Principles and Applications of Science 1 Level 3 Unit 1-2 PDF

T AF R Photo credit: Stephen Coburn/123RF D Principles and Applications o...

T AF R Photo credit: Stephen Coburn/123RF D Principles and Applications of Science I 1 UNIT 1 Getting to know your unit All scientists and technicians need to understand core science concepts. Assessm ent Chemists need to understand atoms and electronic structure to predict You will be assesse d how a range of chemical substances will react to make useful products. Medical professionals need to understand the structure and workings of through a 90-min ute cells when they think about how the body stays healthy as well as when written exam worth 90 diagnosing and treating illness. marks, which is set and marke d by Pearson. Scientists working in the communication industry need a good understanding of waves. How you will be assessed The external paper for this unit will be split into three sections, each worth 30 marks. ▸▸ Section A – Biology (Cell structure and function, Cell specialisation, Tissue structure and function) ▸▸ Section B – Chemistry (Structure and bonding in applications of science, T Production and uses of substances in relation to properties) ▸▸ Section C – Physics (Working with waves, Waves in communication, Use of AF electromagnetic waves in communication) The paper will contain a range of question types, including multiple choice, calculations, short answer and open response. These question types by their very nature generally assess discrete knowledge and understanding of content in this unit. R You need to be able to apply and synthesise knowledge from this unit. The questions on the paper will be contextualised in order for you to show you can do this. There will be two opportunities each year to sit this paper: January and May/June. D Throughout this chapter, you will find assessment practices that will help you prepare for the exam. Completing each of these will give you an insight into the types of questions that will be asked and, importantly, how to answer them. 2 Principles and Applications of Science 1  Getting to know your unit UNIT 1 Unit 1 has four Assessment Outcomes (AO) which will be included in the external Principles and Applications of Science 1 examination. These are: ▸▸ AO1 Demonstrate knowledge of scientific facts, terms definitions and scientific formulae Command words: give, label, name, state Marks: ranges 12 to 18 marks ▸▸ AO2: demonstrate understanding of scientific concepts, procedures, processes and techniques and their application Command words: calculate, compare, discuss, draw, explain, state, write Marks: ranges from 30 to 45 marks ▸▸ AO3: Analyse, interpret and evaluate scientific information to make judgements and reach conclusions Command words: calculate, compare, comment complete, describe, discuss, explain, state Marks: ranges from 18 to 24 marks ▸▸ AO4 Make connections, use and integrate different scientific concepts, procedures, processes or techniques Command words: compare, comment, discuss, explain Marks: ranges from 9 to 12 marks Here are some of the command words. The rest are found in the specifications. Command word Definition – what it is asking you to do T AF Analyse Identify several relevant facts of a topic, demonstrate how they are linked and then explain the importance of each, often in relation to the other facts. Compare Identify the main factors of two or more items and point out their similarities and differences. You may need to say which are the most best or most important. The word Contrast is very similar. Comment R Requires the synthesis of a number of variables from data/information to form a judgement. More than 2 factors need to be synthesised. Define State the meaning of something, using clear and relevant facts. D Describe Give a full account of all the information, including all the relevant details of any features, of a topic. Discuss Write about the topic in detail, taking into account different ideas and opinions. Evaluate Bring all the relevant information you have on a topic together and make a judgment on it (for example on its success or importance). Your judgment should be clearly supported by the information you have gathered. Explain Make an idea, situation or problem clear to your reader, by describing it in detail, including any relevant data or facts. 3  Getting started Scientists working in a hospital laboratory use a range of core scientific principles. Write a list of core scientific principles you think they might need and why they are useful. Remember these may be to do with physics, chemistry or biology. When you have completed this unit, see if you can add any more principles to your list. A Periodicity and properties of elements A1 Structure and bonding in applications in science The electronic structure of atoms You should already know about the structure of an atom. The nucleus contains positive protons and neutral nutrons. Surrounding the nucleus are energy shells T containing negative electrons. You should also know that protons and neutrons both have a relative mass of 1 and that the relative mass of an electron us almost 0. AF Lab technicians need to understand the electronic structure of atoms. They can use this knowledge to predict how chemical substances will behave and react. The protons and the neutrons are found in the nucleus at the centre of an atom. The electrons are in shells or energy levels surrounding the nucleus. Each shell can hold electrons up to a maximum number. When the first shell is full electrons then go into R the second shell and so on. The maximum number of electrons in each shell is shown in Table 1.1. D Sodium ▸▸ Table 1.1: Maximum number of electrons for each electron shell shell 23 Na 11 Electron shell Maximum number of electrons 1 2 2 8 12 n 3 18 11 p 4 32 5 50 A sodium atom containing 11 electrons has an electron arrangement of 2, 8, 1. ▸▸ Figure 1.1: Simple atomic This can be represented by a simple Bohr diagram, as shown in Figure 1.1. structure of sodium This is the simple version of electron structure you will have seen at Key Stage 4. Key term Under Bohr’s theory, an electron’s shells can be imagined as orbiting circles around the nucleus. Orbitals – regions where there is a 95% probability However, it is more complicated than this. Electrons within each shell will not have the of locating an electron. An same amount of energy and so the energy levels or shells are broken down into sub- orbital can hold a maximum shells called orbitals. These are called s, p, d and f orbitals. The orbitals have different of two electrons. energy states. 