Chemistry IGCSE Textbook

NES/Chemistry/IGCSE  Example 15: chlorine gas: A single covalent bond formed between two atoms both with a valency of 1 (e.g. Group VII and hydrogen), has two electrons in the overlap. One from each atom. Dot-Cross Diagram Stick Diagram Cl Cl Cl - Cl Double Bonds and Triple Bonds A double covalent bond formed between two atoms both with a valency of 2 (e.g. oxygen), has four electrons in the overlap. Two from each atom. A triple covalent bond formed between two atoms both with a valency of 3 (e.g. nitrogen), has six electrons in the overlap. Three from each atom. Summary of Covalent Bonds for Non-Metals Atom Valence Electrons Valency Number of Bonds per Atom Hydrogen 1 1 1 Carbon 4 4 4 Nitrogen 5 3 3 Oxygen 6 2 2 Fluorine 7 1 1 Chlorine 7 1 1 Bromine 7 1 1 Iodine 7 1 1 50

NES/Chemistry/IGCSE Examples of Covalently Bonded Molecules Name and Stick Diagrams Dot and Cross Diagram Structure Formulae simple molecular hydrogen gas HH HH (H2) chlorine gas Cl Cl Cl Cl xx simple molecular (Cl2) xxxx simple molecular OO OO simple molecular oxygen gas xx simple molecular (O2) NN xx NN xx simple molecular nitrogen gas H Cl H Cl simple molecular (N2) H H hyrdogen HCH HCH chloride gas H O H (HCl) O HH HH methane gas (CH4) water (H2O) 51

NES/Chemistry/IGCSE Name and Stick Diagrams Dot and Cross Diagram Structure Formulae OCO xx carbon HH xx dioxide gas xx CC xx (CO2) HH xx simple molecular xx simple molecular ethene gas OC O simple molecular (C2H4) simple molecular H H C C H H methanol H H xx (CH3OH) HCOH H C x O xH H H xx ammonia gas N HNH (NH3) HH H H phosphorus P Cl P Cl simple molecular trichloride Cl Cl Cl (PCl3) Cl 52

NES/Chemistry/IGCSE Simple Molecular Water is a small molecule composed of two hydrogen atoms and one oxygen atom covalently bonded. As one molecule is made from a small number of atoms, water is therefore described as simple molecular. Other examples of simple molecules are: carbon dioxide (CO2) hydrogen chloride (HCl) nitrogen dioxide (NO2) carbon monoxide (CO) ammonia (NH3) sulphur dioxide (SO2) Physical Properties Property Reasons Liquids, gases or low melting point solids at room temperature. The molecules are simple molecular with only weak intermolecular van der Waals Low melting and boiling points forces of attraction. Soluble in organic solvents such as Due to weak intermolecular van der Waals benzene. Very slightly soluble or forces of attraction a low amount of energy insoluble in water. has to be supplied to overcome these relatively weak forces. Do not conduct electricity when Covalent molecular (non-polar) substances molten or dissolved dissolve in covalent (non-polar) solvents such as hexane or benzene. Water is a polar solvent and so will not usually dissolve covalent molecular substances. There are no mobile electrons or mobile ions. Intermolecular forces are forces of attraction between molecules e.g. van der Waals. Intramolecular bonds are found inside molecules e.g. covalent bonds. 53

NES/Chemistry/IGCSE 3.2.4 Macromolecules Macromolecules are giant arrangements of covalently bonded atoms. They are: 1. Diamond (an allotrope of carbon) 2. Graphite (an allotrope of carbon) 3. Fullerenes (an allotrope of carbon) 4. Silicon Dioxide (a compound) Allotropes of Carbon 1. Diamond Structure: Each carbon forms 4 strong covalent rigid bonds with other carbon atoms forming a tetrahedral arrangement giving rise to a macromolecular structure. Property Physical Properties High melting point/very hard Poor electrical conductor Reasons All the intramolecular covalent bonds are very strong and rigid, so a large amount of energy has to be supplied to overcome the strong bonds. This is why diamond is hard and has a high melting point. Diamond does not contain any mobile electrons or mobile ions and so does not conduct electricity. 54

NES/Chemistry/IGCSE 2. Graphite Structure: Each carbon atom forms 3 strong van der Waals forces covalent bonds with other carbon atoms forming layers of flat hexagonal rings. There are mobile, delocalised carbon atoms electrons between the flat layers of covalent bonds carbon atoms. This is where carbon's fourth valence electron is found. The layers are only bonded together by weak van der Waals forces of attraction, so the layers can slide over each other. Physical Properties Property Reasons The layers are only bonded together by Lubricant weak van der Waals forces of attraction, and so the layers can slide over each other. Electrical conductor There are mobile electrons between the layers. A lubricant is a substance, which reduces friction between moving parts. Comparing Diamond and Graphite Properties of Diamond Properties of Graphite Does not conduct electricity (no mobile Conducts electricity (has mobile electrons or mobile ions) electrons) A very hard substance Soft and slippery A transparent crystal which sparkles in light A grey-black shiny solid Uses of Diamond Uses of Graphite To make pencil “lead” Used for cutting rocks used as a lubricant Used in glass engraving carbon brushes in motors 55

