. This is often done outdoors, because it results in an exciting little explosion. Iron What happens is: Sodium + 2 Na Water + 2 H2O Hydrogen + Sodium (gas) hydroxide H2 + 2 NaOH Sulfuric acid Fe + H2SO4 Metal + Acid Hydrogen + Magnesium nitrate H2 + Mg(NO3)2 Hydrogen H2 + Iron(II) sulfate + FeSO4 Hydrogen + a Salt It will help you greatly to learn the common laboratory acids Common Name Chem Name Hydrochloric = Hydrogen chloride Sulfuric = Hydrogen Sulfate Nitric = Hydrogen nitrate Once again, some metals react easily and rapidly and form the metal hydroxide, while others react slowly if heated in steam, and form oxides. Hydrogen + Zinc oxide H2 + ZnO Metals like copper and gold do not react at all. There is an “Activity Series” among the metals. Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science + Hydrogen + Zinc chloride H2 + ZnCl2 The ionic compounds formed are collectively known as “salts”, so the general pattern of the reactions is (In fact this is NOT the explosion reaction. The explosion is the reaction of the hydrogen with oxygen, to form water) Zinc + Water Zn + H2O Bubbles of gas are produced. Formula HCl H2SO4 HNO3 Try the WORKSHEET at end of section 6 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Electron Transfer in Metal Reactions The Activity Series of the Metals The chemical reactions that allow us to see the pattern of the Activity Series are just part of an even greater pattern in Chemistry... the process of electron transfer. From these 3 patterns of reaction, it seems there is a further, underlying pattern. Certain metals, like sodium, always seem to react readily and vigorously. Others, like copper, always react slowly or not at all. To understand this, look again at the reaction between a metal and an acid: From this, and other reaction studies, the common laboratory metals can be arranged in an “Activity Series”: Most Active Zinc + Hydrochloric acid K Zn + Hydrogen + Zinc (gas) chloride 2 HCl H2 Na Li Zn + 2H+ + 2Cl- Ca Mg Notice that the chloride ions (Cl ) occur on both sides of the equation unchanged. Nothing has happened to them at all. We say they are “spectator ions”. Like by-standers at a car crash they are not involved, while other atoms and ions undergo serious changes. Zn Fe Sn Pb Since they aren’t actually involved, we can leave the spectators out. This is called a “net equation”. Cu Zn + 2H+ Ag Au H2 + Zn+2 Now we can see what really happened; • a zinc atom became a zinc ion and • 2 hydrogen ions became a (covalent) hydrogen molecule. If you look for these metals on the Periodic Table you will notice a further pattern. 2 6 H2 + Zn+2 + 2Cl- Study this carefully and make sure you understand why there have to be 2 of some ions to agree with the original balanced equation. Al 3 ZnCl2 HCl and ZnCl2 are both ionic compounds. Here is the equation re-written to show the individual ion “species”. Ba Least Active + Positions of the first 6 metals of the Activity Series. To do this, the zinc atom has to lose 2 electrons, and the hydrogen ions must gain a pair of electrons to share. 1 5 4 The highly active metals all lie to the extreme left of the table, AND the higher their activity, the lower down the table they are within each column. Zn+2 + 2e- 2H+ + 2e- H2 Now it should be clear what really happened: the zinc atom gave a pair of electrons to some hydrogen ions. Electrons were transferred from one “species” to another. This is one of many patterns that allows you to use the Periodic Table instead of learning many small facts. For example, instead of memorizing the Activity Series fully, you can remember the pattern above and always be able to figure out the order of the most active metals. Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science Zn The equations above are “Half-Equations” and are often used to describe what is really happening in a reaction. 7 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Oxidation and Reduction First Ionization Energy Although you’re not yet required to know about Oxidation and Reduction, this bit you have to learn. The transfer of electrons from one species to another is one of the most fundamental and important general reactions of Chemistry. Definition The Ionization Energy of an element is the energy required to remove an electron from an atom. The reaction between zinc and acid can be visualized like this: electrons transferred + Zinc atom For technical reasons, the measurement of this energy is carried out for atoms in the gas state. + Zn(g) 2 Hydrogen ions Zn+(g) + e- The energy required for this to happen is the “1st Ionization Energy” Hydrogen molecule We know that zinc atoms normally lose 2 electrons to form the Zn+2 ion. However, the formal definition for this process involves just the loss of 1 electron. +2 Every element has its own characteristic value, even those elements which would not normally lose electrons, such as non-metals like chlorine. Covalent bond (2 electrons being shared) The zinc atom has lost 2 electrons, Cl(g) Zn Zn+2 + 2e- Cl+(g) + e- Normally a chlorine atom forms a negative ion by gaining an electron. Technically though, it is possible for it to lose an electron if energy is added. This energy is the “1st Ionization Energy” For historical reasons, the loss of electrons is called “Oxidation” and the hydrogen ions have gained electrons. 2H+ + 2e- Even the inert gases, which normally do not form ions at all, can be forced to lose an electron if energy is added. They too have a 1st Ionization Energy value. H2 The gain of electrons is called “Reduction” Ionization Energy Determines the Activity Series Now back to the metals and the Activity Series. Neither process can occur alone... they must occur together In order for a metal to begin reacting with an acid, (or with water or oxygen) it must lose an electron. This will require the input of its 1st Ionization Energy. The zinc oxidation allows the hydrogen to be reduced, and the hydrogen reduction allows the zinc to be oxidized. The total reaction is an “Oxidation-Reduction” and is commonly abbreviated to “REDOX”. If the value for 1st Ionization energy is very low, the metal will gain this energy easily and quickly from its surroundings. It will readily enter the reaction, and the reaction will proceed vigorously. Note that the syllabus does NOT require you to know these definitions yet, but it is worth knowing about Redox for future topics. You ARE required to know about electron transfer and its involvement in metal reactions. If its value for 1st Ionization energy is higher, the atom cannot react so readily or vigorously... its activity is lower. Try the WORKSHEET at end of section The ACTIVITY SERIES of the Metals is determined by 1st IONIZATION ENERGY Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 8 Increasing values for 1st Ionization Energy Zinc ion www.keepitsimplescience.com.au K Na Li Ba Ca Mg Al Zn Fe Sn Pb Cu Ag Au keep it simple science TM Emmaus Catholic College SL#802440 Choice of Metals Based on Activity Another example is the choice of metals for water pipes. Sometimes which metal is chosen for a particular application is based on its position in the Activity Series. Steel is cheap, but since iron is about the middle of the Activity Series it will corrode (rust) by contact with water. Is it better to choose a lower activity metal such as copper, which will not corrode as quickly, but is more expensive? Example In critical electronic connections, such as computer network plugs, it is essential that the electric signals get through without loss or distortion. Normally we use copper for electrical wiring, but in a critical connection plug it is worth the extra expense of using gold. The decision is usually to use cheap steel pipes for longer, outdoor uses like your garden taps. Copper is a low activity metal, but can slowly react with oxygen to form a non-conducting oxide layer in the connection. Gold is lower down the activity series and will not react at all, so the plug connection cannot corrode. Brass fittings Copper pipe Indoors, where distances are shorter, and a rusted-out leaking pipe would be a disaster inside a wall or ceiling, copper is chosen, especially for hot water supply. Gold’s extremely low chemical activity (due to a relatively high 1st Ionization Energy) is part of the reason it has always been used for jewellery. Interestingly, sometimes the higher activity metals corrode less. Aluminium and zinc are higher up the Activity Series than iron. They react rapidly when exposed to oxygen, but the surface layer of oxide is airtight and waterproof, and prevents oxygen or water getting to the metal underneath. Therefore, these metals can be used in situations where corrosion needs to be prevented. “Galvanized” steel is coated with a thin layer of zinc to prevent (or slow down) corrosion of steel roofing, fence wires, nails, bolts, etc. Gold’s low activity means it will not tarnish or corrode, so it retains its beautiful colour and lustre. Metals in Art and Religion Photo by Diana Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 9 Photo by Brett Jephson www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Worksheet 2 Part B Practice Problems Part A Fill in the blanks. Check answers at the back. 1. Write a balanced, symbol equation for the reaction of each of the following metals with oxygen. When a metal reacts with oxygen it forms an a)...................... compound. METAL + OXYGEN a) Lead (assume lead(IV) ion forms) b)...................................... b) Iron (Assume iron(III) ion) Some metals will also react with water, forming c)..................................... gas and a d)...................................... compound. METAL + WATER c) Lithium c)..................... + d)................. 2. a) Arrange the metals in Q1 in order of decreasing chemical activity. b) Which one(s), if any, might ignite easily and burn in air with a visible flame? Most metals will react with acids, forming e).......................... gas and an ionic compound called a “f).................................” METAL + ACID e)....................... + f)................. 