1. GCSE Revision Unit C5 The hardest one.

2. C5 Reacting amounts - basics You can count atoms using the mole - “gram-molecular mass” The “molar mass” of substance is its relative formula mass in grams. Need to calculate the molar mass of a substance from its formula (without brackets) by adding the appropriate relative atomic masses. Mass is conserved in a chemical reaction. REMEMBER Carry out simple subtractions, based on the idea that mass is never lost: –mass of gas or water lost during thermal decomposition or heating; –mass of gas gained during reaction; –determine a reacting amount for a simple reaction given all the other reacting amounts. Determine the mass of an element in a known mass of compound given the masses of the other elements present. (i.e. just subtract what you know from the total) E.g. in 100g of CaCO 3, there is 40g of calcium & 12g of carbon. What mass of oxygen is there?

3. C5 More: Add the appropriate relative atomic masses to get molar mass of a substance from its formula (with brackets). Empirical formula: “The simplest whole number ratio of each type of atom in a compound”. Can deduce the empirical formula of a compound from its chemical formula.

4. C5 Higher tier Number of moles = mass ÷ molar mass; –determine the number of moles of an element from the mass of that element; –determine the number of moles of a compound from the mass of that compound; –determine the masses of the different elements present in a given number of moles of a compound. The relative atomic mass of an element is the average mass of an atom of the element compared to the mass of 1/12th of an atom of carbon-12. Calculate mass of products and / or reactants using the mole concept from a given balanced equation and the appropriate relative atomic masses, Calculate empirical formula of a compound given the: –percentage composition by mass; –mass of each element in a sample of the compound.

5. C5 Example NaOH has a mass of 40 (23+16+1) Out of this, 16/40 is oxygen. If I had 100g of NaOH, then 16/40 of this is oxygen, which is 100x16/40 = 40g In this 100g, 100x1/40 is hydrogen, 2.5g The other 57.5g must be sodium.

6. C5 Another example In CaCO 3, the total mass is 100 (40+12+48) So oxygen is 48/100 of the compound Carbon is 12/100 of the compound Potassium is 40/100 of the total So if there is 25g of calcium carbonate, potassium is 25x40/100 = 10g

7. C5 Or: If the calcium carbonate is being heated to make carbon dioxide and calcium oxide: CaCO 3  CaO + CO 2 100 56 44 (relative masses) the CO 2 is 44/100 of the starting materials so it must also be 44/100 of the total at the end. If you start with 200g of calcium carbonate, you must make 200x44/100 grams of CO 2 and 200x56/100g of CaO Therefore you make 88g of CO 2 and 112g of CaO

8. C5 Electrolysis - basics Need to be able to label the apparatus used to electrolyse solutions in a school laboratory: –anode, cathode, d.c. power supply, probably in a beaker. Electrolysis is the decomposition of a compound by passing an electric current through it or a solution of it. Positive ions go to the negative electrode and negative ions go to the positive electrode obviously In the electrolysis of copper(II) sulphate solution using copper electrodes: –the negative electrode is plated with copper; –the positive electrode gets smaller and dissolves. –this is used to purify copper More or less time and current (amps) gives more or less product. Melted Al 2 O 3 (l) gives oxygen and aluminium, Melted PbBr 2 (l) gives lead and bromine, Melted PbI 2 (l) gives lead and iodine Melted KCl(l) gives potassium and chlorine

9. C5 More: Know electrolysis gives: –K 2 SO 4 (aq) – hydrogen and oxygen; –KNO 3 (aq) – hydrogen and oxygen –H 2 SO 4 – hydrogen and oxygen Describe electrolysis in terms of flow of charge and discharge of ions. Describe the changes in mass of the copper electrodes used in the electrolysis of copper(II) sulphate solution: –the negative electrode gains mass; –the positive electrode loses mass; –the gain in mass of the negative electrode is equal to the loss in mass of the positive electrode. Describe the factors that affect the amount of substance produced during electrolysis: –amount made increases as time increases and as current increases. Ionic substances contain ions which are in fixed positions when solid but can move when in solution or when melted.