NES/Chemistry/IGCSE 3. Fullerenes Fullerenes are another allotrope of carbon. They contain 60 carbons in a spherical structure and have the formula, C60. Silicon dioxide Even though silicon and carbon are in the same group, and so have similar chemical properties, their physical properties differ. Carbon dioxide is simple molecular and is a gas at room temperature, however, silicon dioxide is a macromolecule and is a solid at room temperature. Structure: Each silicon atom is covalently bonded to four oxygen atoms, and O O each oxygen is bonded to two silicon atoms by strong rigid covalent Si bonds in a tetrahedral arrangement giving rise to a 3-dimensional Si macromolecular structure, like diamond. The substance is hard as OO all the intramolecular bonds are strong covalent bonds. SiO2 has the same properties as diamond. 56

NES/Chemistry/IGCSE 3.2.5 Metallic Bonding Metals are held together by metallic bonds forming giant metallic structures. Metallic structure is a lattice of positive metal ions surrounded by a mobile „sea‟ of valence electrons in a 3-dimensional structure. The valence electrons of each metal atom in the crystal are delocalised and move around the positive ions. This means that the metal is a giant lattice consisting of regular rows of positive ions surrounded by a “sea of electrons”. A strong non-directional electrostatic force of attraction operates between the delocalised valence electrons and the lattice of positive ions. Structure: e e e e e delocalised mobile e e valence electrons e ee e that have separated from the atoms e ee e lattice of e metal cations e 57

NES/Chemistry/IGCSE Physical Properties Property Reasons Metals usually have high melting points Due to the strong non-directional and electrostatic forces of attraction Metals are usually hard operating between the positive ions and the Metals usually have a high density delocalised mobile valence electrons, a large amount of energy has to be supplied Metals are good electrical conductors to overcome the strong forces holding the ions in fixed positions in the giant lattice. Metals are malleable Large numbers of ions are packed together very closely in a regular arrangement. Metals conduct electricity as they have mobile valence electrons. In a metal the valence electrons move randomly, however when a potential difference is applied across the metal, the „sea‟ of valence electrons move in one direction only. The layers of ions slide over each other due to the continual non-directional electrostatic force of attraction between the lattice of positive ions and the delocalised mobile valence sea of electrons, changing the shape of the metal without breaking. Malleable means that the metal can be bent into shape without breaking. e e e e e e e e e e e force e e e e e 58 e

NES/Chemistry/IGCSE Topic 4.1 - Stoichiometry  Use the symbols of the elements and write the formulae of simple compounds (see Topic 3 notes)  Deduce the formula of a simple compound from the relative numbers of atoms present  Deduce the formula of a simple compound from a model or a diagrammatic representation  Construct word equations and simple balanced chemical equations  Define relative atomic mass, Ar as the average mass of naturally occurring atoms of an element on a scale where the 12C atom has a mass of exactly 12  Define relative molecular mass, Mr as the sum of the relative atomic masses (relative formula mass will be used for ionic compounds)  Determine the formula of an ionic compound from the charges on the ions present  Construct equations with state symbols, including ionic equations  Deduce the balanced equation for a chemical reaction, given relevant information Definitions Ar is the mass of one atom of an element relative to one twelfth of the mass of one atom of 12C Mr is the sum of the relative atomic masses of all the atoms in one molecule of the compound 59

NES/Chemistry/IGCSE Naming Compounds Compounds take their names from the elements they are made up from. The rules for naming compounds are:  The metal, or positive ion, goes first and the non-metal goes second.  If a compound does not have a metal, then the non-metal that is most to the left on the Periodic Table goes first.  The (second) non-metal's name changes its ending. Non-Metal Element Changed Name in a Compound Fluorine Fluoride Chlorine Chloride Bromine Bromide Iodine Iodide Oxygen Oxide Sulphur Sulphide Nitrogen Nitride Phosphorous Phosphide Carbon Carbide Hydride Hydrogen So a compound containing calcium and oxygen would be called calcium oxide and a compound containing lithium and hydrogen would be called lithium hydride and a compound containing phosphorus and chlorine would be called phosphorus chloride. When naming transition metal compounds it is important to include the valency of the transition metal cation. This is because transition metals have variable valencies. Copper sulphate is actually called copper(II) sulphate because the copper has a valency of 2+. Note - organic molecules (see Topic 14) are named by a different system. 60

NES/Chemistry/IGCSE Formula of Compounds To work out the formula of a compound you must know the valency of the elements in the compound. Most elements' valency can be found from the periodic table. The exceptions to this are transition metals, other elements with variable valencies and polyatomic ions. Cations Anions Name Formula Valency Name Formula Valency hydrogen H+ 1+ hydride H- 1- silver(I) Ag+ 1+ nitrate 1- mercury(I) Hg+ 1+ nitrite NO3– 1- copper(I) Cu+ 1+ hydroxide NO2– 1- hydrogen OH– gold Au+ 1+ carbonate 1- manganate(VII) HCO3– ammonium NH4+ 1+ ethanoate 1- MnO4- 1- CH3COO– Name Formula Valency Name Formula Valency zinc Zn2+ 2+ carbonate CO32– 2- mercury(II) Hg2+ 2+ sulphate SO42– 2- copper(II) Cu2+ 2+ sulphite SO32– 2- iron(II) Fe2+ 2+ sulphide S2– 2- lead(II) Pb2+ 2+ silicate SiO32– 2- manganese(II) Mn2+ 2+ chromate(VI) 2- Cr2O72- Name Formula Valency Name Formula Valency aluminium Al3+ 3+ nitride N3- 3- iron(III) Fe3+ 3+ phosphate 3- chromium(III) Cr3+ 3+ PO43– 61

