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TAFE: TPC CHEMISTRY A MODULE: NSWTSCN308A INVESTIGATE THE CHEMICAL NATURE OF MATTER Author: Robyn Krajniak (1995) Revised: Dan Solomon (2013) Graphics: Martin Kent 1 TABLE OF CONTENTS 1. 2. 3. 4. INTRODUCTION – Matter Surrounds Us! 6 CLASSIFICATION SYSTEM 1: STATES OF MATTER 7 1.1 Models of matter Written Exercise: Which state? Activity: Making models Written Exercise: States of matter 7 8 8 10 1.2 Changes of State 11 MODELLING THE SMALLEST PARTICLES OF MATTER 12 2.1 What particles compose the substances around us? 12 2.2 Modelling Atomic Structure Written Exercise: Timeline 14 14 2.3 Isotopes 17 2.4 Quantifying Subatomic Particles in Atoms 18 2.5 How are Electrons organised around the Nucleus? Written Exercise: Drawing structural models of elements Written Exercise: The first 20 elements of the periodic table Activity: Elemental puns 18 20 21 22 CLASSIFICATION SYSTEM 2: ELEMENTS, COMPOUNDS AND MIXTURES 23 3.1 Classifying Everyday Substances Written Exercise: Classifying substances 23 24 3.2 Physical and Chemical Properties Written Exercise: Physical and chemical changes Experiment – Making observations Experiment – Physical versus chemical change 25 26 26 29 3.3 Mixtures and laboratory separation techniques Experiment – Separation techniques 1 Experiment – Separation techniques 2 30 33 36 ELEMENTS AND THE PERIODIC TABLE 39 4.1 Mapping Elements on the Periodic Table 39 4.2 Distribution of the Elements 41 2 5. 6. 4.3 How Atoms can be arranged in Everyday Elements Research Task – Allotropes 42 44 4.4 Classification of the Elements Experiment – Properties of elements 44 44 CHEMICAL BONDING 50 5.1 What makes an Atom Chemically Stable? 50 5.2 Loss and Gain of Electrons – The Ionic Bond 51 5.3 Sharing Electrons – The Covalent Bond Written Exercise: Lewis electron dot formulae 52 52 5.4 Which Chemical Bond Will Form? Written Exercise: Type of chemical bond 53 53 5.5 The Metallic Bond 53 5.6 Summary of Bonding Written Exercise: Formulae and chemical bonding 54 55 5.7 Compound Chemical Formulae Written Exercise: Interpreting chemical formulae 56 58 5.8 Valency 58 5.9 Chemical Formulae and Naming Written Exercise: Writing chemical formulae Written Exercise: Formulae and naming Activity: Modelling molecules and compounds 60 61 62 64 PROPERTIES OF COMPOUNDS 66 6.1 Comparing properties of elements and compounds 66 6.2 Classification of Compounds 67 6.3 Properties of Compounds Written Exercise: Solubility Experiment – Physical properties of compounds Written Exercise: Hydrogen chloride conductor? Written Exercise: Compound properties in summary Experiment – 3 white powders 68 70 71 72 72 73 3 7. 8. CHEMICAL REACTIONS 75 7.1 Energy and Chemical Processes Written Exercise: Chemical reactions (introduction) 75 76 7.2 Writing Chemical Equations - Chemical Shorthand Activity: Modelling a simple chemical reaction Written Exercise: Balancing chemical reactions Written Exercise: Expressing chemical reactions 78 79 81 86 7.3 Ionic Equations Written Exercise: Expressing ionic equations Experiment – Solubility, precipitation and ionic equations 88 89 90 QUANTITIES AND CHEMICAL PROCESSES 92 8.1 An Introduction to Quantitative Chemistry 92 8.2 The Mole Concept Written Exercise: Mole calculations 94 96 8.3 Theoretical Yield 97 8.4 Percentage Yield 98 8.5 Limiting Reagents Written Exercise: Limiting reagent and theoretical yield Yield Experiment 1 – Magnesium Oxide Yield Experiment 2 – Decomposition of a carbonate compound Yield Experiment 3 – Zinc Iodide 99 101 102 103 105 8.6 Moles, Gas Volumes and Yields Written Exercise: Theoretical yield and gas volumes 107 109 8.7 A quick note on Kelvin Yield Experiment 4 – Gas volumes 111 112 8.8 Percentage Composition Written Exercises: Calculating percentage composition 114 114 8.9 Empirical and Molecular Formulae Written Exercises: Percentage composition, empirical and molecular formulae Experiment – Which mercury chloride? 115 Calculating Concentration 120 8.10 4 116 117 9. 8.11 Molarity Written Exercises: Calculating molarity Written Exercises: Solution calculations 121 122 124 8.12 Solution Preparation and Quantitative Dilution in the Laboratory Written Exercise: Dilution scenario calculations Experiment – Solution preparation and serial dilution 125 126 128 CHEMICAL ANALYSIS 130 9.1 Qualitative Vs Quantitative Chemical Analysis 130 9.2 Gravimetric Analysis Practice Questions Experiment – Sulfate content of plant fertiliser 131 133 134 9.3 Volumetric Analysis Practice Questions Experiment 1 – Practice titration and HCl analysis Experiment 2 – HCl standardisation Experiment 3 – NaOH standardisation Experiment 4 – Vinegar analysis Experiment 5 – Miscellaneous titration 136 141 142 144 145 146 147 9.4 Chromatography 149 9.5 Electrophoresis 151 9.6 Colorimetry Experiment – Colorimetry 152 154 9.7 Spectroscopy Experiment 1 – Flame tests Written Exercises: Flame Emission Experiment 2 – Introduction to Flame Emission Experiment 3 – Sodium and potassium in beer Overview: Infra-red Spectroscopy 155 157 160 162 164 168 ANSWERS TO WRITTEN QUESTIONS 169 References 178 5 INTRODUCTION – Matter Surrounds Us ! Chemistry involves the study of matter. All of the different materials around us are made of matter. In this module you will learn how to answer questions such as: - Why do different materials have different characteristics, ie - properties? In what ways is sugar different to salt? How are metals different to plastics or ceramics? - How can a knowledge of the properties of a substance help you to select the best material for a particular purpose? What would be the best material to choose for making a coffee mug or a saucepan? - How can you explain the properties of a substance by understanding its structure? Why are some substances brittle while others are flexible? What makes some substances able to conduct electricity? - How can you predict what will happen if several substances are reacted together? What happens when a substance burns? Why do some substances dissolve in water while other substances don’t? What is Matter? Matter is any material of substance. It has mass and takes up space (volume). Look around you. In any room there are many examples of different types of matter: flooring materials paint metal fittings probably several plastic items water-perhaps in a cup or sink air Is there anything that is not matter? Yes, energy is not matter. All forms of energy (heat, light sound etc) can affect matter but energy has no mass or volume. 6 1. CLASSIFICATION SYSTEM ONE: STATES OF MATTER One way of classifying substances is by grouping them according to whether they are solid, liquid or gas. These are often called the states of matter. In order to understand the different way solids (eg rock), liquids (eg water) and gases (eg air) behave, we set up models which we can use to visualise what is happening deep in the structure of matter. The following models show matter consisting of tiny particles which are in constant motion - even the particles in a solid piece of rock are constantly moving! 1.1 Models of matter In a solid, the particles are just vibrating in set positions. They are packed very close together and attract each other strongly. In a liquid, the particles are further apart. They can slide past one another as they move, colliding with each other and the walls of the container when they reach the edge. The particles are attracted to each other, but not as strongly as in the solid state. In a gas, the particles fly as far apart as the container walls will allow. The particles are only very weakly attracted to each other, and they collide with each other and the walls of the container. 7 Written Exercise: Which State? Many substances are easy to classify as solid, liquid or gas, others can be difficult, particularly if they contain more than one state e.g. an inflated balloon! Place the following substances in their correct category: sand lemonade fog plasticine natural gas balsamic vinegar steel mercury timber paint toothpaste shaving foam milk jelly Add ten more examples including some which may be difficult to classify. Solid Activity: Liquid Gas Difficult to classify? Why? Making models Making models of solids, liquids, gases using marbles or sand in a transparent container. Models help us to understand the behaviour and properties of solids, liquids and gases. The particles of a solid could be represented by gently moving the container so that the objects vibrate (rock backwards and forwards) without changing position. 8 The particles of a liquid could be represented by tilting the container so that the objects slide past one another. The particles of a gas could be represented by shaking the container so that the objects are in constant motion, flying around colliding with all parts of the container. Manipulate these models to demonstrate the following properties. Then explain in words why each is true. (i) Only solids can have a fixed shape. (ii) Only gases can exert a pressure in all directions at the same time. (iii) It would be very difficult for a solid to diffuse. (iv) Increasing the temperature of a substance involves giving the particles more energy (greater movement). Use your models to demonstrate the effect of increasing temperature and corresponding changes of state. 9 Written Exercise: States of matter What are some properties of solids, liquids and gases? 1. Try to complete of the following table. Have a sample of each state of matter in front of you so that you can refer to it! Property Solid Liquid Gas Shape (3-dimensional conformation) Fixed (doesn’t change) Takes the shape of the part of the container it fills Takes the shape of the container Volume (the amount of space taken up) Compressible? (can it be made to occupy a smaller volume?) Diffuses? (can it spread out by itself?) 2. Use the particle models to explain these properties by completing the following descriptions. In a solid the particles are packed closely together. The particles can vibrate but they cannot move around. Therefore the solid has a fixed shape and volume (when at constant temperature). Because the particles are very close together and in fixed positions, solids are virtually incompressible. A solid cannot easily diffuse. Complete similar explanations for liquids and gases. (a) In a liquid the particles .............. (b) In a gas the particles ................ 10 Optional Research: Other States of Matter Are there any other states of matter? You might want to research some information on plasmas as another state. Ref: C1 pp6-7, 261-263 1.2 CC1 pp13-15 Changes of State When a solid is heated, the particles vibrate faster until their energy overcomes the forces (bonds) holding the particles in position. The particles break loose and begin to slide past one another - the solid melts to form a liquid. Similarly heat can provide the energy for particles in a liquid to escape from one another and a gas is formed as the liquid evaporates. Some substances, such as iodine, change directly from a solid to a gas when they are heated - this change is called sublimation. Change Name solid liquid melting, fusion liquid solid freezing, solidification liquid gas boiling, evaporation gas liquid condensing solid gas sublimation gas solid deposition Ref: CC1 pp14-15 Sublimation Melting SOLID Boiling LIQUID Freezing GAS Condensation Deposition 11 2. MODELLING THE SMALLEST PARTICLES OF MATTER 2.1 What particles compose the substances around us? When you use the term particle to explain the different properties of a solid, liquid or gas, you are referring to the simplest, most basic unit of that substance. If you took a sample of that substance and started to divide it into smaller and smaller portions eventually you would reach this basic unit that still has the properties of you substance. The type of basic unit (particle) will depend on how the atoms are arranged in your substance. For example the simplest unit in a sheet of aluminium foil is the aluminium atom. The sheet of foil consists of many billions of atoms arranged in a network structure. Molecules Molecules are formed in many substances when small groups of atoms join together with a unique structural conformation. The atoms can be all the same or there can be several different types. A chemical formula gives information about the types and numbers of atoms present in each of these molecules. Examples: The simplest unit in the substance water is the water molecule, H2O which is a cluster of two hydrogen atoms and one oxygen atom. Water is therefore composed of 2 different elements, classifying it also as a compound. In oxygen gas, O2 there are two oxygen atoms per molecule. 12 Molecules in simple 3-D Chemical molecules are discrete structures representing everyday substances in their most simple form (ie; atomic ratio). You will use a molecular model kit a bit later to represent a variety of different chemical structures. The lines or “sticks” between individual atoms represent . . . . . . . . . . . . . . . . . . . . . . . . . . . Water, H2O Ref: CC1 pp16-22 Ammonia, NH3 Carbon Tetrachloride, CCl4 C1 pp9-13 Other substances are made up of a large network of ions. Ions are atoms or groups of atoms which carry a positive or negative electric charge. Common table salt (which has the chemical name of sodium chloride) is made of a large network of sodium ions, Na+ and chloride ions, Cl - 13 2.2 Modelling Atomic Structure In order to be able to distinguish these different sorts of particles you must understand some fundamental ideas about the structure of an atom. Since atoms are too small to be seen even with the best microscopes, scientists have developed models of atoms to represent them. No model can be exactly like an atom in all respects and so models are being constantly revised as further research reveals more about atoms. In order to explain the properties you are observing, use the simplest possible model. Some models of the atom A Basic Timeline of historical development of models of the atom Written Exercise: Timeline Add a least one more name and date to the Timeline above. Optional Research: An atomic model You may like to research the experimental discoveries that led to the changing of one of the models of the atom. 14 A simple Bohr Model of the Atom This model of the atom will be of particular use to you in this module. Atoms are units composed of much smaller subatomic particles: protons, neutrons and electrons. Electrons: orbiting about the nucleus of the atom Nucleus: A dense mass of protons and neutrons at the centre of the atom KEY: + Proton (mass = 1 a.m.u.) Neutron (mass = 1 a.m.u.) – Electron (mass = 0.00054 a.m.u) These subatomic particles are measured in arbitrary units called atomic mass units, shortened to a.m.u. The periodic table lists the atomic weights of all elements in these units, as individual atoms are far too small and light to weigh individually. Inside the nucleus are found protons (positive charge) and neutrons (neutral charge). Almost all the mass of an atom is in the nucleus with an overall positive charge. Negatively charged electrons orbit the nucleus and define the volume of the atom. This would indicate that atoms are, by far, mostly empty space! A simple calculation based on relative atomic masses above will reveal the nucleus as containing around 99.9% by mass of an entire atom! 15 Data Table of Subatomic Particles: Subatomic Particle Charge Relative mass (a.m.u.) Position Electron: e – –1 0.00054 orbiting the nucleus +1 1 nucleus 0 1 nucleus Proton: p + Neutron: n The electrons form a negatively charged cloud around the nucleus. Atoms are neutral, so that the number of positively charged protons must equal the number to negatively charges electrons. No of protons (p+) = No of electrons (e –) When electrons are lost an atom becomes a positively charged ion or cation. When electrons are gained by an atom it becomes a negatively charged ion or anion. 16 2.3 Isotopes Any element is identified by the number of protons in its nucleus, defined as its Atomic Number. The number of neutrons within an individual element can vary without changing its identity, hence the concept of isotopes. Different isotopes of a given element have the same number of protons, but different numbers of neutrons in the nucleus. Case Study 1: Carbon (Atomic Number 6) All atoms of Carbon will be identified by Atomic Number 6, having this number of protons within its nucleus. The nuclei of most carbon atoms contain also 6 neutrons giving a combines mass number of 12. But statistically, a very small proportion of carbon atoms in nature have 8 neutrons, thus giving a combined mass number of 14. These are two different isotopes of carbon which can be distinguished by their different masses; C-12 and C-14. C-14 is unstable and radioactive. Radioactivity is a feature of many different elemental isotopes. The residual intensity of C-14 radioactivity is used for carbon dating of ancient fossils. So why do many elements appearing on the Periodic Table have mass number values listed with decimal places? Fractions of 1 a.m.u. do not occur in nature! Case Study 2: Chlorine (Atomic Number 17) Chlorine is listed with an atomic mass of 35.45. All Chlorine atoms contain 17 protons in their nucleus. From the atomic mass above it could be concluded that a chlorine nucleus contains 18.45 neutrons; but there is no such thing as 0.45 of a neutron. An atomic mass, as appears on the periodic table is an average mass of all isotopes occurring in nature. This may have decimal places A mass number is a round figure, summing the protons + neutrons of a single atom. Statistically in nature, chlorine atoms are composed of 2 isotopes: Chlorine-35 (having 18 neutrons) accounts for around 76% of chlorine atoms in nature. Chlorine-37 (having 20 neutrons) accounts for around 24% of chlorine atoms in nature. Now, given these statistics of two chlorine isotopes, calculate its atomic mass . . . 17 2.4 Quantifying Subatomic Particles in Atoms The Periodic Table of the Elements (see also page 40) is a complete database of known atomic structures. Atoms are listed in order of atomic number, being the number of protons contained within the nucleus of the atom. An Element is a substance composed of a single specific atom. The Periodic Table lists over 100 different elements providing information on their composition of protons, neutrons and electrons. FEATURES OF ELEMENTS IN THE PERIODIC TABLE: 1. Chemical Symbol: Shortened form of name. 2. Atomic Number: Number of Protons in the Nucleus. 3. Mass Number: Combined mass of Protons and Neutrons. 20 Atomic Number Ca Atomic Mass Chemical Symbol 40.08 Calcium Chemical Name of Element DETERMINING ATOMIC STRUCTURE Number of Protons = Atomic Number Number of Electrons = Number of Protons, ie Atomic Number Number of Neutrons = Atomic Mass (rounded) – Number of Protons 2.5 How are Electrons organised around the Nucleus? Protons and neutrons make up the almost the entire mass of each atom and are located in the nucleus. In our model of the atom, the electrons are arranged in orbits or shells around the nucleus and equal the number of protons in a neutral atom. However, because they may be a long way from the nucleus, outer electrons may be readily removed or added and then an ion is formed. Electrons which are in the same shell are about the same distance from the nucleus and they have about the same energy. Shells are sometimes referred to as energy levels. More complicated models of the atoms refer to subshells, suborbitals, standing waves and so on. Shells are numbered from the nucleus. 18 Electron shells (or orbits) around the nucleus The maximum number of electrons able to “fit” in any given orbital follows what is referred to as the 2n2 rule, where n is the shell number Maximum number of electrons, 2n2 Shell number (n) 1 2 2 8 3 18 4 32 And so on . . . . . . . . . . . . . ...................... Electrons fill the inner shells in order from the nucleus and ever outward. In the third shell, however, eight electrons are placed, then the next two move to the fourth shell before the final 10 are placed in the third shell. This shows a shortfall in this particular model that is solved by using move complicated models of the atom that include subshells. If we consider only the first twenty elements, ie maximum of 20 electrons, then the order of filling the shells follows a somewhat different set of rules: 1st shell 2 nd rd th 3 4 can take up to 2 electrons shell can take up to 8 electrons shell can take up to 8 electrons shell can take remaining 8 electrons When drawing the electrons of a particular atom, the shells are depicted as concentric circles around the nucleus, and the electrons as dashes, dots or “e” symbols. For example, a hydrogen atom with 1 electron 19 Written Exercise: Draw structural models of the following elements: Aluminium Beryllium Calcium Nitrogen 20 THE FIRST 20 ELEMENTS OF THE PERIODIC TABLE Use a periodic table and your knowledge of electron shell structures to complete the following table. Some information has been completed for you. Element Hydrogen Symbol Atomic Number Mass Number Neutron Number 7 5 Beryllium 5 Boron C 2:4 7 Nitrogen 316 Oxygen 10 Fluorine 10 Neon Na 2:8:1 12 Magnesium 2+ 27 Aluminium 14 Silicon 15 Phosphorous 2:8:6 Sulfur 1- Chlorine Argon Potassium Calcium Charge of Ion valency 2 Lithium Sodium Electron Configuration H Helium Carbon Electron Number Ar 19 40 21 Activity: Elemental puns! Are you a creative thinker? Do you like puns? Even if you do not, you might enjoy groaning at these elemental puns. Each of the blanks can be completed with the name of an element. So caesium your pencil and try your luck! 1. 0.5 times holmium: = ……………………… 2. An officer of the constabulary: A .............................................. 3. Funds from your mother’s sister: ............................................... 4. A comical prisoner: A ............................................... 5. What you do with dead cats: You ............................................... 6. A driller’s motto: ............................................... 7. What a good doctor can do for a patient: 8. When you have a hole in the boat: You ............................................... 9. How you mended your clothes: You ............................................... 10. What are you doing with that person’s hair? I’m 11. A nocturnal trojan: A ............................................... 12. A trojan who smokes dope: A ............................................... 13. A man who tried to murder his wife with a razorblade only managed to give her: ............................................... ......................................... …………………………………….. 14. …………………………………………….. Mind your own: Ref: CC1 pp30-34 22 3. 3.1 CLASSIFICATION SYSTEM TWO: ELEMENTS, COMPOUNDS AND MIXTURES. Classifying Everyday Substances The matter around us can be classified into pure and impure substances. Pure substances have a definite composition - they are always the same no matter where the sample comes from. Common table salt is and example of a pure substance since it is always NaCl (sodium chloride) whether the sample comes from Australia or anywhere else on earth. The ratio of sodium ions to chloride ions is 1:1 and it melts at 801 oC. Water is an example of a pure substance since it is always H2O - it contains twice as many hydrogen atoms as oxygen atoms. It melts at 0 oC and boils at 100oC at atmospheric pressure. Impure substances are normally called mixtures since they have no definite composition and the proportion of components will vary between samples. They have no fixed melting and boiling points. Most of the materials surrounding us are mixtures! Air contains oxygen, nitrogen, carbon dioxide, water vapour and other gases. Can you give examples of how air samples may vary? Seawater contains water, dissolved salts and dissolved gases. How do samples of seawater vary? Matter Composed of atoms. Has mass, volume Impure Pure - Atoms are in fixed proportions - Atoms are in variable proportions - Has a definite composition - Has variable composition - Has fixed melting and boiling points - Has no fixed melting or boiling points Elements Atoms are all the same Cannot be decomposed Compounds Mixtures 2 or more different atoms in fixed proportions 2 or more different atoms in variable proportions Can be decomposed into simpler substances Can be physically separated into pure substances 23 Pure substances are further divided into those which cannot be decomposed (broken down into simpler substances) and those which can be decomposed. Elements are our most basic substances - they are made of one type of atom only and therefore cannot be decomposed. Compounds are made of at least two different types of atoms and therefore can be decomposed - although sometimes with great difficulty! Examples : Hydrogen, H2 ; oxygen, O2 and ozone, O3 are all elements. Their formulae show that they each contain only one type of atom. Water, H2O is a compound formed from hydrogen atoms and oxygen atoms when they have chemically bonded together. The formula shows that two different atoms are present in water and that their ratio is fixed – this defines the chemical composition of water. Written Exercise: 1. CLASSIFYING SUBSTANCES Classify the following list of substances below by placing them into their corresponding column. Add further examples to each column if you can. An element is pure substance composed of only one type of atom. A compound is a pure substance composed of 2 or more different atoms. A mixture is a substance that contains 2 or more pure substances mixed together. SUBSTANCES: Elements Compounds Mixtures Water, Air, Copper, Soil, Nitrogen, Gold, Ethanol, Orange Juice, Petrol, Glass, Vinegar, Brass, Sweat, Iodine, Carbon Dioxide. 