4 Principles and Applications of Science 1  Learning aim A UNIT 1 The Aufbau principle states that electrons fill the orbital with the lowest available Principles and Applications of Science 1 Key terms energy state in relation to the proximity to the nucleus before filling orbitals with higher energy states. This gives the most stable electron configuration possible. Electron configuration – the distribution of electrons in an Electrons have the same charge and so repel each other so if there is more than one atom or molecule. orbital in an energy level (sub-shell) they will fill them singly until all the orbitals in that sub-shell have an electron in them and then they will pair up. Spin – electrons have two possible states, ‘spin up’ and Figure 1.2 shows the energy levels of the shells, sub-shells and orbitals for an atom. ‘spin down’. In an orbital, each electron will be in a different Atomic Energy Levels 4f ‘spin state’. 4d n54 4p 3d 4s 3p Energy n53 3s 2p n52 2s Key s n51 n 5 shell ▸▸ Figure 1.2: Energy levels of the shells subshells and orbitals for an atom T Step-by-step: Electron structures 8 Steps When writing out electron structures, you should follow these rules. AF Half arrows are used to represent each electron in the orbitals. They are drawn facing up and down as each electron in an orbital will have a different spin. 1 The electrons sit in orbitals within the shell. Each 5 Electrons fill the lowest energy level orbitals first. orbital can hold up to two electrons. R 2 The first shell can hold two electrons in an s-type 6 Where there are several orbitals of exactly the same D orbital. energy, for example, the three 2p orbitals in the second shell, then the electrons will occupy different orbitals wherever possible. 3 The second shell consists of one s-type orbital and 7 So the electronic structure of nitrogen (which has three p-type orbitals. This diagram represents lithium. 7 electrons) is: 2s1 2p 2p 2p 2s1 2p1 2p1 2p1 1s2 1s2 4 The third shell consists of one s-type orbital, 8 and the electronic structure of a sodium atom three p-type orbitals and five d-type orbitals. (which has 11 electrons) becomes: 2p2 2p2 2p2 3s1 1s2 2s2 5  Assessment practice 1.1 Copy out the following table and complete the electronic structures for the elements. Three have been done for you. Element Number of electrons Electron structure hydrogen 1 1s1 helium lithium boron carbon 6 1s2 2s2 2p2 oxygen 8 1s2 2s2 2p4 magnesium chlorine calcium P ause point T Try explaining what you have learned so far. AF Hint Close the book and write out all the key concepts you have learned so far. What do you know about electronic structure? Could you draw the electronic structure for calcium? Extend What is new compared to what you learned at level 2 about electronic structure? R One of the tasks of a lab technician is to make up solutions ready for experiments or for making products. Different types of compounds dissolve in different types of D solvents depending on what type of bonding is in the compound. The lab technician must know what type of compound they are using in order to use the correct solvent. Ionic bonding Noble gases (elements in group 0 of the periodic table) have a stable electronic configuration. They have full outer shells. This means they do not react easily and most do not react at all. Elements in the other groups do not have full outer shells. Key term This means that they react to gain stable electronic configurations. Ionic bonding – electrostatic Ionic bonding occurs when an atom of an element loses one or more electron and attraction between two donates it to an atom of a different element. The atom that loses electrons becomes oppositely charged ions. positively charged and the atom that gains electron(s) become negatively charged because of the imbalance of protons and electrons. For example, the bonding in sodium chloride is ionic. This means that the sodium atom loses the electron in its outer shell to become the positively charged sodium ion, Na+, with the same electron configuration as neon. Chlorine gains an electron to become the negatively charged chloride ion, Cl−, with the same electron configuration as argon. This means that both the sodium ion and the chloride ion have a full outer shell and become stable. The positive charge on the sodium ion and the negative charge on the chloride ion are attracted. 6 Principles and Applications of Science 1  Learning aim A UNIT 1 Principles and Applications of Science 1 Na Cl Na1 Cl2 Ionic bond ▸▸ Figure 1.3: Electron transfer and bonding in sodium chloride Key terms Electrostatic attraction Figure 1.3 shows bonding using a dot and cross diagram. The dots and crosses – the force experienced by represent electrons in the shells. The electrons fill up the shells singly and then in pairs. oppositely charged particles. Ions containing more than one element can also be formed. For example, in sodium It holds the particles strongly hydroxide, Na+ bonds with the hydroxide ion (OH)–. together. The opposite charges on the ions are what hold them together. This is electrostatic Giant ionic lattice – a regular attraction. arrangement of positive ions and negative ions, for The opposite charged ions in sodium chloride form a giant ionic lattice (see Figure example, in NaCl. 1.4) where the ions are arranged in a regular pattern. sodium ions chloride ions ▸▸ Figure 1.4: Lattice structure of sodium chloride The strength of the electrostatic force and, therefore, of the ionic bond is dependent on the ionic charge and the ionic radii of the ions. The more electrons a positive ion has, the more shells it will have. If an ion has more shells, then its radius will be bigger than an ion with fewer shells. The electrostatic force is stronger when the ionic charge is higher. However, the force becomes weaker if the ionic radii are bigger. This is because, when the ionic radius is bigger, the ionic charge is spread over a larger surface area. 7  Covalent bonding Covalent bonding usually occurs between atoms of two non-metals. A covalent bond forms when an electron is shared between the atoms. These electrons come from the top energy level of the atoms. (a) Chlorine, Cl2 (b) (c) H H O O Cl Cl ▸▸ Figure 1.5: Covalent bonding in (a) a chlorine molecule (b) a hydrogen molecule (c) oxygen moelcule A chlorine molecule has a covalent bond (see Figure 1.5). The highest shell in each chlorine atom contains seven electrons. One electron from the highest shell in each atom is shared to give each chlorine atom the electron configuration of argon with a stable full outer shell. Dative covalent bonding In some covalent molecules, both sharing electrons