NES/Chemistry/IGCSE Symbols and Formula All elements have symbols as a quick way of writing their names Some elements are diatomic and two atoms are found in each molecule: Element Formula of Diatomic Molecule Hydrogen Nitrogen H2 Oxygen N2 Fluorine O2 Chlorine F2 Bromine Cl2 Br2 Iodine I2 Astatine At2 Compounds have a formula that tells you the number and type of atoms in the molecule. The formula is always in a fixed ratio. For example, water is H2O which means that water contains hydrogen and oxygen bonded together. It also means that the H:O ratio is always 2:1. Working Out Formula Step 1 - Write down the symbols of each element Step 2 - Write down the valency of each element (number only, not the sign) Step 3 - Swap the valencies over Step 4 - Simplify the numbers if possible  Example 1: sodium chloride Step 1 Symbol Na Cl 11 Step 2 Valency Na1Cl1 NaCl Step 3 Swap Step 4 Simplify 62

NES/Chemistry/IGCSE  Example 2: magnesium bromide Step 1 Mg Br Step 2 21 Step 3 Mg1Br2 Step 4 MgBr2  Example 3: potassium oxide Step 1 KO Step 2 12 Step 3 K2O1 Step 4 K2O  Example 4: calcium sulphide Step 1 Ca S Step 2 22 Step 3 Ca2S2 Step 4 CaS  Example 5: germanium oxide Step 1 Ge O Step 2 42 Step 3 Ge2O4 Step 4 GeO2 63

NES/Chemistry/IGCSE Polyatomic Ions These are ions made of more than one element, for example SO42–, or CO32–. These ions need (brackets) if they have a number.  Example 6: calcium hydroxide Step 1 Ca OH Step 2 21 Step 3 Ca1(OH)2 Step 4 Ca(OH)2  Example 7: magnesium carbonate Step 1 Mg CO3 Step 2 22 Step 3 Mg2(CO3)2 Step 4 MgCO3 Brackets are not needed as there is no other number following CO3 after being simplified. 64

NES/Chemistry/IGCSE Working Out the Number of Elements Present in Compounds from the Formula If we have the formula of a compound we can work out the number of different elements present in the compound by counting the number of different symbols in the formula. H2SO4 If no subscript is subscript after each present then it is element tells us the taken to be one number of atoms of that type So sulphuric acid, H2SO4 has 2 hydrogen atoms, 1 sulphur atom and 4 oxygen atoms all bonded together in one molecule.  Example 8: Formula Number of Different Total Number of Atoms in Elements Molecule HCl 2 2 HNO3 3 5 H2SO4 3 7 CuSO4. 5H2O 4 21 65

NES/Chemistry/IGCSE Types of Formula 1. Structural formula shows all the atoms and bonds in a molecule. 2. Molecular formula shows how which atoms are in a molecule and how many there are. 3. Empirical formula is the simplest whole number ratio of the molecular formula.  Example 9: Structural Formula Molecular Empirical Name of Formula Formula Compound HH O HCCC C3H6O3 CH2O lactic acid H O HO H H2C CH3 CH limonene H2C CH2 C10H16 C5H8 HC CH2 C CH3 66

NES/Chemistry/IGCSE Relative Atomic Mass (Ar) This is the relative mass of an atom compared to a standard (carbon-12). Ar = 1 mass of one atom of an element . 12 the mass of one atom of carbon- 12 Calculating Average Isotopic Mass Average mass = (mass 1 x % isotope 1) + (mass 2 x % isotope 2) 100 100  Example 10: We know that about 75% of all chlorine is 35 Cl and only 25% 17 is 37 Cl 17 Average mass = (35 x 75/100) + (37 x 25/100) = 35.5 The average mass of the isotopes is 35.5 so the Ar of Cl is 35.5 67

NES/Chemistry/IGCSE Relative Molecular Mass (Mr) This is the relative mass of a compound, or molecule compared to a standard (carbon-12). To do this multiply each Ar by the number of that type of atom present and then add all the masses together.  Example 11: water Molecule Formula Atoms Present Number of Each Ar Total Mass water H2O Type of Atom H 2 1 2 O 1 16 16 Total 18 Mr of water is 18  Example 12: calcium carbonate Molecule Formula Atoms present Number of each Ar Total Mass type of atom calcium CaCO3 Ca 1 40 40 carbonate C 1 12 12 O 3 16 48 Total 100 Mr of calcium carbonate is 100 68

NES/Chemistry/IGCSE Symbol Equations A chemical equation is balanced when there are equal number of atoms and charges on both sides of the equation. Step 1 - Write symbols, or formula for each chemical Step 2 - Count the number of atoms of each element on the left and on the right Step 3 - Balance each element by adding either more reagent, or more product.  Example 13: magnesium + hydrochloric acid  magnesium chloride + hydrogen. Step 1 Mg + HCl  MgCl2 + H2 Step 2 1 magnesium 1 magnesium 1 hydrogen 2 hydrogen 1 chlorine 2 chlorine Step 3 Mg + 2 HCl  MgCl2 + H2 By adding an extra hydrochloric acid, all of the atoms now balance. 1 magnesium 1 magnesium 2 hydrogen 2 hydrogen 2 chlorine 2 chlorine In step 3 we always use numbers before the symbol, or formula to show how many molecules there are of that chemical. 69