3. Write a word equation AND a balanced, symbol equation to describe the reaction of: In all these reactions the various metals react at g)................................... rates, showing an order of chemical h)...................................... From these reactions and others, the “Activity Series” has been determined. a) calcium metal with water (reacts spontaneously at room temperature) Metals such as i)...................................... and ............................. are the most active. These are the elements located in the j)........................... columns of the Periodic Table. b) Tin metal with water (heated in steam) (Assume tin(II)) Some metals such as k)............................. and ......................... have very low activity, and often do not react at all. Other common metals like l)............................................. and .................................... are in the middle of the series. They will react, but generally do so m)............................................ 4. All the following equations are Metal + Acid reactions. Fill in all blank spaces, then re-write in symbols and balance. All these reactions involve the transfer of n)......................... In the case of the Metal + Acid reaction, the metal atoms always o)........................... electron(s) while a pair of p)................................ ions gain 2 electrons (which they share in a q)........................................ bond) and form a r)........................................ molecule with formula s)............... a) Zinc + Sulfuric acid ....................... +......................... b) Calcium + Hydrochloric acid “Oxidation” is the technical term for t)................................... ................................. The opposite is “u)................................... In the Metal + Acid reaction, the metal is always v).............................................. while w).............................. ions are always x).................................................. ........................ +...................... c)....................... +........................... d).................... + ......................... Hydrogen + Barium nitrate Hydrogen + iron(II) chloride The “1st y).................................... Energy” of an element is defined as the energy required to z)......................................... ............................... from atoms in the aa)....................... state. The very active metals are like that because they have very ab)................................... (high/low) values for this. Metals further down the series do not react as vigorously because their values are ac)................................................... 5. For each of the reactions in Q4, which chemical species a) lost electrons? b) gained electrons? c) was a “spectator”? Sometimes the choice of which metal to use is determined by the activity level. An example is ad).................................... ....................................................................................................... WHEN COMPLETED, WORKSHEETS BECOME SECTION SUMMARIES Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 10 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 3. PATTERNS OF THE PERIODIC TABLE History of the Periodic Table Atomic Structure, Number and Mass The modern concept of a chemical element as a substance containing identical atoms was first accepted almost exactly 200 years ago. Here is a quick reminder of some basics about atoms you need to know: In the Nucleus are Protons & Neutrons By 1830 there were about 40 known elements. Even with such a small sample, people began to notice patterns: Dobereiner (German) pointed out that there were several groups of 3 elements with remarkably similar properties: Lithium, sodium & potassium was one “triad”. Chlorine, bromine and iodine formed another “triad”. In orbit around the nucleus are the Electrons By 1860, with over 60 known elements, Newlands (English) proposed a “Law of Octaves”. No.Electrons = No.Protons = “Atomic Number” If the elements were arranged in order of relative weights, Newlands found that every 8th element (an “octave”) was similar in properties. These similar elements included Dobereiner’s triads. Each element’s atoms have a different, characteristic, number of protons and electrons. Therefore, each element has a different Atomic Number. In the Periodic Table the elements are arranged in order of Atomic Number. The system worked well for the first 20 elements, but then became confused. No.Protons + No.Neutrons = “Mass Number” (Electron mass is insignificant) The Mass Number is always a whole number, but in the Periodic Table the “Atomic Weight” is shown instead. (How and why this is different will be explained in a later topic) The basis of the modern Periodic Table was developed by the Russian, Dmitri Mendeleev in 1869. Mendeleev used many physical and chemical properties: • atomic weight • density • melting point • formula of oxide compound • density of oxide and many more, and arranged the elements in order of weight, but with elements with similar properties under each other. The Periodic Table is firstly a list of the elements, arranged in order, and showing all the basic details. Atomic Number Equal to the number of electrons and the number of protons in each atom 18 Ar Argon 39.95 Similar elements placed in vertical columns Inert Gases had NOT been discovered Chemical Symbol Mendeleev’s vertical “families” included Dobereiner’s “triads” and Newland’s “octaves”, but had one big difference... Element Name “Atomic Weight” NOT the “Mass Number” Mendeleev’s genius was to realize that there were probably missing elements that hadn’t been discovered yet. He cleverly left gaps in his table for these undiscovered elements. However, the Periodic Table is far more than a simple list. Why is it such a complicated shape? The most famous case was that of the “missing” element Mendeleev called “eka-silicon”. He used the patterns in his table to predict, very precisely, the properties for ekasilicon. Scientists went looking for such a substance and soon found a new element (which was named “Germanium”) with properties almost exactly as predicted. The shape and arrangement of the Periodic Table is a very clever device to allow many patterns and groupings to be accommodated. You have already learnt one pattern in the position of the most active metals, and their 1st Ionization Energies. There are lots more... Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 11 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Patterns of the Periodic Table In Mendeleev’s day no-one could explain why these patterns existed. However, when scientists see patterns in nature like this, they know there must be underlying “rules” or “laws of nature” causing and controlling the patterns. Perhaps Mendeleev’s great contribution was not just the Periodic Table itself, but the stimulus it gave other scientists to investigate the reasons behind the patterns. Within 40 years Science had unravelled the secrets of atomic structure, the electron energy levels, and more. At this stage, your task is to learn some of the patterns. Melting Point You learned in topic 1 how melting point is determined by the bonding within a substance. Electrical Conductivity As you go across any row (“period”) of the table, you will move through a number of metals, then one or two semimetals, then into the non-metals. Therefore, the conductivity will start out high, but rapidly decrease as you encounter a semi-metal, and become extremely low at the non-metals. Semi-Metals NonMetals Metals Conductivity Moving to the right across a period you enter the “Transition Block” containing typical hard, high melting point metals, held strongly together by “metallic bonding”. Further right you hit the Semi-Metals. These often have very high melting points because of their covalent lattice structure. Then you enter the Non-Metals which have covalent molecular structures and quite low mp’s. At the far right column, each period ends with an Inert Gas which are all single-atom molecules, and have the lowest mp of each period. This pattern repeats itself along each period. 2,000 decreasing At the left side of the table are the very active metals of the Activity Series. They are also usually soft, and have relatively low (for metals) melting points. Melting Points of Elements Periods 3 V (oC) Boiling Points follow a similar pattern to Melting Points Sketch Graph. 0 Melting Point 1,000 Si Valencies are the same down each group Peaks are Transition Metals or Semi-Metals Period 4 Rb K Na Inert Gases Ar Kr Atomic Number Chemical Bonding, Valency & Reactivity What you’ve already learnt about the Activity Series, Ionic and Covalent Bonding and Valency will help you make sense of the following: Group 8 Inert Gases +1 0 +2 +3 4 -3 3 1 -2 2 -1 Activity of Metals Most active at bottom-left. Activity (generally) decreases upwards and to the right. Metals (+ve ions) Bonding Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science Activity of Non-M Metals Semi-Metals (Covalent only) Most active at top-right (Fluorine) Non-Metals (Covalent or (-ve) ions) 12 No chemical reactions, no bonding Activity (generally) decreases downwards and to the left. www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Atomic Radius The size of an atom is the distance across its outer electron shell. You might think that the atoms along each period would be the same size, because it’s the same orbit being added to. However, the increasing amount of positive charge in the nucleus pulls that orbit inwards closer and closer to the centre. H 37 Na 186 K 231 He 50 The numbers given are the atomic radii in picometres. 1 picometre = 1x10-112 metre Radius increasing down a group Li 152 The following diagrams are to scale and show the relative sizes of the first 20 elements Be 112 Mg 160 Ca C B N 77 88 Al 66 P Si 143 O 70 S 110 118 102 Ne F 70 68 Ar Cl 94 99 Radius decreasing across a period 197 Down each group the radius increases. This is because, as you go down a group, you have added an entire electron shell to the outside of the previous layer “Periodic” means “recurring at regular intervals”. This graph shows what a spreadsheet plot gives for the radii of the first 37 elements. Notice how the same graphical pattern keeps recurring... it is a periodic pattern. Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science Rb K De acr creasi oss ng a p Tren erio d d 300 200 Na Li 100 When you do, you can clearly see how the Periodic Table got its name. rend sing T Increa a group down He Ne Ar Kr d ng Tren Increasai group n w do 0 The Syllabus requires that you produce a table and a graph of the changes in a property • across a period, and • down a group Atomic Radius (picometre) Spreadsheet Plot of Atomic Radii 1 10 20 30 Atomic Number There are a number of irregularities and “glitches” apparent on the graph. It is beyond the scope of this course (and way beyond the K.I.S.S. Principle) to attempt an explanation of these. 13 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Ionization Energy Successive Ionization Energies The meaning of the “1st Ionization Energy” was explained previously in relation to the Activity Series of Metals. Having added the energy of 1st I.E. and removed an electron from any atom, it is then possible to add more energy and remove a 2nd electron, and a 3rd, and so on. A+(g) A(g) + e- where “A” stands for any atom in the gas state Any atom can lose an electron if enough energy is supplied... even atoms which do not normally lose electrons. decreasing A+(g) e- 2nd I.E. A+(g) A+2(g) + e- 3rd I.E. A+2(g) A+3(g) + e- + Once the first electron is removed, the remaining electrons are pulled in tighter to the nucleus. Each one experiences increased force of attraction, so it requires more energy to remove the next electron. The trend for the whole Periodic table is: Lowest A(g) ...and so on, according to how many electrons the atom has The Periodic Trend in 1st Ionization Energy You should remember that the very active metals are the ones with low 1st ionization energies. They easily lose their outer electron(s) and so react readily. 1st Ionization Energy 1st I.E. Highest value Each successive ionization requires more energy. Once the entire outer orbit has been stripped away, the next ionization must remove an electron from an underlying orbit, which requires a huge increase in the next ionization energy. This results in an interesting pattern: increasing Patterns in Successive Ionization Energy Data (values shown are energy measurements) Successive Elements on Period 3 Explanations: 1st I.E. increases to the right because each atom across a period has more and more (+ve) nuclear charge attracting and holding electrons in the orbit concerned. Therefore, it requires more energy to remove an electron. 1st I.E. decreases down each group because, at each step down, an extra whole layer of electrons has been added to the outside of the atom. The outer shell is further away from the nucleus, and is partially “shielded” from nuclear attraction by the layers of electrons underneath it. Therefore, it becomes easier and easier to remove an electron. Atoms with a tendency to lose electrons easily (low 1st I.E.) and form (+ve) ions have very low values. Once again, there is a pattern in these values in the Periodic Table. Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 1st I.E. 2nd I.E. 3rd I.E. 4th I.E. Sodium 2.8.1 0.5 4.5 6.9 9.6 Magnesium 2.8.2 0.7 1.4 7.7 10.5 Aluminium 2.8.3 0.6 1.8 2.8 11.6 Highest Value Fluorine (values decrease down) Atoms with a tendency to gain electrons and form negative ions have high values. Electron Config. Notice how the values “jump” (underlined data) as the next ionization has to remove an electron from the next lower orbit. Electronegativity is a value assigned to each element to describe the power of an atom to attract electrons to itself. Element 1.0 1.5 0.9 0.8 Electronegativity Values of selected elements (values decrease to left) Inert gases not included 2.0 2.5 3.0 3.5 4.0 3.0 2.8 0.8 2.5 0.7 2.2 0.7 14 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Worksheet 3 Atomic Radius ae).......................................... across a period because each successive element has af)........................... (more/less) positive charge in the ag).................................. to attract the electron shell and pull it inwards. As you go down a group the radius ah)................................. as each new electron shell is added. Part A Fill in the blanks. Check answers at the back. As early as 1830, the German a)............................................ noticed patterns in the properties of the elements. In 1860, the English scientist b)...................................... proposed a “Law of c).........................................” describing the repeating pattern of properties. First ai)........................ Energies aj)............................ across a period, as the increasing amount of nuclear charge makes it more and more difficult to ak).................................. an electron. The values al).................................... down a group because each extra shell of electrons is am)................. (more/less) strongly held than the previous. It was the Russian d).......................................... who invented the e)........................................................, in more or less its modern form. He realized that there were probably many elements that had not f)............................................................., so he g)................................................... in his table for later additions. By studying the details of known elements, he was able to h)....................................... very precisely the properties of the missing elements. Successive Ionization Energies measure the energy required to an)............................ another, subsequent electron from an atom. The energy required to remove the next electron is always ao).............................................. (higher/lower). When the next electron happens to be in the next lower shell, the value ap)............................................ by a huge amount. Sure enough when discovered, the missing elements were found to have properties i)........................................................ ....................................................................................................... aq)................................................... is a value which describes the power of an atom to ar)................................... electrons. The element with the highest value is as)..............................., and values decrease as you move to the at)......................... and as you move au).................................. the Periodic Table. The patterns in the Periodic Table include: Conductivity, which generally j)......................................... to the right, as you go from metals to k)..................................... and ................................................. Part B Practice Problems Melting Points: tend to l)........................to about the middle of each period, then m)............................. The highest value is usually a n)...................................... metal or one of the o)................................................. elements. The lowest value on each period is always the p)................................ gas member on the extreme q)................................ (right/left) 1. a) Write equations to represent the 1st, 2nd, 3rd & 4th ionisations for a calcium atom. b) Between which two of these successive ionisations would you expect a huge increase in the required energy? 2. On each of the following Periodic Table diagrams label the arrows with the word “increasing” or “decreasing” to correctly describe the trend in the direction shown. Valencies are r)................................. down each vertical group. Bonding follows the pattern of the main categories of elements. s).................................. form t)............................. bonds when they lose electrons and become u).................... ions. The Semi-metal elements form only v)........................... bonds. The Non-metals can bond w)....................................... or can x)................................ electrons to form y).................... ions. ii) (down) b) Electronegativity Also indicate (“H”&“L”) the position of elements with highest & lowest values. Chemical Reactivity is different for metals and nonmetals. The most active metals are located at the left z)................................... (top/bottom) of the table. Generally, activity decreases aa)............................. and to the ab)................................... The Inert Gases show no chemical activity. Apart from them, the most active non-metals are located on the right ac).................................. (top/bottom) of the table. Activity generally decreases as you move ad)........................................... and .......................................... i) ii) i) c) 1st Ionisation Energy WHEN COMPLETED, WORKSHEETS BECOME SECTION SUMMARIES Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science i) (right) a) Atomic Radius Show“H”&”L” ii) 15 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 4. QUANTITY CALCULATIONS & THE MOLE Defining the Mole Quantities in Chemical Calculations For technical reasons, the “atomic standard” used to compare the masses of all atoms is the carbon atom, which contains Atoms, molecules and ions always react with each other in fixed, whole-number ratios. That’s why balancing an equation is so important... it actually brings the equation into line with what is happening at the particle level. 