come from one atom. This is called a dative (coordinate) covalent bond. (see Figure 1.6). O O ▸▸ Figure 1.6: Covalent bonds can be formed when atoms share more than one pair of electrons, e.g. oxygen atoms. The double bonds between the oxygen are formed by two shared pairs of electrons. If three pairs of electrons are shared, then a triple covalent bond is formed. A triple bond is present in a nitrogen molecule (see Figure 1.7). Nitrogen Triple Bond N N ▸▸ Figure 1.7: Bonding in a nitrogen molecule An ammonium ion contains a dative bond (see Figure 1.8). When ammonia reacts Key term with hydrochloric acid, a hydrogen ion from the acid is transferred to the ammonia Lone pair – a non-binding molecule. A lone pair of electrons on the nitrogen atom forms a dative covalent bond pair of electrons. with the hydrogen ion. 8 Principles and Applications of Science 1  Learning aim A UNIT 1 positive charge because negative charge because Principles and Applications of Science 1 only the hydrogen nucleus the hydrogen has left its Image credit: after http://www.chemguide.co.uk/ has moved to the nitrogen electron behind atoms/bonding/dative.html, Jim Clark H H 1 2 H N H Cl H N H Cl H H lone pair dative covalent bond of electrons ▸▸ Figure 1.8: Dative bond formation in reaction between ammonia and hydrochloric acid Single bonds have a greater length than double bonds and double bonds have a greater length than triple bonds. The shorter the length of the bond, the stronger the bond is. Therefore, triple bonds are stronger than double or single bonds. A single bond between carbon atoms has a length of 154 pm and a bond energy of 347 kJ mol−1. A double bond between carbon atoms has a length of 134 pm and a bond Key term energy of 612 kJ mol−1. A triple bond between atoms has a bond length of 120 pm and a bond energy of 820 kJ mol−1. Organic compound – a compound that contains one or more carbons in a carbon Covalent bonding in organic molecules chain. Carbon makes four covalent bonds so it forms many compounds which are called organic compounds. Methane has the formula CH4. Each carbon atom bonds covalently with four hydrogen H atoms. The carbon gains the stable electron structure of neon and hydrogen gains the stable electron structure of helium. These four bonds mean that methane is not a flat molecule. It has a tetrahedral structure C (see Figure 1.9). This is because the bonds are as separated from each other as possible, H because the negative electron pairs repel each other, with each bond angle being 109.5o. H If you were to build a model of a methane molecule, it would have a 3D shape with a H hydrogen pointing down towards you, one pointing down away from you, one pointing ▸▸ Figure 1.9: Tetrahedral structure down to the side and one pointing up, all connected to the carbon in the centre. of methane Step by step: Building models of organic compounds 3 Steps 1 Use molecular model kits to build models of the following organic compounds. methane CH4 ethane CH3CH3 propane CH3CH2CH3. 2 Write down what you notice about the structure of these molecules. 3 Look at one of the carbons in each molecule and the atoms bonded to it. Write down what you notice about the shape. 9  Organic compounds with three or more carbons in a chain cannot be linear because of the tetrahedral structure around each central carbon (see Figure 1.10). H H H C C H H C C H H H H H Butane ▸▸ Figure 1.10: A butane model Metallic bonding Metals are giant structures of atoms held together by metallic bonds. The metal Key term structure is a regular lattice (see Figure 1.11). Delocalised electron – Metallic bonding is caused because the electrons in the highest energy level of a metal electrons that are free to atom has the ability to become delocalised. They are free to move through the metal in move. They are present in a ‘sea’ of electrons. This gives the metal nuclei a positive charge which is attracted to the metals and are not associated negative charge on the delocalised electrons. There is a very strong force of attraction with a single atom or covalent between the positive metal nuclei and the negative delocalised electrons. However, the bond. forces in metallic bonding are not as strong as in covalent or ionic bonding. ▸▸ Figure 1.11: Metallic structure The metal structure is a lattice of positive ions with electrons flowing between these ions. P ause point What have you learned about bonding? Hint Describe the differences between ionic, covalent and metallic bonding. Extend Give two examples of elements, compounds or molecules with each type of bond. The electronegativity of two atoms will determine what type of bond will form between them. Key term Electronegativity– This is the tendency of an atom to attract a bonding pair of electrons. 10 Principles and Applications of Science 1  Learning aim A UNIT 1 Atoms that have similar electronegativities form covalent bonds. Principles and Applications of Science 1 There is a strong electrostatic attraction between the two nuclei and the shared pair(s) of electrons between them. This is the covalent bond. Both chlorine atoms have the same electronegativity, and so the electrons are equally shared. The chlorine molecule is non-polar (see Figure 1.6). Hydrogen only has one shell containing one electron. This electron from each hydrogen is shared to give each atom the electronic configuration of helium. Oxygen only has 6 electrons in its highest energy shell. Each oxygen atom shares two of its electrons with another oxygen atom, giving both 8 electrons in their outer shell. This makes the atoms in the oxygen molecule stable. ▸▸ Figure 1.12 : Non-polar covalent bond In most covalent compounds, the bonding is polar covalent (see Figure 1.7). The shared electrons are attracted more to one nucleus in the molecule than the other. The atom with the higher electronegativity will attract the electrons more strongly. This gives the atom a slight negative charge. The other atom in the molecule will have a slight positive charge. Key terms Non-polar molecule – a molecule where the electrons are distributed evenly throughout the molecule. Polar molecule – a molecule with partial positive charge in one part of the molecule and similar negative charge in another part due to an uneven electron distribution. ▸▸ Figure 1.13: Polar covalent bond 11  As the difference in electronegativity between the atoms increases, the bond will become more polar. See Figure 1.4. ▸▸ Figure 1.14: Electronegativity spectrum The electronegativities of some of the common elements you will use are shown in Table 1.2. ▸▸ Table 1.2: Electronegativities of elements Element Electronegativity