NES/Chemistry/IGCSE  Example 14: methane burning in oxygen to make carbon dioxide and water. Step 1 CH4 + O2  CO2 + H2O Step 2 1 carbon 1 carbon 4 hydrogen 2 hydrogen 2 oxygen 3 oxygen Step 3 CH4 + 2O2  CO2 + 2H2O By adding an extra oxygen and an extra water, all of the atoms now balance. 1 carbon 1 carbon 4 hydrogen 4 hydrogen 4 oxygen 4 oxygen Use of Fractions in Equations  Example 15: ethane burning in oxygen to make carbon dioxide and water. Step 1 C2H6 + O2  CO2 + H2O Step 2 2 carbon 1 carbon 6 hydrogen 2 hydrogen 2 oxygen 3 oxygen Step 3 2 C2H6 + 7 O2  4 CO2 + 6 H2O We can also write the equation like this C2H6 + 7 O2  2 CO2 + 3 H2O 2 70

NES/Chemistry/IGCSE State Symbols When writing equations, it is sometimes required to write state symbols after each compound, or element: (s) solid (l) liquid (g) gas (aq) aqueous, which means dissolved in water Ionic Equations  Example 16: Write an ionic equation for the displacement / redox / exothermic reaction between aluminium powder and copper(II) sulphate solution Step 1 Write a symbol equation 2Al(s) + 3CuSO4(aq)  Al2(SO4)3(aq) + 3Cu(s) Step 2 Write the equation with all aqueous solutions ionised, or dissociated. To do this split each aqueous compound into its ions: 2Al(s) + 3Cu2+(aq) + 3SO42–(aq)  2Al3+(aq) + 3SO42–(aq) + 3Cu(s) Step 3 Cancel out any species on both sides of the equation which have not changed their valency, or state of matter. These are called spectator ions. 2Al(s) + 3Cu2+(aq) + 3SO42–(aq)  2Al3+(aq) + 3SO42–(aq) + 3Cu(s) Step 4 Write out what is left 2Al(s) + 3Cu2+(aq)  2Al3+(aq) + 3Cu(s) 71

NES/Chemistry/IGCSE Balancing Redox Equations  Example 17: Balance the redox equation Al + Cu2+  Al3+ + Cu Step 1 Split the equation into two ionic half-equations - one for oxidation and one for reduction. Al  Al3+ + 3e– Cu2+ + 2e–  Cu Step 2 Balance each half-equation so there are the same number of electrons in each equation. 2Al  2Al3+ + 6e– 3Cu2+ + 6e–  3Cu Step 3 Cancel out the electrons Step 4 2Al  2Al3+ + 6e– 3Cu2+ + 6e–  3Cu Add the two ionic half-equations together 2Al + 3Cu2+  2Al3+ + 3Cu 72

NES/Chemistry/IGCSE Topic 4.2 - The Mole Concept  Define the mole and the Avogadro constant  Use the molar gas volume, taken as 24dm3 at room temperature and pressure  Calculate stoichiometric reacting masses, volumes of gases and solutions, and concentrations of solutions expressed in g/dm3 and mol/dm3 (Calculations involving the idea of limiting reactants may be set. Questions on the gas laws and the conversion of gaseous volumes to different temperatures and pressures will not be set.)  Calculate empirical formulae and molecular formulae  Calculate percentage yield and percentage purity Definitions A mole is the amount of substance which contains 6 x 1023 atoms, ions or molecules The limiting reactant in a particular experiment is the reactant that governs the maximum amount of product that can be formed Empirical formula shows the simplest whole number ratio of atoms present Molecular formula shows the actual number of atoms of each element present 73

NES/Chemistry/IGCSE Moles It was discovered that if you have 6 x 1023 particles of any chemical, then the mass in grams will be the same as the chemicals mass number (Ar or Mr)  Example 1: Calculate the mass of 6 x 1023 particles of calcium. Calcium has a mass number of 40, so the mass is 40g  Example 2: Calculate the mass of 6 x 1023 particles of water. Water has a mass number of (16 + 1 + 1), so the mass is 18g The number 6 x 1023 is the Avogadro constant. 1 mole = 6 x 1023 particles Using Relative Atomic Mass (Ar) Ar is the same as the element's mass number and can be found on the Periodic Table. It has the units g/mol, which just means how many grams there are in 1 mole. We can use the Ar to work out the mass of one mole of an element. The Ar of sodium is 23g/mol. So if we wanted to weigh out 1 mole of sodium atoms then we simply weigh out 23g. This will contain 6 x 1023 atoms (Avogadro's constant). The relative atomic mass (Ar), in grams, of any element contains 1 mole of atoms. Number of moles = mass used mass used relative atomicmass mole Ar 74

NES/Chemistry/IGCSE  Example 3: How many moles are there in 46g of sodium? Number of moles = 46 = 2 moles of sodium atoms 23 Using Relative Molecular Mass (Mr) The relative molecular mass (Mr) in grams, of any compound contains 1 mole of molecules Number of moles = mass used mass used relative molecular mass mole Mr  Example 4: How many moles are there in 1.8g of water? Number of moles = 1.8 = 0.1 moles of water molecules 18 75