2. Which laboratory methods would you use to separate the following mixtures? (a) Charcoal and salt (b) Water and oil (c) Alcohol and water (d) Sugar from a sugar solution ................................................................................. (e) Water from muddy river water ................................................................................... ................................................................................... ................................................................................... ................................................................................... Differences in physical or chemical properties decide how we classify matter. 24 3.2 Physical and Chemical Properties Physical changes are those which affect only the physical form or state of a substance. They do not affect the chemical composition of a substance. A physical property of a substance can usually be observed without a change in chemical composition taking place. Physical changes can easily be reversed in many instances. Examples: Physical Change Physical Property The boiling of water Boiling point The dissolving of salt in water Solubility of salt In a physical change, the elements and / or compounds remain chemically the same! Chemical changes are those which cause the chemical composition of the substance to be altered, that is a new substance is formed. A chemical property can be observed as a substance chemically reacts with another, or breaks down (decomposes), forming new substance(s). Chemical changes are often difficult, even impossible, to reverse. Examples: Chemical Change Chemical Property Combustion of fuel Flammability Converting water to hydrogen and oxygen Ease of decomposition In a chemical change new substances are formed, but the number of atoms remains the same - they have just been rearranged ! 25 Written Exercise: Physical and Chemical Changes Classify the following as chemical (C) or physical (P) changes and determine which property is concerned. 1. Water freezing to form ice. P Melting point (freezing point) 2. An iron bar rusting in the air. ............. ............................................... 3. Petrol combusting in a car engine. ............. ............................................... 4. Moulding plastic into a new shape. ............. ............................................... 5. Setting off some fireworks. ............. ............................................... Experiment – MAKING OBSERVATIONS QUALITATIVE VERSUS QUANTITATIVE OBSERVATION: Qualitative refers to any observation identifying any characteristic or process which cannot easily be expressed as an amount or quantity. Quantitative refers to any observation which can, in theory be easily measured and therefore be expressed as a number or quantity, for example: Mass, Volume, Concentration, Temperature etc…… PHYSICAL AND CHEMICAL PROPERTIES: A Physical Property is any characteristic of a pure substance which: A Chemical Property is any characteristic of a pure substance which: PART A: Observing a burning candle Aim – to make as many observations as possible on a burning candle. ** See over page for task and results page 26 Task: Light a candle and begin immediately listing your observations in table 1 below. Classify each observation according to the criteria listed above. Extinguish the candle and continue making observations in table 2 according to the same criteria. TABLE 1 – BURNING CANDLE Observation Physical or Chemical Change? Physical or Chemical Property Qualitative or Quantitative? TABLE 2 – EXTINGUISHED CANDLE Observation Physical or Chemical Change? Physical or Chemical Property 27 Qualitative or Quantitative? PART B: Observing and interpreting everyday phenomena Aim: To interpret the science behind everyday phenomena by observation Equipment: - Ice (Large Beaker) Bunsen burner Forks / coin demo Volumetric flask + dye solution - Match / pin demo Coke / Diet Coke Bucket Method: - Observe each station carefully. Record your observations and any other required explanation in the table below. Results: STATION 1: STATION 2: Matchstick and Bunsen arrangement Warm coloured water in a sealed container Observation: …………… Observation (s) ……………….. Interpretation: Hint - Identify at least one process you think might be occurring Interpretation: ……………… ……………………. STATION 3: STATION 4: Arrangement of forks, coin and a beaker Regular and diet soft drink cans in water Observation: …………… Observation: ………………….. Interpretation: ……………… Interpretation: …………………. STATION 5: Ice Observation: The ice is turning from a ____________ state into a ______________ state. Interpretation: ……………… Why do you think this is happening? ……………………. 28 Experiment: PHYSICAL VERSUS CHEMICAL CHANGE Aim: To distinguish a physical from a chemical change based on observation. Task: (Teacher demonstration) - You will be required to record observations at each station set up and identify each as either a physical or chemical change. Summarise the process(es) taking place at each station, expressing each in terms of starting and finishing substances. Expressing physical states as subscripts next to each substance will assist in this regard. ! Risk Assessment You will be required to complete a 5 min risk assessment form, that that outlines any form of risk encountered at each station. Results: Station 1: BOILING WATER Station 2: ELECTROLYSIS OF WATER: VOLTAMETER APPARATUS Station 3: HEATING IODINE IN A TEST TUBE. Station 4: HEATING SUGAR IN A TEST TUBE Station 5: A BUNSEN BURNER FLAME Station 6: SALT IN WATER Station 7: EVAPORATING A SALT SOLUTION Station 8: SODIUM METAL IN WATER Station 9: MAGNESIUM METAL IN HYDROCHLORIC ACID Station 10: HEATING MAGNESIUM METAL IN A BUNSEN FLAME Station 11: ADDITION OF 2 CLEAR / COLOURLESS SOLUTIONS: POTASSIUM IODIDE AND LEAD NITRATE Station 12: ADDITION OF SODIUM CARBONATE POWDER TO A DILUTE HYDROCHLORIC ACID SOLUTION Discussion: Identify one station you felt with confidence demonstrated a physical change. Why was this obvious? Identify one station you felt with confidence demonstrated a chemical change. Why was this obvious? Was there a station you found difficult to classify as a physical / chemical change? Did any station demonstrate both a physical and chemical change? Emphasise the difference between physical and chemical change by defining both. 29 3.3 Mixtures and laboratory separation techniques Most substances in nature are not found pure but as mixtures. Mixtures can be separated into pure substances using methods based on differences in their physical properties. Examples: A magnet can be used to separate iron (and some other substances) from other solids. physical property? differences in magnetic properties Solids and liquids can be separated by decanting, filtering or centrifuging. physical property? differences in density or particle size Two immiscible liquids can be separated using a separating funnel. physical property? differences in density Example - Separating a mixture of chalk powder and salt If you were given a mixture of chalk powder and salt to separate you could use differences in their physical properties to prepare a plan - flowchart - of the steps needed. Mixture of chalk powder and salt Add water and stir Filter to separate liquid and undissolved solid Chalk powder (wet) allow to dry Salt solution evaporate water by heating Chalk powder Salt In this separation we were using differences in solubility of the two solids in water. However the water also had to be separated by evaporation. Some industrial examples of separating mixtures: Gold panning Oil refining Sugar refining Separating rutile etc from beach sands Recovering salt from seawater Recovering pure water from contaminated water 30 Laboratory techniques for separating mixtures METHOD (property differences) Decanting (density) Separating funnel MIXTURE TYPE Used to separate undissolved sediment from a liquid or immiscible liquids. used to separate immiscible liquids (density) Filtration Gravity and Vacuum techniques used to separate an undissolved solid from a liquid. (particle size) 31 DIAGRAM METHOD (property differences) Centrifuge (density) Evaporation / Crystallisation Distillation MIXTURE TYPE DIAGRAM used to separate undissolved substances (some liquid) of differing densities used to separate a dissolved solid from a solvent (liquid) used to separate mixtures of liquids (boiling point) Chromatography used to separate miscible (partition – differences in liquids attraction of sample matrix for mobile phase Vs stationary phase) Note: Solutions are mixtures - solutions will be considered in detail in a later module! 32 Experiment – SEPARATION TECHNIQUES 1 Aim: To devise a method of separating a known sand / salt mixture, and express the percentage of salt in the mixture. Equipment: Method: Accurately weigh and record the mass of approximately 8 g of sand and 2 g of salt into a small beaker. Stir gently with a spatula until mixture is homogeneous in composition. Now devise a series of steps to achieve complete separation of the sand and salt components below. Flow Chart: 33 Results: TRUE VALUES: what mass of sand and salt did you weigh out? Mass: sand = ________________________ g Mass: salt = ________________________ g Total mass of mixture = _________________________ g % Salt = MEASURED VALUE: what mass of salt did you separate and weigh back? SALT Mass: Clean Dry Container (g) Mass: Container + Dry Salt (g) Mass: Salt (g) % Salt in mixture = Mass Salt obtained Total Mixture Mass x 100 HOW CLOSE WAS YOUR MEASURED RESULT TO THE TRUE RESULT? - Calculate the % error based on the percentage of salt you calculated after the separation, as follows. A result of 3% or less is considered acceptable at this level. Step 1: ABSOLUTE ERROR = Difference between TRUE and MEASURED result Step 2: RELATIVE ERROR (%) = ABSOLUTE ERROR TRUE VALUE 34 x 100 SALT Absolute Error = Relative Error = Questions: 1. What is meant by the term quantitative transfer? Give an example of one step in this task that used this technique. 2. Identify one physical property of both the sand and salt that allowed for their separation to take place. 3. Identify any difficulties or errors you experienced whilst carrying this separation out. 35 Experiment – SEPARATION TECHNIQUES 2 Aim: To devise and undertake a method for the separation and recovery of a mixture containing sand, salt and sawdust. Equipment: Method: Accurately weigh and record the mass of an unknown mixture containing sand, salt and sawdust. Devise a series of steps to achieve complete separation of the sand and salt components below. 36 Results: SAMPLE MASS: _________________________ g i) SAND COMPONENT Description Quantity / Result % Sand = ii) SALT COMPONENT Description Quantity / Result % Salt = iii) SAWDUST COMPONENT Description Quantity / Result % Sawdust = 37 Discussion: Compare your results with the true values as supplied by the lab staff. - Optional: Calculation of corresponding Relative Errors for each separated substance will assist you in making a more informed comparison. Identify how individual separations were possible based on your knowledge of physical properties as they apply to each separated component. What sources of error can you attribute to this separation task? Hindsight: can you suggest any improvements to your experimental set up or technique that might increase your experimental success? Conclusion: What were your findings? How successful overall was this exercise? 38 4. Elements and the Periodic Table Elements are pure substances that contain only one type of atom. Elements cannot be decomposed to give simpler substances. A listing of elements can be obtained from Chemical Data books or many other texts, however a very useful arrangement of the elements can be found in a Periodic Table which lists the elements in order of atomic number. 4.1 Mapping Elements on The Periodic Table See Periodic Table over page and notice where the various types of elements are placed: (a) (b) (c) (d) metals and non-metals solids, liquids and gases at room temperature natural and man-made elements radioactive elements Periods Use a key to mark them on you copy of the Periodic Table. Groups The horizontal rows on the Periodic Table are called periods. The first period contains just two elements, hydrogen and helium. (Remember that the first shell of electrons could hold 2 electrons). Periods 2 and 3 contain eight elements - the end of each period marks the filling of an electron shell. Periods 4 and 5 contain 18 elements. A section of period 6 is removed and printed below the bottom of the table. This section is named the lanthanides. Similarly a section of period 7 is also printed below the main table. It is called the actinides. (The removal of both sections is for ease of printing, so that the table did not become too wide). The vertical columns of the table are called groups. They are families of elements with closely related characteristics. The main group elements are numbered with Roman numerals: Group I is headed by Lithium (Li) – referred to as the Alkali Metals. Group II is headed by Beryllium (Be) – referred to as the Alkaline Earths. Then skip a few across the central “valley” of transition elements to arrive at . . . Group III headed by Boron (B). Group IV headed by Carbon (C). Group V headed by Nitrogen (N). Group VI headed by Oxygen (O) Group VII headed by Fluorine (F) – referred to as the Halogens. Group VIII headed by Neon (Ne) – referred to as the Noble Gases. The central block of elements constituting the rectangular “valley” as described above are referred to as the Transition Metals. 39 THE PERIODIC TABLE OF THE ELEMENTS 1 H 1.008 Hydrogen KEY 3 Li 6.941 Lithium 4 Be 9.012 Beryllium 11 Na 22.99 Sodium 12 Mg 24.31 Magnesium 19 K 39.10 20 Ca 40.08 Calcium 21 Sc 44.96 Scandium 22 Ti 47.87 Titanium 23 V 50.94 Vanadium Rubidium 38 Sr 87.62 Strontium 39 Y 88.91 Yttrium 40 Zr 91.22 Zirconium 41 Nb 92.91 Niobium 55 Cs 132.9 Cesium 56 Ba 137.3 Barium 57–71 Lanthanoids 72 Hf 178.5 Hafnium 73 Ta 180.9 Tantalum 87 Fr [223.0] Francium 88 Ra [226.0] Radium 89–103 Actinoids Potassium 37 Rb 85.47 2 He 4.003 Helium Lanthanoids Actinoids Atomic Number Relative Atomic Mass 57 La 138.9 Lanthanum 89 Ac [227.0] Actinium 24 25 Cr Mn 52.00 54.94 Chromium Manganese 26 Fe 55.85 Iron 43 44 Tc Ru [97.91] 101.1 Molybdenum Technetium Ruthenium 42 Mo 95.94 74 W 183.8 Tungsten 105 106 Db Sg [262.1] [166.1] Rutherfordium Dubnium Seaborgium 104 Rf [261.1] 58 Ce 140.1 Cerium 13 Al 26.98 Aluminium 75 Re 186.2 Rhenium 76 Os 190.2 Osmium 107 Bh [264.1] Bohrium 108 Hs [277] Hassium 60 61 62 Nd Pm Sm 144.1 [144.9] 150.4 Praseodymium Neodymium Promethium Samarium 59 Pr 140.9 90 91 92 Th Pa U 232.0 231.0 238.0 Thorium Protactinium Uranium Symbol of element Name of element 5 B 10.81 Boron 6 C 12.01 Carbon 7 N 14.01 Nitrogen 8 O 16.00 Oxygen 9 F Fluorine 19.00 10 Ne 20.18 Neon 13 Al 26.98 Aluminium 14 Si 28.09 Silicon 15 P 30.97 Phosphorus 16 S 32.07 Sulphur 17 Cl 35.45 Chlorine 18 Ar 39.95 Argon 27 Co 58.93 Cobalt 28 Ni 58.69 Nickel 29 Cu 63.55 Copper 30 Zn 65.41 Zinc 31 Ga 69.72 Gallium 32 Ge 72.64 Germanium 33 As 74.92 Arsenic 34 Se 78.96 Selenium 35 Br 79.90 Bromine 36 Kr 83.80 Krypton 45 Rh 102.9 Rhodium 46 Pd 106.4 Palladium 47 Ag 107.9 Silver 48 Cd 112.4 Cadmium 49 In 114.8 Indium 50 Sn 118.7 Tin 51 Sb 121.8 Antimony 52 Te 127.6 Tellurium 53 I 126.9 Iodine 54 Xe 131.3 Xenon 77 Ir 192.2 Iridium 78 Pt 195.1 Platinum 79 Au 197.0 Gold 80 Hg 200.6 Mercury 81 Tl 204.4 Thallium 82 Pb 207.2 Lead 83 Bi 209.0 Bismuth 84 Po [209.0] Polonium 85 At [210.0] Astatine 86 Rn [222.0] Radon 68 Er 167.3 Erbium 69 Tm 168.9 Thulium 70 Yb 173.0 Ytterbium 71 Lu 175.0 Lutetium 109 110 111 112 Mt Ds Rg Cn [268] [271] [272] [285] Meitnerium Darmstadtium Roentgenium Copernicium 63 64 Eu Gd 152.0 157.1 Europium Gadolinium 93 94 95 Np Pu Am [237.0] [244.1] [243.1] Neptunium Plutonium Americium 96 Cm [247.1] Curium 65 Tb 158.9 Terbium 66 67 Dy Ho 162.5 164.9 Dysprosium Holmium 97 98 99 Bk Cf Es [247.1] [251.1] [252.1] Berkelium Californium Einsteinium 100 Fm [257.1] Fermium For some elements, the relative atomic mass of their most common radioactive isotope is shown in brackets. 40 102 103 No Lr [259.1] [262.1] Mendelevium Nobelium Lawrencium 101 Md [258.1] 4.2 Distribution of the elements Approximately 109 elements are known (90 are natural to earth, the rest are man-made). Throughout the universe, hydrogen (83%) and helium (16%) are by far the most abundant. On the earth’s crust and atmosphere the distribution by mass of elements is: In the human body, the distribution of elements is different again . . . 41 Occurrence of the elements Only a few elements actually exist in a pure form on earth; most are in the form of compounds. Activity: Earth Elements What are some of the elements that exist uncombined on earth? Two are gold (as nuggets etc) and oxygen (as gas in the atmosphere). List three others. 1. .................................................................... 2. .................................................................... 3. ................................................................... Why do the others only exist naturally in compounds? Because most other elements are too chemically unstable to exist by themselves. They instead react with substances in the natural environment such as water, oxygen and carbon dioxide, forming various compounds which have much greater chemical stability. 4.3 How atoms can be arranged in everyday elements Some elements exist as single free atoms. They are called MONATOMIC elements and the most common examples are in the group at the far right of the Periodic Table called the rare or noble gases: He, Ne, Ar, Xe, Rn. These gaseous atoms are extremely unreactive (ie; chemically stable). They have a full outer shell of electrons. The other gaseous elements form molecules of two or more atoms. DIATOMIC elements form molecules composed of 2 like atoms. Examples are oxygen, O2 ; hydrogen, H2 ; nitrogen N2 ; fluorine, F2 and chlorine, Cl2. Bromine, Br2 is a liquid at room temperature and iodine, I2 is a solid at room temperature. POLYATOMIC elements have 3 or more atoms in their molecules. Examples are phosphorus, P4 and sulphur, S8 although they can also exist in molecules of differing numbers of atoms, or allotropes. Metal elements have many atoms grouped together in a regular structure called a crystal lattice. They do not form separate “molecules”, nor are they free atoms. Some nonmetals such as carbon also have a crystal lattice structure. 42 Allotropes Some elements can exist in a number of different pure forms, where different numbers of atoms exist in a molecule, or the atoms are joined in different ways. Examples include Oxygen which can exist as O2 (oxygen gas) or as O3 (ozone). Carbon can exist as graphite, diamond or bucky balls! These examples and some others are illustrated below: OXYGEN O3 O2 Ozone C60 Oxygen gas Graphite Diamond CARBON Buckminsterfullerene SULFUR S6 S8 S4 43 Research: Allotropes You may wish to research one or more of these allotropes, or some others of your choosing. This would be suitable for including in your portfolio. Ref: CC1 pp23-25 C1 pp13-16 4.4 Classification of the elements Experiment – Properties of elements This practical activity must be done in a science laboratory under the supervision of a chemistry teacher. The sections marked as demonstrations must only be done by the teacher with the student observing and recording carefully. Normal laboratory safety precautions will need to be observed at all times. In this practical exercise you will move to various workstations. Make and record careful observations as you proceed. Do not remove samples from their containers. 1. Complete the following table by observing samples of elements on display. All the elements are contained in glass sample tubes. Some may be stored in a liquid for safety reasons. Element Name / Symbol Colour and Appearance 44 Metal or Non Metal? 2. A range of different elements are available for density calculation: Lead (Pb), Iron (Fe), Carbon (C), Aluminium (Al), Copper (Cu), Zinc (Zn), Sulfur (S) and Iodine (I2) Calculate the density of each element ** via the following simple method: ** Calculate the density of Sulfur and Iodine (teacher demonstration) in a clean, dry 25 mL measuring cylinder without water Record the mass of each element. Fill a measuring cylinder of appropriate size with a volume of tap water enough to submerge the entire object. Record this volume. Submerge the object in the water and record the new volume of water. Calculate the actual volume of water displaced by the object. Calculate the density of the object, where: Density = Mass (g) Volume (mL) Consult an SI Chemical Data book and record the literature density values of each element: (ρ) g / cm3 Element Vol 1 (mL) Meas. Cyl Vol 2 (mL) Meas. Cyl Sulfur, S N/A N/A Iodine, I2 N/A N/A ID Mass (g) 45 Element Volume (mL) Density (g / mL) Literature Density (ρ) g / cm3 3. (Teacher Demonstration). The Melting Point of available elements will be determined qualitatively using a bunsen burner on the blue flame setting. Use the temperature probe to measure the temperature of the blue bunsen flame. Record this value. Observe the ease (or otherwise) to which each element melts. Rate each melting point using word descriptions: very low low medium medium-high high very high. Consult an SI Chemical Data book and record the literature melting point values of each element. Bunsen flame temperature: _________________________ Element ID Observed Melting Point 46 Literature Melting Point o C 4. The Electrical Conductivity of each element can be easily tested by touching live electrical probes onto the side of each. Predict the electrical conductivity (good fair poor) of each element. Observe the reading of an ammeter connected in series within the circuit. Record these observations using similar descriptions. Element ID Predicted Electrical Conductivity Observed Electrical Conductivity Describe on what basis you are making your predictions of electrical conductivity: 47 5. (Teacher Demonstration). Observe the malleability / ductility of available elemental samples. Malleability can be observed by hammering or attempting to bend each sample. Ductility is best observed when sodium metal is placed in a laboratory device called a Sodium Press. Briefly describe the meaning of the terms: Malleable: …………………………………………………………………….. …………………………………………………………………………………………………….. Ductile: …………………………………………………………………….. …………………………………………………………………………………………………….. Element ID Malleable / Ductile ? Yes / No What class of elements generally exhibit characteristics of malleability / ductility? ……………………………………………………………………………………………….. Ref: Metals C1 pp23-26 Non metals C1 pp51-56 Metals CC1 pp49-60 48 6. (Teacher Demonstration). Elements can be made to undergo physical and chemical changes under certain conditions. Complete the table below after observing each scenario. Scenario Observations Zinc + Hydrochloric Acid Copper + Hydrochloric Acid Magnesium + Hydrochloric Acid Iron + Hydrochloric Acid Sodium + Water, Δ Sulfur heated with a bunsen burner in a gas jar Addition of Potassium Iodide, KI to a solution containing Sodium Hypochlorite bleach, NaClO Explain how Physical Changes are different from Chemical Changes . 49 5. CHEMICAL BONDING All particles are held together by chemical bonds of some kind. Water molecules are made up to two hydrogen atoms and one oxygen atom always. The three atoms are held together by chemical bonds, which are strong and therefore not easily broken. How then do ice, liquid water and steam differ? They all contain water molecules. Gas particles move far away from each other. Liquid particles are closer together. Solid particles are very close together, held in a fixed position. The main reason for the difference is that bonds can occur between molecules, called intermolecular forces, and they increase in strength as the molecules get closer together. These bonds are much weaker in strength than chemical bonds. Chemical bonds form between atoms to form stable partnerships, since most atoms are not stable alone. Chemical bonds exist between the hydrogen and oxygen atoms in the stable water molecule. Chemical bonds involve the outer electron shells of atoms. 5.1 What makes an Atom Chemically Stable? The elements in the group at the right-hand end of the periodic table - He, Ne, Ar, Kr, Xe are called the rare, noble or inert gases. They are inert because they are extremely unreactive. This means that they don’t form compounds except under rare circumstances. They are extremely stable as monatomic elements - as single, separate atoms. Electron configuration : (2) (2,8) (2,8,8) In each of these atoms all electron shells present are full. The outer shell of electrons is called the valence shell. Chemical stability is obtained when an atom has a complete (full) outer shell of electrons. 