Fluorine 3.98 Oxygen 3.44 Nitrogen 3.04 Carbon 2.55 Chlorine 3.16 Hydrogen 2.20 Lithium 0.98 Sodium 0.82 Key terms Intermolecular forces Intermolecular forces – Intermolecular forces also affect how chemical substances behave. A laboratory the attraction or repulsion technician must know where these are present and understand how they will affect between neighbouring the behaviour and reactions of chemical substances they are working with. molecules. London dispersion forces Dipole – separation of One type of intermolecular force is called London dispersion forces (also called charges within a covalent temporary dipole – induced dipole forces). They are weak forces present between molecule. non-polar covalent molecules. They are less than 1% of the force of a covalent bond (see Figure 1.15). When the electron distribution in a molecule becomes non-symmetrical (i.e. there are more electrons at one end of the molecule than the other) then one end of the molecule can become more positive and one end can become more negative. This causes a temporary dipole. The positive and negative charge in the dipole can disturb the electrons in a nearby molecule, repelling the electrons and so causing (inducing) a dipole in that molecule. The molecule with the temporary dipole and the molecule with the induced dipole attract each other and pull the molecules together. The forces are temporary because the electrons are constantly moving so electron density in any part of a molecule is constantly changing. Larger molecules have more electrons which can move further so more temporary dipoles can form, meaning the force is bigger. more electrons → more movement → bigger dipoles → stronger attraction 12 Principles and Applications of Science 1  Learning aim A UNIT 1 Principles and Applications of Science 1 H H H H Molecule A Molecule B even distribution of electrons throughout both molecules. H H H H Molecule A Molecule B Uneven distribution of electrons in Molecule A causes a temporary dipole in the molecule. This will induce a dipole in molecule B as the electrons in Molecule B will be attracted to the positive end of Molecule A H H H H Molecule A Molecule B This forms a temporary dipole – induced dipole ▸▸ Figure 1.15: London Dispersion forces London dispersion forces are the only forces that exist between noble gases and non‑polar molecules. Assessment practice 1.2 Pentane (C5H12) boils at 309 K and ethane (C2H6) boils at 185 K. This means that pentane is a liquid at room temperature (293 K) and ethane is a gas. Explain why pentane is a liquid at room temperature but ethane is a gas. Dipole-dipole forces Another form of van der Waals forces are dipole-dipole forces. These are permanent Key term forces between polar molecules (see Figure 1.16). Polar molecules have a permanent negative end and a permanent positive end. These oppositely charged end attract each Van der Waals forces – All other. Dipole-dipole forces are slightly stronger than London dispersion forces but are intermolecular attractions are still weak in comparison to a covalent bond. The force is about 1% the strength of a van der Waals forces. covalent bond. Molecules that have permanent dipole-dipole forces include hydrogen chloride, HCl, and iodine monochloride, ICl. In both cases, the chlorine atom in the molecule is slightly negative. The hydrogen and iodine atoms are slightly positive. d1 d2 d1 d2 attraction ▸▸ Figure 1.16: Dipole-dipole forces There are dipole-dipole forces between molecules of iodine monochloride (ICl). 13  Hydrogen bonding The strongest form of intermolecular force is a hydrogen bond. These are a special type of dipole-dipole bond and are forces that are about 10% the strength of a covalent bond. Hydrogen bonds will form when compounds have hydrogen directly bonded to fluorine, oxygen or nitrogen. This is because there is a large difference in electronegativity between hydrogen and and of these three atoms. This large difference means that very polar bonds are formed so the molecules have permanent dipoles. When two of these Hd1 molecules are close together, there will be an attraction between the positive end of one d2 N d1 and the lone pair of electrons of the other. This is a hydrogen bond. d2 d1 H N H Hd1 This is different to other dipole-dipole forces because there are inner bonding electrons. d1 H d1 The single electron in the hydrogen atom is drawn to the nitrogen (see Figure 1.17), H hydrogen bond oxygen or fluorine atom. There are no non-bonding electrons shielding the nucleus of ▸▸ Figure 1.17: Hydrogen bond in the hydrogen. The hydrogen proton is strongly attracted to the lone pair of electrons on ammonia the nitrogen atom of another molecule. Discussion Hydrogen bonding in water is the reason why water has such unusual properties, e.g. solid water is less dense than liquid water, it has a higher boiling point than expected, it is a good solvent for many chemical substances. Research how hydrogen bonding is caused in a water molecule. Work in pairs to list properties of water due to the hydrogen bonding. In groups, explain the properties to other pairs of learners. P ause point Try to describe all the different types of intermolecular forces to a partner. Hint Draw a table showing the different types of intermolecular bonding and their properties. Extend Explain how each type of intermolecular bond affects the properties of the molecules. Quantities used in chemical reactions Balancing equations All chemical reactions can be written as a balanced equation using the chemical formulae for the reactants and the products involved in the reaction. Symbols for elements can be found in the periodic table. The numbers in the formulae show how many atoms of each element there are. You can use the periodic table to predict whether the compound is covalent or ionic. The group numbers will show you how many electrons the atom needs to lose or gain or share to form a bond. The equation must balance like a maths equation. There should be the same number and types of atoms on both sides of the equation. Step by step: Writing a balanced equation 5 Steps 1 Write the equation as a word equation including all the reactants and all the products. 2 Write out the formulae for each substance in the reaction. Note that gaseous elements (except those in group 0) like hydrogen and oxygen are diatomic (molecules with two atoms) so they must be written as H2 and O2. Metal elements and the noble gases are monatomic (one atom) 14 Principles and Applications of Science 1  Learning aim A UNIT 1 Principles and Applications of Science 1 3 Write out the number of each element on both sides. 