NES/Chemistry/IGCSE Using the Mass of Reacting Material This combines using the number of moles formula above, with an equation. Step 1 Write a balanced equation Step 2 Find the number of moles of 1 of the chemicals Step 3 Use the mole ratio to find the number of moles of the other chemicals Step 4 Calculate the mass of the other chemicals  Example 5: 20g of sodium hydroxide reacts with an excess of hydrochloric acid to form sodium chloride and water. Calculate the mass of salt made. Step 1 NaOH + HCl  NaCl + H2O Step 2 Number of moles NaOH = 20 = 0.5 moles 40 Step 3 1 1 1 1 mole ratio 0.5 0.5 0.5 0.5 moles Step 4 Mass sodium chloride = Mole x Ar Mass sodium chloride = 0.5 x 58.5 = 29.25g  Example 6: Iron(III) oxide is reduced to iron using carbon monoxide in the blast furnace. How much iron will be produced from 100g of iron(III) oxide? Step 1 Fe2O3 + 3CO  2Fe + 3CO2 Step 2 100 Number of moles Fe2O3 = 160 = 0.625 moles Step 3 1 3 2 3 mole ratio 0.625 1.875 1.25 1.875 moles Step 4 Mass iron = Mole x Ar Mass iron = 1.25 x 56 = 70g 76

NES/Chemistry/IGCSE  Example 7: Insoluble salts are made by precipitation. An equation for the preparation of barium sulfate is: FeSO4(aq) + BaCl2(aq)  BaSO4(s) + FeCl2(aq) This reaction can be used in an experiment to find the value of x in the formula for hydrated iron(II) sulfate crystals (FeSO4.xH2O) A known mass of the crystals of hydrated iron (II) sulfate was dissolved in water. Excess barium chloride solution was added. The precipitated barium sulfate was filtered, washed and dried. Finally it was weighed. Mass of hydrated iron (II) sulfate crystals = 1.390g Mass of barium sulfate formed = 1.165g Calculate the value for x in FeSO4.xH2O. 77

Step 1 NES/Chemistry/IGCSE FeSO4 + BaCl2  BaSO4 + FeCl2 Step 2 1.165 Number of moles BaSO4 = 233 = 5 x10–3 moles Step 3 11 1 1 mole ratio 5 x10–3 5 x10–3 5 x10–3 5 x10–3 moles Step 4 rearranging mass FeSO4.xH2O = Mole x Mr Mr FeSO4.xH2O = mass used mole 1.390 Mr FeSO4.xH2O = = 278 g/mol 0.005 And we know that Mr FeSO4 = 152 g/mol So the mass of water is 278g – 152g = 126 g So Number of moles of water is mass used 126 = =7 Mr 18 Formula of hydrated iron(II) sulfate is FeSO4.7H2O 78

NES/Chemistry/IGCSE Limiting Reagent This is used when we have different numbers of moles of reagents. The reagent with the least number of moles is the limiting reagent and will all react (there will be none left after the reaction has finished). The other reagent will be in excess (there will be some left over after the reaction has finished). To calculate the limiting reagent you must calculate the number of moles of each reagent and use the mole ratio to compare values. Step 1 Write the equation for the reaction Step 2 Calculate the number of moles of a reagent Step 3 Calculate the number of moles of the other reagent Step 4 Use the mole ratio to compare which reagent has fewer moles - this is the limiting reagent  Example 8: 10g of calcium carbonate reacts with 29.4g of sulphuric acid. Which reagent is in excess? Step 1 CaCO3 + H2SO4  CaSO4 + H2O + CO2 Step 2 10 Number of moles CaCO3 = 100 = 0.1 moles Step 3 29.4 Number of moles H2SO4 = 98 = 0.3 moles Step 4 As the mole ratio is 1:1 then CaCO3 is the limiting reagent as it has fewer moles The limiting reagent is used to calculate the number of moles of the other reagents as it is all used up in the reaction. 79

NES/Chemistry/IGCSE Empirical Formula and Molecular Formula  Example 9: Glucose molecular formula C6H12O6 empirical formula CH2O So glucose, C6H12O6 is made up from 6 empirical units of CH2O. To calculate the empirical formula of a compound you must use the following steps: Step 1 Write down the elements in the compound Step 2 Write down the % or mass in grams of each element Step 3 Divide by the Ar to get the ratio of atoms Step 4 Divide by the smallest number to turn the atom ratio into a whole number for the formula And then to calculate the molecular formula you will also need: Step 5 number of empirical units = molecular mass empirical mass Step 6 molecular formula = empirical formula x number of units 80

NES/Chemistry/IGCSE  Example 10: A carbohydrate of molar mass 180g/mol contains by weight 40.0% carbon; 6.7% hydrogen and 53.3% oxygen. What is the empirical formula and molecular formula? Step 1 carbon hydrogen oxygen Step 2 40.0 6.7 53.3 Step 3 40 6.7 53.3 12 1 16 3.3 6.7 3.3 Step 4 3.3 6.7 3.3 3.3 3.3 3.3 1 21 So the empirical formula = CH2O with a mass of (12+2+16) = 30 Step 5 180 Step 6 number of empirical units = 30 molecular formula = CH2O x 6 so the molecular mass = C6H12O6 81