10. C5 Higher tier In the electrolysis of aqueous solutions it may be easier to discharge ions from the water rather than from the solute. Half equations for the electrode processes that happen during the electrolysis of the following given the formula of the ions in the electrolyte: K 2 SO 4 (aq) and KNO 3 (aq) Half equations for electrode processes that happen during the electrolysis of CuSO4(aq) using copper electrodes: –(Anode) positive electrode Cu - 2e -  Cu 2+ –(Cathode) negative electrode Cu 2+ + 2e -  Cu Relationship between charge transfer, current and time: Q = It Perform simple calculations based on current, charge and the amount of substance produced in electrolysis. They might expect half-equations for what happens at each electrode during the electrolysis of: Al 2 O 3 (l), PbBr2(l), PbI 2 (l), KCl(l)

11. C5 Concentration - basics Concentration of solutions may be measured in g/dm 3 (g per dm 3 ). Concentration of solutions may be measured in mol/dm 3 (mol per dm 3 ). Volume is measured in dm 3 (litre) or cm 3 1000 cm 3 equals 1 dm 3 (litre) Describe how to dilute a concentrated solution You must get concentration right by diluting food preparations, medicine and baby milk. Dangers of getting the dilution wrong (obvious ones): food consistency wrong; overdoses, very poorly babies Food packaging, recommended daily allowances.

12. C5 More: The more concentrated a solution the more crowded the solute particles. Volume in cm 3 is 1000 times the volume in dm 3 Volume in dm 3 is 1000 th of the volume in cm 3 Perform calculations involving concentration for Simple dilutions of solutions e.g. how to dilute a 1.0 mol/dm 3 solution into a 0.1 mol/dm 3 solution

13. C5 Higher tier To convert concentration in g/dm 3 into mol/dm 3 divide by the molecular mass To do the opposite, multiply by molecular mass Calculate the concentration of a solution given appropriate information about the mass or number of moles of solute in a particular volume of solution. Perform simple calculations involving concentration, number of moles and volume of solution. Interpret more complex food packaging information and its limitations e.g: convert amounts of sodium to amounts of salt; understand that an ion may come from several sources, so the above conversion will be inaccurate.

14. C5 Titrations - basics Remember changes in pH when an acid reacts with an alkali: –pH decreases when acid is added to alkali; –pH increases when alkali is added to a acid; –They expect you to interpret a simple pH curve, find the pH at a particular volume added or read volume at a pH – major grid lines (whole numbers). You can estimate the pH of a solution with universal indicator. Apparatus used in an acid-base titration: –burette and conical flask; pipette and pipette filler. Describe the procedure for carrying out a simple acid-base titration: –acid in burette, alkali in pipette (or vice versa); –acid slowly added to alkali (or vice versa) until end point is reached; –end point detected by the change in colour of an indicator. Calculate the titre given appropriate information from tables or diagrams. Indicators in acids and alkalis: –universal indicator, - range from red/yellow/green/turquoise/dark blue –Litmus – red in acid, blue in alkali –Phenolphthalein – colourless in acid, pink in alkali, changes around 9 –screened methyl orange – red in acid, yellow in alkali, changes/orange around pH 5

15. C5 More: There is a sudden change in pH at the “end point” of a titration, as the “balance” tips the indicator –neutralisation; –acid + alkali  salt + water Interpret a simple pH curve: –determine the volume of acid or alkali at neutralisation; –determine the pH when a particular volume added or vice versa. Need several consistent titre readings in titrations, otherwise you can’t be sure.... A single indicator such as litmus - sudden colour change Mixed indicator (like universal) has a continuous colour change – less useful in titration

16. C5 Higher tier Sketch a pH titration curve for the titration of an acid or an alkali. Exam questions will be a one to one molar ratio (one mole acid : one mole alkali). Use equation concentration 1 x volume 1 = concentration 2 x volume 2 – you know three of these, just change the subject. State and use the relationship between moles, concentration and volume: –moles = concentration × volume in dm 3 –concentration = moles ÷ volume in dm 3 –volume in dm 3 = moles ÷ concentration –Think of the unit for concentration – moles per litre, means conc. is moles divided by volume then make a triangle Why is an acid-base titration better with a single indicator rather than a mixed indicator? moles concentration solvent volume (dm 3 )

17. C5 Gas volumes - basics Possible apparatus to collect and measure the volume of a gas from a reaction: –gas syringe; –upturned measuring cylinder; –upturned burette. If you can measure the mass of the reaction mixture, it will get lighter as gas is lost to the air. A reaction stops when one of the reactants is all used up. The more amount of reactant you have, the greater volume of gas is made. Interpret graphs or tables of data about the volume of gas produced during the course of a reaction (limited to major grid lines on graphs): – deduce total volume of gas produced; – deduce when the reaction has stopped; – deduce volume of gas at a particular time and vice versa; – deduce when the reaction is at its fastest..