4 Make the number of each atom equal on each side. Remember that you cannot change the formula of the compounds. To increase the number of atoms of a particular element, you must place a number in front of the compound it is in. This will affect the number of atoms of all the other elements in the compound. 5 Check that there is the same number of atoms of each element on both sides. Worked Example 1 Write a balanced equation for the following reaction. Ethanol + oxygen → carbon dioxide +water Step 1: Write out the formulae for each substance in the reaction. C2H5OH + O2 → CO2 + H2O Step 2: Write out the number of each element on both sides. left-hand side right-hand side C2 C1 H6 H2 O3 O3 Step 3: Make the number of each atom equal on each side. In this case, start by putting a 2 in front of the carbon dioxide to equal out the carbons. This will also add two more oxygens to the right-hand side C2H5OH + O2 → 2CO2 + H2O left-hand side right-hand side C2 C12 H6 H2 O3 O35 Put a 3 in front of the water to balance the hydrogens. Remember to add to the oxygens again C2H5OH + O2 → 2CO2 + 3H2O left-hand side right-hand side C2 C12 H6 H26 O3 O357 The carbons and hydrogens are now equal on both sides so you must multiply the oxygens on the left-hand side to finish balancing the equation C2H5OH + 3O2 → 2CO2 + 3H2O left-hand side right-hand side C2 C12 H6 H26 O37 O357 This equation is now balanced. 15  Write a balanced equation for the following reaction: P ause point butanol (C4H9OH) + water → carbon dioxide + water. Hint Remember you can only change the number of moles of each substance, you cannot change the formula. Extend Now write a balanced equation for: magnesium carbonate + hydrochloric acid → m  agnesium chloride + water + carbon dioxide. Assessment practice 1.3 Write balanced equations for the following reactions. 1 methane (CH4) + oxygen → carbon dioxide + water 2 calcium carbonate (CaCo3) + hydrochloric acid (HCl) → calcium chloride + carbon dioxide + water 3 calcium hydroxide (CaOH) + hydrochloric acid → calcium chloride + water Moles, molar masses and molarities Chemical equations allow you to work out the masses of the reactants you need to use in order to get a specific mass of product. Chemists never use one molecule of a substance because that would be too small. Even 0.1 g of hydrochloric acid will contain millions of molecules of the acid. These numbers are very big and difficult to work with so chemists use a quantity called a mole with the symbol mol. Do not let the idea of a mole confuse you. It is just a number. One mole of a chemical means there are 6.023 × 1023 particles (Avogadro’s constant). 6.023 × 1023 is a number is standard form. This is a simple way of showing a very large number. 1 × 103 is how you would write 1000 in standard form. The 103 means that if you write the number out in full, it will have 3 zeroes at the end. So 6.023 × 1023 is a simple way to write 6023 with 20 zeroes at the end. A mole is the amount of a substance which has the same number of particles as there are atoms in 12 g of carbon-12. So one mole of carbon dioxide has the same number of particles as one mole of gold. The molar mass of a substance is equal to the mass of one mole of a substance. It is useful to be able to convert masses into moles and moles into masses. Key terms Mass (g) = molar mass × number of moles ▸▸ The relative atomic mass (Ar) of an element on the periodic table tells you how Mole – a unit of substance equivalent to the number of much mass there is in one mole of the element. The relative atomic mass is the atoms in 12 g of carbon-12. average mass of an atom of an element compared to one twelfth of the mass of an 1 mole of a compound has atom of carbon–12. The relative atomic mass of hydrogen is 1.0. The relative atomic a mass equal to its relative mass of oxygen is 16.0. atomic mass expressed in ▸▸ The relative formula mass is the sum of all the relative atomic masses of all the grams. atoms in the empirical formula (simplest formula) of a compound (Mr). Molar mass – the mass of The relative formula mass of water, H2O, is (1 × 2) + 16 = 18 one mole of a substance. Relative atomic and formula masses do not have any units as they are only relative to carbon–12. 16 Principles and Applications of Science 1  Learning aim A UNIT 1 Principles and Applications of Science 1 Assessment practice 1.4 What is the relative formula mass for these molecules? 1 CO2 2 NaOH 3 H2SO4 4 Ca(OH)2 5 Fe2O3 The following worked examples show how to convert masses to moles. Worked Example 1 What is the number of moles in 136.5 g of potassium? Number of moles of an element = mass/Ar For potassium Ar =39 136.5 Number of moles = ______ ​ 39 = 3.5 moles 2 What is the number of moles in 20 g of sodium hydroxide, NaOH? Number of moles = mass/Mr For sodium hydroxide Mr = 23 + 16 + 1 = 40 20 Number of moles = ___ ​ 40 = 0.5 moles Empirical formula This shows the ratio between elements in a chemical compound. It is useful when discussing giant structures such a sodium chloride. The empirical formula of a compound can be calculated from the masses of each element in the compound. These masses are worked out through experimental analysis of the compound. Step by step: Empirical formula 3 Steps 1 Divide the mass of each element present in the compound by its molar mass to get its molar ratio. 2 Divide the answer for each element by the smallest molar ratio calculated. This gives you a ratio of 1:x for each element present. 3 If the answers are not all whole numbers, multiply them all by the same number to get whole numbers. e.g. if the ratio is 1:1.5:3 then multiplying all the numbers by 2 will give you an answer with all whole numbers 2:3:6 17  Molecular formula Molecular formulae are used for simple molecules. To work out the molecular formula you need to know the empirical formula and the relative molecular mass. e.g. a compound has the empirical formula CH4. This has an empirical formula mass of 12 + (1 × 2) It has a relative molecular mass of 42. To work out its molecular formula you first divide its relative molecular mass by the empirical mass. 42/14 = 3 You write out the formula multiplying each part of the CH2 unit by 3. This gives C3H6. This is the molecular formula. Reacting quantities When carrying out titrations, a chemist has to use solutions of a known concentration. These are called standard solutions. They have been prepared and tested to ensure they are of the specific concentration needed. The number of moles of solute in a given volume of solvent tells you how concentrated the solution is. When 1 mole of solute is dissolved in 1 cubic decimetre of