NES/Chemistry/IGCSE  Example 11: An oxide of phosphorus has a mass of 0.89g, which formed from 0.39g of phosphorus reacting with oxygen. Calculate the empirical formula and molecular formula? Step 1 phosphorus oxygen Step 2 0.39 (0.89-0.39) = 0.5 Step 3 0.39 0.5 31 16 0.0126 0.0313 Step 4 0.0126 0.0313 0.0126 0.0126 1 2.5 As the empirical formula must be whole numbers, we must double the ratio 25 So the empirical formula = P2O5 For ionic compounds the molecular formula = empirical formula So the molecular formula = P2O5 82

NES/Chemistry/IGCSE Mole Calculations for Gases 1 mole of any gas occupies a volume of 24dm3 at a temperature of 25oC and a pressure of one atmosphere (this is at r.t.p. - room temperature and pressure). If the question gives the volume of gas in dm3: number of moles of gas = volumeof gas(dm3 ) volume (dm3) 24 mole 24 If the question gives the volume of gas in cm3: Gas volume conversion 1dm3 = 1000cm3 number of moles of gas = volumeof gas(cm3 ) 24000  Example 12: Find the number of moles there are of ammonia gas at r.t.p. in a 200cm3 gas jar. number of moles = 200 24000 = 0.00833 moles of ammonia gas 83

NES/Chemistry/IGCSE  Example 13: A 10cm3 sample of alkene gas X needed 45cm3 of O2 for complete combustion and 30cm3 of CO2 were produced. By completing the following equation, work out the formula of gas X, which is a hydrocarbon. Because 1 mole of any gas at room temperature and pressure (r.t.p.) occupies a volume of 24dm3, we can compare volumes of gas without having to calculate the number of moles. C…..H….. + …..O2  …..CO2 + …..H2O 10 cm3 45 cm3 30 cm3 Using the volumes to balance the equation gives us 1 C…..H….. + 4½ O2  3 CO2 + .....H2O We know that there are 3 carbons in the hydrocarbon, and using the alkene general formula CnH2n we also know there are 6 hydrogens So the formula of the hydrocarbon is C3H6 84

NES/Chemistry/IGCSE Mole Calculations for Solutions  Molar concentration (molarity) is measured in mol/dm3  Mass concentration (molality) is measured in g/dm3 This can be used to work out the Mr of a solute: massconcentration Mr = molarconcentration Or the number of moles in a solution: Number of moles = volume of solution (in dm3) x molar concentration Number of moles = volumeof solution(cm3 ) x molar concentration 1000 mole volume conc (dm3) To calculate the concentration of a solution in a reaction you must use the following steps: Step 1 Write down the chemical equation Step 2 Write down the mole ratio Step 3 Calculate the number of moles of reagent Step 4 Use the mole ratio to calculate the number of moles of the other reagent Step 5 Calculate the concentration of the other reagent 85

NES/Chemistry/IGCSE  Example 14: 25.0cm3 of hydrochloric acid is neutralised by 30.0cm3 of a solution of sodium hydroxide of concentration 0.25mol/dm3. Find the concentration of the hydrochloric acid. Step 1 HCl + NaOH  NaCl + H2O Step 2 Step 3 11 11 Moles NaOH = volume x concentration 1000 Moles NaOH = 30.0 x 0.25 1000 = 7.5 x 10–3 moles Step 4 Moles HCl = 7.5 x 10–3 moles Step 5 Concentration HCl = mole = 7.5x 10–3 volume 0.025 Concentration HCl = 0.3mol/dm3 86

NES/Chemistry/IGCSE Percentage Yield The percentage yield of a product is the percentage of its theoretical yield achieved in practice. The theoretical yield of a product is the maximum calculated mass that can be obtained from a given mass of specified reactant. Percentage yield = mass producedx 100% theoretical yield Step 1 Write the equation Step 2 Calculate the number of moles of reagent Step 3 Use the mole ratio to work out the number of moles of the other chemicals Step 4 Calculate the theoretical mass of product Step 5 Calculate the percentage yield  Example 15: Copper(II) sulfate-5-water was prepared by the following reactions. CuO + H2SO4  CuSO4 + H2O CuSO4 + 5H2O  CuSO4.5H2O In an experiment, 25cm3 of 2.0mol/dm3 sulfuric acid was neutralised with an excess of copper(II) oxide. The yield of crystals, CuSO4.5H2O, was 7.3g. Calculate the percentage yield. Step 1 CuO + H2SO4  CuSO4 + H2O CuSO4 + 5H2O  CuSO4.5H2O Step 2 Number of mole of H2SO4 = 0.025 x 2.0 = 0.05 mol Step 3 11 11 0.05 mol 0.05 mol 0.05 mol 0.05 mol Step 4 Theoretical mass CuSO4.5H2O = 0.05 x 250 = 12.5 g Step 5 Percentage yield = 7.3 x 100 = 58.4% 12.5 87