50 5.2 Loss and Gain of Electrons – The Ionic Bond Sodium and chlorine are extremely reactive as pure elements, but the compound they form, sodium chloride, is extremely stable. The sodium atom has one electron in its valence shell, while the chlorine atom has seven electrons in its valence shell (which is one short of a full shell). A transfer of this one electron from the sodium atom to the chlorine atom results in the formation of more stable particles, now charged ions, which have a full outer electron shells. The sodium atom loses an electron to become a positive sodium ion, while the chlorine atom gains the electron to become a negative chloride ion. Both the sodium ion and the chloride ion have full outer shells of electrons. Since only the outer electron shells are involved in chemical bonding between atoms, only the outer shell of electrons is often shown as an Electron Dot Formula. This concept is also referred to as a Lewis Structure Sodium atom Sodium ion + e - Chlorine atom + e - Chlorine ion 51 5.3 Sharing Electrons – The Covalent Bond The other way for an atom to fill its outer shell of electrons is to share electrons with other atoms. When sharing electrons each atom involved contributes an equal number of electrons - one, two or three - to form a single, double or triple covalent bond. A pair of shared electrons is represented as a dash ( – ) in a structural formula, ie a single bond. Some elements have their atoms covalently bonded in simple molecules (such as oxygen) or even giant networks (such as carbon) to form stable units. The following common elements have stable diatomic molecules: H2, N2, O2, F2, Cl2 Br2, I2. Written Exercise: Lewis Electron Dot Formulae Use Lewis Structures to illustrate covalent bonds for the following molecular elements: H2 O2 N2 F2 Structural formula: H–H Structural formula: O=O Simple molecular compounds contain covalent bonds within their molecules. Complete the following examples: H2O NH3 CH4 Structural formula: CO2 HCl 52 HCN 5.4 Which Chemical Bond Will Form? When atoms of one element need to gain electrons to become stable, and atoms of another element need to lose electrons to become stable then an ionic bond will form. Ionic bonds usually form between metals and non-metals. Metals will not bond chemically with other metals to form compounds but they can form mixtures called alloys. When the atoms of both elements need to gain electrons to become stable then a covalent bond is formed. This situation arises when compounds form between nonmetals. Exception: The ammonium ion NH4+ is a positive ion containing non-metals, however it bonds ionically with any negative ion. Written Exercise: Type of chemical bond? Identify the nature of chemical bond that will form between the following elements. Mg and Cl ............................... K and S ............................... C and Br ............................... O and O ............................... H and Cl ............................... Na and Ca ............................... 5.5 take care ! The Metallic Bond A Model of a Metallic Lattice Metal atoms generally have few electron shells (1 3) occupying outer valence shells. In theory these electrons need to be lost in order for the metal to form a stable ion. However in the pure element there are no atoms to gain them since all are identical in electron configuration (arrangement). 53 The outer electrons are freed from the metal atoms and form a negative cloud which is the glue that holds the resultant positive ions (ie nucleus + inner electron orbitals) together in a regular lattice structure. The freedom of these electrons to move (even in the solid state) is the reason why metals are good conductors of electricity. It also helps to explain properties of malleability and ductility as the free electrons can flow to bind a distortion in the lattice. 5.6 Summary of Bonding Atoms are held together in compounds by Chemical Bonds. Chemical bonds result from the sharing or transfer of valence electrons between pairs of atoms. Bonded atoms attain the stable electron configuration of a noble gas. The noble gases themselves exist as isolated atoms because that is their most stable condition. The transfer of one or more electrons between atoms produce positively and negatively charged ions: cations and anions. The attraction between a cation and an anion is an ionic bond. A substance with ionic bonds is an ionic compound. Nearly all ionic compounds are crystalline solids at room temperature, having high melting points. These solids consist of positive and negative ions packed in an orderly arrangement. The total positive charge is balanced by the total negative charge, and therefore the ionic compound (also called a salt) is electrically neutral. When atoms share electrons to gain stable electron configuration of a noble gas, they form covalent bonds. A shared pair of electrons constitutes a single covalent bond. Sometimes two or three pairs of electrons are shared to give double and triple covalent bonds. Metals consist of positively charged ions packed together and surrounded by a sea of their valence electrons. This arrangement constitutes the metallic bond. The valence accounts for the excellent electrical conductivity in metals and helps explain why metals are malleable and ductile. Ref: CC1 pp48-51; 63-67 CC1 pp82-85 C1 pp23-25; 50-58 C1 pp69-70 54 Written Exercises: 1. Write electron dot structures for each of the following elements. (a) 2. Cl 5. 6. 8. (c) Al (d) Li (e) C (b) Al (c) Na (d) Li (e) Ba (f) Mg (b) S (c) N (d) Cl (e) P (f) O Write the formula for the stable ion formed from each of the following elements. (a) aluminium (b) lithium (c) barium (d) potassium (e) calcium (f) strontium (g) bromine (h) oxygen (i) arsenic (j) nitrogen (k) selenium (l) sulfur Which of the following pairs of elements are likely to form ionic compounds? (a) magnesium and bromine (b) chlorine an iodine (c) potassium and helium (d) nitrogen and sulfur (e) lithium and fluorine (f) carbon and oxygen Draw Lewis Structures of the compounds formed from these pairs of ions. (a) K+ ; S2- 7. S How many electrons must be gained by each of the following atoms to attain a noble gas configuration (full electron shell)? (a) I 4. (b) How many electrons must be lost by each of the following atoms to form an ion? (a) Ca 3. Formulae and Chemical Bonding (b) Ca2+ ; O2- (c) Al3+ ; Cl- Write formulae for the ions in the following compounds. (a) KCl (b) NaNO3 (c) BaSO4 (d) MgBr2 (e) KOH (f) Li2CO3 Draw Lewis electron structures of the following. Classify each as ionic or covalent. (a) H2O (b) Na2O (c) CS2 (d) CaS (e) NH3 (f) SO2 (g) Al2O3 55 5.7 Compound Chemical Formulae A chemical formula represents one molecule or formula unit of an element or compound. A chemical formula can feature: FEATURE DEFINITION EXAMPLES Symbols identify element(s) present. C Subscripts i) indicate number ratios of elements or polyatomic formula units CO2 or C6H12O6 ii) states of matter Mg NaCl or CaO etc O2 (g) H2O (l) AgCl (s) or HCl (aq) Superscripts indicate charge of cation or anion Coefficient Numbers represent mole quantity ratios in balanced chemical formulae and equations Na + or CO3 2- CuSO4.3H2O N2 (g) + 3H2 (g) 2NH3 (g) Brackets express multiple polyatomic ion units in a chemical formula. Al (NO3)3 (aq) Roman Numerals clarify valency of transition metal elements in word formulae Iron III Oxide, Prefix Names indicate number ratios of non metal elements in covalent molecular compounds Dinitrogen Pentoxide Fe2O3 N2O5 A molecule is a discreet group of atoms which are bonded together in a fixed ratio defined by chemical formula. Examples: Elements Compounds H2 - Hydrogen H2O - Water P4 - Phosphorus C6H12O6 - Glucose (a sugar) S8 - Sulfur NH3 - Ammonia 56 A formula unit represents the simplest ratio of atoms in a substance which may be a giant lattice, ie network structure. Examples: Each of these substances is present as a giant lattice structure. Substance Type Formula Diamond element C Sodium Chloride compound (ionic) NaCl Silica compound (network covalent) SiO2 Ions (charged particles) use superscripts to show the electrical charge. Examples: Mg2+ means a magnesium ion having a charge of +2. CO3 2- means one carbon atom and three oxygen atoms forming an ion (called a carbonate ion) with a charge of -2. Further Examples: 1. 2 atoms of hydrogen H2SO4 1 atom of sulfur 2. 4 atoms of oxygen Ca(OH)2 2 groups of OH 1 atom of calcium 2 atoms of oxygen 57 2 atoms of hydrogen Written Exercise – Interpreting chemical formulae: 1. (NH4)2SO4 ( ......... groups of .................... ) ............ atoms of ............. atoms of .............. atoms of ............ atoms of .......................... ............................ ............................. ........................... 2. CH4 represents ...................................................................................................... 3. Sn3(PO4)4 represents .............................................................................................. 4. CuSO4.5H2O represents.......................................................................................... 5. Cl- represents ........................................................................................................ 6. OH- represents ........................................................................................................................... 5.8 Valency Valency is a numerical measure of combining power in determining correct compound formula. Cations have positive valency, Anions negative. Atoms combine together according to their values of valency to form stable compounds, commonly of neutral charge, but not always. . . Where an element can have more than one valency, it is given as a Roman Numeral in the name of the compound. This commonly occurs with transition elements. 58 VALENCY TABLE Polyatomic Ions – contain 2 or more elements making up their charged arrangement. CATIONS 1+ 2+ 3+ 4+/- Name Formula Name Formula Name Formula Name Formula Hydrogen H+ Calcium Ca2+ Iron (III) Fe3+ Carbon C 4+/- Lithium Li+ Magnesium Mg2+ Aluminium Al3+ Silicon Si 4+/- Sodium Na+ Copper II Cu2+ Potassium K+ Cobalt II Co2+ * * Carbon and Silicon Ammonium NH4+ Zinc II Zn2+ do not form ions, Silver I Ag+ Iron II Fe2+ rather covalent bonds Copper I Cu+ Lead II Pb2+ ANIONS 1Name 2Formula – Name 3Formula 2– Name Formula Nitride N3– Fluoride F Oxide O Chloride Cl– Sulfide S2– Phosphide P3– Bromide Br– Carbonate CO32– Phosphate PO43– Iodide I– Sulfate SO42– Hydroxide OH– Sulfite SO32– Ethanoate CH3COO– Oxalate C2O42– Permanganate MnO4– Chromate CrO42- Hydrogen carbonate HCO3– Dichromate Cr2O72- Nitrate NO3– Nitrite NO2– Cyanide CN– 59 5.9 Chemical formulae and naming The valency of an atom or polyatomic ion can be viewed as a bonding position which must be filled to form a stable compound. In order to find the formula of a compound you must recognise the parts or constituents from the name. Since the name is in two parts this recognition is easy as long as you realise that the ending of the name of the second atom is often changed to -ide. Oxygen becomes oxide; sulfur becomes sulfide; chlorine becomes chloride etc. Na Na will join the ratio 1:1 Cl Cl Sodium’s valency is 1+ Chloride’s valency is 1- 2. Formula is Na1Cl1 or NaCl (1 is automatically assumed) Aluminium nitrate is a compound made from aluminium and nitrate polyatomic ion NO3 Al NO3 will join in the ratio 1:3 Al NO3 NO3 Aluminium has a valency of 3+ Nitrate has a valency of 1- Formula will be Al(NO3)3 3. Lead IV oxide is a compound made from lead (valency 4) and oxygen. Pb O will join in the ratio 1:2 O Pb O Lead has a valency of 4+ Oxygen has a valency of 2- The formula will be PbO2 60 Cross Multiply Method Step 1. Write the symbols which represent each part of the name. Step 2. Write the valencies as a superscript to each part. Step 3. Cancel by dividing the valencies by any common factor Step 4. Cross over the numbers to form subscripts, using brackets where necessary. Examples: 1+ 1. Na 2. 3. Sodium Chloride Aluminium Nitrate Lead (IV) Oxide Written Exercises: 1Cl 3+ 1- Al NO3 Formula is Na1Cl1 or NaCl Formula is Al1(NO3)3 or Al(NO3)3 4+ 2- Pb O 2 1 4 2 Formula is Pb1O2 or PbO2 Write chemical formulae for the following compounds. (a) Hydrogen bromide ……………………… (g) Ammonium nitride ……………………… (b) Calcium chloride ……………………….. (h) Iron (III) hydroxide ……………………… (c) Zinc (II) sulfate ……………………….. (i) Potassium sulfite ……………………….. (d) Magnesium oxide ……………………… (j) Nickel (II) carbonate ……………………. (e) Lead (IV) sulfide ……………………… (k) Silver (I) phosphate …………………….. (f) Barium phosphide ……………………… (l) 61 Tin (II) nitrate …………………………. Prefix Numbering When two non-metals combine to form a compound, they can often form several different compounds depending on the prevailing conditions. For example sulfur and oxygen can form SO2 or SO3. These compounds are named so that the number of each atom present is indicated by a prefix. SO2 is sulfur dioxide SO3 is sulfur trioxide The prefixes used are : mono 1 hexa 6 di 2 hepta 7 tri 3 octa 8 tetra 4 nona 9 penta 5 deca 10 The prefix mono is not used for the first atom, but is used for the second. For example, CO is carbon monoxide - notice also that the final “o” from the prefix name is dropped so that it is easier to pronounce. Written Exercises: 1. Formulae and Naming (also termed “Nomenclature”) Write formulae for the following compounds: (a) potassium nitrate ………………. (b) sodium carbonate ………………………… (c) cobalt (II) sulfate ……………… (d) ammonium carbonate ……………………. (e) calcium phosphate ……………… (f) disulfur dichloride .....………………………. (g) nitrogen triodide ......................... (h) iodine heptafluoride ................................. (i) silicon dioxide ............................ (j) selenium dibromide ................................... 2. Name the following compounds: (a) SiF4 ............................................. (b) CaSO4 …………………………………….… (c) FeO ………………………………… (d) CuCl …………………………………………. 62 (e) AgCl ……………………………….. (f) NaHCO3 …………………………………….. (g) NO .................................................. (h) P2O5 ............................................................ (i) PCl3 ................................................. (j) CS2 ............................................................... 3. Complete the following table: Formula Name Name MgO sodium oxide CaS calcium carbonate CO ammonia CO2 aluminium hydroxide Csl dinitrogen tetroxide CaBr2 ammonium sulfate H2O aluminium oxide NO2 copper (I) bromide N2O4 chromium (III) fluoride AsCl3 manganese (IV) oxide AgNO3 mercury (II) sulfide KBr oxygen difluoride Ca(OH)2 iron (II) chloride CoCl2 iron (III) iodide HI copper (II) sulfate SnO2 carbon tetrachloride CCl4 strontium chloride SnCl2 potassium nitrate BaF2 calcium phosphate 63 Formula Activity – MODELLING MOLECULES AND COMPOUNDS Aim: To classify and construct models to represent some common substances. Task: 1. Correspond a chemical formula to each chemical name. 2. Classify each substance according to chemical bond type. 3. Draw a structural representation of each molecular model. CHEMICAL NAME / FORMULA CLASSIFICATION (Ionic / Covalent) Chlorine gas ………….……………. Oxygen gas …………………………… Nitrogen gas …………………………… Ammonia …………………………… Sodium Chloride …………………………… Lithium Oxide …………………………… 64 STRUCTURE Sodium Hydroxide …………………………… Carbon Dioxide …………………………… Potassium Sulfate …………………………… Methane …………………………… Ethanol ……… C2H5OH …… Ethanoic Acid ……… CH3COOH …… Question – Consider and write down some of the advantages and disadvantages of using molecular model kits to understand the nature of chemical substances. 65 6. Properties of Compounds 6.1 Comparing properties of elements and compounds Compounds are formed when elements combine together in fixed proportions. The compound formed will often have properties that are very different from its constituent elements. The formation of the compound will involve an absorption or a release of energy. Compounds can be decomposed back into its constituent elements, sometimes with great difficulty. Example: When a spark is introduced into a mixture of hydrogen gas and oxygen gas an explosion occurs and droplets of liquid are formed: Hydrogen (g) (element) + Oxygen (g) (element) Water (I) (compound) + Energy Comparison of some properties - elements and a compound Property Physical Chemical Hydrogen Oxygen Water Gas at room temp - little attraction between molecules, low density. Gas at room temp - little attraction between molecules, low density Liquid at room temp - attraction between molecules evident. Density approx 1.0 g / mL. Good solvent for many substances. Reacts vigorously with many substances. Necessary for combustion burning. Does not support combustion. Burns explosively in air. 66 6.2 Classification of Compounds Compounds can be classified in several different ways. The following classifications rely on differences in both chemical and physical properties. 1. Organic / Inorganic Compounds Inorganic Organic Compounds composed of elements other than carbon and hydrogen. Exceptions include oxides of carbon: 2CO, CO2, CO3 , HCO3 , cyanides, CN- and strong acids, HX. Composed predominately of the elements carbon and hydrogen. Can contain small quantities of oxygen, sulfur, nitrogen, phosphorous and halogen elements. Found in living things, (plants & animals), foods, fossil fuels, fibres, plastics etc . . . Many organic compounds undergo charring when they are heated. They turn black, showing a carbon residue. CARE! Always heat small samples in a well ventilated area. 2. Ionic / Simple Molecular / Network Covalent Compounds Ionic Contains charged Particles Ions. Cation is a metal ion or ammonium, NH4+ Network arranged. Covalent contain non-metals only Simple Molecular Small, discrete groups of atoms 67 Network Covalent Giant, covalently bonded lattice 6.3 Properties of Compounds Melting Point A substance has a high melting point if it requires lots of energy (heat) to break apart the forces holding the individual particles together. Ionic compounds (eg NaCl, common salt) usually have high melting points, 900oC +. This means that the forces holding the ions together in ionic solids are strong. Ionic bonds are strong. Simple molecular compounds (eg H2O water) have low melting points, 0oC. This means that the forces holding the molecules together in simple molecular solids are weak. Intermolecular bonds are weak. Network covalent solids (eg SiO2 silica or quartz) have very high melting points, 1500oC +. This means that the forces holding the atoms together in network covalent solids are very strong. Covalent bonds are very strong. Solubility A substance will dissolve in a liquid if it has similar properties to that liquid. “Like dissolves like . . .” Polar inorganic liquids such as water dissolve many ionic solids, like Sodium Chloride, NaCl but not Silver Chloride, AgCl (!! Solubility rules apply). Non-polar liquids such as petrol and kerosene will dissolve some simple covalent molecular solids. Ionic compounds will NOT dissolve in organic solvents. Polar organic liquids such as alcohols, (ethanol is a good example) are miscible with a diverse range of organic and inorganic liquids, like petrol and water respectively. Ionic compounds will commonly dissolve in water: This means that the water molecules are attracted strongly to the positive and negative ions such that the energy holding the ions in the crystal lattice is overcome. Cations and anions randomly move about in aqueous solution (aq). There are still many examples of insoluble ionic compounds, and therefore there is a general guide that can be referred to when determining solubility of such compounds. See Solubility Rules table over page 68 SOLUBILITY RULES Use this table as a guide to determine the solubility of ionic compounds in water SOLUBILITY RATING ION EXCEPTIONS Lithium, Li+ Completely Soluble Sodium, Na+ None Potassium, K + Ammonium, NH4+ Nitrate, NO3Ethanoate, CH3COO – Mostly Soluble Iodide, I - Insoluble Exceptions: - Silver I, Ag+ Bromide, Br - Lead II, Pb2+ Chloride, Cl Mercury II, Hg2+ Mostly Soluble Sulfate, SO42- Insoluble Exceptions: Calcium, Ca2+ Barium, Ba2+ Lead II, Pb2+ Mostly Insoluble Hydroxide, OH - Soluble Exceptions: Carbonate, CO32- Lithium, Li+ Phosphate, PO43- Sodium, Na+ Potassium, K + Ammonium, NH4+ 69 Written Exercise: Identify each compound below as either Precipitate or Soluble according to the solubility rules. Mercury II Nitrate: _______________ Copper II Hydroxide: _______________ Magnesium Sulfate: ________________ Calcium Carbonate: _______________ Ammonium Phosphate: ______________ Lead II Ethanoate: _________________ Electrical Conductivity A compound will conduct electricity if it possesses charged particles which are free to move, ie mobile. Solutions of ionic compounds are good conductors because they contain ions which can move in the liquid. Solid ionic compounds are poor conductors because their ions are not free to move out of the solid lattice. Molten (liquid) ionic compounds are good conductors because their ions can move. Solutions of molecular compounds are poor conductors because they contain no ions. Solid molecular compounds are poor conductors because they contain no ions. Molten (liquid) molecular compounds are poor conductors because they contain no ions. 70 Experiment – PHYSICAL PROPERTIES OF COMPOUNDS Complete the table below carefully. Reference information where necessary from the SI Chemical Data manual. Test compounds available using appropriate laboratory apparatus. Compound Name Compound Classification Compound Type Include a Chemical Formula Organic / Inorganic Ionic, Covalent, Network etc.. Appearance Melting Point OBSERVED High, Med, Low, Char. 71 Literature Value Solubility Water Other Name it Conductivity Solution Molten State Written Exercise: Hydrogen chloride conductor? Hydrogen chloride is a simple molecular compound (gas) and a non-conductor. It is very soluble in water, forming a solution which is an excellent electrical conductor. Can you explain this? ............................................................................................................................................... ............................................................................................................................................... Ref: CC1 pp92-93 C1 pp37-45 Written Exercise: COMPOUND PROPERTIES IN SUMMARY Compound Type COVALENT MOLECULAR IONIC CHEMICAL EXAMPLE(S) CHEMICAL BOND TYPE SOLID, LIQUID & GAS PHYSICAL STATE @ Room Temp MELTING POINT (Low, High, Very High) WATER SOLUBILITY (High, Low, Nil) SEE SOLUBILITY RULES LOW - MEDIUM ELECTRICAL CONDUCTIVITY - Solid State ELECTRICAL CONDUCTIVITY - Molten State ELECTRICAL CONDUCTIVITY - Aqueous Sol’n NO Hydrogen salt exceptions 72 COVALENT NETWORK Experiment: 3 WHITE POWDERS Background: 3 different unknown compounds resembling white powders will be issued for positive identification. You will perform a range of physical tests to classify each compound based on its unique set of physical properties. You will then confirm the identity of each white substance based on a simple chemical diagnostic test. Glucose, Sodium Chloride and Starch are the chemical substances you must positively identify. Perform all chemical / physical tests in boiling tubes PART A – CHEMICAL DIAGNOSTIC TESTS Complete the following table, classifying the type of chemical compound and providing results / observations of diagnostic tests: Glucose Sodium Chloride COMPOUND CLASSIFICATION ? Ionic ? Covalent Molecular ? Covalent Network ? Organic / Inorganic IODINE (Observation) SILVER NITRATE (Observation) BENEDICTS REAGENT + HEAT (Observation) 73 Starch PART B – RESULTS OBSERVATION TABLE UNKNOWN SUBSTANCE UNKNOWN: UNKNOWN: UNKNOWN: _______ _______ _______ PHYSICAL PROPERTY SOLUBILITY (soluble / insoluble) CONDUCTIVITY of aqueous solution (yes / no) MELTING POINT (low / high) CHAR (yes / no) PART C – SUBSTANCE ID CLASSIFICATION: ? Ionic ? Covalent Molecular ? Covalent Network CLASSIFICATION: ? Organic ? Inorganic SUBSTANCE ID ? Glucose ? Sodium Chloride ? Starch DIAGNOSTIC TEST (Confirmed / Incorrect) 74 7. CHEMICAL REACTIONS Chemical reactions occur when a change leads to the formation of at least one NEW substance. The starting substances in a reaction are called reactants. The substances formed during a reaction are called products. During a chemical reaction, atoms are rearranged and energy is absorbed or released in the process. A reaction has occurred when one or more of the following can be observed: evolution of a gas (not the vapour of a reactant) formation of a precipitate (a solid formed from two solutions) a permanent colour change a noticeable temperature change (heat) or other energies released/absorbed disappearance of a solid that is known not to be soluble. A chemical equation is a shorthand way of describing the chemical change that has occurred. Reactants Products In this module we will use word equations to describe the reactions observed. This concept will then be extended to include balanced chemical equations and use them to calculate quantities of reactants and products involved. Chemical reactions are occurring inside us and around us everyday. Our body’s metabolism involves hundreds of thousands of different chemical reactions for the production of energy and growth. Combustion is an important chemical reaction that is also used for energy production (light, heat) and transport amongst other uses. Photosynthesis is an essential process for the existence of life on earth. Essential sugars and oxygen are produced when carbon dioxide and water react in the presence of sunlight within the green leaves of plants. Most metal objects have been produced through industrial chemical processes. They would otherwise remain existing mostly as an Earth based compounds distributed in rich crustal deposits called ores. We can look more closely at specific chemical changes: 7.1 Energy and Chemical Processes When a spark ignites a mixture of hydrogen gas and oxygen gas, an explosion occurs. A large amount of energy is released. The product formed is water. Hydrogen (g) + Oxygen (g) Water (l) + energy The subscripts used in this equation indicate the states of reactants and products. (s) = solid (l) = liquid (g) = gas (aq) = aqueous (dissolved in water) 75 Chemical reactions which release energy are called exothermic, hence “Energy is lost to the surrounding environment”. Reactions which absorb energy are called endothermic, hence “energy is absorbed from the surrounding environment”. Photosynthesis is an example of an endothermic chemical process. Plants absorb energy in the form of sunlight to change carbon dioxide gas and water to complex organic molecules such as glucose, releasing oxygen gas. This process is endothermic. Carbon + Dioxide (g) Water(l) + Energy Glucose (aq) + Oxygen (g) Many chemical reactions require some energy to start the reaction. However endothermic reactions require energy to be available continuously, so that the reaction can continue. Ref: C1 pp70-71 Written Exercise & Research: Chemical Reactions 1. Find two more examples of both exothermic and endothermic reactions. 