6 protons 6 neutrons 6 electrons For example, when hydrogen and oxygen react to form water, the balanced equation is 2H2 + O2 Atomic Number = 6 Mass Number = 12 2H2O This is a true description of what is happening to the molecules: 2 Molecules of H2 + 1 Molecule of O2 The mass of this atom is defined to be exactly 12.000000 atomic mass units (a.m.u.) and all other atoms are given masses relative to this one. Since this is the standard of comparison, the formal definition of the mole is: “the number of atoms contained in exactly 12 grams of carbon-12” 2 Molecules of H2O However, when we carry out chemical reactions in the laboratory or in Chemical Industry, we cannot see or count the molecules. Instead, we measure the mass or volume of substances. Note: In Topic 1 it was pointed out that the Mass Number for any atom is a whole number. It has still not been explained why the “Atomic Weights” in the Periodic Table are mostly not whole numbers. To measure out the correct numbers of particles for a reaction we need a simple way to convert masses and volumes to numbers of molecules, and vice-versa. That’s the purpose of This WILL be explained in a later topic. If you cannot wait, go find out about “Isotopes”. The Mole Avogadro’s Number Just how many atoms are in 1 mole? 1 mole is a quantity of a chemical substance. Obviously, it is a very large number. We now know that it is about 6,000 billion trillion. 1 mole of any element or compound contains exactly the same number of particles. Avogadro’s Number 1 mole of each substance has a different mass, because the atoms and molecules all weigh differently. 6.022 x 1023 particles in 1 mole of anything The really clever and convenient thing about the mole is its link to the Periodic Table and the “Atomic Weights” shown. 6 Carbon C Argon Ar Pb 12.01 39.95 207.2 1 mole = 12.01 grams 1 mole = 39.95 grams This number is named in honour of an Italian scientist who you will learn about soon. 82 18 6p+ 6n0 207.2 grams of Lead contains 6.022 x 1023 Lead atoms Lead 1 mole = 207.2 grams 39.95 grams of Argon contains 6.022 x 1023 Argon atoms 12.01 grams of Carbon contains 6.022 x 1023 Carbon atoms EACH OF THESE HAS THE SAME NUMBER OF ATOMS Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 16 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Calculating Mole Quantities Moles and Numbers of Particles You need to be able to calculate mole quantities in terms of both mass and number of particles. Since one mole of any substance contains Avogadro’s Number of particles: Molar Mass The “Molar Mass” of any chemical species is the mass (in grams) of 1 mole of the substance. No. of moles = No. of particles you have Avogadro’s Number n= You need to add up all the Atomic Weights of all the atoms given in the formula. Examples: Name Argon Sodium Formula Ar Na Example Calculations 1. How many moles are present in a sample of lead containing 7.88 x 1024 atoms? Molar Mass (g) 39.95 22.99 Solution (for elements like these just use Atomic Weight) Oxygen Chlorine O2 Cl2 (16.00 x 2) (35.45 x 2) 32.00 70.90 Water H2O (1.008x2 + 16.00) 18.016 Carbon Dioxide CO2 (12.01 + (16.00x2) 44.01 Sodium chloride NaCl (22.99 + 35.45) 58.44 b) m = n x MM = 0.0250 x 207.2 (molar mass of Pb) = 5.18 g Try the Worksheet at the end of this section Number of Moles in a Given Mass Try the Worksheet at the end of this section When you weigh a chemical sample you then need to be able to calculate how many moles this contains. Mole Quantities in Chemical Equations When you consider an equation like No. of moles = mass of substance you have molar mass Solution 2H2 + O2 m MM n = m = 5.23 = 0.215 mol MM 24.31 b) n = m = 96.7 MM (2x1.008 + 16.00) = 96.7/18.016 = 5.37 mol 2 Molecules of H2 1 Molecule of O2 2 Molecules of H2O 2 million H2 + 1 million O2 2 million H2O or, 200 zillion H2 + 100 zillion O2 200 zillion H2O or, (let’s use Avagadro’s number) (2 x NA) H2 + NA O2 so m = n x MM = 1.50 x (22.99 + 35.45) = 1.50 x 58.44 = 87.7 g = 2 moles H2 + 1 mole O2 (2 x NA) H2O 2 moles H2O The Balancing Coefficients in a Chemical Equation May be Interpreted as Mole Ratios Try the Worksheet at the end of this section Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science + However, the number of molecules reacting is really just a ratio. The actual numbers might be 2. What mass is needed if you want to have 1.50 moles of salt (sodium chloride)? n= m MM 2H2O you know it means a) 5.23g of magnesium? b) 96.7g of water? a) = 7.88x102423 6.022x10 = 13.1 mol Solution a) n = N so N = n x NA = 0.0250 x 6.022x1023 NA = 1.51 x 1022 atoms (add up At.weights of all atoms in the formula) Example Calculations 1. How many moles in n= N NA 2. a) How many atoms of lead are needed to make 0.0250 mole? b) What would be the mass of this quantity? (these elements are diatomic molecules... 2 atoms each) n= N NA 17 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Mole Quantities in Chemical Equations (cont.) Calculating Mass Quantities in Reactions Mole calculations allow you to calculate the mass of products and reactants involved in a reaction. The balancing coefficients of an equation can be interpreted as the mole ratio of reactants and products. 2 H2 So, + O2 Example Problem Aluminium burns to form aluminium oxide. If 4.29g of aluminium was burned, a) what mass of oxygen would be consumed? b) what mass of aluminium oxide would be formed? 2 H2O means 2 mol. reacts with 1 mol. to form 2 mol. or, 4 mol. reacts with 2 mol. to form 4 mol. or, 100 mol. reacts with 50 mol. to form 100 mol. or any other proportional quantities. Solution Always start with the balanced equation: Example Problem a) If 0.50 mol of sodium reacted completely with hydrochloric acid, how many moles of products would be formed? mole ratio b) What mass of each product would be formed? No. of moles of Aluminium: n = m = 4.29 MM 26.98 = 0.159 mol Solution a) The balanced equation is mole ratio so, 2 Na + 2 HCl H2 2 mol : 2 mol : 0.50 mol : 0.50 mol : : 2 mol. 0.25 mol : 0.50 mol b) Mass of Hydrogen: m = n x MM = 0.25 x 2.016 = 0.50 g Mass of salt: m = n x MM = 0.50 x 58.44 = 29 g 3 : 2 ∴ mass of Al2O3: m = n x MM = 0.0795 x 101.96 = 8.11 g Try the Worksheet at the end of this section Try the Worksheet at the end of this section Practical Work: Using Mass & Mole Ratios to Determine a Formula A common experiment is to burn a piece of magnesium in a crucible, as suggested by the diagram. The difficulty is to open the lid enough to admit oxygen for complete combustion, but to limit the loss of powdery product. Magnesium + Oxygen : 2 Al2O3 b) Mass Al2O3 produced: mole ratio Al : Al2O3 = 4: 2 (i.e. 2:1) ∴ moles of Al2O3 = 1/2 x 0.159 = 0.0795 mol Answer: 0.25 mol of H2 and 0.5 mol of NaCl The reaction is 4 + 3 O2 a) Mass O2 consumed: mole ratio Al : O2 = 4 : 3 ∴ moles of O2 = 0.159 x 3 = 0.119 mol 4 ∴ mass of O2: m = n x MM = 0.119 x 32.00 = 3.81 g + 2 NaCl 1 mol 4 Al ceramic crucible Magnesium oxide Careful measurement of mass allows the empirical formula for magnesium oxide to be determined. Analysis of Results Remember that to convert any Typical Measurements mass to moles: n = m / MM Mass of empty crucible = 42.74 g Mass of magnesium = 2.05 g Mass of crucible Elements Magnesium : Oxygen + product after burning = 46.22 g Ratio of masses: 2.05 g : 1.07 g Ratio of moles: 2.05 / 24.31 : 1.07 / 16.00 (divide by Atomic Weight) ∴ Mass of magnesium oxide = 0.0843 mol : 0.0669 mol formed = 3.48 g Simplified ratio = 0.0843/0.0669 : 0.0669/0.0669 (divide both by the ∴ Mass of oxygen in = 1.26 : 1 smaller) compound = 1.07 g Nearest whole number ratio 1 : 1 There is often a large error due to incomplete burning ∴ Empirical Formula is MgO Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 18 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Comparing Mass Changes When Metals Burn A Little History... How the Mole was Invented Atoms always react in simple whole-number mole ratios, but atoms have different masses, and compounds have various formulas, so the result is that chemicals do NOT react in simple ratios by mass. The “mole” as a measure of chemical quantities, is a mathematically convenient device (a “trick”) to help chemical calculations. Here’s how it was figured out... That’s why we need the mole... we measure quantities by their mass, but this makes no sense until moles are calculated. Gay-Lussac’s Law Joseph Gay-Lussac was a French scientist with an unfortunate name, by modern standards. He lived 200 years ago, and was very interested in flight using balloons, so he investigated the way gases react chemically. The syllabus requires that you should consider the mass changes involved when various metals combine with oxygen to form their oxide compound. After a series of clever experiments, in which the volumes of reacting gases were measured, in 1808 he proposed the “Law of Combining Volumes”: The following table shows the mass changes for 100g of the metal in each case: 100g of Metal Formula of oxide Mass O2 needed(g) When measured at constant temperature and pressure, the volumes of gases in a chemical reaction show simple, whole-number ratios to each other. Mass of Oxide formed Lithium Li2O 115 215 Iron Fe2O3 43 143 Zinc ZnO 49 149 Lead PbO2 15 115 The volume of a gas is easily changed by temperature and pressure, so it is very important that the volumes are all measured at the same conditions. Examples of Gay-Lussac’s Law: Empirical Formulas v Molecular Formulas You are reminded that a molecular formula really does describe the atoms present in a molecule. Hydrogen(g) + Chlorine(g) 1 litre 1 litre Hydrogen chloride(g) 2 litres The molecular compound methane, has formula CH4, because that’s exactly what each molecule contains... 1 carbon atom and 4 hydrogen atoms. Hydrogen(g) + Oxygen(g) 2 litres 1 litre Water(g) (vapour) 2 litres Hydrogen(g) + Nitrogen(g) 3 litres 1 litre Ammonia(g) 2 litres Lattice structures, either ionic or covalent are NOT molecular. Example: sodium chloride, NaCl Notice that in every case, that the volumes are always in a simple, whole number ratio to each other. The formula does NOT describe a molecule, but only gives the simplest ratio between the bonded atoms... this is an empirical formula. Now consider the balanced equations for these three example reactions: H2(g) + 2 H2(g) + On the previous page was an example of how formulas are determined by analysing the mass composition of a compound. Cl2(g) 2 HCl(g) O2(g) 2 H2O(g) 3 H2(g) + N2(g) 2 NH3(g) The mole ratios are the same as the volume ratios discovered by Gay-Lussac! You should note that this method can only produce an empirical formula. (In fact, the word “empirical” means something determined by experiment, not by theory.) Why should this be? If a molecular compound, with molecular formula X2Y6 was analysed by mass measurements, its empirical formula would be calculated to be XY3... simplest ratio of atoms. Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science ... enter Avogadro! 19 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Avogadro’s Hypothesis Molar Volume of a Gas The Italian, Amadeo Avogadro (1776-1856) was trained in Law, but became very interested in Science. If 1 mole of any chemical species contains the same number of particles (Avogadro’s Number) AND if equal volumes of gases contain equal number of particles (Avogadro’s Hypothesis), then it follows that 1 mole of any gas must occupy the same volume, if measured at the same temperature and pressure. In 1811, he noticed the similarity between Gay-Lussac’s Law (an empirical “law” based on experiment) and the concept that atoms must combine in simple, whole number ratios to form compounds. This volume is the “Molar Volume” and is the same for every gas. It is measured at the standard set of conditions known as Standard Laboratory Conditions (SLC); 25oC and 1 standard atmosphere of pressure. This led him to make an hypothesis: Equal Volumes of all Gases Contain Equal Numbers of Molecules 1 mole of any gas = 24.8 litres at SLC (when measured at the same conditions of temperature and pressure) Mole Calculations Involving Gases This additional knowledge opens up the opportunity to carry out quantity calculations which involve mass and volumes of gases. This was a vital breakthrough in the history of Chemistry. For example, consider the reaction: Hydrogen(g) + Chlorine(g) Example Problems 1. If 15.65g of calcium carbonate (CaCO3) was completely decomposed by heat, what volume of carbon dioxide gas would be produced (if measured at SLC)? Hydrogen chloride(g) Prior to Avogadro, it was assumed that the the reaction involved single atoms, like this: H(g) + Cl(g) HCl(g) Solution Always begin with the balanced equation for the reaction. CaCO3(s) CO2(g) + CaO(s) mole ratio = 1 : 1 : 1 Moles of CaCO3: n = m = 15.65 = 0.1564 mol MM 100.09 Mole ratio is 1 : 1, so moles of CO2 formed = 0.1564 but the combining volumes (discovered by experiment) were Hydrogen(g) + Chlorine(g) 1 volume : 1 volume : Hydrogen chloride(g) 2 volumes Now, reasoned Avogadro, gases react in simple, wholenumber volume ratios because each litre of gas has the same number of molecules in it. Therefore, to get the volume ratios shown above, each hydrogen molecule, and each chlorine molecule, must have 2 atoms! ∴ Volume of CO2 = 0.1564 x 24.8 = 3.88 L (at SLC) 2. What volume of hydrogen gas (at SLC) would be produced if 10.00g of lithium metal was reacted with sulfuric acid? i.e. Hydrogen is H2(g) and Chlorine is Cl2(g), and the correct equation is H2(g) + Cl2(g) 2 HCl(g) Solution 2 Li(s) + H2SO4(aq) 2 : 1 Then, for the same reaction, scientists could measure the masses of these gases as well as volumes. This showed that chlorine atoms must be about 35 times heavier than hydrogen... chemists were on the way to figuring out the relative atomic weights of elements, and being able to calculate chemical quantities. : H2(g) + Li2SO4(aq) 1 : 1 Moles of lithium: n = m = 10.00 = 1.441 mol MM 6.941 Mole ratio is 2:1, so moles of H2 = 1/2 x 1.441=0.7204 ∴ Volume of H2 = 0.7204 x 24.8 = 17.9 L (at SLC) Although he did not invent the concept of the mole, we name the number of particles in 1 mole in Avogadro’s honour... 23 Avogadro’s Number, NA = 6.022 x 10 Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science Molar Vol. of all gases at SLC Try the Worksheet at the end of this section 20 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Worksheet 4 4. Mole Ratios in Equations Sodium reacts with water as follows: H2 + 2Na + 2H2O Part A Practice Problems a) If 1 mole of sodium reacted, how many moles of i) hydrogen formed? ii) water consumed? b) If 0.25 mol of NaOH formed, how many moles of i) sodium consumed? ii) hydrogen formed? c) If 0.75 mole of hydrogen formed, how many moles of i) sodium consumed? ii) NaOH produced? 1. Molar Masses Calculate the molar mass of: a) potassium b) krypton c) tin d) bromine (Br2) e) nitrogen gas f) magnesium oxide g) sodium iodide h) iron(III) sulfide i) ammonia j) copper(II) sulfate k) aluminium oxide l) glucose (C6H12O6) Aluminium reacts with oxygen: 4 Al + 3 O2 2 Al2O3 2. No. of Moles in a Given Mass How many moles in: a) 100.0g of lead? b) 100.0g of zinc? c) 100.0g of water? d) 100.0g of copper(II) nitrate? e) 38.55g of magnesium fluoride? f) 60.00g of carbon dioxide? g) 1.000g of zinc oxide? h) 500.0g of glucose (C6H12O6)? i) 3.258 x 10-3g of salt (sodium chloride)? j) 128.6g of ammonium carbonate? d) If 0.5 mole of Al used, how many moles of i) Alum.oxide formed? ii) oxygen used? e) If 0.1 mole of alum.oxide formed, how many moles of i) aluminium used? ii) oxygen used? 5. Mass Quantities in Reactions a) Calcium burns in oxygen to form calcium oxide: 2Ca + O2 2CaO If 8.50g of calcium reacted, what mass of calcium oxide would be formed? b) Silver carbonate decomposes when heated: 2Ag2CO3 2CO2 + 4Ag 3. Moles and Number of Particles a) How many particles (atoms/molecules) in: i) 3 moles of water? ii) 2.478 mol of CO2? iii) 5 mol of salt? iv) 0.007862 mol of copper v) 1/1000 mol of helium + O2 If 20.0g of silver carbonate was decomposed i) what mass of silver metal would form? ii) what mass of CO2 would be produced? iii) what mass of O2 would be formed? c) Aluminium reacts with hydrochloric acid: 2Al + 6HCl 3H2 + 2AlCl3 b) Convert between mass, moles and no.of particles. 25 i) If there are 8.800x10 atoms of tin, how many moles is this, and what would be the mass? If 6.50g of aluminium reacted i) what mass of HCl would be consumed? ii) what mass of hydrogen gas produced? iii) what mass of aluminium chloride produced? ii) You have a sample containing 2.575x1024 molecules of water. How many moles is this, and what is its mass? d) Tin reacts with steam as follows: 2H2(g) + Sn(s) + 2H2O(g) iii) If you weigh out 400.0g of water, how many moles is this, and how many molecules are present? SnO2(s) If 14.8g of tin reacted i) what mass of tin(IV) oxide would be formed? ii) What mass of steam would be consumed? iii) what mass of hydrogen would be produced? iv) If you have 2.569g of pure nickel, how many atoms are there? v) What mass of sulfur would contain 2.500x1023 atoms? FOR MAXIMUM MARKS SHOW FORMULAS & WORKING, APPROPRIATE PRECISION & UNITS IN ALL CHEMICAL PROBLEMS Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 2NaOH Worksheet continues next page... 21 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 6. Empirical Formulas from Mass Composition a) A compound containing only copper and chlorine is decomposed, and the masses measured to find the mass composition: Mass of copper present = 12.84g Mass of chlorine present = 7.16g i) Find the empirical formula. ii) Name the compound. 8. Mass & Gas Volume Calculations a) To “scrub” the air and remove poisonous CO2 on board the Space Shuttle, the air is continually pumped through canisters containing 5.0kg of lithium oxide. The reaction is Li2O(s) + CO2(g) Li2CO3(s) b) i) Find the empirical formula of a compound containing carbon and hydrogen; a sample was found to contain 1.5g of carbon and 0.5g of hydrogen. ii) Name the compound, given that its empirical and molecular formulas are the same. b) Iron reacts with oxygen: 4Fe(s) + 3O2(g) i) How many moles of lithium oxide in each canister? ii) How many moles of CO2 can this amount absorb? iii) What volume of CO2(g) is this? (measured at SLC) i) If 10.0L of O2 at SLC reacted, what mass of iron(III) oxide would be formed? ii) If 100g of iron reacted, what volume of oxygen (at SLC) would be needed? c) A compound was found to contain 30% nitrogen and 70% oxygen by mass. i) Find the empirical formula. ii) It is later found that its molecular formula is a 2 times multiple of the empirical. Write the molecular formula. iii) Name the compound. c) The electrolysis of water causes decomposition: 2H2O(l) 2H2(g) + O2(g) i) If 1.00g of water was decomposed, what volumes of each gas (measured at SLC) would be formed? In an electrolysis experiment, 50mL (0.050 L) of oxygen was produced. (measured at SLC) ii) What volume of hydrogen (at SLC) was produced? iii) What mass of water must have been decomposed? 7. Volumes of Reacting Gases ( Assume all are measured at same temperature & pressure) 2 H2(g) + O2(g) 2 H2O(g) Part B Fill in the blanks. Check answers at the back a) If you used 5 litres of hydrogen, how many litres i) of oxygen consumed? ii) of water vapour formed? The formal definition of the mole is “the a).......................... of atoms in 12.00 grams of b).................................................” One mole of any substance contains the same number of c).................................. The mass of 1 mole of any substance is equal to its d)........................................................... in grams. The actual number of particles in one mole of anything is known as “e).................................................’s Number” and has a value of f).................................................. b) If you used 0.25 litres of oxygen, how many litres of i) water vapour formed? ii) hydrogen consumed? c) If this reaction produced 20 litres of steam, how many litres of i) hydrogen consumed? ii) oxygen consumed? In a balanced chemical equation, the “balancing numbers” (coefficients) may be interpreted as being g)........................... .............................. of reactants and products. By converting between the h).......................... of substances and the number of i)..............................., it becomes possible to calculate all the quantity relationships during a chemical j).................................... From the mass composition it is also possible to determine the k).................................. formula of compounds. Ammonia gas is formed by reaction of hydrogen with nitrogen 3 H2(g) + N2(g) 2 NH3(g) d) In order to make 9 litres of ammonia, what volume i) of hydrogen needed? ii) of nitrogen needed? Historically, the mole concept arose from the work of 2 men: The Frenchman l).................