NES/Chemistry/IGCSE Percentage Purity The percentage purity of a reactant is the actual mass reacted divided by the initial mass used expressed as a percentage. Percentage purity is used when a reagent is impure. It contains impurities that do not react. Percentage purity = actual mass (pure)reacted x 100% initial mass(impure)used  Example 16: An excess (more than is needed) of hydrochloric acid was added to 1.570g of impure barium carbonate. The volume of carbon dioxide gas collected was 0.120dm3. The impurities did not react with the acid. Calculate the percentage purity of the barium carbonate. Molar gas volume at r.t.p. is 24dm3 Step 1 2HCl + BaCO3  BaCl2 + H2O + CO2 Step 2 moles of CO2 = 0.120 = 5 x 10–3 moles 24 Step 3 21 1 11 Step 4 1x10–2 5x10–3 5x10–3 5x10–3 5x10–3 Step 5 actual mass BaCO3 = 5 x 10–3 x 197 = 0.985 g Percentage purity = 0.985 x 100 1.570 = 62.7% pure 88

NES/Chemistry/IGCSE Topic 5 - Electricity and Chemistry  Define electrolysis as the breakdown of an ionic compound, molten or in aqueous solution, by the passage of electricity  Describe the electrode products and the observations made during the electrolysis of: 1. molten lead(II) bromide 2. concentrated hydrochloric acid 3. concentrated aqueous sodium chloride 4. dilute sulfuric acid  between inert electrodes (platinum or carbon)  State the general principle that metals or hydrogen are formed at the negative electrode (cathode), and that non-metals (other than hydrogen) are formed at the positive electrode (anode)  Predict the products of the electrolysis of a specified binary compound in the molten state  Describe the electroplating of metals  Outline the uses of electroplating  Describe the reasons for the use of copper and (steel cored) aluminium in cables, and why plastics and ceramics are used as insulators  Relate the products of electrolysis to the electrolyte and electrodes used, exemplified by the specific examples in the Core together with aqueous copper(II) sulfate using carbon electrodes and using copper electrodes (as used in the refining of copper)  Describe electrolysis in terms of the ions present and reactions at the electrodes in the examples given  Predict the products of electrolysis of a specified halide in dilute or concentrated aqueous solution  Construct ionic half-equations for reactions at the cathode  Describe the transfer of charge during electrolysis to include: 1. the movement of electrons in the metallic conductor 2. the removal or addition of electrons from the external circuit at the electrodes 3. the movement of ions in the electrolyte  Describe the production of electrical energy from simple cells, i.e. two electrodes in an electrolyte. (This should be linked with the reactivity series in section 10.2 and redox in section 7.4.) 89

NES/Chemistry/IGCSE  Describe, in outline, the manufacture of: 1. aluminium from pure aluminium oxide in molten cryolite (refer to section 10.3) 2. chlorine, hydrogen and sodium hydroxide from concentrated aqueous sodium chloride (Starting materials and essential conditions should be given but not technical details or diagrams.) Definitions Electrolysis is the decomposition of an electrolyte using electricity An electrolyte is a substance, which when molten or dissolved in solution conducts electricity and is decomposed (broken down) by it 90

NES/Chemistry/IGCSE Electrolysis Before starting this topic a review of bonding (see Topic 3) would be helpful. Electrolysis involves splitting up compounds by using electricity. The compound will turn back into atoms usually. It is a bit like bonding in reverse. Diagram of an Electrolytic Cell external circuit battery anode (+) electron flow bulb cathode (-) anions release cations remove e to the anode e from the cathode forming atoms anion cation electrolyte (molten or aqueous) The external circuit is made up of the battery, bulb, electrodes and wires. This is where the electricity is conducted by electrons. The other part of the circuit is the electrolyte, where the electricity is conducted by mobile ions. During electrolysis, ions are changed back to being atoms by losing, or gaining electrons. It is a bit like bonding in reverse. There are two types of electrolyte molten (liquid) and aqueous (dissolved in water). 91

NES/Chemistry/IGCSE Electrolytic Cell An electrolytic cell converts electrical energy into chemical energy and consists of two electrodes connected to a power supply by an external circuit and an electrolyte. The battery pushes electrons around the circuit. The electrons travel around an electrolytic cell from anode to cathode. The ions in the electrolyte move to the oppositely charged electrodes. A negatively charged ion is called an anion - this gets attracted to the positively charged electrode, called an anode. A positively charged ion is called a cation - this gets attracted to the negatively charged electrode, called a cathode. This means electrons are moving in the external circuit and ions are moving in the electrolyte. Electrons enter the external circuit at the anode (anions lose their extra electrons to the electrode forming atoms/molecules; oxidation). Electrons are removed from the external circuit at the cathode (cations gain electrons and are reduced to form atoms; reduction). A bulb may be added to the circuit to show that the circuit is complete and working. 92

NES/Chemistry/IGCSE Molten Electrolysis For molten (liquid) salts, metals are formed at the cathode, and non-metals at the anode. molten electrolyte cathode (-) metals anode (+) non-metals (except hydrogen gas)  Example 1: electrolysis of molten lead (II) bromide Carry out in a fume cupboard as lead is poisonous and bromine is toxic. battery +- Use inert electrodes of graphite or platinum. When the circuit is complete the bulb does not initially light. This is because when the lead (II) bromide is a solid the ions are not free to move and so does not conduct electricity. The solid is heated with a Bunsen burner, and electrodes when enough energy has been supplied to overcome the electrostatic forces of attraction holding the ions in the lattice, the solid melts and the bulb lights due to the bromine molten presence of mobile ions and so the molten gas (Br2) PbBr2 lead (II) bromide conducts electricity. molten lead HEAT 93