2. Do each of the following involve chemical reactions? State what you would observe in each case to justify this conclusion. (a) A piece of wood burns leaving grey ashes (b) A piece of sodium metal disappears after fizzing as it zips and releases sparks over the surface of the water. (c) A tablespoon of salt dissolves when mixed in water. (d) A headache tablet fizzes as it dissolves in the water. Practical Activities at Home: The following chemical reactions can be carried out safely in the home or in the laboratory. As a safety precaution you should always have long hair tied back, your eyes protected by safety glasses and your feet fully protected by shoes. Protective clothing such as laboratory coats are worn in laboratories together with other safety wear as required. Always use small quantities of reactants and carry out the reactions in open glass containers. 1. The acid in vinegar is ethanoic (acetic) acid, CH3COOH. Bicarb of soda or sodium hydrogencarbonate, NaHCO3 is used in cooking, usually because it will release carbon dioxide gas to help foods to rise. Sodium hydrogencarbonate is also found in baking powder. When ethanoic acid is added to sodium hydrogencarbonate, the result is the formation of carbon dioxide, water and a salt called sodium ethanoate, NaCH3COO. 76 Observations : You may like to include a simple labelled diagram Word Equation : ……………………………………………………………………………………………………… Conclusions: Summarise your evidence to show that a chemical reaction occurred. You may like to substitute other acids in the home for the vinegar and test their reactions with bicarb of soda or baking powder. Try carbonic acid, H2CO3 in soft drink or citric acid in lemon juice. Other carbonates like calcium carbonate in limestone and egg shells can be substituted for sodium hydrogencarbonate. 2. When two different metals dip into a conducting solution (electrolyte), electrical energy can be produced. Your metals could be copper and iron or any available. Make sure the surfaces are clean by rubbing with emery paper. The electrolyte (solution) could be an acid such as lemon juice, or salty water. A small torch light bulb or a multimeter can detect the electric current produced. Electrodes Connecting wires Electrolyte Light Globe If you used copper and iron, the iron metal will slowly dissolve, forming Fe 2+ and release electrons (electric current). You may notice a fizzing at the surface of the copper as hydrogen gas is produced. Word Equation: Iron(s) + Acid(aq) Iron + Ion (aq) Is this reaction exothermic or endothermic? Report on your results. 77 Hydrogen + Energy 7.2 Writing Chemical Equations - Chemical Shorthand A chemical equation is a shorthand method of communicating information about reactants and products, quantities involved and sometimes how much energy is needed or released. It accounts for all the atoms involved in the re-arrangement of bonds. Reactants (Ingredients) (form) Products (new substances) During a chemical reaction, bonds are broken in the reactants and new bonds are formed making the products. The atoms are re-arranged during a reaction, but the numbers of the different types of atoms remain constant. The law of Conservation of Matter (or mass) states that during a chemical reaction the total mass of reactants is the same as the total mass of products formed. During a chemical reaction, matter is neither created nor destroyed, it is only changed from one form into another. Law of Conservation of Mass (Matter) Total Mass of reactants = Total Mass of products Note: The total number of atoms of each element is unchanged! Example 1: When a spark ignites a mixture of hydrogen gas and oxygen gas, an explosion occurs - a large amount of energy is released. The product formed is water. Chemical reactions which release energy are called exothermic. Word Equation : Symbol Equation: (unbalanced) Hydrogen(g) + H2(g) + Oxygen(g) Water(l) + energy O2(g) H2O(l) + energy The correct chemical formula is written under each reactant and product. The extra subscripts indicate the state of that substance. The equation above is said to be unbalanced because the numbers of various atoms represented on the reactants’ side (left of the arrow) is not the same as those on the products’ side (right of the arrow). This would mean that mass is not conserved! 78 Reactants ( H2 + O2 ) Products ( H2O ) H=2 H=2 O=2 O=1 The numbers of oxygen atoms are not balanced! We must adjust the numbers of molecules of each type present to reach a balance! This involves placing numbers where required in front of each formula. The number will multiply each atom in the formula. This was referred to previously as a coefficient number. A “2” in front of the H2O product will provide the required 2 oxygen atoms on the right. H2(g) + O2(g) 2 H2O(l) + energy Reactants ( H2 + O2 ) Products ( 2 H2O ) H=2 H=4 O=2 O=2 Whilst oxygen has now been balanced, Hydrogen is now unbalanced! In order to balance the hydrogen atoms it is necessary to place a “2” in front of the H2 on the left. This provides 4 hydrogen atoms on the left also. Balanced Equation: 2 H2(g) + O2(g) 2 H2O(l) + energy Reactants ( 2 H2 + O2 ) Products ( 2 H2O ) H=4 H=4 O=2 O=2 Diagram: Activity: Use your atomic models to show the re-arrangement of atoms which takes place in this reaction. Notice that you will need to use the numbers of molecules as indicated in the equation, ie; 2 x hydrogen molecules for each oxygen molecule. 79 Example 2. When an emergency flare lights up, magnesium (or aluminium) metal combines with oxygen gas releasing a bright light. This is also an exothermic reaction. The substance formed is a metal oxide. Word equation: Magnesium(s) Symbols: Mg(s) + Oxygen(g) + O2(g) Magnesium + energy oxide(s) MgO(s) + energy Reactants ( Mg + O2 ) Products ( MgO ) Mg = 1 Mg = 1 O=2 O=1 The numbers of oxygen atoms are not balanced! Balance the oxygen atoms by placing a coefficient “2” in front of MgO. This will also produce 2 magnesium atoms. Mg(s) + O2(g ) 2MgO(s) + energy Reactants ( Mg + O2 ) Products ( 2 MgO ) Mg = 1 Mg = 2 O=2 O=2 Then balance the magnesium atoms by placing a coefficient “2” in front of Mg. Balanced equation: 2 Mg(s) + O2(g) 2 MgO(s) + energy Reactants ( 2 Mg + O2 ) Products ( 2 MgO ) Mg = 2 Mg = 2 O=2 O=2 Ref: C1 pp70-71 80 Written Exercises 1. Balance the following chemical equations. (a) CH4(g) + O2(g) CO2(g) + H2O(l) (b) Na(s) + H2O(l ) NaOH(aq) + H2(g) (c) Zn(s) + HCl(aq ) (d) P + O2 (e) NH3 + H2SO4 (f) CuO + (g) H2O2 H2O + O2 (h) H2CO3 H2O + CO2 (i) Fe + O2 Fe2O3 (j) C8H18 + O2 CO2 ZnCl2(aq) + H2(g) P2O5 (NH4)2SO4 HCl CuCl2 + + H2O H2O 2. You may have earlier experimented at home the reaction between ethanoic (acetic) acid, CH3COOH with sodium hydrogencarbonate, NaHCO3 to form carbon dioxide, water and a salt called sodium ethanoate (acetate), NaCH3COO. Write a word and balanced equation for this reaction. Word equation: ……………………………………………………………………………………………………….. Balanced equation: ……………………………………………………………………………………………………….. 81 EXPRESSING BALANCED CHEMICAL REACTIONS: A reaction occurs because atoms are REARRANGED. Atoms are not formed or destroyed. Mass of Reactants and products remains the SAME. 1. Identify Reactants of Chemical Reaction, for example; Sodium Hydroxide + Sulfuric Acid ? 2. Try to identify the Nature of reactants. Then correspond with a typical reaction type, ie Sodium Hydroxide + Sulfuric Acid ? BASE + ACID SALT + WATER 3. Write down a WORD EQUATION: Sodium Hydroxide + Sulfuric Acid Sodium Sulfate + Water 4. Write Symbols for compounds NaOH + H2SO4 NaSO4 + H2O. 5. Balance atoms in compounds according to Valency: 1+ 2- NaOH + H2SO4 Na2SO4 + H2O. 6. Reactant Atoms = Product Atoms Balance atoms either side of arrow. - Balance Na (2NaOH) 2NaOH + H2SO4 Na2SO4 + H2O. - Balance O (2H2O) 2NaOH + H2SO4 Na2SO4 + 2H2O. Elements in equation now BALANCED. 82 Some Typical Chemical Reactions - Atoms can be rearranged in a variety of ways…. - Write a word and balanced chemical equation of each reaction type listed below in the corresponding space at the right. REACTION TYPE EXAMPLES ACID + BASE SALT + WATER NEUTRALISATION An ACID mixed with a BASE (Alkali) produces a Neutral Salt and Water. ACID + METAL An ACID added to a METAL will produce a Metal Salt and Hydrogen Gas, H2 ACID + METAL SALT + HYDROGEN ACID + CARBONATE SALT + CARBON DIOXIDE + WATER ACID + CARBONATE An ACID added to a Metal CARBONATE compound will produce a Salt, Carbon Dioxide, CO2 and Water, H2O. ORGANIC FUEL + OXYGEN CARBON DIOXIDE + WATER ORGANIC COMBUSTION Organic Fuels BURNT in the presence of OXYGEN produce Carbon Dioxide and Water ** ** assuming complete combustion. 83 REACTION TYPE EXAMPLES ELEMENT + OXYGEN ELEMENT OXIDE INORGANIC COMBUSTION Some elements will readily burn in the presence of oxygen to form an oxide compound. PRECIPITATION Two SOLUBLE Compounds may React and Rearrange to produce an INSOLUBLE (solid) Compound, ie Precipitate. AB (aq) + CD (aq) AD (s) + BC (aq) A B+C+... DECOMPOSITION A single compound can break down to form 2 or more smaller compounds or elements. HYDROLYSIS Reaction of any chemical with WATER 84 TESTING FOR GASES - Hydrogen, Carbon Dioxide and Oxygen gases can be identified via simple diagnostic chemical tests. TEST TYPE DESCRIPTION POP TEST A diagnostic test for Hydrogen gas. SPLINT TEST A diagnostic test for Carbon Dioxide gas. LIMEWATER TEST A second diagnostic test for Carbon Dioxide gas. GLOWING SPLINT TEST A diagnostic test for Oxygen gas. 85 CHEMICAL EQUATIONS – Exercise 1. Carefully read the following chemical reaction scenarios. Firstly decide which substances are reactants and which are products, then: i) Write down a word equation of reactants and products. ii) Write a balanced chemical equation. 1. Aqueous Magnesium Chloride and Hydrogen gas are produced when Magnesium ribbon is placed into a solution of Hydrochloric acid. 2. Aluminium metal reacts with Nitric acid to produce an aqueous solution of Aluminium Nitrate and Hydrogen gas. 3. Iron metal rusts in the presence of Oxygen to produce the compound Iron III Oxide. 4. Lithium metal, a reactive group 1 alkali metal, reacts violently when placed in water to produce a strongly alkaline solution of Lithium Hydroxide and Hydrogen gas. 5. Solid Silver I Carbonate decomposes when heated to produce solid Silver Oxide and Carbon Dioxide 6. Solid Silver I Chloride and an aqueous solution of Sodium Nitrate are produced when solutions of Silver I Nitrate and Sodium Chloride are combined. 7. A neutral solution of Sodium Sulfate and water are produced at equivalence point when Sulfuric Acid is fully neutralised by the addition of Sodium Hydroxide. 86 8. A bright yellow precipitate of Lead II Iodide and aqueous solution of Potassium Nitrate are produced when aqueous solutions of Lead Nitrate and Potassium Iodide are combined. 9. When Hexane, C6H14 (a colourless liquid hydrocarbon) combusts in a plentiful supply of oxygen, the compounds of carbon dioxide and water are formed. Heat is given off indicating that the process is exothermic. CHEMICAL EQUATIONS - Exercise 2 Use your knowledge of Chemical Formulae, Valencies and Reactions to complete the following Exercises. Complete the WORD and BALANCED CHEMICAL EQUATIONS below. 1. Magnesium + Nitric Acid 2. Copper + Sulfuric Acid 3. Potassium + Water 4. Zinc (II) Carbonate + Nitric Acid 5. Copper II Carbonate (s) + HEAT 6. Aluminium + Sulfuric Acid 7. Sodium Hydroxide (aq) + Copper II Sulfate (aq) 8. Sodium Hydroxide + Sulfuric Acid 87 7.3 Ionic Equations When solutions are involved in a reaction, only some of the ions present are usually involved. Other ions may be present, but they are still in the solution at the end of the reaction, unchanged by the chemical process. These ions are called spectator ions and are best left out of the balanced equation. When spectator ions are left out of an equation an ionic equation results. Ionic equations are the best representation of the chemical reaction, since they show only those species which have undergone a change. Example 1. When a solution of lead (II) nitrate is mixed with a solution of potassium iodide, a brightly coloured precipitate forms. This precipitate is solid lead(II) iodide. Write an ionic equation for this chemical change. General Equation (balanced): Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq) Aqueous compounds split into their respective ions as they are dissolved in water. We can expand out the general equation as follows: Expanded Equation Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) PbI2(s) + 2K+(aq) + 2NO3-(aq) Note: PbI2 (s) on the right does not form ions, as it is in the solid state. In this “expanded” form, the equation shows the “whole picture” of the chemical process. If we look a bit closer at the reactants and products however, we can see that some of the chemical components have remained unchanged in the process. Can you see this ??? If you can, you will have identified 2NO3-(aq) and 2K+(aq) as remaining identical, hence unchanged on the left and right side of the equation. They are merely “spectator ions” but can be mistaken as being part of the chemical process when referring back to the balanced general equation. If they are deleted from the left and right of the equation, we will arrive at the ionic equation, ie: Ionic Equation: Pb2+(aq) + 2I-(aq) PbI2(s) Example 2. When magnesium metal reacts with dilute hydrochloric acid, a solution of magnesium chloride remains and hydrogen gas is evolved. Write a balanced, general chemical equation and an ionic equation for this chemical change. Magnesium(s) + hydrochloric acid (aq) magnesium + chloride (aq) hydrogen (g) MgCl2(aq) H2(g) General Equation (balanced): Mg(s) + 2HCl(aq) 88 + Expanded Equation Mg(s) + + (aq) 2H + - 2Cl (aq) 2+ (aq) Mg - + 2Cl (aq) + H2(g) Note: Mg (s) on the left and H2 (g) on the right do not form ions. They are in the solid and gaseous state respectively. The spectator ions in this case are the chloride ions, which are present in the dilute hydrochloric acid as Cl-(aq) and remain unchanged on the right as a component of aqueous magnesium chloride. When Cl-(aq) are deleted from the left and right of the expanded equation, we arrive at the net ionic equation: Ionic Equation: Mg(s) + 2H+(aq) Mg2+(aq) + H2(g) An Ionic Equation should also appear as a balanced equation. If not, a mistake has been made and you will need to go back and check where something went wrong. Ionic equations must also be balanced in terms of their overall charge on the left and right. This can be done as a simple calculation Reactants ( Mg(s) + 2H +(aq) ) Products (Mg2+(aq) + H2 (g) ) 2 x 1+ = 2+ 1 x 2 + = 2+ Total charge on left hand side of equation = 2+ Total charge on right hand side of equation = 2+ You will find it useful to recognise certain common types of chemical reactions so that you can correctly predict the products which will form. Ionic equations will be used where appropriate in the examples following. More work on ionic equations will be done in another module! Written Exercises: Write out the net ionic equation of each full equation below. 1. AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) 2. Na2CO3 (aq) + H2SO4 (aq) Na2SO4 (aq) + CO2 (g) + H2O (l) 89 Experiment: SOLUBILITY, PRECIPITATION REACTIONS AND IONIC EQUATIONS Aim: To observe precipitation reactions and express ionic equations Method: Combine pairs of aqueous compounds using a spotting tray as per the table below. Report “yes” or “no” to indicate formation of a precipitate. Record precipitate colour if observed. Try naming each precipitate compound formed. Results: Copper II Sulfate Barium Chloride Lead II Nitrate Calcium Chloride Potassium Iodide Sodium Hydroxide Sodium Carbonate Silver I Nitrate Precipitate? Precipitate? Precipitate? Precipitate? Colour: Colour: Colour: Colour: Name: Name: Name: Name: Precipitate? Precipitate? Precipitate? Precipitate? Colour: Colour: Colour: Colour: Name: Name: Name: Name: Precipitate? Precipitate? Precipitate? Precipitate? Colour: Colour: Colour: Colour: Name: Name: Name: Name: Precipitate? Precipitate? Precipitate? Precipitate? Colour: Colour: Colour: Colour: Name: Name: Name: Name: Written Exercise: Choose one precipitate formed from each row above. Express each process as a word equation, balanced chemical equation and finally, an ionic equation. See over page for working space and structure: 90 1. Copper II Sulfate Word Equation: Copper II Sulfate + ________________ ________________ + _______________ Balanced Chemical Equation: CuSO4 (aq) + ________________ _________________ + _________________ Ionic Equation: ………………………………………………………………………………………………………… 2. Barium Chloride Word Equation: Barium Chloride + ________________ ________________ + _______________ Balanced Chemical Equation: BaCl2 (aq) + _________________ __________________ + ________________ Ionic Equation: ………………………………………………………………………………………………………… 3. Lead II Nitrate Word Equation: Lead II Nitrate + ________________ ________________ + _______________ Balanced Chemical Equation: Pb(NO3)2 (aq) + ________________ _________________ + _______________ Ionic Equation: ………………………………………………………………………………………………………… 4. Calcium Chloride Word Equation: Calcium Chloride + ________________ ________________ + _______________ Balanced Chemical Equation: CaCl2 (aq) + _________________ __________________ + __________________ Ionic Equation: ………………………………………………………………………………………………………… 91 8. QUANTITIES AND CHEMICAL PROCESSES How can you determine how much of each substance is needed to make your product from a chemical reaction? How can you work out the amount of a chemical substance present in a product from the supermarket? In this module you will learn how to work out quantities from balanced chemical equations. You will later encounter practical techniques which are used to analyse samples for particular chemicals. 8.1 An introduction to Quantitative Chemistry Read the following 2 scenarios below. Have a go answering each of the questions. Scenario 1: If we combined 10 g of Silver Nitrate and 10 g of Sodium Chloride (both in aqueous state), a chemical reaction occurs and a white solid powder of Silver Chloride forms. If we evaporate the water after completion of the reaction, what mass of material will be obtained? Write down a corresponding chemical equation. Name which chemical substance(s) would make up this remaining material. What issues must we consider? Scenario 2: If we burn 5 g of magnesium in an open porcelain dish, a bright white flame is observed and 8.3 g of magnesium oxide (a white powder) is obtained. Write down a corresponding chemical equation. What mass of oxygen gas was consumed? Are equal masses of reactants always consumed in a chemical reaction? Which reactant in this case will there be an excess of? (ie; Excess Reagent) Which reactant would run out first, (ie; Limiting Reagent)? Ref: BFK p70 92 All chemical reactions occur in simple ratios of particles. The quantities required are given by the balancing numbers in the chemical equation. Examples: 1. Electricity can be used to decompose water molecules into its elements of hydrogen and oxygen. RATIO: 2H2O(l) 2H2(g) 2 molecules of water 2 molecules of hydrogen 1 molecule of oxygen 2 2 1 : + : O2(g) Two molecules of water decompose to form two molecules of hydrogen and one molecule of oxygen. Of course we do not work with single molecules - they are much too small! However, the simplest whole number ratio of molecules involved in this process will always be 2 x H2O : 2 x H2 : 1 x O2 2. Sodium and chlorine react together violently to produce sodium chloride. 2Na(s) RATIO: + Cl2(g) 2NaCl(s) 2 atoms of sodium 1 molecule of chlorine 2 units of sodium chloride 2 1 2 : : Two atoms of sodium react with 1 molecule of chlorine to produce 2 formula units of sodium chloride. 2 x Na : 1 x Cl2 : 2 x NaCl Again, we cannot see molecules and so making sure that we have the right quantities for a chemical reaction could be difficult! Simplest whole number ratios are referred to as coefficient numbers in a balanced chemical equation. These numbers represent mole quantities, a unit of scientific measurement which can be reported with accuracy in the laboratory. 93 8.2 The Mole Concept We need to “scale up” our quantities from the molecular level to an amount which we can see and measure in the laboratory. The quantity of particles chosen is called the mole which is approximately 6.02 x 1023 particles of any pure substance (ie; element or compound). Scientific notation is expressed in this number as this is a gigantic quantity! The Periodic Table measures the mass of each element in atomic mass units (a.m.u.) One mole of any substance contains 6.02 x 1023 particles and has a mass measured in grams equal to its formula weight (a.m.u.) One mole of water, H2O contains 6.02 x 1023 molecules. The mass of 1 mole of water is equal to its formula weight (FW), expressed in grams; FW (Hydrogen) = 1.008 a.m.u. FW (Oxygen) = 16.00 a.m.u. (2 x 1.008) + 16.00 = 18.016 g. Likewise, one mole of hydrogen (gas) contains 6.02 x 1023 molecules of hydrogen and has a mass of 2 x 1.008 = 2.016 g. And - one mole of oxygen (gas) contains 6.02 x 1023 molecules of oxygen and has a mass of 2 x 16.00 = 32.00 g. We could, if we wanted, determine the ratio, by mass of the substances involved in the decomposition of water. 2H2O(l) 2H2(g) + O2(g) 2 molecules of water 2 molecules of hydrogen 1 molecule of oxygen 2 moles of water 2 moles of hydrogen 1 mole of oxygen 2 x 18.016 = 36.032 g 2 x 2.016 = 4.032 g 1 x 32.00 = 32.00 g If we started with 1 mole (ie 36.032 g) of water, we would obtain 4.032 g of hydrogen and 32.00 g of oxygen. What mass of hydrogen and oxygen would we obtain if we started with 1.00 g of water? H2O (l) H2 (g) O2 (g) 36.032 36.032 4.032 36.032 32.00 36.032 1.00 g 0.11 g 94 + 0.89 g If we started with 1.00 g of water we would obtain 0.11 g of hydrogen and 0.89 g of oxygen. Notice that the balancing numbers do not give us masses directly, that is we cannot say that 2 g of water will give 2 g of hydrogen and 1 g of oxygen! When we use the mole concept we are taking into account that the various atoms have different masses, and the balancing numbers give us the right ratios of atoms involved. It makes life much easier when interpreting number coefficients of a balanced chemical equation as mole quantities, because that’s exactly what they are ! Take special note that the total mass of reactants equals the total mass of products. No atoms are lost or gained, as understood from our work on writing and balancing chemical equations. Atoms are just re-arranged in a chemical reaction. We could continue our calculations to account for any mass of water, or we could calculate the quantity of water required to produce a particular amount of hydrogen or oxygen. The mole ratios of reactants to products are determined from the balanced chemical equation and a knowledge of the mole concept. We can determine the number of moles present in a sample of a substance by using one or more of the following formulae: No of moles = No of particles 6.02 x 1023 This mole equation will not be immediately useful as it deals with a known number of chemical particles, which in reality is mostly unrealistic. You may, however be asked to calculate a mole quantity given such data. Mole Equation 1. No of moles = Mass (g) Formula Weight This equation is useful in the real world, as mass and Formula Weight are measurable quantities in the real world. Memorise it !!! Shorthand: n = m FW 95 Written Exercises: Mole calculations 1. Calculate the Formula Weight, (FW) for each of the following substances. (a) Sodium Chloride, NaCl (b) Ammonium Sulfate, (NH4)2SO4 (c) Glucose, C6H12O6 (d) Potassium Hydrogen Phthalate, KC8H5O4 2. Calculate the number of moles of each substance (a) 58 g of NaCl (b) 13.2 g of (NH4)2SO4 (c) 90 g of C6H12O6 (d) 5.0 g of KC8H5O4 3. What mass of each substance is present? (a) 2.5 moles of NaCl (b) 0.5 moles of (NH4)2SO4 (c) 4 moles of C6H12O6 (d) 0.015 moles of KC8H5O4 Ref: CC1 p205 C1 pp170-173; 213-218 DB pp20-22 96 8.3 Theoretical Yield (also Chemical Stoichiometry) A Theoretical Yield is the calculated prediction of an amount of substance produced (product) in a chemical reaction. It is a very useful application of the mole concept, and is an integral part of the chemical industry. This concept can alternatively be used to calculate the amount of chemical substance required (reactant) in a chemical process. Scenario 1: Predicting the mass of a product obtained in a chemical reaction. Calculate the mass of Zinc Chloride, ZnCl2 produced when 5.0g of Zinc metal is dissolved in an excess of Hydrochloric Acid, HCl. Step 1. Calculate moles of Zinc metal reacted Step 2. Write down a balanced chemical reaction Step 3. Express the reaction ratio between the reactant Zn and product ZnCl2 chemicals of interest. ________ : ________ Step 4. Calculate the number of moles of ZnCl2 produced: apply the reaction ratio (ie mole ratio) to the calculated number of moles in Step 1. Step 5. Calculate the Theoretical Yield mass of ZnCl2 using the mole equation (rearranged..) 