-............................. discovered that “the m)........................... of gases in chemical reactions always show simple, n)............................................ ratios to each other”. Soon after, the Italian o)................................................. suggested that “Equal p)................................... of all gases contain q)....................... numbers of r)................................... (when measured at the same conditions of s).............................. and ..........................) The standard conditions usually used are known as t)...................... (abbreviation) and is a pressure of o u)........................................... and temperature v)............... C. e) If 0.6 litre of hydrogen reacted, what volume i) of ammonia formed? ii) of nitrogen need? WHEN COMPLETED, WORKSHEETS BECOME SECTION SUMMARIES Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 2Fe2O3(s) 22 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 5. METALS FROM THEIR ORES The Importance of Predicting Yield from an Ore Ores and Minerals ... and now back to the metals. The whole situation of economic feasibility makes the science of Analytical Chemistry vital in the mining and metals industry. Minerals are naturally occurring compounds. “Rocks” are mixtures of various minerals. Most minerals are lattice structures, both ionic and covalent. Some very common minerals include: Mining operations cost millions of dollars to set up. To do so, the operators need to be sure that the ore contains enough metal to be profitable. Chemical analysis in the laboratory is used to measure the mineral content of the ore body, to predict the final yield of the metal. • Silica, which is chemically silicon dioxide (SiO2) and is the most common mineral on Earth. Other compounds are often included in the silica lattice to make “silicate” minerals. These occur in virtually all rocks. • Calcite, which is calcium carbonate (CaCO3) is the main mineral in limestone and marble. Photo: Thanks Ulrik Some minerals contain significant quantities of metal(s), chemically combined in the compound. Ores are rocks and/or minerals from which it is economically worthwhile to extract a desired metal. It is the economic part of this definition which is critical. For example, there are many rocks and minerals that contain significant amounts of iron and aluminium. These are not “iron ore” or “aluminium ore” unless it is economically worthwhile to mine and process them to get the metal. Ores are Non-Renewable Resources Minerals and ores have been formed over millions and billions of years of geological processes on Earth. Photo courtesy of Comalco Aluminium Ltd Because of that time-frame, the ores are non-renewable in the sense that once we use them up, they cannot be replaced. There is no immediate concern for running out of the most important ores, but unlimited exploitation of any non-renewable resource is: • irresponsible, to future generations. • unsustainable, because all non-renewable things must eventually run out. • economically stupid, because it may be cheaper to re-use and recycle, than to constantly extract “new” materials. • environmentally damaging, because mining and metal smelting have a history of pollution and ecosystem destruction. What Makes It Economically Worthwhile? Basically, economic feasibility is the balance between: • the Commercial Price for which the metal can be sold and • the Production Costs of mining and transporting the ore, and chemically extracting and purifying of the metal. Another factor is the abundance of the metal and its ores on Earth. For example, iron is relatively cheap because it is very common in huge deposits of iron ores. Platinum is very rare, so it commands a high price. This makes it worthwhile to mine even very low-grade ores. A low-grade iron ore would not be worth mining! Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science In the not-too-distant future it may become economically worthwhile to begin “mining” the old rubbish dumps around our cities, to recover the discarded metals in society’s garbage. 23 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Case Study: Extraction of Copper from its Ores Copper Ores include a variety of compounds of copper, including: Froth Flotation to Concentrate the Ore • copper(I) sulfide, Cu2S • copper(II) hydroxide mixed with copper(II) carbonate, Cu(OH)2.CuCO3 The ore is crushed into a powder and the copper minerals are separated from the silicates by a process of “Froth Flotation” which relies on differences in “wettability” and density. These compounds usually occur as thin “veins” of bluegreen minerals embedded in masses of worthless silicate rock. Froth overflows for collection The copper content of the entire ore body might be only 3% or less. Therefore, the first step after mining is to separate the copper minerals from the “rock”. Compressed air creates froth bubbles Compressed air creates a froth of bubbles in a detergent solution. Copper minerals, sprayed with a special oil, cling to the bubbles and are carried upwards to overflow with the froth. Chemistry of Smelting The concentrated copper minerals now undergo Decomposition Reactions. Silicate minerals are wetted by the water and, being denser, sink to the bottom. In Australia, the main copper ores contain copper(I) sulfide. If this is heated in a furnace supplied with plenty of air the reaction is: Copper(I) sulfide + oxygen Cu2S + O2 The collected froth is then treated to separate the oil and detergent for re-use. Copper + Sulfur dioxide 2Cu + SO2 The ore concentrate is now about 30% copper. The copper collected is about 98% pure. Sulfur dioxide is a serious pollutant if released from the smelter. These days it is collected and used to manufacture sulfuric acid... a useful by-product. Final Purification by Electrolysis The major use of copper is for electrical wiring. For this it needs to be 99.9% pure. Copper is purified by electrolysis: The impure copper is immersed in CuSO4 solution and electrified: Cu Cu +2 - + 2e The copper dissolves into the solution, but impurities do not. - + Impure Copper dissolves into solution Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science Cu+2 ions migrate through CuSO4 solution Pure Copper deposits on electrode After migrating through the solution, the ions are redeposited as pure copper metal on the other electrode: Cu+2 + 2e- Cu Impurities 24 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 The Case for Recycling Producing the electricity usually involves the burning of coal at a power station. The burning of fossil fuels like coal is a major contributor to the “Greenhouse Effect” which many scientists are now convinced is causing massive climate changes to the entire Earth. The point that mineral ores are non-renewable has already been made. Eventually, any non-renewable resource must run out, so recycling is inevitable. There is also a strong environmental case for recycling of metals, especially aluminium. Recycling aluminium requires about 7kJ of energy, a saving of about 96% in energy and environmental impact! Extracting aluminium from its ore requires about 200kJ (kilojoules) of energy per kg of metal. This energy is mainly in the form of electricity, which is needed in huge quantities for the electrolytic smelting process. Most local councils now operate “Recycling Centres” which can sort out paper, glass, plastic, etc from our garbage, as long as we remember to put recyclables in the correct bin. Aluminium (mainly drink cans) collected this way is returned to scrap-metal businesses which clean and re-melt the metal to return it to manufacturing industry for re-use. Photo by Griszka Worksheet 5 Fill in the blanks. Check answers at the back. “Minerals” are naturally occurring a)........................................ which are mixed together in rocks. An “ore” is a b).................................. from which it is c)......................................... worthwhile to extract a desired d).............................. Whether it is worthwhile (or not) to mine an ore depends on the balance between the e)........................................................... and the f)........................................................... of mining, transporting and g)..................................... the metal. h)..................................... analysis of an ore deposit is vital to predict the i)..................................... from the ore, to determine if it is worth mining. Ores are j)............................................... resources because once used, they cannot k)........................................................ due to the immense time it takes for l)....................................... processes to form them. WHEN COMPLETED, WORKSHEETS BECOME SECTION SUMMARIES Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 25 Copper ores contain compounds such as m)........................... ............................ and ..................................................................... After mining, the ore is crushed, then concentrated by “n)........................... .......................................”. This process uses a froth of bubbles to separate the o)................................... density copper compounds from the worthless rock which is mainly p)........................................ minerals. The “smelting” process involves q)............................................ reactions. For a sulfide ore, it reacts with r)............................. to form s).............................. metal and t).................................... gas. The final step is to u)................................... the copper by a process of v)....................................................... There are many good reasons to w)......................................... metals, especially x)........................................ which requires large amounts of y)................................... energy to extract from its ore. Producing the electricity required is often done by burning z).......................... fuels such as aa)...................... This contributes to the “ab).................................................... Effect”, responsible for global climate changes. Recycling aluminium requires only a fraction of this energy. www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 CONCEPT DIAGRAM (“Mind Map”) OF TOPIC Some students find that memorizing the OUTLINE of a topic helps them learn and remember the concepts and important facts. Practise on this blank version. METALS Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 26 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Practice Questions 7. The scientist most responsible for the development of the Periodic Table was: A. Avogadro B. Newlands C. Gay-Lussac D. Mendeleev When you have confidently mastered this level, it is strongly recommended you work on questions from past exam papers. 8. Element “X” is in Group 2 and element “Y” in Group 7. These are not intended to be "HSC style" questions, but to challenge your basic knowledge and understanding of the topic, and remind you of what you NEED to know at the K.I.S.S. Principle level. Part A Multiple Choice If X & Y formed a compound, you would expect it to be 1. Which list shows metals used by humans in the correct chronological order of their history of usage? A. bronze, aluminium, iron B. copper, bronze, iron C. gold, iron, bronze D. copper, steel, bronze A. ionic, with formula X2Y B. covalent, with formula X2Y C. ionic, with formula XY2 D. covalent, with formula Y2X 2. Which list correctly identifies an alloy, and the elements it contains? A. Steel; iron and tin B. Bronze; tin and zinc C. Solder; copper and lead D. Brass; copper and zinc 9. If the elements “X” & “Y” in Q8 lie in the same period of the table, you would expect: A. X to have a smaller radius than Y. B. Y to have a higher electronegativity than X. C. X to have a higher 1st ionization energy than Y. D. Y to have a higher melting point than X. 3. The metals used by humans have changed over the course of history. The availability of new metals has often been dependent on the: A. availablity of energy to extract metals from ores. B. discovery of new minerals as people explored the world. C. invention of new alloys. D. development of new technologies to use the metals. 10. The reason for the trend in atomic radius as you move across a period to the right, is: A. increasing nuclear charge. B. addition of extra electron shells. C. decreasing attraction of electrons to the nucleus. D. increasing mass of the atoms. 4. A metal which reacts readily and vigorously with oxygen, water and dilute acids would probably: A. have a high value for 1st ionization energy. B. be from the “Transition” block of the Periodic Table. C. have a very low 1st ionization energy. D. be located at extreme right of the Periodic Table. 11. An atom of argon is about twice as heavy as an atom of neon. You would expect: A. a mole of argon to contain about half as many atoms as a mole of neon. B. equal masses of each element to contain about the same number of atoms. C. 2g of argon to contain about the same number of atoms as 1g of neon. D. the molar mass of neon to be about twice the molar mass of argon. 5. If nickel reacted with sulfuric acid, the products of the reaction would be: A. hydrogen gas and nickel sulfate B. carbon dioxide gas and nickel sulfate. C. nickel sulfide and hydrogen gas. D. sulfur dioxide gas and nickel hydroxide. 6. During the reaction in Q5, the basic underlying change occurring is: A. the breaking covalent bonds. B. the transfer of electron(s) from one species to another. C. chemical changes in “spectator ions”. D. physical dissolving of metal in the acid. Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 12. Which line shows correctly the molar mass (to the nearest gram) of the named substance? A. water, 18g B. carbon dioxide, 28g C. oxygen gas, 16g D. helium gas, 8g 27 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 13. Aluminium reacts with oxygen to form aluminium oxide. 4 Al + 3 O2 2 Al2O3 20. (5 marks) Give an outline of an experiment you have done to investigate the relative chemical activity of some metals. Include the observation(s) you made to assess metal activity, and state the conclusion(s) reached. If 1 mole of aluminium (about 27g) was to be reacted, you would need how many moles of oxygen gas? A. 0.75 mol B. 3 mol C. 1 mol D. 1.3 mol 21. (6 marks) Write a balanced symbol equation for the reaction of: a) magnesium with hydrochloric acid. b) calcium with water (reacts at room temperature). c) potassium with oxygen. 14. Avogadro’s number can be described by the abbreviation NA. If you had 2 moles of methane (CH4), then the number of hydrogen atoms present is: B. 4 x NA A. 2 x NA C. 8 x NA D. 10 x NA 22. (4 marks) When barium metal reacts with an acid there is an exchange of electrons such that hydrogen gas and barium ions are formed. Write 2 “half-equations” to show clearly the species gaining, and the species losing, electrons. 15. Carbon monoxide gas reacts with oxygen gas to form carbon dioxide gas as follows: 2CO(g) + O2(g) 2CO2(g) 23. (4 marks) a) Write an equation (including states) for the first ionization of i) magnesium ii) oxygen b) Describe how the Activity Series of Metals is related to the values of 1st Ionisation Energy. If 100mL of carbon dioxide was produced, then the total volume of reactants (all measured at the same temp. & pressure) before the reaction would have been: A. 100mL B. 150mL C. 50mL D. 250mL 24. (5 marks) a) Sketch a graph (values are not required) to show the general changes in melting points of the elements across one period of the Periodic Table. b) Briefly explain the general trend shown in your graph. 16. The “smelting” of a metal ore always involves: A. separating the metal-containing mineral from the rock. B. decomposing a compound of the metal. C. purifying the extracted metal by electrolysis. D. all of the above. 25. ( marks) a) Write equations (including states) to describe the i) 1st ii) 2nd and iii) 3rd ionisations of potassium. b) At which ionization would you expect a huge increase in the ionization energy? Explain your answer. Longer Response Questions Mark values shown are suggestions only, and are to give you an idea of how detailed an answer is appropriate. 17. (5 marks) Give an example of a) a metal used in its elemental state, and b) a non-ferrous alloy (naming its components) in common use. For each, relate the properties of the metal to its particular use(s). 26. (6 marks) a) Write a balanced equation for the reaction of aluminium metal with hydrochloric acid. b) If 6.58g of aluminium reacted fully, calculate: i) the number of aluminium atoms involved. ii) the mass of aluminium chloride formed. iii) the volume of hydrogen gas (at SLC) 18. (3 marks) Give a reason why a) metal tools are superior to stone tools. b) iron replaced bronze in the history of metallurgy. c) aluminium did not come into common use until the 20th century. 27. (4 marks) It was found by experiment that a compound containing only tin and oxygen, contained 88% tin, by mass. Showing your working, determine the empirical formula for this compound, and give its correct chemical name. 19. (6 marks) The most common metal in use today is steel, which comes in a variety of forms, with different properties and uses. Compare 3 different types of steel, stating the composition of each and relating its properties to a common use. Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 28 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 28. (4 marks) In the reaction of nitrogen and hydrogen gases to form ammonia gas, it was found by experiment that 300mL of hydrogen reacted completely with 100mL of nitrogen. 200mL of ammonia gas was produced. All the gas volumes were measured at a pressure of 10 standard atmospheres and 150oC. a) Write a balanced equation for the reaction. b) Explain how the experimental measurements are in agreement with Gay-Lussac’s Law. 30. (8 marks) a) Give the name and formula for a compound commonly found in copper ores. b) Name, and briefly describe the process by which a copper ore is concentrated and separated from the surrounding “rock”. c) Write a chemical equation to describe the reaction which occurs in the smelting of the ore. (Involving the compound you named in part (a)) 29. (5 marks) a) Differentiate between a “mineral” and an “ore”. b) Outline the role of Chemical Science in assessing the economic feasibility of mining a mineral resource. c) Briefly discuss the sustainability of using the Earth’s mineral resources, and outline a strategy for conservation. d) Name the process by which the smelted copper is purified, and relate the need for purification to a common use of the metal. FOR MAXIMUM MARKS SHOW FORMULAS & WORKING, APPROPRIATE PRECISION & UNITS IN ALL CHEMICAL PROBLEMS NOTICE ANY ERRORS? Our material is carefully proof-read but we’re only human If you notice any errors, please let us know Need to contact us? PO Box 2575 PORT MACQUARIE NSW 2444 (02) 6583 4333 keep it simple science ABN 54 406 994 557 Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science FAX (02) 6583 9467 www.keepitsimplescience.com.au [email protected] TM 29 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Answer Section 4. a) Worksheet 1 b) a) stone or wood/bone b) gold c) decoration/jewellery d) copper e) roasted by fire f) copper and tin g) bronze h) hard and strong i) brittle j) a saw blade k) iron l) harder and stronger m) temperature/energy n) coal o) steel p) carbon q) aluminium r) energy s) ores t) iron u) steel v) cheap/economical w) hard and strong x) mild y) stainless z) copper and zinc aa) solder ab) tin and lead ac) plumbing and electronics ad) lightweight ae) corrosion af) drink cans/window frames/aircraft construction ag) copper ah) conductivity ai) ductile aj) decomposition ak) endothermic al) copper am) aluminium an) electricity ao) metal ap) energy c) 4Li + O2 Part B Practice Problems 1. Ca+(g) a) 1st Ca(g) + FeCl2 + e- 2nd Ca+(g) Ca+2(g) + e- 3rd Ca+2(g) Ca+3(g) + e- 4th Ca+3(g) Ca+4(g) + eb) Between 2nd and 3rd, because 3rd ionisation takes an electron from an inner orbit. 