solution, its concentration is written as: 1 mol dm−3. This can be written as 1M for short. This is the molarity of the solution. I mole of HCl has a mass of 1 + 35.5 = 36.5 g 36.5 g of HCl in 1 dm3 of solution has a concentration of 1 mol dm−3 or 1M or 36.5 g dm−3. Key terms Titration – a method of volumetric analysis used to calculate the concentration of a solution. Solution –a liquid mixture where a solute is dissolved in a solvent Standard solution – a solution of known concentration used in volumetric analysis. Solute – the substance dissolved in a solvent to form a solution. Solvent - a liquid that dissolves another substance. Worked Example 1 How many moles of hydrochloric acid are there in 100 cm 3 of 1M hydrochloric acid solution? Number of moles (N) = molarity (C) × volume of solution (V) (dm3) N = CV The volume is given in cm3 so this needs to be converted into dm3 by dividing by 1000. (Remember 1 dm3 = 1000 cm3) 100 number of moles = ______ ​ ​× 1 1000 = 0.1 mol 18 Principles and Applications of Science 1  Learning aim A UNIT 1 Principles and Applications of Science 1 2 What is the concentration of a sample of sodium hydroxide solution if 10 dm3 contains 0.5 mol? Number of moles (N) = molarity (C) × volume of solution (V) (dm3) N = CV 0.5 = C × 10 0.5 C = ​____​= 0.05M 10 3 What volume in cm3 of 2M sulfuric acid solution would you need to ensure you had a sample containing 0.05 mol? Number of moles (N) = molarity (C) × volume of solution (V) (dm3) N = CV 0.05 = 2 × V 0.05 V = _____ ​ ​= 0.025 dm3 2 Multiply by 1000 to give answer in cm3 0.025 × 1000 = 25 cm3 4 Calculate the number of moles of HCl in 20 cm3 of a 2 mol dm–3 solution of HCl(aq). Convert 20 cm3 to dm3 by dividing by 1000. 20 ______ ​ ​= 0.02 dm3 1000 Use the equation Number of moles (N) = molarity (C) × volume of solution (V) (dm3) N = CV 0.02 × 2 = 0.04 mol of HCl in solution Using a chemical equation to calculate the quantities of reactants and products Chemical equations can be used to calculate the quantities of reactants and products. Here is an example. Calcium chloride can be produced by reacting calcium carbonate with hydrochloric acid. This is the equation for the reaction. CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l) Note that the equation includes state symbols. A solid substance is indicated by (s), a solution is indicated by (aq), a liquid is indicated by (l) and a gas is indicated by (g). The equation shows that one mole of calcium carbonate reacts with two moles of hydrochloric acid. One mole of calcium chloride is produced as well as one mole each of carbon dioxide and water. This is an example of stoichiometry. Key term Stiochiometry – involves using the relationships between the reactants and the products in a chemical reaction to work out how much product will be produced from given amounts of reactants. 19  Worked Example 1 Calculate the expected mass of calcium chloride produced when 50 g of calcium carbonate is reacted with excess hydrochloric acid. Ar (H) = 1, Ar (C) =12, Ar (0) = 16, Ar (Cl) = 35.5, Ar (Ca) = 40 One mole of CaCO3 produces one mole of CaCl2. You know this from the balanced equation CaCO3+ 2HCl → CaCl2+ CO2+H2O. This shows a one to one (1:1) ratio. Add up the relative atomic masses for each compound. 40 +12 + (3 × 16) g = 100 g of CaCO3 produces 40 + (35.5 × 2) g = 111 g of CaCl2. As one mole of CaCO3 produces one mole of CaCl2 then 100 g CaCO3 produces 111 g CaCl2. In this case, only 50 g of CaCO3 was used so 111 50 g CaCO3 produces _____​× 50 g CaCl2. 100 50 g CaCO3 produces 55.5 g CaCl2. You could say that only ​__12 ​a mole of CaCO3 was used so therefore only half the amount of CaCl2 would be produced and this would give the same answer of 55.5 g. This is the theoretical mass. 2 Calculate the expected mass of water if 10 g of oxygen is reacted with excess hydrogen. Ar (H) = 1, Ar (O) = 16 Use a balanced equation to find out the ratio between oxygen and water. 2H2 +O2 → 2H2O So one mole of oxygen gives 2 moles of water. This is a 1:2 ratio. Add up the relative atomic masses for each substance. Remember there will be two lots of water. 2 × 16 g = 32 g of O2 produces 2 × (2 × 1) +16 g = 36 g of H2O 32 g O2 produces 36 g H2O So 36 10 g O2 produces ___​× 10 g of H2O 32 10 g O2 produces 11.25 g H2O Assessment practice 1.5 Silver iodide is used in the manufacture of photographic paper. Calculate the theoretical yield of silver iodide for 34 g of silver nitrate reacting with excess sodium iodide. The equation for the reaction is as follows. AgNO3 (aq)+ NaI (aq) → AgI (s) + NaNO3 (aq) Ar (N) = 14, Ar (0) = 16, Ar (Ag) = 108, Ar (I) = 127 Percentage yields The theoretical mass is the amount of product you can produce in a reaction. In most reactions it is unlikely that the total amount of product possible is made. 20 Principles and Applications of Science 1  Learning aim A UNIT 1 Some may be lost in transferring product from one vessel to another. Some of the Principles and Applications of Science 1 Key terms reactants or products may react with impurities. In reversible reactions, products react to become the reactants and so are not all extracted from the reaction system. Theoretical mass – the Chemists need to know how efficient their reaction process is so they calculate the expected amount of product percentage yield. from a reaction calculated from the balanced equation. The percentage yield is the actual mass compared to the theoretical mass. An efficient process would give a percentage yield as close to 100% as possible. Percentage yield – the actual amount of yield The formula for calculating percentage yield is: worked out as a percentage actual number of moles of the theoretical yield. Percentage yield = _____________________________     ​× 100% expected number of moles Reversible reaction – a It can also be calculated as: reaction where the reactants react to form products and actual mass Percentage yield = _________________ ​× 100% the products simultaneously theoretical mass react to re form the reactants. The first step in working out percentage yield is to measure accurately the mass of s, for example, in NaCl. product that you have obtained. How accurate your measurements are may depend on the equipment you have, but the mass should be measured to at least two decimal places. You should be able to use a top pan balance for this. If you are using small quantities, or if you want more accurate measurements, you may use a chemical balance which measures to 3 decimal places. Once you have measured the mass of your product, you can work out how many moles you have produced. You can then divide this by the number of