NES/Chemistry/IGCSE At the Anode: 2Br –(l)  Br2(g)+ 2e– The Br– anions are attracted to the anode where they lose their extra electron to the electrode forming atoms. Bubbles of brown gas (Br2) are observed. At the Cathode: Pb2+(l) + 2e–  Pb(l) The Pb2+ cations are attracted to the cathode where they gain electrons from the cathode to form lead atoms. A silvery liquid (molten lead) is collected at the bottom of the boiling-tube. Adding the two half-ionic equations together we get the following full ionic equation: Pb2+(l) + 2Br–(l)  Pb(l) + Br2(g) Why can the bulb continue to glow, even after the battery is switched off? After 30 minutes many lead cations (Pb2+) have been reduced to molten lead atoms which fill the bottom of the test tube. Eventually there is so much lead metal that it is in contact with both electrodes, so even if the heat is removed and the lead solidifies the circuit is still complete as solid metal conducts electricity. solid lead 94

NES/Chemistry/IGCSE Aqueous Electrolysis Aqueous solutions are more difficult to predict than molten. This is because as well as the cation and anion from the salt, there is a cation (H+) and anion (OH–) from the water. Remember water ionises very slightly to form ions H2O ⇌ H+ + OH– Predicting the Products of Aqueous Solutions At the Cathode (-) At the Anode (+)  If the metal is Dilute solutions Concentrated solutions more reactive than hydrogen, Oxygen gas is the only  If the anions fluoride, then hydrogen product formed by the chloride, bromide or gas is produced. equation: iodide are present then fluorine,  If the metal is less chlorine, bromine or iodine is produced. reactive than 4OH–  O2 + 2H2O + 4e– hydrogen, then  If there is no halide anion present, then the metal is oxygen gas is produced. deposited. The most unreactive cation in the reactivity series (see Topic 10.1) will be the product halogen concentrated oxygen aqueous anode (+) dilute oxygen electrolyte cathode (-) metal hydrogen gas 95

NES/Chemistry/IGCSE  Example 2: electrolysis of aqueous copper(II) sulfate solution Platinum or graphite electrodes are used because they are inert (unreactive) and so do not affect the products made at each electrode. electrolytic cell The four ions present in the solution test tube are: electrode cations Cu2+ H+ anions SO42– OH– Ions attracted to the cathode Cu2+ and H+ Product formed at the cathode is copper metal because it is less reactive than hydrogen. Cu2+ + 2e–  Cu Product formed at the anode is oxygen gas because there is no concentrated halide ion present. 4OH–  O2 + 2H2O + 4e– The solution left over at the end is sulphuric acid (H2SO4) made from the H+ and SO42- ions left over. The other ions have been discharged. 96

NES/Chemistry/IGCSE Products and Observations of Some Aqueous Electrolytes The table shows the products and observations of the most common aqueous electrolytes: Electrolyte Products Observations Cathode - Hydrogen gas Cathode - bubbles of Concentrated aqueous 2H+(aq) + 2e–  H2(g) colourless gas hydrochloric acid Anode - Chlorine gas Anode - bubbles of 2Cl–(aq)  Cl2(g) + 2e– green gas Concentrated Cathode - Hydrogen gas aqueous Cathode - bubbles of sodium 2H+(aq) + 2e–  H2(g) colourless gas chloride Anode - Chlorine gas Anode - bubbles of Dilute 2Cl–(aq)  Cl2(g) + 2e– green gas aqueous Cathode - Hydrogen gas sulphuric acid Cathode - bubbles of 2H+(aq) + 2e–  H2(g) colourless gas Anode - Oxygen gas Anode - bubbles of 4OH–(aq)  2H2O(l) + O2(g) + 4e– colourless gas 97

NES/Chemistry/IGCSE Purification or Refining of Copper using Active Metal Electrodes This is carried out using a pure copper cathode and an impure copper anode. The electrolyte is blue acidified copper(II) sulfate solution. Active metal electrodes mean that the anode decreases in mass/size as the copper atoms are oxidised to form Cu2+ ions and enter the electrolyte and the cathode increases in mass/size as the Cu2+ ions are reduced to copper atoms so Cu gets deposited. Therefore the anode decreases in mass/size and the cathode increases in mass/size. start later impure pure electron copper copper flow anode cathode pure Cu impure copper anode CuSO4(aq) cathode decreases in mass/size electrolyte increases Cu2+ ions are formed in which enter the electrolyte anode sludge/impurities mass/size Anode At the anode copper atoms on the impure copper plates are oxidised, each losing two valence electrons to form copper(II) ions (Cu2+) which enter the electrolyte. The anode decreases in mass/size. The electrons travel around the external circuit to the cathode. Cu(s)  Cu2+(aq) + 2e– The anode sludge/impurities are left on the bottom of the beaker and can be removed by filtration. 98

NES/Chemistry/IGCSE Cathode At the cathode the copper(II) ions are reduced, each gaining 2 electrons, forming pink brown copper atoms which are deposited on the cathode surface. The cathode increases in mass/size and is pure copper metal. Cu2+(aq) + 2e–  Cu(s) The concentration of the electrolyte stays constant throughout electrolysis because for every one copper atom oxidised to Cu2+ at the anode, only one Cu2+ ion is reduced to a copper atom at the cathode, therefore the number of Cu2+ ions per unit volume remains constant, so the colour remains the same - blue. 99