97 Scenario 2: Calculating the required mass of a reactant chemical to produce a desired known quantity of a given product. Iron metal, Fe is produced from one of its ore compounds Iron III Oxide, Fe2O3 by heating in the presence of excess Carbon Monoxide, CO gas. Carbon dioxide is given off as a by-product. - Calculate the mass of Iron III oxide, Fe2O3 required to produce 500 g of Iron metal. Step 1. Calculate the moles of desired Iron metal. Step 2. Write down a balanced chemical reaction. Step 3. Express the reaction ratio between the reactant / product chemicals of interest. Fe2O3 Fe _______ : _______ Step 4. Apply reaction ratio to calculate the moles of Iron III Oxide (Fe 2O3) required. Step 5. Convert moles of Iron III Oxide (Fe2O3) into a mass (via mole equation again…) 8.4 Percentage Yield In reality, it is very unlikely that a 100% yield of product is obtained. Experimental errors, transfer losses and incomplete reaction often tend to contribute to a lower than expected yield. (Note: yields greater than 100% indicate that an error in calculation, or perhaps moisture in the final weighed product might have been encountered). % Yield = obtained yield theoretical yield x 100 This percentage calculation is quite easy when comparing your obtained yield with the theoretical (ie expected yield). 98 Scenario 2 Revisited: Calculating a percentage yield The required mass of Iron III oxide, Fe2O3 calculated in Scenario 2 previously was carefully weighed out, and chemically converted into Iron via the same chemical process. A mass of 462.2 g of Iron was recovered. i) Calculate the % Yield of this process ii) Suggest reasons for this obtained % 8.5 Limiting Reagents As we have seen, it is possible to calculate the exact quantities of reactants that would in theory react chemically right down to the very last molecule. This, in reality does not occur – there is always an excess of one substance (quite possibly verrry verrry small or large depending on the situation). The chemical substance totally consumed in a chemical process is referred to as the Limiting Reagent – it is the substance that determines when the reaction will cease and the amount of product that will be obtained. As an example – the pop test is a chemical reaction between Hydrogen Gas, H2 and Oxygen Gas, O2. Water, H2O is the resulting product. The amount of water produced in this case is limited by the volume (and hence mass) of hydrogen gas. Atmospheric oxygen is in plentiful supply (21%) and is therefore not particularly likely to be entirely consumed…… Oxygen is referred to here as the Excess Reagent. Identifying the Limiting Reagent is yet another application of the mole concept. The mole quantities of each chemical reactant present is applied to the reaction ratio according to the balanced chemical equation. A decision can be made from here via simple calculation as to which one is in limited supply, (ie; Limiting Reagent identification) Scenario 1 Revisited: 10.0g of Zn (s) is chemically reacted with 10.0g of HCl (aq). PART A: Identify the Limiting Reagent Step 1. Calculate moles Zn Step 2. Calculate moles HCl 99 Zn HCl Step 3. Express reaction ratio: _______ : _______ (from Step 2, Scenario 1) Step 4. Apply reaction ratio to moles of Zn (Step 1.) to calculate the actual mole requirement of HCl Moles HCl Required: ___________________________ Moles HCl Available: ___________________________ (from Step 2) Step 5. Are there sufficient moles of HCl available ?? - If yes, Zn will be the Limiting Reagent (it will be consumed first) - If no, then HCl will be the Limiting Reagent. There is insufficient HCl to fully react with the mass of Zn available, hence HCl will be consumed first. Limiting Reagent ID: ____________________________________ PART B: Calculate the Theoretical Yield of ZnCl2 produced in this scenario. - The amount of ZnCl2 is governed by the available amount of Limiting Reagent. Step 1. Identify the limiting reagent (as per Step 5, Part A): _____________________ Step 2. Calculate moles of Limiting Reagent (as per Step 2, Part A): ______________________ Step 3. Express the reaction ratio between the limiting reagent and the product ZnCl2 Step 4. L.R. ZnCl2 _______ : _______ Apply this reaction ratio in calculating the moles of ZnCl2 produced. Moles ZnCl2 = _____________________ Step 5. Calculate the Theoretical Yield mass of ZnCl2 (using the mole equation…) 100 PART C: Calculate the mass of Excess Reagent remaining unreacted. Step 1. Identify the excess reagent: _________________________ Step 2. Calculate the moles of excess reagent available (as per Step 1, Part A): Step 3. Calculate moles of limiting reagent (as per Step 2, Part A): Step 4. Express the reaction ratio between the excess and limiting reagent: Zn Step 5. : HCl = _______ : _______ Apply reaction ratio and calculate the moles of excess reagent consumed - Moles Zn consumed in reaction = _________________________ Step 6. Convert moles of excess reagent into mass of excess reagent consumed. Step 7. Calculate the mass (g) of excess reagent remaining. Written Exercise: Hydrazine, N2H4 is used as a rocket fuel. It reacts with oxygen to form nitrogen gas and water vapour. (a) Write a balanced chemical equation for this reaction. (b) 1 kg of hydrazine and 1 kg of oxygen are available on the rocket. Identify the Limiting Reagent in this scenario. (c) What mass of water vapour is produced during the flight? 101 Yield Experiment 1 – MAGNESIUM OXIDE Aim: To predict a theoretical yield when magnesium metal is combusted. Method: (record all results in the table below) Obtain and record accurately the mass of a clean empty crucible + lid. Accurately weigh approximately 0.2g of Mg metal into your crucible. Record the combined mass. Heat crucible and contents with a bunsen flame. Use tongs to lift crucible lid periodically and observe until all Mg has combusted. Allow 10 minutes to cool. Reweigh Crucible and combusted contents. Results: Mass: empty crucible + lid (g) Mass: crucible + lid + Mg (g) Calculate mass: Mg metal (g) Mass: crucible + lid + MgO (g) Calculate mass: MgO (g) Calculations (Theoretical Yield): 1. Moles Mg: Moles = Mass / FW ……………………………………………………………………………………………………… 2. Write a word and balanced chemical equation for the combustion of Mg (s). i) Word Equation: ……………………………………………………………………………………………………… ii) Balanced Chemical Equation: ……………………………………………………………………………………………………… Mg MgO 3. Reaction Ratio between Mg (s) and MgO (s) = _____ : 4. Moles MgO = ……………………………………….. 5. Mass MgO (g) (theoretical yield): Mass = Moles x FW (MgO) ______ ………………………………………………………………………………………………… Calculations; Percentage Yield: 6. Calculate this % based on the mass of MgO obtained experimentally. 102 Yield Experiment 2 – DECOMPOSITION OF A CARBONATE COMPOUND Background: Decomposition Reactions involve the break down of large compounds into 2 or more smaller components. Heat is a large factor of many decompositions. Bicarb Soda, a carbonate used in baking, will decompose into simpler substances – most notably Carbon Dioxide. This is what causes bread to rise. Procedure: Obtain and record accurately the mass of a clean empty porcelain crucible + lid. Accurately weigh approximately 1.5g of Copper Carbonate Powder into the crucible. Record the combined mass of crucible, lid and Copper Carbonate powder. Set up a heatproof mat, Bunsen burner, tripod and pipeclay triangle. Carefully heat the Carbonate substance with a blue Bunsen flame for between 5 and 10 minutes. Use tongs to occasionally lift the crucible lid and observe the change in colour of the compound. Allow 10 minutes for the crucible and lid to cool. Reweigh and record the new mass. Results: Mass: Crucible + Lid (g) Mass: Crucible + Lid + CuCO3 (g) Mass: CuCO3 (g) After heating with Bunsen Burner: Mass: Crucible + Lid + CuO (g) Mass: CuO (g) Calculations 1. Theoretical Yield: Moles CuCO3 : Moles = Mass ÷ FW ……………………………………………………………………………………………………… 103 2. Write a word and balanced chemical equation for the decomposition of CuCO3 (s) i) Word Equation: …………………………………………………………………………………… ii) Chemical Equation: ……………………………………………………………………..…………… CuCO3 3. Reaction (mole) Ratio between CuCO3 and CuO = 4. Moles CuO = ……………………………………………………………….. 5. Mass CuO (g) Mass = moles x FW (CuO) ______ : CuO ______ (this value is the theoretical yield) ……………………………………………………………………………………………………… CALCULATIONS 6. Percentage Yield: Calculate the Percentage Yield of CuO, based on the mass you obtained experimentally. % Yield = Obtained Yield x 100 Theoretical Yield Questions: 1. Atoms are neither created nor destroyed in chemical reactions – rather they are rearranged. How can we therefore account for the loss of mass in the reaction? 2. Explain why it is not necessary to periodically remove the crucible lid to ensure completion of the chemical reaction. 3. Yield percentages of over 100% are commonly encountered in this task. Explain why. 104 Yield Experiment 3 – ZINC II IODIDE Background: Zinc metal, Zn reacts with iodine, I2 to produce the compound zinc II iodide, ZnI2. You will react an accurately known mass of iodine crystals with excess zinc metal, so that all the iodine can react but some zinc will remain. Ethanol will be used as a solvent for the iodine to speed up the reaction. Zinc iodide is highly hygroscopic - it will absorb water from the air. ! Risk Assessment: Iodine vapour is mildly corrosive and an irritant. Iodine will stain skin and clothing easily. The reaction between zinc and iodine is exothermic and iodine vaporises easily. Method: 1. 2. Collect approximately 2g of zinc pieces and record their mass accurately. Obtain a sample tube and lid. Tare sample tube and accurately weigh approximately 1g of iodine. Record this value. 3. Carefully add the zinc pieces to the tube containing the iodine. 4. Collect about 20 mL of ethanol and add it to the tube a few drops at a time while gently swirling the contents of the tube. The ethanol should be added very slowly and the tube should become very warm between additions. If the ethanol is added too quickly the reaction will become very slow. 5. Continue to add the ethanol and swirl the contents until all reaction has ceased. The brown colour of dissolved iodine should have almost disappeared. Allow the tube to stand covered until all colour has been lost. 6. Carefully weigh a clean, dry evaporating basin. Decant the liquid from the sample tube into the basin, retaining unreacted zinc in the sample tube. 7. Wash the zinc and the insides of the tube with several small quantities of ethanol. Add washings to the evaporating basin. 8. Evaporate the ethanol from the evaporating basin using an infra-red lamp if possible. Otherwise use an electric hotplate on a low setting. 9. Allow the dry evaporating basin to cool in a desiccator and weigh the basin and contents quickly. From your measurements you should now be able to calculate the percentage yield of zinc iodide. Results: Mass: Zinc = __________________ g Mass: Iodine = __________________ g Mass: Clean, dry evaporating basin = _________________________ g Mass: Evaporating basin + dry Zinc Iodide = _______________________ g Mass: Zinc Iodide = _________________________ g 105 Calculations: 1. Calculate to show that the Iodine was the Limiting Reagent 2. Calculate the Theoretical Yield of Zinc Iodide 3. Calculate the Percentage Yield of Zinc Iodide 106 8.6 Moles, Gas Volumes and Yields It is often inconvenient to have to weigh a gas in order to determine the quantity present. Why? Try weighing an inflated balloon in order to determine the mass of gas inside! Various problems such as buoyancy effects and the mass of the container become apparent. When handling gases, it is preferable to measure volume. Of course we must use set conditions of temperature and pressure in order to compare quantities, as differences in both temperature and pressure change volume! There are 2 conditions of Molar Volume chosen by scientists universally. Each is a measure of the volume of 1 mole of any gas. They are as follows: Standard Temperature and Pressure, S.T.P. Temperature Pressure Molar Volume 0oC (273K) 100 kPa 22.71 L Standard Laboratory Conditions, S.L.C. Temperature Pressure Molar Volume 25oC (298K) 100 kPa 24.79 L 100kPa, for interests sake is very close to that of atmospheric air pressure It has been found that 1 mole of any gas will occupy the corresponding volume according to the respective conditions outlined in the tables above. We can use this information to therefore easily determine the number of moles present. No of moles = volume (L) molar volume (L) Example: Carbon disulfide, CS2 is an important industrial solvent. It is prepared by the reaction of coke (almost pure carbon) with sulfur dioxide gas. What mass of coke and volume of sulfur dioxide (at 25oC and 100 kPa) would be required to produce 1 kg of carbon disulfide? Equation: 5C(s) + 2SO2 (g) Mole ratio: 5 moles : 2 moles CS2 (l) + : 1 mole 107 4CO (g) : 4 moles i) Calculate required mole quantities: The balancing numbers tell us that if we wanted 1 mole of carbon disulfide we would require 5 moles of carbon (coke) and 2 moles of sulfur dioxide. How many moles of carbon disulfide do we require? Using the formula: No of moles = (of CS2) Mass (g) FW = 1000 g ( 1 kg = 1000 g) 12.01 + (2 x 32.07) = 13.1 Therefore we must produce 13.1 moles of carbon disulfide. For each mole of carbon disulfide we need 5 moles of coke. No of moles (of C) = 5 x 13.1 = 65.5 For each mole of carbon disulfide we need 2 moles of sulfur dioxide. No of moles (of SO2) = ii) 2 x 13.1 = 26.2 Calculate corresponding mass (C) and volume (SO2) quantities The question requires answers as mass (C) and volume (SO2) and so we will need to use the corresponding mole equations, covered previously to calculate these. 1. Calculating Mass of Carbon, C Moles C = 65.5 Mass C = ? FW (C) = 12.01 No of moles = Mass (g) Formula Weight rearrange to make “Mass” the subject of the formula Mass (g) = No of moles x FW = 65.5 x 12.01 = 787 g (rounded) See over page for calculation of SO2 volume 108 2. Calculating Volume of SO2 (g) Moles SO2 = 26.2 Volume SO2 = ? Molar Volume (SLC) = 24.79 L No of moles = Volume (L) 24.79 rearrange to make “Volume” the subject of the formula Volume (L) = No of moles x 24.79 = 26.2 x 24.79 = 650 L (rounded) If the gases present were not under S.L.C. (or S.T.P.) conditions, you would need to use another mathematical formula: PxV = nxRxT where; P = Gas pressure (kPa) V = Volume of gas (L) n = No of moles of gas present R = the Gas Constant T = Temperature (Kelvin) Ref: C1 pp276-279 Written Exercise: Theoretical Yield and Gas Volumes 1. Limestone and shells are made mostly of calcium carbonate. When they are heated to produce “quicklime” or calcium oxide, carbon dioxide gas is released. What volume of carbon dioxide (measured at 25oC and 100 kPa) is released in the production of 112 kg of quicklime? (i) Write a balanced equation for the reaction, and hence indicate mole ratios. (ii) How many moles of quicklime are produced? (iii) How many moles of carbon dioxide are produced? (iv) What volume of carbon dioxide is produced? 109 2. When hydrogen gas, H2 is needed in the laboratory it is commonly produced by the reaction of dilute hydrochloric acid on zinc metal. What volume of hydrogen gas o could be produced at 25 C and 100 kPa, when 13.0 g of zinc is reacted with excess hydrochloric acid. (i) Write a balanced equation for the reaction, and hence indicate mole ratios. (ii) How many moles of zinc metal are used? (iii) How many moles of hydrogen will be produced? (iv) What volume of hydrogen will be produced? 3. Silver metal can be reclaimed from wastes containing silver chloride by reacting them with hydrogen gas. What volume of hydrogen (measured at 25 oC and 100 kPa) would be needed to change 28.7 g of AgCl to silver metal? AgCl(s) + H2(g) Ag(s) (i) Balance the equation above. (ii) Write the mole ratio under the balanced equation. (iii) How many moles of AgCl need to be reacted? (iv) How many moles of H2 will be required? (v) What is this volume of H2? + HCl(g) 4. Acetylene, C2H2 is an important fuel which is used in welding. We can generate acetylene in the laboratory by reacting calcium carbide, CaC2 with water. Calculate the volume of acetylene released at 25oC and 100 kPa if 0.450 mol of calcium carbide is reacted with excess water. CaC2(s) + H2O(l) Ca(OH)2(aq) + C2H2(g) See question structure over page 110 (i) Balance the equation. (ii) Write the mole ratio under the balanced equation. (iii) How many moles of acetylene will one mole of CaC2 produce? (iv) How many moles of acetylene will 0.450 moles of CaC2 produce? (v) What is this volume of acetylene? Another bad joke: Did you hear about the cat which accidently consumed a meal of calcium carbide after falling pregnant? It later had acetylene kittens! 8.7 A Quick note on Kelvin . . . Temperature is measured on a scale of degrees Celsius, (oC) as we are all familiar with. Many calculations encountered in chemistry (and physics) often require values of temperature to be measured not in oC, rather in Kelvin, (K). Where does temperature start and finish? Cold temperatures could be measured in the order of –10 oC in a very cold environment, or maybe even –150 oC when some gases liquefy. Air temperature is around 25 oC, a volcano might be 1000 oC 5000 oC, whereas the core temperature of a star might be in the order of 1,000,000 oC or more! In short, temperature has no maximum value, but does have a minimum value, referred to as Absolute Zero. Absolute Zero has been scientifically calculated to be –273.16 oC. This is normally rounded to –273 oC. It is not possible to achieve a temperature colder than this! Kelvin is referred to as an absolute measure of temperature, and therefore has a minimum value of 0 Kelvin, or 0 K. This is equivalent to the Celsius value identified above as –273 oC. Hence: 0 K = –273 oC Any calculation requiring conversion of oC K is simply a matter of adding 273 to the value of oC. Kelvin = oC + 273 111 Yield Experiment 4 – GAS VOLUMES Background: Predicting mass yields in chemical processes via mole based quantities is by now a familiar concept. Gases produced in chemical reactions have until now been accounted for as a mass loss. However there is a directly proportional relationship between the amount of any gas produced and its volume. The molar volume of any gas is identical whilst temperature and pressure remain unchanged. Most laboratories operate in an environment close to Standard Laboratory Conditions (S.L.C.) Prework: Part A – Carbon Dioxide Volume Yield: Accurately weigh approximately 0.2 g of Calcium Carbonate and record in the table over page. Based on the mass you weighed out, calculate the volume yield of Carbon Dioxide expected at S.L.C. Part B – Hydrogen Gas Volume Yield Accurately weigh approximately 0.05 g of Aluminium metal and record in the table over page. Based on the mass you weighed out, calculate the volume yield of Hydrogen gas expected at S.L.C. PROCEDURE: Perform the general procedure below separately for both Parts A and B. Set up equipment required for transferring and collecting the gas produced into an inverted 100 mL measuring cylinder containing tap water. Record the initial volume of air in the inverted measuring cylinder. 112 Measure around 20 mL of 5.0 M HCl. Transfer into the conical flask and immediately attach bored stopper. Ensure the gas displaced is successfully transferred into the inverted measuring cylinder. Record the final volume of gas. Report the volume of gas displaced, and % Yield based on this result. RESULTS: PART A: Calcium Carbonate + HCl PART B: Aluminium + HCl Mass weighed (g) Initial Volume in Measuring Cylinder (mL) Final Volume in Measuring Cylinder (mL) Gas Volume Displaced (mL) CALCULATIONS % Yield = Obtained Yield x 100 Theoretical Yield Percentage Yield: Calculate the Percentage Yield of obtained gas, based on the volume you obtained experimentally. PART A: Calcium Carbonate + HCl PART B: Aluminium + HCl QUESTIONS: 1. Identify possible errors in the design and execution of this experiment. 2. The gas collected in the measuring is a mixture of displaced air and carbon dioxide. Does this contribute errors in your volume measurements? Explain. 3. Summarise the conditions of S.T.P. and S.L.C. Identify what both conditions stand for and their respective conditions of pressure and temperature. 113 8.8 Percentage Composition Any chemical compound has a set chemical formula and therefore set percentages of elements by mass in it. If we can determine the percentage composition by mass of a compound in a laboratory, we can often identify it. A chemical formula makes it possible to calculate the percentage total of each element. % element in compound = total mass component of element formula weight of compound x 100 Consider water as an example: H2O contains two elements, hydrogen and oxygen. From the formula, there are twice as many hydrogen atoms as oxygen atoms. However, hydrogen atoms are much lighter than oxygen atoms, so the fraction of the mass due to oxygen will presumably be much greater than one third! Formula weight H2O = % hydrogen = % oxygen (2 x 1.008) + (1 x 16.00) 2 x 1.008 18.016 x 100 x 100 = 11.2 % = 1 x 16.00 18.016 = 88.8 % = 18.016 Of course, we did not need to calculate % oxygen once we knew % hydrogen! Since oxygen is the only remaining element in water: % oxygen Written exercise: = 100% – 11.2% = 88.8% Percentage composition 1. Calculate the percentage composition by mass of each element in the following compounds. (a) CO2 (b) H2SO4 (c) C6H12O6 114 2. Zinc and iodine react to form a compound, zinc iodide, ZnI2. Calculate the percentage by mass of each element in zinc iodide. 3. Mercury reacts to form two different chlorides; mercury (I) chloride, Hg2Cl2 and mercury (II) chloride HgCl2. Mercury (II) chloride is more toxic although both must be handled with care. Calculate the percentage composition by mass of Hg and Cl in each compound. 8.9 Empirical and Molecular Formulae The Empirical Formula of a compound is the simplest whole number ratio of elements composing a compound. The Molecular Formula is the true chemical formula of the compound. Octane is an organic compound having a molecular formula of C8H18. Its empirical formula would be C4H9. Propane is another organic compound having a molecular formula of C 3H8. Its empirical formula is also C3H8 as it cannot be simplified any further. You can calculate the molecular formula for a compound if you are given its elemental percentage composition and its formula weight. Consider the following: Example: A compound in white wine has been analysed to give the following percentage composition. C = 52.3% H = 13.1% C O = 34.6% : H : O Ratio by mass 52.3 13.1 34.6 Ratio by number of atoms 52.3 12.01 13.1 1.008 34.6 16.00 4.36 13.1 2.16 4.36 2.16 13.1 2.16 2.16 2.16 2 : 6 : ( atomic mass) ( smallest number ratio) 1 Following these steps will calculate the empirical (simplest ratio) formula, ie C2H6O. 115 Its formula weight has been is known to be approximately 46. What is its molecular formula? The empirical formula weight, C2H6O = = (2 x 12.01) + (6 x 1.008) + 16.00 46.068 This is very close to the given formula mass and therefore the molecular formula is the same as the empirical formula C2H6O. If the molecular FW was larger than the empirical formula weight, then a multiplying factor would apply. If the molecular FW was say 92, which is 2x the empirical FW, then this factor would apply to the empirical formula. It would hence be C4H12O2. Written Exercises: Percentage Composition, Empirical and Molecular Formulae 1. The percentage composition of an organic compound is 48.6% carbon, 8.10% hydrogen and 43.2% oxygen. What is its empirical formula? 2. A compound is analysed to give the following results: H = 5.00% F = 95.00% Its formula weight was determined to be 40.0. Find its empirical and molecular formula. 