moles you were expecting to obtain and multiply by 100. If you are using solutions, then you will need to calculate the number of moles for the volume of solution used. Calculating concentration times volume, CV, will give you the number of moles in the volume of solution used. This equation can be rearranged to find out what volume of a known concentration of solution is needed in a reaction. Worked Example 1 When 50 g of calcium carbonate is reacted with excess hydrochloric acid solution to make calcium chloride, the theoretical yield is 55 g.When the reaction was carried out, only 44 g of calcium chloride was produced. Calculate the percentage yield of calcium chloride. actual yield Percentage yield = ​_________________    ​× 100%    theoretical yield 44 ___ Percentage yield = ​× 100% 55 Percentage yield is 80% 2 When I mole of oxygen reacts with excess hydrogen, 2 moles of water should be produced. When this reaction was carried the actual yield was 1.8 moles of water. Calculate the percentage yield. actual number of moles Percentage yield = _____________________________     ​× 100% expected number of moles 1.8 Percentage yield = ____ ​ ​× 100% 2 Percentage yield is 90% 21  22 Principles and Applications of Science 1 0 1 4 H He hydrogen helium Ke y 1 2 7 9 mass number 11 12 14 16 19 20 Li Be B C N O F Ne The periodic table atomic symbol lithium beryllium name boron carbon nitrogen oxygen fluorine neon 3 4 atomic (proton) number 5 6 7 8 9 10 23 24 27 28 31 32 35.5 40 Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon 11 12 13 14 15 16 17 18 39 40 45 48 51 52 55 56 59 59 64 65 70 73 75 79 80 84 ▸▸ Figure 1.18: A section from the periodic table K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 relation to properties 85 88 89 91 93 96 101 103 106 108 112 115 119 122 128 127 131 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 133 137 139 178 181 184 186 190 192 195 197 201 204 207 209 Cs Ba La* Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium lanthanum hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Fr Ra Ac* francium radium actinium 87 88 89 A2 Production and uses of substances in  Learning aim A UNIT 1 Periods 1, 2, 3 and 4 Principles and Applications of Science 1 Key term The periodic table (see Figure 1.18) shows all the chemical elements arranged in Atomic number – the order of increasing atomic number. Chemists can use it to predict how elements will number of protons in an behave, or what the physical or chemical properties of the element may be. atom. (This is the same as the A laboratory technician needs to be very familiar with the periodic table. It is an number of electrons in the information sheet on all the elements and their properties. atom.) The elements on the periodic table are organised into groups (vertical columns) and periods (horizontal rows). Chemical properties are similar for elements in the same group. The atomic number increases as you move from left to right across a period. This is because each successive element has one more proton than the one before. ▸ Table 1.3: Characteristics of each period Discussion Period Characteristics Look at the periodic table and 1 Contains hydrogen and helium. Both are gases. The electrons in these two write down five key features elements fill the 1s orbital. Helium only has two electrons and, chemically, of the periodic table. Work in helium is unreactive. Hydrogen readily loses or gains an electron, and so can pairs and try to list the names behave chemically as both a group 1 and a group 7 element. Hydrogen can form compounds with most elements and is the most abundant chemical of any groups in the periodic element in the universe. table as you can. Discuss any 2 Contains eight elements, lithium, beryllium, boron, carbon, nitrogen, oxygen, facts you know about the fluorine and neon. The outer electrons in these elements fill the 2s and 2p elements in the groups you orbitals. Nitrogen, oxygen and fluorine can all form diatomic molecules. Neon have listed. These may be is a noble gas. Carbon is a giant molecular structurre. properties of the elements or 3 Contains eight elements, sodium, magnesium, aluminium, silicon, trends within groups. phosphorus, sulfur, chlorine and argon. The outer electrons in these elements fill the 3s and 3p orbitals. 4 Contains 18 elements, from potassium to krypton. The first row of the transition elements is in this period. The outer electrons on these elements fill the 4s, 4p and 3d orbitals. Groups – s block, p block, d block The periodic table is also organised by element blocks. An element block is a set of elements in groups that are next to each other. Element blocks are named for the orbital the highest energy electrons are in for that set of elements. Groups 1 and 2 of the periodic table are in s block. Groups 3 to 7 and group 0 make up p block. This block contains all the non-metals except for hydrogen and helium. The transition metals are in the d block. For example, carbon had electronic structure of 1s2 2s2 2p2. The highest energy electron in carbon is in a p orbital and therefore carbon is a p block element. Assessment practice 1.6 Explain why calcium is an s block element. P ause point Summarise what you have learned about the periodic table. Hint Consider what you know about groups, periods and trends. Extend Choose three elements in different areas of the table, and explain why their atomic structure and properties means they are in the position they are in. 23  Physical properties of elements Atomic radius The radius of an atom changes depending on what is around it. The only way to measure the radius is to measure the distance between the nuclei of two touching atoms and divide by two. The atomic radius decreases across the period from left to right. Across the group, more protons and electrons are added. However, the extra electrons are added to the same s and p sub-shells and so the size does not increase. The extra protons increase nuclear charge. The increased nuclear charge attracts the extra electrons and pulls them closer to the nucleus. This leads to a decrease in atomic radius. As you go down a group the atomic radii increases. This is because the extra electrons are added to additional shells and so the radius increases. Although nuclear charge increases, the number of inner shells increases and so the nuclear charge is shielded more. This means that the atomic radius increases. The trend is slightly different for the transition metals. The atomic radii get slightly smaller as you go across the start of the transition metals but then the radii stay very similar. This is because the additional nuclear