2. a) i) decreasing ii) increasing b) i) increasing ii) decreasing H= fluorine L= francium c) i) increasing ii) decreasing H = helium L = francium 2Fe2O3 2Li2O 2. a) Li, Fe, Pb b) Lithium 3. a) calcium + water Ca + 2H2O b) Tin + water Sn + H2O Ba(NO3)2 a) Dobereiner b) Newlands c) Octaves d) Mendeleev e) Periodic Table f) been discovered g) left gaps h) predict i) almost identical to the predictions j) decreases k) semi-metals & non-metals l) increase m) decrease n) Transition o) semi-metal p) inert q) right r) identical s) metals t) ionic u) positive v) covalent w) covalently x) gain y) negative z) bottom aa) upwards ab) right ac) top ad) down and left ae) decreases af) more ag) nucleus ah) increases ai) Ionisation aj) increase ak) remove al) decrease am) less an) remove ao) higher ap) increases aq) Electronegativity ar) attract as) fluorine at) left au) down Part B Practice Problems 1. a) Pb + O2 PbO2 3O2 + Worksheet 3 Part A a) oxide b) METAL OXIDE c) hydrogen d) hydroxide (or oxide) e) hydrogen f) salt g) different h) activity i) potassium and sodium j) left k) copper and gold l) iron and tin/lead/zinc m) slowly n) electrons o) lose p) hydrogen q) covalent r) hydrogen s) H2 t) loss of electrons u) Reduction v) oxidized w) hydrogen x) reduced y) Ionisation z) remove one electron aa) gas ab) low ac) higher ad) gold used in electronics, because it will not corrode. + hydrogen + zinc sulfate H2 + ZnSO4 hydrogen + calcium chloride H2 + CaCl2 Ca + 2HCl c) Barium + nitric acid Ba + 2HNO3 H2 d) Iron + hydrochloric acid Fe + 2HCl H2 5. a) the metals: Zn, Ca, Ba, Fe b) hydrogen ions (from the acid) c) sulfate, chloride and nitrate ions. Worksheet 2 Part A b) 2Fe Zn + H2SO4 hydrogen + calcium hydroxide H2 + Ca(OH)2 hydrogen + tin(II) oxide H2 + SnO Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science 30 www.keepitsimplescience.com.au keep it simple science TM Emmaus Catholic College SL#802440 Worksheet 4 Part A 1. Molar Masses a) 39.10g d) 159.8g g) NaI= 149.9 j) CuSO4=159.6g b) 83.80g e) N2 = 28.02g h) Fe2S3=207.9g k) Al2O3=102.0g d) n(Sn) = m / MM = 14.8 / 118.7 = 0.125 mol i) ∴ n(SnO2) = 0.125 mol m(SnO2) = n x MM = 0.125 x 150.7 = 18.8g ii) ∴ n(H2O) = 0.125 x 2 = 0.250 mol m(H2O) = n x MM = 0.250 x 18.016 = 4.50g iii) ∴ n(H2) = 0.125 x 2 = 0.250 mol m(H2) = n x MM = 0.250 x 2.016 = 0.504g c) 118.7g f) MgO = 40.31g i) NH3=17.03g l) 180.2g 6. Empirical Formulas a) i) Copper masses = 12.84g No moles = 12.84/63.55 = 0.2020 ratio = 1 ∴ emp. formula is CuCl ii) Copper(I) chloride b) i) Carbon masses = 1.5g No moles = 1.5/12.01 = 0.125 = 0.125/0.125 = 1 ratio ≅ 1 ∴ emp. formula is CH4 ii) methane c) i) Nitrogen masses = 30% No moles = 30/14.01 = 2.14 = 2.14/2.14 ratio = 1 ≅ 1 ∴ emp. formula is NO2 ii) 2 x (NO2) = N2O4 iii) dinitrogen tetra-oxide 2. Moles in a Given Mass use n = m/MM a) n= 100.0/207.2 = 0.4826 mol b) n = 100.0/ 65.39 = 1.529 mol c) n = 100.0 / 18.016 = 5.551 mol d) n= 100.0 / 251.12 = 0.3982 mol e) n = 38.55 / 62.31 = 0.6187 mol f) n = 60.00 / 44.01 = 1.363 mol g) n = 1.000/ 81.39 = 0.01229 mol h) n = 500.0 / 180.2 = 2.775 mol i) n = 3.258x10-3/ 58.44 = 5.575 x 10-5 mol j) n = 128.6 / 96.094 = 1.338 mol 3. Moles & Particles a) use n = N/NA and N= n x NA i) N = 3 x 6.022x1023 = 1.807x1024 molecules ii) N = 2.478 x 6.022x1023 = 1.492x1024 molecules iii) N = 5 x 6.022x1023 = 3.011x1024 “sets” of ions. iv) N = 0.007862 x 6.022x1023 = 4.734x1021 atoms v) N = 1/1000 x 6.022x1023 = 6.022x1020 atoms b) i) n = N/NA = 8.800x1025/6.022x1023 = 146.1 mol m = n x MM = 146.1x118.7 = 17,340g (=17.34kg) ii) n = N/NA = 2.575x1024/6.022x1023 = 4.276 mol m = n x MM = 4.276 x 18.016 = 77.04g iii) n = m/MM = 400.0/ 18.016 = 22.20 mol N = n x NA = 22.20 x 6.022x1023 = 1.337x1025 molecules iv) n = m/MM = 2.569/58.69 = 0.04377 mol N = n x NA = 0.04377 x 6.022x1023 = 2.636x1022 atoms v) n = N/NA = 2.500x1023/6.022x1023 = 0.4151 mol m = n x MM = 0.4151 x 32.07 = 13.31g : Hydrogen : 0.5g : 0.5/1.008 : 0.496 : 0.496/0.125 : 3.96 : 4 : Oxygen : 70% : 70/16 : 4.375 : 4.375/2.14 : 2.04 : 2 7. Volumes of Reacting Gases a) i) 2.5L ii) 5 L b) i) 0.5 L ii) 0.5 L c) i) 20 L ii) 10 L d) i) 13.5 L ii) 4.5 L e) i) 0.4 L ii) 0.2 L 4. Mole ratios in Equations a) i) 0.5 mol ii) 1 mol b) i) 0.25 mol ii) 0.125 mol c) i) 1.5 mol ii) 1.5 mol d) i) 0.25 mol ii) 0.375 mol e) i) 0.2 mol ii) 0.15 mol 8. Mass & Gas Volume a) i) n(Li2O) = m/MM = 5,000/29.882 = 167 mol ii) n(CO2) = 167 mol iii) v(CO2) = 167 x 24.8 = 4.14x103 L (>4,000L !) b) i) n(O2) = 10.0/24.8 = 0.403 mol ∴ n(Fe2O3) = 0.403 x2/3 = 0.269 mol m(Fe2O3) = n x MM = 0.269 x 159.7 = 42.9g ii) n(Fe) = m/MM = 100/55.85 = 1.79 mol ∴ n( O2) = 1.79 x 3/4 = 1.34 mol V(O2) = 1.34 x 24.8 = 33.2 L c) i) n(H2O) = m /MM = 1.00/ 18.016 = 0.0555 mol ∴n(H2) = 0.0555, v(H2) = 0.0555 x 24.8 = 1.38 L and n(O2) = 0.0555/2, v(O2) = (0.0555/2) x24.8=0.688L ii) use Gay-Lussac’s Law: v(H2) = 100mL (0.10 L) iii) n(H2) = 0.10 / 24.8 = 0.00403 mol ∴ n(H2O) = 0.00403 mol m(H2O) = n x MM = 0.00403 x 18.016 = 0.073g 5. Mass Quantities in Reactions a) n(Ca) = m/MM = 8.50/40.08 = 0.212 mol ∴ n(CaO) = 0.212 mol m(CaO) = n x MM = 0.212 x 56.08 = 11.9g b) n(Ag2CO3) = m/MM = 20.0/275.81 = 0.0725 mol i) ∴n(Ag) = 0.0725 x 2 = 0.145 mol m(Ag) = n x MM = 0.145 x 107.9 = 15.6g ii) ∴ n(CO2) = 0.0725 mol m(CO2) = n x MM = 0.0725 x 44.01 = 3.19g iii) ∴ n(O2) = 0.0725 / 2 = 0.03625 mol m(O2) = n x MM = 0.03625 x 32.00 = 1.16g c) n(Al) = m / MM = 6.50/26.98 = 0.241 mol i) ∴ n(HCl) = 0.241 x 3 = 0.723 mol m(HCl) = n x MM = 0.723 x 36.458 = 26.4g ii) ∴ n(H2) = 0.241 x 3/2 = 0.3615 mol m(H2) = n x MM = 0.3615 x 2.016 = 0.729g iii) ∴ n(AlCl3) = 0.241 mol m(AlCl3) = n x MM = 0.241 x 133.33 = 32.1g Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science : Chlorine : 7.16g : 7.16/35.45 : 0.2019 : 1 31 www.keepitsimplescience.com.au keep it simple science Worksheet 4 Part B a) number c) particles e) Avogadro’s g) ratio of moles i) moles (particles) k) empirical m) volume o) Avogadro q) equal s) temperature & pressure u) 1 standard atmosphere TM Emmaus Catholic College SL#802440 21. a) Mg + 2HCl b) carbon (carbon-12) d) formula mass f) 6.022 x 1023 h) mass j) reactions l) Gay-Lussac n) whole-number p) volumes r) molecules/particles t) SLC v) 25 c) 4K 22. 2K2O + O2 Ca(OH)2 Melting Point Practice Questions 13. A 14. C 15. B 16. B Part B Longer Response In some cases there may be more than one correct answer possible. The following “model” answers are correct, but not necessarily perfect. 17. a) Copper. Used for electrical wiring, due to its excellent conductivity and high ductility. b) Solder, an alloy of tin & lead. Used for joining pipes in plumbing, and joining wires in electronics, because of its very low melting point. 18. a) not brittle/ can be re-sharpened/shape possiblities (saw) b) Iron is stronger and harder... tools are superior. c) Needs electricity for smelting. 19. Mild Steel (0.2% carbon). Used for car bodies & sheet metal, because it is strong but very malleable. Tool Steel (1.5% carbon). Used for hammers, drills, etc because it is very hard and strong. Stainless Steel (20% nickel & chromium). Used for food utensils and medical equipment because it resists corrosion and is very hygenic. 20. Small pieces of metal added to dilute acid in test tubes. (To keep expt. fair, the acid must be same strength, and metal pieces same size.) Observe the rate of gas production to assess reactivity. Conclusion: order of activity is: Mg > Zn > Fe > Pb > Cu Preliminary Chemistry Topic 2 Copyright © 2005-2007 keep it simple science H2 + Ba Ba+2 + 2e2H + 2e H2 (Barium lost, hydrogen ions gained) 23. a) i) Mg(g) Mg+(g) - + e+ ii) O(g) O (g)+ e b) The lower the ist Ionisation Energy the more active the metal, because the metal readily loses electron(s) to enter a reaction. 24 a) graph b) At the beginning of a period (left) the elements are soft metals with moderate to low mp’s. Moving right, the mp’s rise to a maximum at a transition metal, or semi-metal. Then mp’s fall rapidly at the nonmetals. Lowest values are the inert gases on far right. 25. K+(g) -+ ea) 1st: K(g) + +2 2nd: K (g) +2 K (g) + e 3rd: K (g) K+3(g) + eb) Between 1st & 2nd, because the 2nd ionization involves an electron from an inner orbit, which will require a big increase in energy to remove. 26. a) 2Al + 6HCl 3H2 + 2AlCl3 b) n(Al) = m / MM = 6.58 / 26.98 =2 0.244 mol23 i) N(Al)= n x NA=0.244x6.022x10 = 1.47x10 atoms ii) n(AlCl3) = 0.244 mol m(AlCl3) = n x MM = 0.244 x 133.33 = 32.5g iii) n(H2) = 0.244 x 3/2 = 0.366 mol V(H2) = 0.366 x 24.8 = 9.08 L 27. Tin : Oxygen % mass 88 : 12 moles = 88/118.7 : 12/16.00 = 0.74 : 0.75 ≅ 1 : 1 ∴ empirical formula is SnO. Tin(II) oxide 28.a) N2 + 3H2 2NH3 a) compounds b) mineral c) economically d) metal e) commercial price f) production cost g) extracting (smelting) h) Chemical i) yield j) non-renewable k) be replaced l) geological m) copper(I) sulfide & copper carbonate/hydroxide n) Froth-flotation o) lower p) silicate q) decomposition r) oxygen s) copper t) sulfur dioxide u) purify v) electrolysis w) recycle x) aluminium y) electrical z) fossil aa) coal ab) Greenhouse Multiple Choice 5. A 9. B 6. B 10. A 7. D 11. C 8. C 12. A b) Ca + 2H2O + Worksheet 5 Part A 1. B 2. D 3. A 4. C H2 + MgCl2 b) volumes = 100mL 300mL 200mL Vol. ratio = 1 : 3 : 2 The volumes of the gases are in a simple, whole number ratio to each other. This is Gay-Lussac’s Law. 29. a) A mineral is a naturally-occurring crystalline compound. An ore is a mineral which is economically worth mining to extract a metal from. All ores are minerals; not all minerals are ores. b) Chemical analysis allows an ore body to be analysed to predict the yield of metal. c) Ores are non-renewable resources, and once used cannot be replaced. Therefore, it is wise to conserve these resources by recycling metals wherever possible. 30. a) Copper(I) sulfide, Cu2S. b) Crushed ore is separated by “froth flotation”. Low density ore is carried in a detergent froth, while silicates fall to the bottom. c) Cu2S + O2 2Cu + SO2 d) Electrolysis. Copper needs to be very pure for its main use in electrical wires. If impure, conductivity is lower. 32 www.keepitsimplescience.com.au