3. The mass of phosphorus in an 18.3 g sample of a compound containing only phosphorus and oxygen is 10.3 g. The formula weight of the compound is 220. Find the empirical and molecular formula of the compound. Ref: C1 pp211-213; CC1 p202 116 Experiment: Which Mercury Chloride? Task: You will be given a sample powder (approximately 3g) in a specimen tube. Your task is to identify your given compound as mercury (I) chloride, Hg2Cl2 or mercury (II) chloride, HgCl2. Both of these compounds are white powders. Use the percentage composition calculated in the previous exercise to help you make your decision. You will be able to determine the percentage of mercury in your compound by determining the mass of the white powder (the compound) and then decomposing it to obtain the mercury metal which can be weighed also. % Hg = mass of mercury metal mass of compound x 100 Decomposition of the mercury chloride is achieved by heating the substance with a mixture of sodium hypophosphite and hydrochloric acid until the element mercury is produced. Do not attempt to write a balanced equation for this reaction at this stage. ! Risk Assessment: All compounds of mercury should be considered to be highly poisonous. Don't breathe in the fumes from the solution – carry experiment out in a fume cupboard and make sure that the mercury is covered by solution during heating. Wash your hands thoroughly at the completion of the exercise and make sure all spillages are cleaned up! All mercury residues must be disposed of in the labelled waste collection containers. Chemicals required: Unknown sample tube containing approximately 3g of mercury(I) or mercury(II) chloride. Sodium hypophosphite, 2g in 5 mL H2O + conc HCl 3 mL methylated spirit 5 mL Procedure: 1. 2. 3. 4. 5. 6. 7. 8. 9. Obtain a sample tube containing a mercury chloride unknown and record the identification code on the tube. Weigh and record the mass of the sample tube, lid and contents. Obtain a clean, dry evaporating basin. Label with your name. Weigh and record its mass. Carefully add the mercury chloride to the evaporating basin. Collect 20 mL of distilled water and add the water to the evaporating basin. Re-weigh the empty sample tube. Add the sodium hypophosphite / hydrochloric acid solution to the contents of the evaporating basin. Place over a steam bath in a fume cupboard and stir to mix the reactants. ! Care - liquid is corrosive. Continue heating and stirring until mercury collects in large silvery globules. ! Avoid inhaling vapour while heating. When all the mercury globules have coalesced, remove the basin from the steam bath and decant the colourless liquid. 117 10. 11. 12. 13. Wash the contents of the basin twice with water and once with small amounts of methylated spirits. Collect all washings in the waste container provided. Dry the mercury by pulling pieces of filter paper through it as demonstrated. Re-weigh the basin containing the dry mercury. Return the mercury to the class collection bottle. Results MERCURY COMPOUND A MERCURY COMPOUND B ID Code: Mass: Sample Tube + compound Mass: Sample Tube (empty) Mass: Mercury Compound (g) Mass: Evaporating Basin (empty) Mass: Evaporating Basin + Mercury Mass: Mercury (g) Mass: Chloride (g) Calculations MERCURY COMPOUND A % Mercury = % Chloride = Mole Ratio Mercury = Mole Ratio Chloride = Empirical Compound Formula Molecular Compound Formula 118 MERCURY COMPOUND B Questions 1. Why is it unlikely that your experimental composition was exactly the same as the calculated values from the formula? 2. Is your mercury chloride a covalent or ionic compound? Why? 3. This exercise began with a mercury chloride, and resulted in mercury metal. Why did we not need to measure the amount of chlorine produced? 4. Name two (2) safety procedures which you needed to observe while completing this experiment. 5. Mercury is classified as a heavy metal and the term “heavy metal residues” often appears in pollution reports. You may wish to find out about pollution by mercury compounds, where they are found and how they are detected. 119 8.10 Calculating Concentration Concentration is a measure of the quantity of one substance within another. Concentration can be measured as a function of all 3 states of matter: solids, liquids and gases. Liquids and gases are normally expressed as volume based concentrations. A liquid containing one or more dissolved substances becomes a mixture by definition. A liquid mixture is more commonly referred to as a solution. A common example of a solution: seawater Seawater, simplistically speaking contains salt dissolved in water, ie; Seawater = Water + Solution = Solvent + Salt Solute The Solvent is the main liquid component. The solute is normally the “ingredient of interest” dissolved within the solvent and may be in the form of a solid, liquid or gas. As soon as any substance becomes dissolved in a liquid, it is recognised as being aqueous, (aq) in its physical form. Solution concentration is more conventionally measured as the amount of solute dissolved within a set volume of solution, (mL, 100 mL, or L) The solute component is often expressed as a mass (g, mg, or ug) but can also be measured as a volume where a liquid solute, eg: ethanol applies. Mass based concentrations can be calculated using the following general equation; Concentration = Amount of Solute Volume of Solvent Examples: Calculate the concentration of substances in the scenarios below. 1. 4.8 grams of salt dissolved in 2 litres of water Concentration 2. = 4.8 g 2L = 2.4 g / L 456 milligrams of lead dissolved in 80 litres of water Concentration = 456 mg 80 L = 5.7 mg / L 120 3. 18 mL of alcohol dissolved in 375 mL of beer Concentration = 18 mL 375 mL = 0.048 mL / mL This concentration might not sound overly meaningful, or even confusing as it is a figure comparing mL with mL . . . . Normally alcohol concentration is reported as a volume based percentage (%). In simple terms, if there is 0.048 mL of alcohol in 1 mL of beer, then there would be 100 x as much alcohol in 100 mL of beer . . . makes sense ? This is essentially a unit conversion, but in short if you multiply the calculated concentration above by 100, then a more meaningful alcohol concentration of 4.8 mL / 100 mL will be arrived at. This could also be reported as 4.8% v/v. In conclusion, you will notice that the units of concentration corresponding with the calculated number are determined by the individual measurement units of the solute and solvent. 8.11 Molarity A term not associated whatsoever with Morality! We saw in 8.10 that solution concentration can be calculated and expressed in a variety of different units of measurement. In one example, concentration was calculated from a mass based solute (grams) and volume based solvent (measured in Litres), ie: Concentration (g / L) = Mass (g) Volume (L) We have also explored the mole concept, whereby a given mass (g) of pure substance can be expressed as a quantity in moles. It follows logic that if convert a mass (g) of solute instead into moles, then it would modify our units of concentration, ie; Concentration (moles / L) = Moles Volume (L) This measurement of concentration is more conventionally referred to as a MOLARITY, and is calculated using the following equation: Mole Equation 2. Molarity = Moles Volume (L) NOTE ! ! Volume is always measured in litres (L) Shorthand: M = n V (L) 121 We can see that Molarity is a measurement of concentration in units of Moles / L. These units can be further shortened to the symbol M, meaning Molar. Either way, Molarity, Moles / L, Molar and M are all different representations of exactly the same measurement! To further clarify, a 1M concentration of any chemical solution will contain 1 mole of chemical per 1 litre of solution. For example, a 1M solution of Sodium Hydroxide, NaOH would contain 1 mole of NaOH per litre of solution. This 1 mole of NaOH (FW 40.08 a.m.u.) per litre could be expressed back to a mass based concentration of 40.08 g/L. Practice Examples: Use mole equation 2 to calculate the molarity of solution in the following examples: i) 6 moles in 2 L ii) 0.5 moles in 200 mL iii) 0.0025 moles in 25 mL The examples above were fairly straight forward, except for the requirement to convert units of mL L where identified. The following 3 examples deal with a specific mass of compound dissolved in a given volume of solution. Each mass will firstly need to be converted to moles (mole equation 1), before calculating the molarity of solution. iv) 58.5g of NaCl in 2L of v) 4.0 g of NaOH in 200 mL vi) 0.5 g of Na2CO3 in 50 mL The six practice examples above all used the mole equation 2 in its basic form to calculate the concentration of solution. However, another important application of this equation is to assist us in preparing a solution of known concentration. For this to happen, a laboratory technician will need to know 2 things about the solution to be prepared: 1. 2. its desired Concentration (M), and the Volume required The task therefore becomes calculating the (3.) quantity of solute that will need to be weighed and dissolved in solution. 122 These 3 pieces of data will also fit into the mole equation 2: Molarity = moles ? Volume (L) The values of Molarity and Volume are known, making the value of moles as our unknown quantity in this scenario. It will be necessary to rearrange this equation so that moles becomes the subject of the formula, ie; moles = Molarity x Volume (L) This intermediate calculation gives us a quantity of moles that are required to be weighed and dissolved, however we cannot weigh mole quantities of any substance! We therefore refer back to our original mole equation 1 to convert our intermediate calculation of moles into a mass, ie; Moles = mass (g) ? FW You will become familiar with solution preparation scenarios as requiring a 2 step calculation. Practice examples help! Example: Explain how to make 500 mL of a 0.10 M NaCl solution. Volume = 500 mL = 0.5 L Concentration = 0.10 M No of moles of NaCl = = 0.10 x 0.5 0.050 Mass of NaCl No of moles x FW 0.050 x 58.44 2.9 g = = = Therefore to make the desired solution we would need to accurately weigh 2.9 g of NaCl, add water until the volume reaches 500 mL and mix well to dissolve. Normally a 500 mL volumetric flask would be used for this procedure. 123 Written Exercises: Solution calculations 1. Explain how to make the following solutions. (a) 250 mL of a 0.100 M AgNO3 solution. (b) 100 mL of a 0.33 M K2CO3 solution. 2. What is the resulting molarity of the following solutions? (a) 9.20 g of C2H6O in 500 mL of solution. (b) 1.06 g of LiCl in 10.0 L of solution. (c) 1.31 g of Ba(NO3)2 in 250 mL of solution. 3. What mass of solute is contained in each of the following solutions? (a) 250 mL of 0.010 M Na2CrO4 solution. (b) 1000 L of 1.0 x 10-4 M AuCl3 solution. (c) 1.0 mL of 18 M H2SO4. 4. What volume of a 0.200 M solution contains these masses of solute? (a) 8.40 g of NaHCO3 (b) 5.95 g of KBr (c) 1.17 g of NaCl BONUS QUESTION: Carbon dioxide reacts with limewater, Ca(OH)2(aq) as follows: Ca(OH)2(aq) + CO2(g) CaCO3(s) + H2O(l) 100 mL of a 0.10 M Ca(OH)2 solution was used as a test for carbon dioxide. Assuming an excess of CO2 gas is passed through the solution, what is the maximum mass of CaCO3 precipitate which could form? 124 8.12 Solution Preparation and Quantitative Dilution in the Laboratory Written calculation and explanation of solution preparation in the classroom is one thing, however achieving this skill in a laboratory practical sense is another thing altogether. Knowledge of appropriate laboratory glassware, equipment and technique is another thing completely! Quantitative (ie accurate) dilution of an existing solution is another important role of a laboratory chemist. Some methods of chemical analysis can require solutions to be extremely diluted down, sometimes to the extent of being diluted down to one thousandth of their original concentration or less! This cannot be achieved in a single dilution step alone, so the technique of serial dilution is employed. An upcoming experiment will get you to put into practice the theoretical and practical concepts of solution preparation and dilution. So far, the calculation concepts of solution preparation have been covered – so lets look at what quantitative dilution is all about! The process of diluting a solution achieves accuracy through the careful use of quantitative glassware including pipettes and volumetric flasks. The extent of a single dilution can typically be done accurately in the range of 1:2 down to 1:100. The dilution equation relates the mathematical relationship between the original and diluted solution: Concentration 1 x Volume 1 = Concentration 2 x Volume 2 OR: C1.V1 = C2.V2 This equation has a simple message – adding water to an existing solution (C1.V1) gives you a higher volume of solution (V2) at a lower concentration (C2). The amount of solute in the original solution in reality remains the same. A dilution factor is a number multiplier that expresses the magnitude to which the solution has been diluted. For example, if you quantitatively add 50.0 mL of a solution to a 100.0 mL volumetric flask and fill to volume with distilled water, then the new solution is ½ the original concentration. It is therefore twice as dilute, otherwise reported as 2x. Scenario: A student wished to make 100 mL of a potassium chloride solution of 5 g/L. 1. Calculate the mass of KCl required. Hint: 125 Concentration = Mass (g) Volume (L) 2. The student then pipetted 5 mL of this solution into a second 100 mL volumetric flask and made it up to volume. a) What was the dilution factor applied here? b) Calculate the new concentration of the solution. 3. The student was required to dilute the original solution (5 g/L) down to a concentration of 1 g/L a) What dilution factor is required here? b) Briefly outline a means of achieving this dilution (ie; what glassware will you use?) 4. A further 1:20 dilution of the 1 g/L solution in 3. above was performed. a) Briefly outline a means of achieving this dilution. b) Calculate: Written Exercises: 1. i) the new solution concentration. ii) the overall dilution factor applied Dilution Scenario Calculations Calculate the dilution factor involved in the following scenarios: i) 5 mL of solution is diluted to 50 mL ii) 20 mL of solution is diluted to 500 mL iii) 15 mL of solution is diluted to 250 mL 126 2. Calculate the concentration of the diluted solution in the following: i) 25 mL of 2.5 g/100 mL is diluted to 100 mL ii) 15 mL of 2.5 mole/L is diluted to 2.5 L iii) 5 mL of 400 mg/L is diluted to 200 mL 3. Calculate the volume of original solution required to perform the following dilutions: i) How much 2000 mg/L solution is required to make 500 mL of 100 mg/L? ii) How much 25 %v/v solution is required to make 250 mL of 5 %v/v? iii) How much 0.05 M solution is required to make 200 mL of 0.01 M? Ref: DB pp27-29 127 Experiment – Solution Preparation and Serial Dilution A: Sodium Chloride Prework: In the table below, briefly outline a method of solution preparation. Test by Conductivity Solution Method Your Sample Comparison Standard 50 g / L NaCl Next perform a series of 1:10 dilutions from the initial solution. Serial Dilutions Aliquot Volume Volumetric Flask Size Diluted NaCl concentration NaCl Test by Conductivity Your Sample Comparison Standard Dilution 1 Dilution 2 Dilution 3 B: Potassium Permanganate Prework: In the table below, briefly outline a method of solution preparation. Test by Absorbance Prepare a Solution Method Your Sample 1000 mg / L KMnO4 Next perform a series of specified dilutions from the initial solution (over page) 128 Comparison Standard Serial Dilutions Aliquot Volume Volumetric Flask Size Diluted KMnO4 Concentration KMnO4 Test by Absorbance Your Sample Comparison Standard Dilution 1 1:10 Dilution 1:5 Dilution 1:4 Dilution 1:2 C: Hydrochloric Acid Perform a series of 1:10 dilutions on an existing HCl solution. The concentration of the original HCl solution is: ________________ M Serial Dilutions Aliquot Volume Volumetric Flask Size Diluted Concentration Original HCl Solution N/A N/A N/A Dilution 1 Dilution 2 Dilution 3 Dilution 4 129 pH Yours N/A Comparison Standard 9. CHEMICAL ANALYSIS Chemical analysis is a necessary requirement of many industrial processes, particularly in satisfying government regulations. It is necessary for chemists to be able to identify the composition of raw materials, manufactured products and substances which are discharged as wastes. An integral component of a typical chemist’s role is in Quality Control. 9.1 Qualitative Vs Quantitative Chemical Analysis When determining the composition of a substance we need to find out: What chemicals are present? This identification process is called qualitative analysis. Sometimes this is all that is required, but often we also need to know: how much of each chemical is present. This measurement process is called quantitative analysis. In the mercury chloride laboratory task you determined whether your unknown sample contained mercury (I) or mercury (II) chloride. The methods you used involved quantitative analysis since you measured masses of reactants and products in order to analyse your sample. Consider the following examples: A jar of peanut butter has listed as ingredients: peanuts, vegetable oils, sugar salt. This represents a qualitative analysis only. Normally ingredients are listed in order of decreasing mass but no actual quantities have been given. On a packet of pain relief capsules the following ingredients are listed: Paracetamol: 500 mg ; Codeine phosphate: 8 mg This represents a quantitative analysis. Note that there may be other ingredients also which are unlisted, since the manufacturers are only obliged to list active ingredients, not fillers, binders etc. Written Exercise / Research 1. Examine the labels of other containers for examples of qualitative and quantitative analysis. List three further examples of each. 2. Care must be taken with some ingredients to make sure that their concentrations stay within set limits. For example, the allowable pesticide residue limits in foods is carefully controlled. Find another example and find out how this product is analysed. What methods can we use in chemical analysis? Chemists employ many methods in order to analyse the millions of different substances recognised today. We will look at some examples from which you may be able to choose for your major investigation in this module - some involve simple laboratory apparatus 130 while others will involve sophisticated instruments. You will need to adapt any procedure given to make use of the resources available to you. Consult a chemistry teacher once you have made your choice. You should practice as many techniques as possible. Consult references to find out about other techniques. 9.2 Gravimetric Analysis Gravimetric analysis involves the separation and weighing of a substance obtained from a sample. This form of analysis separates the substance of interest from the sample based on its physical or chemical properties. The concentration of separated substance is calculated and reported as a percentage by mass. Often there are three main steps to this procedure: 1. Obtain and weigh sample mass 2. Employ means of physical or chemical separation of the analyte 3. Reweigh sample (or separated analyte component) Gravimetric analysis in its most simple form could be the analysis of the moisture content of say, a food substance like cereal. A suitable sample container needs to be chosen as it must not change mass during the analysis. An accurately weighed mass of cereal is placed into the container and heated in air oven at around 120 oC for a minimum time period to evaporate the entire moisture content. The container is removed, cooled and reweighed. The difference in mass (confidently assuming it became lighter !) is calculated and expressed as a percentage of the sample mass. All calculations in moisture determinations are purely mass based, but there are further examples of gravimetric analysis involving mole based calculations . . . If we were to determine the salt content of a food sample we must first dissolve the food in water. Any undissolved substance will need to be removed using a separation method such as filtration. The food solution can be analysed for salt, as NaCl by adding an excess of AgNO3 solution. This precipitates out the entire chloride ion, Cl- (aq) content as AgCl (s) according to the following chemical equation: and then filtering and weighing the precipitate which forms: Ag+(aq) + Cl-(aq) AgCl(s) The precipitate is separated by filtration. After the precipitate is dried, it is then weighed and its mass recorded. The number of moles of AgCl formed can be calculated. From the equation, this is the same as the number of moles of Cl- in the food solution. This is therefore equal to the number of moles of salt, NaCl in the food sample analysed. We can convert this number of moles to a mass and express it as a percentage of the mass of the original sample. 131 Example: A 25.0 mL sample of soup was analysed for its salt content, as NaCl. The sample was filtered to remove any suspended solids. An excess of Silver nitrate solution was added to the filtrate to precipitate out the available salt component. After standing for a few minutes a further few drops of silver nitrate were added to make sure that precipitation was complete. The precipitate was collected by filtration, then dried and weighed. A mass of 0.23g of precipitate was obtained. Find the percentage (g / 100 mL) of salt in the soup. No of moles of AgCl produced Chemical equation: + (aq) Ag + From the equation above, moles Cl- = mass of AgCl FW = 0.23 143.4 = 1.60 x 10-3 - Cl (aq) AgCl(s) = moles of AgCl = 1.60 x 10-3 We are assuming that all the chloride ions present came from dissolved salt - NaCl. This is a reasonable assumption in most food samples. Therefore, moles of NaCl in the soup sample = mass of NaCl in the soup sample = % of NaCl in soup sample 1.60 x 10-3 moles x FW = 1.60 x 10-3 x 58.45 = 0.0938 = 0.0938 x 100 25.0 = 0.38 % w/v You should note the following important points: solid samples need to be dissolved so that the substance to be detected is in solution. Any undissolved material should be removed by filtration. the precipitate formed needs to be a highly insoluble substance, so that the proportion of soluble ions remaining in solution for analysis is negligible. Check the table of solubility rules, under the chapter “properties of compounds” – your teacher will assist you with this! the filtered precipitate needs to be washed carefully with small volumes of water to remove other soluble substances present in the solution. the precipitate should be dried in an oven at 110 oC until constant mass is obtained and allowed to cool before weighing. 132 Practice Questions: 1. A student analysed a popular breakfast cereal for its moisture content . The standard procedure was followed and results were listed in the student's logbook. Determine the moisture content: Empty container mass: 25.3403g Mass container + cereal: 29.3489g Mass container + cereal (2hr @ 110 oC): 29.2897g 2. The Limestone, CaCO3 content of a rock sample was analysed by dissolving 8.3917g of rock into a solution of HCl. The CO2 evolved was collected in a special sodaasbestos tube. It gained in mass by 2.4918g. Calculate the %w/w of limestone in the sample. BONUS QUESTION 3. Aluminium can be analysed by precipitation with the organic compound 8-hydroxyquinoline, C9H6NO. A 0.3271g of a bauxite ore was analysed for its aluminium content. A mass of 1.4827g of the aluminium precipitate, Al(C9H6NO)3 was recovered. Determine the percentage of Aluminium in the ore. 133 Experiment – Sulfate Content of Plant Fertilizer Background: Sulfate is an important nutrient for plant growth. It is added to many fertilizer materials, present in soluble forms, most notably as Ammonium Sulfate. This method aims to dissolve the sulfate component of a fertilizer sample into a slightly acidified aqueous solution. The available Sulfate, SO4 2- (aq) will be precipitated, filtered and weighed as BaSO4 (s) by addition of a soluble Barium Chloride solution, BaCl2 (aq) Method: 1. Obtain a 600 mL beaker, place onto a 3 decimal place balance and tare. 2. Accurately weigh approximately 0.75g of fertiliser and record this mass into the results table below. 3. Dissolve the weighed material in around 100 mL of distilled water. 4. Add around 5 mL of the supplied 2M HCl and heat to boiling on a hotplate. 5. Remove from heat. Filter solution through #4 (fast) filter paper. Ensure quantitative transfer and rinse final residue in filter paper with distilled water. 6. Place filtered solution onto a hotplate and heat until just boiling. 7. Remove again from heat and add 25 mL of 5% BaCl2. 8. Cover beaker with a watchglass and transfer solution onto a hotplate (! low heat). 9. Add a further 2 mL of the BaCl2 solution after 10 min. 10. Continue to digest your solution on low heat for a further 20 min. 11. Obtain and record the mass of a sintered glass crucible in the results table. 12. Remove from hotplate and cool in an ice bath for 10 min. 13. Set up vacuum filtration apparatus and filter your solution through the pre-weighed sintered glass crucible. Wash remaining residue with a small volume of distilled water. 14. Place into an air oven to dry and leave until next class to weigh back. 15. Remove from oven. Allow 10min to cool and reweigh. Results: Mass (g): Fertiliser Mass (g): Sintered Glass crucible Mass (g): Sintered Glass Crucible + BaSO4 (s) Mass (g): BaSO4 (s) 134 Calculations: Moles BaSO4 (s) Balanced Chemical Equation (precipitation process): Reaction Ratio BaSO4 : SO4 2- = _________ : __________ Moles SO4 2Mass SO4 2- % SO4 2- % SO4 2- = 2- Mass SO4 Sample Mass x 100 Questions: 1 a) Why is it important to initially filter and remove any undissolved fertilizer in step 5 ? 1 b) Why would this undesirable solid residue be washed with distilled water? 2) Explain the purpose of cooling the precipitate prior to filtering in step 12. 3) Explain the importance of washing the solid BaSO4 compound with distilled water prior to drying in step 13 4) Why was the measured volume of distilled water (ie 100 mL) instructed as approximate and not accurate? 