18. C5 More: “Describe an experimental method to measure the volume of gas produced in a reaction given appropriate details about the reaction” “Describe an experimental method to measure the mass of gas produced in a reaction given appropriate details about the reaction” State that the limiting reactant is the one that is used up first of all. Explain why a reaction stops in terms of the limiting reactant present given appropriate qualitative (no numbers) information. Describe that the amount of gas produced is directly proportional to the amount of the limiting reactant present. Interpret graphs as above, but also: –read from minor or estimate from major lines; –deduce the volume of gas produced with different amounts of limiting reactant.

19. C5 Higher tier One mole of gas molecules occupy a volume of 24 dm 3 at room temperature and pressure So we can calculate the volumes of samples of gases. E.g. 5 moles has a volume of 120dm 3. Don’t get cm 3 and dm 3 confused, for example, 250cm 3 is 0.25 dm 3 and 1.2 dm 3 is 1200 cm 3 You may be asked to sketch a graph to show how the volume of gas produced during the course of a reaction changes, given appropriate details: –change in the rate of the reaction; –total volume of gas collected; –when the reaction stops.

20. C5 Equilibria - basics Many chemical reactions are easily reversed, giving a forward and a backward reaction. It is possible for both forward and backward reactions to be proceeding at the same time. The Ý symbol is used to show that a reaction is reversible, e,g: N 2 + 3H 2 Ý 2NH 3 Interpret data in the form of tables or graphs (using major grid-lines) about the equilibrium composition: – composition at particular temperatures; – composition at particular pressures; – effect of temperature and pressure on composition. The raw materials to make sulphuric acid by the Contact Process: – sulphur; – air; – water. State that the production of sulphuric acid by the Contact Process involves the reversible reaction between sulphur dioxide and oxygen: sulphur dioxide + oxygen Ý sulphur trioxide

21. C5 More: Some reversible reactions may reach an equilibrium where: –the rate of the forward reaction equals the rate of the backward reaction; –the concentrations of the reactants and the products do not change and have no reason to be equal. If the concentration of product is greater than the concentration of reactant, we say the position of equilibrium is “to the right” And vice-versa A change in temperature, pressure or concentrations of reactant (LHS) or product (RHS) may change the position of equilibrium. “Interpret data in the form of tables or graphs about the equilibrium composition”: –composition at particular temperatures; –composition at particular pressures; –effect of temperature and pressure on composition. Conditions used in the Contact Process: –V 2 O 5 catalyst; –around 450°C; –atmospheric pressure. Recall that sulphur dioxide needed for the Contact Process often comes from burning sulphur: –sulphur + oxygen Ý sulphur dioxide

22. C5 Higher tier Explain why a reversible reaction may reach an equilibrium: –a closed system; –rate of forward reaction decreases; –rate of backward reaction increases; –eventually rate of forward equals rate of backward reaction. Explain in simple qualitative terms factors that affect the position of equilibrium: –removing a product moves the position of equilibrium to the right or vice versa; –adding extra reactant moves the position of equilibrium to the right or vice versa; –increasing the pressure moves the position of equilibrium to the side with the least number of moles of gas molecules Equations for the manufacture of sulphuric acid by the Contact Process: S + O 2  SO 2 2SO 2 + O 2 Ý 2SO 3 SO 3 + H 2 O  H 2 SO 4 Conditions used in the Contact Process: –high temperature decreases yield and increases rate of reaction so an optimum is used; –Catalysts increase rate but do not change position of equilibrium; “same amounts of products but faster” –Position of equilibrium is already on right so high pressure is expensive and is not needed.

23. C5 Acid strength - basics Ethanoic acid is an example of a weak acid. Hydrochloric, nitric and sulphuric acids are strong acids. Strong acids have a lower pH (about 1-3) than weak acids (about 3-6) of the same concentration. Any acid will react with magnesium to give hydrogen. Any acid will react with calcium carbonate to give carbon dioxide. These two reactions are slower with weak acids like ethanoic acid than with hydrochloric acid of the same concentration. The same amount of same concentration strong and weak acids produce the same volume of gas in these reactions. Ethanoic acid has a lower electrical conductivity than hydrochloric acid of the same concentration. Electrolysis of both ethanoic acid and hydrochloric acid makes hydrogen at the negative electrode. Weak acids are sometimes used as descalers, some occur in foods.