charge is balanced by the extra shielding by the 3d electrons of the outer 4s sub-shell. Periodic Trends in Atomic Radii 300 Cs Fr Rb 250 K Atomic Radius (om) Es 200 La Ac Na Y Tl Sc In Hf 150 Li Rf Xe Rn Hs Ga Kr 100 Ar H Ne 50 He 0 0 10 20 30 40 50 60 70 80 90 100 110 Atomic number ▸ Figure 1.19 Ionic radius The trends in ionic radius down a group follow a similar pattern to the trend for atomic Key terms radius down a group. This is because extra electrons are added to extra shells as you go down the group therefore giving a larger size. Isoelectronic – having the Cations have a smaller radius than their corresponding atom. As you go across a same numbers of electrons. period, the cations all have the same electronic structure. They are isoelectronic, Cations – ions with a positive therefore although number of electrons remains the same, the nuclear charge charge. increases, for example, Na+, Mg2+, A3+. However, the number of protons increases Anions – ions with a negative across the period. This pulls the electrons more strongly to the centre of the ion so the charge. ionic radii of the cations decreases as you go across the period. 24 Principles and Applications of Science 1  Learning aim A UNIT 1 Anions have a larger radius than the corresponding atom because there is more Principles and Applications of Science 1 repulsion between the extra electrons. As you go across the period, the anions are all isoelectronic, for example, N3−, O2−, F−. They have more electrons not fewer. The number of protons still increases as you go across the period whilst the number of shells and electrons stays the same so the ionic radius of the anions also decreases as you go across the period. Electronegativity Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. It increases as you go across a period. It decreases as you go down a group. This means that fluorine is the most electronegative element. The Group 0 gases such as argon that do not form bonds do not have electronegativity that can be reliably determined, because they do not form compounds/bonds. Electronegativity depends on the number of protons in the nucleus, the distance from the nucleus of the bonding pair of electrons and how much shielding there is from inner electrons. As you go across the period, the bonding pair of electrons will be shielded by the same number of electrons. However, the number of protons will increase, so the group 7 element will be more electronegative than the group 1 element. As you go down a group, there is more shielding from inner electrons and the bonding pair of electrons are further from the nucleus. This adds up to less pull on the bonding pair from the positive charge of the nucleus and so electronegativity decreases. Periodic Trends in Electronegativity 4.00 F Electronegativity (Pauling) 3.50 Cl Kr 3.00 Xe 2.50 H At 2.00 Al Ga In Tl Pa 1.50 La Lr Sc Y Hf 1.00 Li Na K Rb Cs Fr 0.50 0.00 0 10 20 30 40 50 60 70 80 90 100 Atomic number ▸ Figure 1.20 Trends in the periodic table are usually identified across periods or down groups. However, there are often similarities between elements that are diagonal to each other. For example, beryllium and aluminium have identical electronegativity. There is an increase across the period from group 2 to group 3 but then as electronegativity decreases down a group, this increase is balanced out. Other diagonal pairs also have similar electronegativity, for example, lithium and magnesium. These similarities mean they form similar bonds and may show similar chemistry. P ause point Explain how electron affinity affects the reactivity of atoms. Hint You will need to use the terms nuclear charge, shells, and shielding. Extend Use this information to explain why potassium is a very reactive metal. 25  First ionisation energy and reasons for trends Key terms First ionisation energy is the minimum energy needed for one mole of the outermost electrons to be removed from one mole of atoms in a gaseous state. One mole of Periodicity – the repeating positively charged ions is formed. First ionisation energies of the elements in a period show pattern seen by the elements periodicity. There is an overall trend of first ionisation energy increasing across the period. in the periodic table. This trend is shown in the graph in Figure 1.21. First ionisation energy – the energy needed for one mole It takes more energy to remove an electron as you go across the period. This is of electrons to be removed because the number of protons increase across the period so the positive charge from one gaseous atoms. For on the nucleus increases. This means that the force of attraction pulling on the example, the equation shows outer electron increases. However, you can see there is not a steady increase in first potassium losing one electron ionisation energy. There is a pattern in the dips and increases for each period. to become a positive ion. Across period 2, it dips at group 3 and group 6 elements. You can see the same pattern in K(g) → K+(g) + e−. period 3. Period four is a little different because it also contains the transition elements. Periodicity of 1st Ionisation Energy for Periods 1-4 (and start of Period 5) 2500 2250 2000 1750 1st IE in kJ/mol 1500 1250 1000 750 500 250 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Atomic number and symbol ▸ Figure 1.21: First ionisation energies of elements in periods 1–4 For periods 2 and 3, there is a pattern which suggests that electrons removed from the third energy level are arranged in different sub-levels. Across period 2, the first electrons are removed from the 2s sub-level. The value for beryllium is higher than for lithium because beryllium has one more proton in its nucleus. There is a decrease for boron where the electron is taken from the 2p sub-level. This is a higher energy level than the 2s sub-level and so the electron is easier to remove i.e. the 2s sub-shell shields the 2p sub-shell, making it easier to remove an electron. Carbon and nitrogen show the expected increase because the electron removed from each element is in the same 2p sub-level. These electrons occupy orbital on their own, they are all unpaired. There is a second dip in the first ionisation energy at oxygen. Here the electron removed is also in the 2p sub-level, but it is paired with another electron in that level. The electrostatic repulsion between the two electrons in the orbital means that it is easier to remove this electron. The first ionisation energy increases then for fluorine and neon because they have increasing positive charge. A similar pattern is seen for period 3 where the electrons are removed from the 3