135 9.3 Volumetric Analysis Unknown concentrations of soluble chemicals in aqueous solution can be determined by a technique referred to as Volumetric Analysis, or Titration. A carefully measured volume of the unknown solution is transferred to a conical flask by pipette. This volume is referred to as an aliquot. A second chemical solution is chosen that is known to chemically react with the unknown solution in a predictable manner, via a balanced chemical reaction. This solution is must have a known concentration as it is referred to as a standard solution. A burette is filled with the standard solution where it is then slowly delivered into the aliquot of unknown solution, allowing the chemical reaction to progressively “consume” the chemical analyte within the unknown solution aliquot. This method of analysis continues to deliver standard solution into the unknown solution aliquot (in the conical flask) until all of the chemical analyte in the unknown solution is consumed. At this point the chemical analysis is complete, and addition of further solution is ceased. The volume of standard solution delivered to the unknown solution by burette is carefully measured and recorded. This is referred to as an endpoint volume. Diagram of basic Titration apparatus: Bulb Pipette Apparatus The fundamental basis of all volumetric analyses is the chemical reaction. It is essential that a balanced chemical equation be expressed as part of the calculation procedure. Generic Equation: HX(aq) + 1 YOH(aq) : XY (aq) + H2O (l) 1 In this example the reaction ratio, and hence mole ratio is 1:1, however different chemical reactants may have mole ratios like 1:2, 1:3, 3:2 and so on. This ratio is an integral part of the calculation procedure in determining the unknown solution concentration. 136 Volumetric analysis (titration) is commonly employed to determine the concentrations of dilute acids and bases. As covered previously, acids and bases react together predictably in what is referred to as a neutralisation reaction, ie; Acid + Base Salt + Water Avoid potential contamination – rinse your glassware correctly prior to use: Burettes require a small sacrificial rinse with the solution it will be filled with. Pipettes require a small sacrificial rinse with the solution it will deliver. Conical flasks require a rinse with distilled water only. Volumetric Flasks (used sometimes for titration) require a rinse with distilled water only. The general steps involved in a titration, in summary are: A measured pipette volume (aliquot) of the unknown solution is added to a conical flask. A few drops of a pH sensitive indicator are normally added to the conical flask to produce a noticeable colour change at endpoint. The burette is filled with the standard solution and volume adjusted to the top volume reading of 0 mL. The standard solution is carefully delivered to the flask. As the colour change becomes more apparent, the delivery rate from the burette is slowed from a steady stream to smaller “squirts” and ultimately dropwise until a permanent colour change is observed. Cease delivering standard solution from burette. Record burette volume. This is the volume of standard solution that was required to consume the last of the analyte in the standard solution. There are 4 measurable quantities recorded in a single titration analysis. 3 of them are known quantities and 1 is unknown initially. Place a tick in the boxes below which correspond with known quantities. Place a cross in the box corresponding with the value to be calculated (unknown). Burette Conical flask Standard solution: Unknown solution aliquot Volume Volume Concentration Concentration 137 The titration technique requires considerable practice in order to obtain reliable results. Volumetric analysis is commonly performed in triplicate. It is desirable that your 3 titration volume results are within 0.2 mL 0.3 mL. This can be difficult to achieve initially! Titrations are used in many situations such as when we want to find the proportion of particular pollutants present in industrial waste water, or the proportion of vitamin C in commercially produced fruit juices. There are a variety of other chemical reaction types that can be used for the technique of volumetric analysis including: Reduction – Oxidation (Redox: Covered in TPC Chemistry B) Complexometric , and Precipitation (TAFE: Cert IV / Diploma in Laboratory Techniques) The following examples use acid / base reactions although many other types of chemical reactions can be used in volumetric analysis. Example 1 A 20.0 mL aliquot of hydrochloric acid, HCl was titrated to endpoint with 15.5 mL of a standardised 0.112 M solution of sodium carbonate, Na2CO3. Determine the concentration of the HCl. moles of Na2CO3 titrated = Concentration x Volume (L) = 0.112 = Equation: Mole ratio: 2HCl(aq) 2 + : 1.74 x 10 x 0.0155 -3 Na2CO3(aq) 2NaCl(aq) + H2O(l) + CO2(g) 1 The mole ratio above indicates there will be twice as many moles of HCl than there will be of Na2CO3. Therefore, moles of HCl reacted Concentration of HCl 138 = 2 x 1.74 x 10-3 = 3.47 x 10-3 = moles HCl Volume (L) = 3.47 x 10 0.02 L = 0.174 M -3 (ie; moles / L) Example 2 How much ammonia (a base) is in ‘Cloudy Ammonia’ ? A 25.0 mL aliquot of ‘Cloudy Ammonia’ was diluted to 250.0 mL in a volumetric flask. A 25.0 mL aliquot of this diluted solution was transferred by pipette into a conical flask. A few drops of methyl orange indicator were added (a yellow solution results). A standardised 0.099 M HCl solution was delivered by burette with constant swirling of the flask. The titration was completed when the colour of the solution just changed to a permanent tinge of apricot-pink. The volume of HCl added was noted. The procedure was repeated three times until consistent results were obtained. Results: Titration No Volume HCl added (mL) 1 28.90 2 27.60 3 27.60 4 27.65 Titration 1 result is ignored as it is not consistent with the following titrations. The results from titrations 2, 3 and 4 are used to find an average titre. Average titre = 27.60 + 27.60 + 27.65 3 moles of HCl titrated Equation: Mole ratio: Therefore: HCl(aq) + 1 : = = = = 27.62 mL concentration x volume (L) 0.0991 x 0.02762 L -3 2.74 x 10 NH3(aq) NH4Cl(aq) 1 moles of NH3 = 2.74 x 10-3 (since reaction ratio is 1:1) These 2.74 x 10-3 moles of NH3 are contained in the 25.0 mL titrated. Concentration of NH3 in the diluted solution = moles NH3 Volume (L) = 2.74 x 10-3 0.025 L = 0.110 M ** ** The original sample had been diluted by 10x in the 250 mL volumetric flask. Therefore, the concentration of NH3 in the ‘Cloudy Ammonia’ = 0.110 M x 10 = 1.10 M 139 TITRATION CALCULATION STEPS Remember - a titration is always a chemical reaction between 2 reacting solutions: A Solution of KNOWN concentration (Standard Reagent) or quantity (Primary Standard) is fully titrated with your solution of UNKNOWN concentration. REMEMBER MOLE EQUATIONS: Equation 1. Moles = Mass (g) FW Equation 2. Molarity = Moles Volume (L) 1. You must firstly read the scenario carefully and identify your KNOWN reagent and UNKNOWN reagent. 2. Calculate moles of Known Reagent 3. Write down a BALANCED CHEMICAL EQUATION between the 2 reagents: Acid + Base Salt + H2O Acid + Carbonate Salt + CO2 + H2O 4. Determine and write down the REACTION RATIO between your Known and Unknown reagent. 5. Calculate MOLES of your Unknown Reagent: Multiply or Divide your Moles of KNOWN reagent (step 1) by the Reaction Ratio (step 3). 6. Calculate the MOLARITY of your Unknown Reagent: MOLARITY = Moles Unknown Reagent Volume (L) Unknown Reagent 140 Practice Questions: 1. Determine the concentration of the unknown solution given the following data: a) 25.0 mL of NaOH is titrated to endpoint with 18.3 mL of 0.1050M HCl. b) 10.0 mL of H2SO4 is titrated to endpoint with 24.8 mL of 0.0978M NaOH. 2. A solution of approximately 0.1M HCl was standardised via sodium carbonate, Na2CO3 as the primary standard. A mass of 0.1472 g of the primary standard was weighed and required 23.7 mL of the HCl to reach endpoint. Calculate its concentration. 3. A sample of vinegar, thought to contain around 4% Ethanoic Acid, CH3COOH was analysed by titration with a standardised 0.1082M NaOH solution. A 10 mL aliquot of the vinegar was diluted to 100 mL in a volumetric flask. Triplicate 25 mL aliquots of this diluted solution were titrated to endpoint by an average of 18.3 mL of the standardised NaOH. Determine the Ethanoic Acid content (%w/v) of the vinegar. 141 Experiment 1 – PRACTICE TITRATION AND HCl ANALYSIS Part A: Indicator Colour Record colours of the following indicators as they apply to a typical acid – base titration. Screened Methyl Orange Methyl Orange Phenolphthalein Acid (HCl) Base (NaOH) Endpoint Part B: Practice Titration ! Be sure to rinse all glassware appropriately prior to use . . . Obtain triplicate 25.0 mL volumes of HCl (0.1M) by pipette and place into 3 conical flasks. Add 3-5 drops of phenolphthalein indicator. Fill your burette with with NaOH (0.1M) to the 0mL mark. (don’t forget sacrificial rinsings where required….) Titrate your solutions to endpoint placing emphasis on the precision of your results. Aim to titrate each set of 3 analyses within 0.2 mL 0.3 mL. Repeat titration steps above using screened methyl orange indicator and if time, methyl orange indicator. RESULTS NaOH concentration: ________________ M NaOH Titration Volume (mL) HCl Aliquot Volume (mL) 1. Phenolphthalein 2. Screened MO 25.0 25.0 25.0 AVERAGE 142 3. MO Part C: HCl Analysis – A Titration Calculation Choose the best set of triplicate titration results from Part B above. This would be the indicator which achieved the smallest titration volume range. Use the average titration value of this indicator to calculate the molarity of the hydrochloric acid, following steps 1 – 5 below: Calculations: 1. Moles NaOH Titrated: = Molarity x Volume (L) 2. Chemical Equation: HCl + NaOH _______________ + _______________ 3. Reaction Ratio: ____ : ____ 4. Moles HCl = 5. Molarity HCl: = Moles Volume(L) QUESTIONS 1. Summarise the rinsing procedure(s) for the following pieces of laboratory equipment: Burette Conical Flask Pipette 2. How is endpoint different to equivalence point when performing volumetric analysis (ie titration)? 143 Experiment 2 – HCl STANDARDISATION Aim: To standardise a solution of hydrochloric acid using the primary standard Na2CO3. Prework: Calculate the mass of Sodium Carbonate, Na2CO3 required to react with 25 mL of an approximately 0.1M HCl solution. Method: HCl Standardisation Weigh accurately, approximately the mass of Na2CO3 calculated above into 3 conical flasks. ! Label 1 – 3. Dissolve in a small volume of water. Add 5 drops of the methyl orange indicator. Fill your burette with the HCl solution to be standardised. Titrate your standard Na2CO3 solution with the HCl until endpoint is reached. Titrate your standard Na2CO3 solution with the HCl until endpoint is reached. Record this volume and repeat for flasks 2 and 3 Results: Mass Na2CO3 (g) HCl Volume (mL) Calculations: Perform separately for each titration result. i) Calculate moles of sodium carbonate ii) Write a balanced chemical reaction, and hence determine the reaction ratio. iii) Calculate moles of HCl iv) Calculate the molarity of HCl Questions: 1. Why are titration volumes between 15 – 30 mL the most desirable for titrations? 2. Why is it OK to average the titration volumes performed on triplicate analyses of solution aliquots, but NOT OK for 3 separate masses of Sodium Carbonate carefully weighed and dissolved within 3 separate conical flasks? 144 Experiment 3 – NaOH STANDARDISATION Aim: To standardise a solution of sodium hydroxide using the primary standard KC8H5O4 (Potassium Hydrogen Phthalate, KHP) Prework: Calculate the mass of Potassium Hydrogen Phthalate, KC8H5O4 required to react with 25 mL of an approximately 0.1M NaOH solution. Method: Weigh accurately, approximately the mass of KC8H5O4 calculated above into 3 conical flasks. ! Label 1 – 3. Dissolve in a small volume of water. Add 5 drops of the phenolphthalein indicator. Fill your burette with the NaOH solution to be standardised. Titrate your standard KC8H5O4 solution with the NaOH until endpoint is reached. Record this volume and repeat for flasks 2 and 3 Results: Mass KC8H5O4 (g) NaOH Volume (mL) Calculations: Perform separately for each titration result. i) Calculate moles of KHP ii) Write a balanced chemical reaction, and hence determine the reaction ratio. iii) Calculate moles of NaOH iv) Calculate the molarity of NaOH 145 Experiment 4 – VINEGAR ANALYSIS Aim: To determine the concentration of Ethanoic Acid, CH3COOH in Vinegar. Method: 1. Perform a 10x dilution on a vinegar sample by transferring a 25.0 mL aliquot of vinegar sample into a 250.0 mL volumetric flask and making up to volume. 2. Transfer 25.0 mL aliquots of the diluted vinegar in triplicate by pipette into 3 conical flasks. 3. Add 3 – 5 drops of phenolphthalein indicator to each flask and titrate with standardised sodium hydroxide, NaOH reagent. Results Table: NaOH = ____________________ M Aliquot volume (mL) Volume NaOH (mL) 25.0 25.0 25.0 Volume NaOH (Average) Calculate the concentration of the vinegar in %w/v…. 1. Moles NaOH titrated = 2. Chemical Reaction (and hence reaction ratio): 3. Moles CH3COOH present in 25.0 mL (diluted) aliquot = 4. Molarity CH3COOH (diluted) = 5. Dilution Factor = 6. Molarity CH3COOH (undiluted solution) = 7. % w/v, (ie g/100 mL vinegar) = …………….. i) Moles / L = ii) g / L = iii) g /100 mL = 146 Experiment 5 – MISCELLANEOUS TITRATION: RESULTS PROFORMA 1. Sample ID: 4. Total sample volume prepared: _________________________________ ________________________________ 2. Analyte ID: 5. Titrant ID: _________________________________ ________________________________ 3. Sample amount taken: 6. Titrant molarity: _________________________________ ________________________________ Aliquot (mL) Titrant Volume (mL) 1. 2. 3. Average Titrant Volume 1. Calculation of Analyte Concentration (M) i) Moles of titrant: ii) Balanced chemical reaction and reaction ratio: iii) Moles of analyte in sample solution: Calculations continued over page 147 iv) Molarity of sample solution: v) Dilution factor (if applicable) 2. Report your analyte concentration as a percentage, ie; Liquid samples as % w/v, (ie g/100 mL) i) Moles / L = ii) g/L= iii) g /100 mL = _______________ OR Solid samples % w/w, (ie g/100 g) i) Moles / L = ii) g/L= iii) g / total sample volume (volumetric flask) iv) g / sample amount taken v) % w/w = ________________ 148 9.4 Chromatography Chromatography is a procedure for separating mixtures containing several dissolved substances. Each dissolved substance has a difference in attraction for the solvent, referred to as the mobile phase, versus the blotting paper, referred to as the stationary phase. By comparing the distance moved by one substance in the mixture with that of a known substance, we can identify the presence of that substance in the mixture. You can separate some food colourings or inks using this method. Concentrated dots of colour are placed along one edge of a piece of blotting or other absorbent paper as shown. In this example, three coloured inks A, B and C and a known substance S are spotted. The paper is suspended in a suitable solvent in a covered glass container for several hours. At the beginning ` cover covered glass container blotting paper spots are small and concentrated pencil line marking origin solvent level is below the spots on the paper A B C S Known reference substance (S) can be spotted next to the substances to be tested. After several hours ---------------- solvent front distance solvent front has moved known reference substance (S) A B C S 149 Substance A contains substance S and at least one other substance (since this solvent may not have separated all the substances present). B and C do not contain S because there are no spots at this height on the chromatogram. The components of the inks can be identified by the distance they travel through the paper compared to the distance that the solvent front has travelled. This is expressed as an R f value for each component. Rf = distance component spot has moved from the origin distance solvent front has moved from the origin solvent front component P component Q Distance moved by Q Distance moved by P Distance moved by solvent = = = 5.2 cm 11.8 cm 15.0 cm Rf for component P = 11.8 15.0 = 0.79 Rf for component Q = 5.2 = 0.35 The solvent that is best for a particular separation is found by experimentation. You might try various mixtures of water, alcohol (colourless eg vodka, white rum), white vinegar or ammonia solutions. Other absorbent materials such cotton or even a stick of chalk can be used for the stationary phase of the chromatogram. Thin layer chromatography makes use of a fine layer of silica or aluminium oxide on a glass plate. The absorbent material is packed in to a glass column for column chromatography and a solvent is dripped through after the sample is placed on top. Practical Activity / research: 1. 2. 3. Chromatography Use chromatography to separate some pigments of your choice. In your report comment on the effectiveness of your procedure and suggest improvements. Research instrumental methods such as high performance liquid chromatography and gas-liquid chromatography. Observe demonstrations of these methods if possible. Chromatography is used extensively in industry. Use some modern sources to investigate one application. Ref: CC1 pp172-174 C2 pp13-14 CC2 pp25-29; 58-61 150 9.5 Electrophoresis Anything which conducts an electric current must contain mobile electric charge carriers. Conduction can arise in two different ways: by movement of electrons or by movement of ions. In metal wires, only electrons move through the rigid crystal lattice but this is not the case in solutions where ions move. Solutions which contain ions and therefore conduct electricity are called electrolytic solutions while any substance which produces ions in a solution is an electrolyte. Electrolytic solutions conduct because they contain charge-carriers (ions) which are absent from solutions of non-electrolytes and virtually absent from water itself. The electrically charged ions in the compound sodium chloride, Na+ & Cl- will conduct an electric current if they are free to move. The ions can move if the compound is molten (liquid) or dissolved in water. Electrophoresis uses this movement of ions to separate them in a process which is, in some ways is similar to chromatography. The positive ions are attracted to the wire and metal clip attached to the negative terminal of the electricity source. The negative ions are attracted to the oppositely charged positive electrode. Smaller ions can move faster than larger ones and so a separation and identification can take place. If the ions are not coloured they can be made visible by spraying with a suitable dye or other means. Diagram 151 9.6 Colorimetry The intensity of a coloured solution is directly related to its concentration. The more intense the colour, the more concentrated the solution. A copper(II) sulfate solution of unknown concentration is mid-blue. We can estimate its concentration by comparing its colour intensity with other copper sulfate solutions of known concentrations. This analysis technique is called colorimetry. The colour intensity of solutions can be more reliably and accurately compared using an instrument called a colorimeter than using our eyes. The instrument measures the intensity of light which passes through the solution of unknown concentration. A series of measurements is first carried out on the standard solutions and a calibration graph is drawn. The concentration of the unknown solution can then be read from the graph. Colourless substances need to be converted first to coloured compounds before simple colorimetry can be used. Phosphates are converted to a blue complex by reacting them with ammonium molybdate. In this way colorimetry can be used to calculate the amount of phosphates in detergent samples. Example: The concentration of a solution of dye orange-X needed to be determined. Standard solutions of orange-X were made up using distilled water to the concentrations shown in the table. A blank (plain distilled water) was measured for absorbance in a colorimeter as was each of the standards. Solution concentration (mg / L) Absorption 0 (blank) 0.00 0.10 0.16 0.20 0.30 0.30 0.45 0.40 0.61 0.50 0.76 Unknown 0.37 A calibration graph has been plotted. Your task is to interpolate from the graph a corresponding for the unknown solution. See over page 152 Orange - X Analysis by Colorimetry 0.8 0.7 Absorbance 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 Concentration of Orange - X (mg / L) From the calibration graph the concentration of the orange-X solution was determined to be ................ mg / L. Ref: CC2 pp38-39 C2 pp49-50 153 0.6 Experiment – COLORIMETRY Aim: To analyse the relationship between colour intensity and concentration of an analyte in aqueous solution. ! Graph Paper required as part of this exercise Method: 1. Devise a simple method for the preparation of a series of standard copper sulfate, CuSO4 solutions from a 1.0 M stock solution: 2. Analyse standard solutions through the colorimeter. Record each standard absorbance value in the results table. 3. Record the absorbance of your unknown(s). 4. Make a calibration graph from your CuSO4 standards 5. Determine the concentration of your unknown solutions. Results: Standard Solution Method of Preparation Absorbance 0M 0.2 M 0.4 M 0.6 M 0.8 M Unknown Solution Unknown Solution Concentration: _________________________ Questions: 1. Consider the accuracy of this method. Suggest ways in which it could be improved. 2. Discuss the use of a calibration graph as a mathematical model in determining the concentration of your unknown. 154 9.7 Spectroscopy Chemicals are capable of emitting or absorbing energy in the form of visible light, ultraviolet and infrared radiation. By measuring the amount and types of energy absorbed or emitted we have a method of identifying the chemical (qualitative analysis) and in some cases determining the amount present (quantitative analysis). Visible and U.V. spectra An electron cannot move in an orbit of just any radius - there are in fact very few orbits whose radius values are “legal”, and an electron must be in one or another of these. If an electron in a “legal” orbit near the nucleus absorbs enough energy it can move to an orbit which is further away. The electron is now said to be in an “excited” state and will usually fall back fairly promptly into an empty “legal” orbit of lower energy closer to the nucleus. When this happens the energy the electron no longer carries is released as a pulse of radiant energy such as light. This process can only be studied properly in a collection of isolated atoms, for in polyatomic molecules or in condensed phases (solids or liquids) adjacent atoms interfere with the free movement of outer electrons. Ways of obtaining isolated atoms include: 1) 2) vaporising a substance at high temperatures (eg in a bunsen flame); subjecting a gas at low pressure to an electric discharge (eg in a fluorescent tube). Both processes result in a collection of isolated gaseous atoms in excited states which will release light or other radiant energies as they revert to less energetic states. Common applications of the principle are the red neon advertising signs and both mercury (blue) and sodium (orange-yellow) streetlights. Each element produces energies of a characteristic pattern which makes this phenomenon a powerful tool in chemical analysis. The instrument for examining the wavelengths of these energies is a spectroscope. An incandescent liquid (eg molten iron) or solid (eg the tungsten filament of an electric light globe) when viewed through a spectroscope shows a “continuous” spectrum, ie; a rainbow of colours with no gaps, ranging from red to violet. Incandescent gaseous atoms however produce light at only certain sharply-defined positions in the spectrum (only at specific wavelengths), with darkness in between. Through the spectroscope discrete bright coloured lines appear across a dark background. It is the number and positions (colours) of these lines which constitute the unique “fingerprint” of each element called its emission spectrum. Continuous spectrum 155 Sodium spectrum Hydrogen spectrum It is also possible to analyse which frequencies of light from the continuous spectrum are absorbed by a particular element - an absorption spectrum. The spectrum appears as the continuous spectrum with black lines across it corresponding to those frequencies which would be emitted by the same element (ie. a reverse of the emission spectrum). Flame Emission Spectrophotometry Many metal compounds at the high temperature of a bunsen flame or similar produce excited metal atoms whose emission spectra are analytically useful. The technique is very simple; the only apparatus required is: · a burner which can be adjusted to give an almost non-luminous flame · a platinum wire holder or · a plastic squeeze bottle with a very fine nozzle from which a fine spray of a solution of the compound may be delivered. Either method will result in a coloured flame characteristic of a particular element. Not all metals emit energy in the visible light region of the electromagnetic spectrum. Magnesium, for example, does not. However for those that do, the colour may be used to identify the metal either directly by observation or by spectroscopic analysis. Non-metals do not emit in this way. Each metal gives off its own unique spectrum even if other metals are present in the specimen as well. The spectrum produced will contain all the lines from each individual metal’s spectrum. Potassium alone Sodium alone If the colour of one metal masks that of another because of overwhelming brightness, spectroscopic examination will still reveal the presence of the fainter one. Sodium, for example, generally outshines all others, and a mixed sodium-potassium specimen will give a flame colour which looks yellow like the sodium alone to the naked eye. 156 Spectroscopically however the flame appears like this: Mixture (sodium-potassium) Flame spectroscopy is the most convenient way of detecting the various “alkali metals” (Group I) in their compounds. They have very similar properties and often occur naturally together. Written exercises: Visible spectra 1. Why does salty water boiling over in a saucepan turn a gas flame yellow? 2. How can hydrogen with only one electron produce multiple lines in its spectrum? Why are some lines brighter than others? 3. The following spectra are provided so that you can answer the questions below. (a) (b) Identify the metallic elements in the unknown specimen. Which alkali metal is most abundant in the atmosphere of the sun? Experiment 1 – FLAME TESTS The colour of light emitted when different metal ions are sprayed into a flame are characteristic for each metal element, as it appears on the Periodic Table. PART A - METAL ION ID Metal Ion Colour 157 PART B - UNKNOWN METAL ION IDENTIFICATION Try matching an element to the colour of the unknown metal ion upon being aspirated into the flame. Unknown Metal Ion ID Colour Metal Ion PART C - COLOUR INTENSITY AND METAL ION CONCENTRATION 1. Prepare a solution containing an approximately 10% concentration of Sodium Chloride. 2. Aspirate this through the flame. 3. Dilute 10 mL of this solution up to 100 mL with distilled water. 4. Aspirate the diluted solution through the flame. 5. Repeat steps 1 – 4 another 5 times. Summarise your observations of this task below. Ref: CC1 p27 (flame tests) C2 pp10-13 CC2 pp29; 383 158 Analysis of Sodium and Potassium by Flame Emission Spectrophotometry Potassium (K+) and Sodium (Na+) are soluble metal ions commonly found in natural water bodies. The ocean is a salt water environment. This salt is composed predominantly of table salt, known chemically as Sodium Chloride (NaCl). Sodium ions (Na+) exist in concentrations of around 1.2%, or 1.2 g/100 mL Potassium (K+) ions exist at lower concentration - around 0.4%, or 0.4 g/100mL We have seen in previous lessons how various metal ions release different colours of light when placed into a hot gas flame. This is referred to as Light Emission. A relationship exists between the concentration of soluble metal ion and colour intensity. As concentration increases, flame colour emitted becomes brighter, or more INTENSE. Flame colour intensity therefore relates directly to metal ion concentration. An instrument called a FLAME EMISSION SPECTROPHOTOMETER is designed to aspirate (or spray) small volumes of solution into a gas flame. The instrument has a detector designed to measure the intensity of coloured light being emitted by the metal ion. If we can prepare a series of different known concentrations of a given metal ion (eg Sodium or Potassium), we will see that the colour intensity of light emitted changes according to the relationship described above. If we make a graph of metal ion concentration versus colour intensity (vertical axis), we will see on graph paper that colour intensity increases as soluble metal ion concentration increases (see graph). An unknown concentration of the same metal ion will give a particular flame intensity when aspirated through the flame photometer. We can use our graph to convert this intensity to a concentration. Flame Emission Spectrophotometers are designed to analyse for the elements Sodium (Na) and Potassium (K). The instrument will accurately analyse concentrations of these elements between 1mg/L and 100 mg/L. Outside of these concentrations the light emitted is too weak or too strong to analyse accurately. Samples which are too concentrated must be accurately diluted until the Sodium / or Potassium concentration is lowered back within this range. 159 Graph of Metal Ion Concentration versus Colour Intensity Colour Intensity 120 100 80 60 40 20 0 0 20 40 60 80 100 Concentration (mg / L) Written Exercises: 1. What is the chemical name of the salt found predominantly in sea water? 2. Name 2 soluble metal ions commonly found in sea water. 3. What do the following symbols stand for: Na+ ______________________ mL: K+ ______________________ g/100mL: ___________________________ g: ______________________ g/L: __________________________ mg: ______________________ mg/L: __________________________ L: ______________________ __________________________ 4. What is the name of the instrument used to analyse for potassium and sodium ion concentration? 160 5. A small volume of solution is aspirated into the photometer flame to measure analyte concentration. What is meant by the term "aspirate"? 6. When metal ions release different colours of light in a gas flame, this referred to as Fl_________________ E___________________. 7. What is the purpose of the detector within the photometer machine? 8. According to the Flame Emission principle, what is light intensity dependant on? 9. How might we be able to still analyse a sample for both Na+ and K+ when their concentrations are too high to be measured? 161 Experiment 2 – INTRODUCTION TO FLAME EMISSION Background: We can see the relationship between metal ion concentration and colour intensity through the method of Flame Emission. A series of known concentrations (or standards) of the sodium metal ion will be aspirated through a Flame Emission Spectrophotometer for this purpose. This experiment aims to develop the following skills: 1. Preparation of standard solutions through serial dilution. 2. Tabulating and graphing results. 3. Determining the concentration of an unknown solution. Method: - Complete the information below summarising how you will prepare your standard solutions. Run your prepared standards and unknown sample through the spectrophotometer. Record the intensity readings of each solution in the table at right of page. Results: Sodium (Na+) ion Stock Solution: ___________________ mg/L Intermediate solution ___________________ mg/L Preparation: __________ mL (Pipette) __________ mL (Volumetric flask) STANDARD SOLUTION PREPARATION Intermediate (mL) - Pipette Total Volume - Volumetric Flask Standard 1 0 mg/L Standard 2 20 mg/L Standard 3 40 mg/L Standard 4 60 mg/L Unknown 162 ANALYSIS: FLAME EMISSION SPECTROPHOTOMETER Sodium (Na+) Intensity Analysis: 1. Construct a calibration graph of Concentration (axis ) versus + Colour Intensity (axis ) for your Sodium (Na ) standard solutions. PASTE CALIBRATION GRAPH HERE 2. From your graph, describe how the colour intensity changes with sodium concentration. 3. Use your graph to calculate the concentration of Na+ in the unknown solution. 163 Experiment 3 – SODIUM AND POTASSIUM IN BEER Background: We have seen there is a relationship between metal ion concentration and colour intensity through the method of Flame Emission. The concentration of Sodium (Na+) and Potassium (K+) ions in beer can be determined via a similar method. Method: - Complete the information below summarising how you will prepare your standard Na + and K+ solutions. De-gas a beer sample by pouring from one beaker to another several times. Prepare and aspirate your prepared standards through the spectrophotometer and record your results in the table below. Quantitatively dilute beer sample until Na+ / K+ intensities fall within the range of standard concentrations. Aspirate diluted beer sample and record this diluted Na + / K+ intensity. Draw a calibration graph for both Na+ and K+ + + Calculate the K and Na in your original beer sample. Results: PART A: Sodium Na+ Stock Solution: ___________________ mg/L STANDARD SOLUTION PREPARATION Na+ Stock Na+ (mL) - Pipette Total Volume - Volumetric Flask Standard 1 0 mg/L Standard 2 10 mg/L Standard 3 20 mg/L Standard 4 30 mg/L DILUTED BEER 164 ANALYSIS: FLAME EMISSION SPECTROPHOTOMETER Sodium (Na+) Intensity Dilution Factor - Beer: Beer Volume PIPETTE = ____________ mL PART B: Total Volume VOL FLASK = ____________ mL DILUTION FACTOR x __________ Potassium K+ ion Stock Solution: ___________________ mg/L STANDARD SOLUTION PREPARATION + K Stock K+ (mL) - Pipette Total Volume - Volumetric Flask ANALYSIS: FLAME EMISSION SPECTROPHOTOMETER Potassium (K+) Intensity Standard 1 0 mg/L Standard 2 20 mg/L Standard 3 40 mg/L Standard 4 60 mg/L DILUTED BEER Dilution Factor - Beer: Beer Volume PIPETTE = ____________ mL Total Volume VOL FLASK = ____________ mL Analysis - Over Page 165 DILUTION FACTOR x __________ Analysis: PART A - Sodium Construct a calibration graph of Concentration (axis ) versus Colour Intensity (axis ) for your Sodium (Na+) standard solutions. PASTE CALIBRATION GRAPH HERE Calculations: Sodium Ion Concentration: 1. Na+ intensity (diluted beer) = _________________ 2. Na+ concentration (diluted beer) = ________________ mg/L 3. Dilution Factor (beer) = x ______________ Na+ concentration (beer sample) ___________________ mg/L 166 Analysis: PART B - Potassium Construct a calibration graph of Concentration (axis ) versus + Colour Intensity (axis ) for your Potassium (K ) standard solutions. PASTE CALIBRATION GRAPH HERE Calculations: Potassium Ion Concentration: 1. K+ intensity (diluted beer) = _________________ 2. K+ concentration (diluted beer) = ________________ mg/L 3. Dilution Factor (beer) = x ______________ K+ concentration (beer sample) ___________________ mg/L 167 Infra-Red (I.R.) Spectroscopy Other radiations such as infra-red can also be absorbed or emitted by chemicals in a spectrum which allows them to be identified and even measured. You may be able to observe infra-red spectroscopy being used. It is particularly useful in identifying covalently bonded molecules and it is the type of bond which absorbs the energy enabling it to be identified. X-rays are another radiation which is often used in this identification process. Figure 1: An infrared spectrum of the compound butanoic acid Figure 2: A general infrared spectrum. Fingerprint region is unique to each compound 168 SOLUTIONS TO WRITTEN EXERCISES Page 8 Solid sand plasticine steel timber Liquid lemonade mercury milk Gas natural gas Difficult to classify? Why? fog balsamic vinegar paint toothpaste shaving foam jelly Page 10 Property Shape Solid fixed (own) shape Gas takes the shape of the container fixed at constant temperature Liquid takes the shape of the part of the container it fills fixed at constant temperature Volume (the amount of space taken up) Compressible? (can it squash into a smaller volume?) no no yes Diffuses? (can it spread out by itself?) no yes yes fills whatever container it is in and has its volume (b) In a liquid the particles can slide past one another but they are still very close. Their sliding movement means that they will diffuse until they meet the walls of the container and the liquid will take part of its shape. The particles are still very close and so the volume of the liquid is fixed at constant temperature and it is incompressible. (c) In a gas the particles can fly apart until they collide with the walls of their container. Therefore the gas takes the shape and volume of the container. The particles have a lot of space in between them and therefore a gas is easily compressible. Gases diffuse quickly because of the flying movement of their particles. Page 21 Element Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium Aluminium Symbol Atomic Number Mass Number Neutron Number Electron Number Electron Configuration H He Li Be B C N O F Ne Na Mg Al 1 2 3 4 5 6 7 8 9 10 11 12 13 1 4 7 9 11 12 14 16 19 20 23 24 27 0 2 4 5 6 4 7 8 10 10 12 12 14 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 2:1 2:2 2:3 2:4 2:5 2:6 2:7 2:8 2:8:1 2:8:2 2:8:3 169 Charge of Ion valency 1+ 0 1+ 2+ 3+ 4 +/3210 1+ 2+ 3+ Silicon Phosphorous Sulfur Chlorine Argon Potassium Calcium Page 22 Si P S Cl Ar K Ca 14 15 16 17 18 19 20 28 31 32 35 40 39 40 14 16 16 18 22 20 20 14 15 16 17 18 19 20 2:8:4 2:8:5 2:8:6 2:8:7 2:8:8 2:8:8:1 2:8:8:2 Elemental puns 1. 0.5 times holmium: = ......hafnium................. 2. An officer of the constabulary: A .....copper............................... 3. Funds from your mother’s sister: 4. A comical prisoner: A ......silicon............................... 5. What you do with dead cats: You .......barium............................. 6. A driller’s motto: .......boron.................. 7. What a good doctor can do for a patient: .......helium.......or curium........ 8. When you have a hole in the boat: You .......zinc.................................. 9. How you mended your clothes: You ........sodium............................ 10. What are you doing with that person’s hair? I’m ........platinum........................ 11. A nocturnal trojan: A ........nitrogen.......................... 12. A trojan who smokes dope: A ........hydrogen........................ 13. A man who tried to murder his wife with a razorblade only managed to give her: ….arsenic… 14. Mind your own: Page 24 1. 4+/3210 1+ 2+ Elements Copper Nitrogen Gold Iodine ......antimony........................... ….bismuth……. Compounds Water Ethanol Vinegar or Carbon Dioxide Mixtures Air Soil Orange Juice Petrol Glass Vinegar Brass Sweat 2. (a) Charcoal and salt ...........dissolution and filtration................... (b) Water and oil ......separation funnel or distillation................ (c) Alcohol and water ..........distillation........................................ (d) Sugar from a sugar solution ..............evaporation or distillation ........... (e) Water from muddy river water .............filtration or distillation...................... Page 26 Classify the following as chemical (C) or physical (P) changes and determine which property is concerned. 2. An iron bar rusting in the air. ........C..... ...........Corrosion / reaction with oxygen.... 3. Petrol combusting in a car engine. ........C..... ...........Combustion / reaction with oxygen.. 170 4. Moulding plastic into a new shape. .......P...... ...........Malleable................... 5. Setting off some fireworks. .......C...... ...........Combustion. / reaction with oxygen.. Page 53 Mg, Cl ....ionic.......................... O, O ....covalent.................... K, S ....ionic.......................... H, Cl ....covalent.................... C, Br ....covalent.................. Na, Ca ....metallic….... Pages 55 .. :S: c) : Al . a) .. : Cl : b) ‘ Ca - 2 b) Al - 3 c) 3) a) I-1 S-2 c) 4) (a) aluminium Al (d) potassium K (g) bromine Br (j) nitrogen N 1) a) 2) 5) b) 3+ d) Li . e) :C: Na - 1 d) Li - 1 e) Ba - 2 f) Mg - 2 N-3 Cl - 1 e) P-3 O-2 d) + (b) lithium Li + (e) calcium Ca - (h) oxygen O (k) selenium Se (a) magnesium and bromine yes (b) chlorine an iodine no (c) potassium and helium no (d) nitrogen and sulfur no (e) lithium and fluorine yes (f) carbon and oxygen no + 2- 3- 2+ 2- 2- 3+ Sr (i) arsenic As (l) sulphur S - 7) (a) KCl K and Cl (d) MgBr2 Mg (a) H2O covalent (b) Na2O ionic (c) CS2 covalent (d) CaS ionic (e) NH3 covalent (f) SO2 covalent (g) Al2O3 ionic 2+ and Br - (c) Al (f) strontium K2S - CaO Ba (a) K ; S + ; O 2- + - (b) NaNO3 Na and NO3 (e) KOH K and OH + - 2+ c) barium 6) 8) (b) Ca 2+ f) ; Cl AlCl3 2+ Ba and SO4 (f) Li2CO3 Li and CO3 + 2- (NH4)2SO4 of ........ NH4.... ) ......2.... atoms of ......8.... atoms of .......1..... atom of .....4..... atoms of .......Nitrogen..... .........Hydrogen.... .............Sulfur..... ......Oxygen....... CH4 represents .........Carbon x 1 and Hydrogen x 4....................... 171 2- (c) BaSO4 ( ...2.... groups 2. 3- 2- Page 58 1. 2+ 3. Sn3(PO4)4 represents .....Tin x 3, Phosphorus x 4 and Oxygen x 16..... 4. CuSO4.5H2O represents.....Cu x 1, Sulfur x 1, Oxygen x 9, and Hydrogen x 10.... 5. Cl - represents ....... Chloride anion...................................................................... - 6. OH represents ...Hydrogen x 1 and Oxygen x 1. Net charge of -1 (or: hydroxide polyatomic ion)... Page 61 Written Exercises: Write formulae for the following compounds. (a) Hydrogen bromide ……HBr………… (g) Ammonium nitride ………(NH4)3N…… (b) Calcium chloride ……CaCl2……….. (h) Iron (III) hydroxide ……Fe(OH)3………. (c) Zinc (II) sulfate …………ZnSO4……… (i) Potassium sulfite ………K2SO3…….. (d) Magnesium oxide ……MgO……… (j) Nickel (II) carbonate ……NiCO3…. (e) Lead(IV) sulphide ……PbS2……… (k) Silver (I) phosphate ……Ag3PO4…….. (f) Barium phosphide …Ba3P2………… (l) Tin (II) nitrate ………Sn(NO3)2……. Pages 62-63 Formulae and Naming (also termed “Nomenclature”) Written Exercises: 1. Write formulae for the following: (a) potassium nitrate …KNO3…. (b) sodium carbonate …..Na2CO3…… (c) cobalt II sulfate …..CoSO4…. (d) ammonium carbonate …..(NH4)2CO3 (e) calcium phosphate ....Ca3(PO4)2... (f) disulfur dichloride .....S2Cl2...... (g) nitrogen triodide ....NI3.......... (h) iodine heptafluoride .........IF7............ (i) silicon dioxide ........SiO2.......... (j) selenium dibromide ....SeBr2................. 2. Name the following compounds: (a) SiF4 ......Silicon tetrafluoride......... (b) CaSO4 ......Calcium sulfate......... (c) FeO ......Iron II oxide......... (d) CuCl ......Copper I chloride......... (e) AgCl......Silver I chloride......... (f) NaHCO3 ......Sodium hydrogen carbonate......... (g) NO .......Nitrogen monoxide.... (h) P2O5 ......Diphosphorous pentoxide......... (i) PCl3 ......Phosphorous trichloride.... (j) CS2 ..........Carbon disulfide........... 3. Formula Name Name Formula MgO magnesium oxide sodium oxide Na2O CaS calcium sulfide calcium carbonate CaCO3 CO carbon monoxide ammonia NH3 CO2 carbon dioxide aluminium hydroxide Al(OH)3 172 Csl caesium iodide dinitrogen tetroxide N2O4 CaBr2 calcium bromide ammonium sulfate (NH4)2SO4 H2O dihydrogen monoxide aluminium oxide Al2O3 NO2 nitrogen dioxide copper (I) bromide CuBr N2O4 dinitrogen tetroxide chromium (III) fluoride CrF3 AsCl3 arsenic trichloride manganese (IV) oxide MnO2 AgNO3 Silver I nitrate mercury (II) sulfide HgS KBr potassium bromide oxygen difluoride OF2 Ca(OH)2 calcium hydroxide iron (II) chloride FeCl2 CoCl2 cobalt (II) chloride iron (III) iodide Fel3 HI hydrogen iodide copper (II) sulfate CuSO4 SnO2 tin (IV) oxide carbon tetrafluoride CF4 CCl4 carbon tetrachloride strontium chloride SrCl2 SnCl2 tin (II) chloride potassium nitrate KNO3 BaF2 barium fluoride calcium phosphate Ca3(PO4)2 Page 72 Written Exercise: Hydrogen chloride conductor? .......As a solid or gas hydrogen chloride assumes a covalent molecular form. It is unique as having the properties of a strong acid, releasing hydrogen ions and also chloride ions when dissolved in water. In this form it is an excellent conductor of electricity.... Page 72 Compound Type CHEMICAL EXAMPLE(S) CHEMICAL BOND TYPE PHYSICAL STATE @ Room Temp MELTING POINT (Low, High, Very High) WATER SOLUBILITY (High, Low, Nil) COMPOUND PROPERTIES IN SUMMARY IONIC COVALENT MOLECULAR COVALENT NETWORK Sodium Chloride Carbon Dioxide Silica Ionic Covalent Covalent Solid SOLID, LIQUID & GAS Solid High Low Very High SEE SOLUBILITY RULES LOW - MEDIUM Insoluble 173 No No No Yes No No ELECTRICAL CONDUCTIVITY - Solid State ELECTRICAL CONDUCTIVITY - Molten State Yes ELECTRICAL CONDUCTIVITY - Aqueous Sol’n NO Hydrogen salt exceptions Page 81 1. 2. (a) CH4(g) + 2O2(g) (b) 2Na(s) + (c) Zn(s) (d) CO2(g) + 2H2O(l) 2H2O(l) 2NaOH(aq) + H2(g) + 2HCl(aq ) ZnCl2(aq) + H2(g) 4P + 5O2 (e) 2NH3 + H2SO4 (f) CuO + 2HCl CuCl2 (g) 2H2O2 2H2O + O2 (h) H2CO3 H2O + CO2 (i) 4Fe 3O2 2Fe2O3 (j) 2C8H18 + 25O2 16CO2 + 18H2O + Sodium hydrogen + carbonate NaHCO3 + 2P2O5 (NH4)2SO4 + ethanoic Sodium + acid ethanoate H2O water CH3COOH NaCH3COO + H2O + carbon dioxide + CO2 Page 86-87 1. Magnesium + hydrochloric acid magnesium chloride + hydrogen gas Mg (s) + 2HCl (aq) MgCl2 (aq) + H2 (g) 2. Aluminium + nitric acid aluminium nitrate + hydrogen gas 2Al (s) + 6HNO3(aq) 2Al(NO3)3 (aq) + 3H2 (g) 3. Iron + oxygen iron III oxide 4Fe (s) + 3O2 (g) 2Fe2O3 (s) 4. Lithium + water lithium hydroxide + hydrogen gas 2Li (s) + 2H2O (l) 2LiOH (aq) + H2 (g) 5. Silver I carbonate silver oxide + carbon dioxide Ag2CO3 (aq) Ag2O (s) + CO2 (g) 6. Silver I nitrate + sodium chloride silver chloride + sodium nitrate AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) 7. Sulfuric acid + sodium hydroxide sodium sulfate + water H2SO4 (aq) + 2NaOH (aq) Na2SO4 (aq) + 2H2O (l) 174 No 8. Lead II nitrate + potassium iodide lead II iodide + potassium nitrate Pb(NO3)2 (aq) + 2KI (aq) PbI2 (s) + 2KNO3 (aq) 9. Hexane + oxygen carbon dioxide + water (+ energy) C6H14 (l) + 9½O2 (g) 6CO2 (g) + 7H2O (l) or 2C6H14 (l) + 19O2 (g) 12CO2 (g) + 14H2O (l) Page 87 1. Magnesium + Nitric Acid Magnesium nitrate + hydrogen gas Mg (s) + 2HNO3 (aq) Mg(NO3)2 (aq) + H2 (g) 2. Copper + Sulfuric Acid Copper II sulfate + hydrogen gas Cu (aq) + H2SO4 (aq) CuSO4 (aq) + H2 (aq) 3. Potassium + Water Potassium hydroxide + hydrogen gas 2K (s) + 2H2O (l) 2KOH (aq) + H2 (g) 4. Zinc II Carbonate + Nitric Acid Zinc II nitrate + carbon dioxide + water ZnCO3 (aq) + 2HNO3 (aq) Zn(NO3)2 (aq) + CO2 (g) + H2O (l) 5. Copper II Carbonate (s) + HEAT copper II oxide + carbon dioxide CuCO3 (s) CuO (s) + CO2 (g) 6. Aluminium + Sulfuric Acid Aluminium sulfate + hydrogen gas 2Al (s) + 3H2SO4 (aq) Al2(SO4)3 (aq) + 3H2 (g) 7. Sodium Hydroxide + Copper II Sulfate Copper II hydroxide + sodium sulfate 2NaOH (aq) + CuSO4 (aq) Cu(OH)2 (aq) + Na2SO4 (aq) 8. Sodium Hydroxide + Sulfuric Acid Sodium sulfate + water 2NaOH (aq) + H2SO4 (aq) Na2SO4 (aq) + 2H2O (l) Page 89 + - 1. Ag 2. 2CO3 (aq) (aq) + Cl (aq) + 2H AgCl (s) + (aq) CO2 (g) + H2O (l) Page 96 1. a) 58.44 b) 132.15 c) 180.16 d) 204.22 2. a) 0.99 mol b) 0.10 mol c) 0.50 mol d) 0.024 mol 3. a) 146.1 g b) 66.08 c) 720.64 g d) 3.06 g Page 101 a) N2H4 (g) + O2 (g) N2 (g) + 2H2O (g) b) L.R. = Hydrazine (just!) c) 1124 g Pages 109-111 1. i) ii) iii) iv) CaCO3 (s) CaO (s) + CO2 (g) 1997.1 mol 1997.1 mol 49 508 L 175 2. i) ii) iii) iv) Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g) 0.20 mol 0.20 mol 5.0 L 3. i) ii) iii) iv) v) 2AgCl (s) + H2 (g) 2 Ag (s) + 2HCl (aq) 2 : 1 : 2 : 2 0.20 mol 0.10 mol 2.48 L 4. i) ii) iii) iv) v) CaC2 (s) + 2H2O (l) Ca(OH)2 (aq) + C2H2 (g) 1 : 2 : 1 : 1 1 0.450 mol 11.16 L Pages 114-115 1. a) C = 27.3% ; O = 72.7% b) H = 2.1% ; S = 32.7% ; O = 65.2% c) C = 40.0% ; H = 6.7% ; O = 53.3% 2. Zn = 20.5% ; I = 79.5% 3. Hg2Cl2: Hg = 85.0% ; Cl = 15.0% HgCl2: Hg = 73.9% ; Cl = 26.1% Page 116 1. C3H6O2 2. Emp Formula: HF ; Molec Formula: H2F2 3. Emp Formula: P2O3 ; Molec Formula: P4O6 Page 122 i) 3M iv) 0.5M ii) 2.5M v) 0.5M iii) 0.1M vi) 0.09M Page 124 1. a) Dissolve 4.248 g of AgNO3. Make up to 250.0 mL b) Dissolve 4.561 g of K2CO3. Make up to 100.0 mL 2. a) 0.40M b) 0.0025M c) 0.02M 3. a) 0.405 g b) 30.335 g c) 1.766 g 4. a) 500 mL b) 250 mL c) 100 mL BONUS QUESTION: 1.001 g Pages 126-127 1. i) 10x ii) 25x 2. i) 0.63 g/100 mL 3. i) 25 mL iii) 16.67x ii) 0.015M ii) 50 mL iii) 10 mg/L iii) 40 mL 176 Page 133 1. 1.5% w/w 2. 67.5% w/w 3. 26.6% w/w 1.b) 0.1213M 2. 0.1172M Page 141 1.a) 0.0769M 3. 4.76% w/v Page 160-161 1. sodium chloride, NaCl 2. sodium Na + 3. potassium K + magnesium Mg 2+ calcium Ca 2+ What do the following symbols stand for: + Na + K g: mg: L: __sodium ion____ __potassium ion_____ ____grams__________ ___milligrams_____ ____litres_________ mL: _______millilitres________ g/100mL: ___grams per 100 millitres___ g/L: _____grams per litre_____ mg/L: ____milligrams per litre_____ 4. flame emission spectrophotometer 5. to spray a finely dispersed mist of solution 6. flame emission 7. detector measures light intensity emitted 8. analyte concentration 9. quantitative dilution of the sample 177 REFERENCE LIST Some of these references are referred to in the module text by abbreviation: CC1 James et al (1991) Chemical Connections. Book One. Jacaranda ISBN O 7016 2750 6 C1 Elvins et al. (1995) Chemistry One. Materials: Chemistry in Everyday Life. Second Edition Heinemann. ISBN 0 85859 708 X Elvins et al. (19xx). Chemistry One. Teachers’ Resource Book. Heinemann. ISBN 0 85859 xxx x C2 Commons et al. (1994) Chemistry Two. Chemistry and the Marketplace; Energy and Matter. Heinemann. ISBN 0 85859 755 1 Commons et al. (1995). Chemistry Two. Teachers’ Resource Book. Heinemann. ISBN 085859 758 6 CC2 James et al. (1991) Chemical Connections. Book One. Jacaranda. ISBN 0 7016 2750 6 James et al. (1992) Chemical Connections. Book Two. Jacaranda. ISBN 0 7016 3062 0 Selinger, B. (1994). Chemistry in the Marketplace. Fourth Edition. Harcourt Brace & Co. ISBN 0 7295 0334 8 Elvins et al. (1995). Chemistry One. Materials; Chemistry in Everyday Life. Second Edition. Heinemann. ISBN 0 85859 708 X Commons et al. (1994). Chemical Connections. Book Two. Chemistry and the Marketplace; Energy and Matter. Heinemann. ISBN 0 85859 755 1 DB Barker, D. (2001). 6849AA Calibration and Data Handling BFK Barker, D. Fullick, G and Krajniak, E. (2003). Chemistry For Technicians. Harcourt Brace & Co. ISBN 0 7295 3291 7 Laboratory Manuals Deretic, G. & Ware, G. (1995). Senior Chemistry Practical Manual. Heinemann. ISBN 0 85859 789 6. Burke, S. (1992). Chemistry Works. Science Press. ISBN 0 85583 170 7 Further References Smith, A. and Dwyer, C. (1991). Key Chemistry. Book One. Materials and Everyday Life. ISBN 0 522 844502 Smith, A. et al. (1992). Key Chemistry. Book Two. Energy, Matter and the Marketplace. ISBN 0 522 84461 8 178