24. C5 More: All acids ionise in water to produce H + ions. A strong acid completely ionises in water. HCl  H + + Cl - The ionisation of a weak acid is an example of a reversible reaction. For a weak acid HA: HA Ý H + + A - This gives an equilibrium mixture. Ethanoic acid reacts slower than hydrochloric acid because there are fewer hydrogen ions and so fewer collisions with hydrogen ions. But the volume of hydrogen gas formed in a metal reaction is determined by the amounts of reactants present not the acid strength. Ethanoic acid is less conductive than hydrochloric acid of the same concentration because there are fewer hydrogen ions available to move. Hydrogen is produced during the electrolysis of ethanoic acid and of hydrochloric acid. Hydrogen ion + electron  hydrogen gas Strong acids are a bad idea for descaling agents, as they would corrode metals.

25. C5 Higher The pH of a weak acid is much higher than the pH of a strong acid of the same concentration because there are fewer hydrogen ions per unit volume The difference between acid strength and acid concentration: –acid strength (strong or weak) is a measure of the degree of ionisation of the acid; –concentration of an acid is a measure of how many moles of acid in one dm 3. That is, strong acids separate to give more H + (aq) ions than weak ones do Equations for examples of the ionisation of weak and strong acids: CH 3 COOH Ý CH 3 COO - + H + HCl  H + + Cl - (these are the only ones they will ask for) Ethanoic acid reacts more slowly than hydrochloric acid of the same concentration because: –ethanoic acid is weak and hydrochloric acid is strong; –greater concentration of hydrogen ions; –greater collision frequency with hydrogen ions. ethanoic acid is less conductive than hydrochloric acid of the same concentration because: –ethanoic acid is weak and hydrochloric acid is strong; –greater concentration of hydrogen ions to carry the charge. “Explain why a weak acid may be more useful than the more dilute strong acid.”

26. C5 Ionic reactions - basic A precipitation reaction involves two solutions reacting together to make an insoluble substance. Most precipitation reactions involve ions from one solution reacting with ions from another solution. A nitrate solution (usually silver nitrate) can be used to test for halide ions, giving: –white precipitate with Cl- –cream precipitate with Br- –yellow precipitate with I Describe that barium nitrate or barium chloride solution can be used to test for sulphate ions (both form a white precipitate). You need to be able to identify the reactants and the products in an ionic equation. You need to be able to recognise and use the state symbols (aq), (s), (g) and (l). You need to be able to label apparatus used during the preparation of an insoluble compound by precipitation (basically, a beaker for the reaction, followed by a filter and funnel to get the solid.)

27. C5 More: Ionic substances contain ions which are in fixed positions in the solid but can move when in solution or when melted. Ions must collide with other ions if they are to react. This happens when solutions are mixed. Use word equations for simple precipitation reactions e.g. for the reaction between solutions of barium chloride and sodium sulphate. Preparation of a dry sample of an insoluble compound by precipitation, (you will be given the names of the reactants): –mix solutions of reactants; –filter; –wash and dry the residue.

28. C5 Highest level Most precipitation reactions are extremely fast reactions between ions. They occur because the forces attracting the ions to make a precipitate are stronger than those holding the ions to water molecules. Heat energy is released as the ionic lattice forms, but you can’t usually detect it. Ionic equations involve just the ions that react. ‘Spectator ions’ are the ions that stay in solution and haven’t formed a precipitate. E.g. in NaBr + AgNO 3  AgBr + NaNO 3 Solubility rules tell us what should happen, because Na + and NO 3 - ions always stay dissolved, so the Ag + and Br - ions can be the only precipitate: Ag + (aq) + Br - (aq)  AgBr (s)

29. C5 Simplified solubility rules Group 1 metal and ammonium (NH 4 + ) compounds always dissolve Nitrates (NO 3 - ) always dissolve Chlorides, bromides and iodides dissolve, except those with silver, mercury and lead Metallic sulphates usually dissolve, but not those of calcium, strontium, barium, silver, mercury and lead Metallic oxides, carbonates and hydroxides are insoluble except where the first rule applies. All metallic hydroxides (OH - ) are insoluble except where the first rule applies and calcium, strontium and barium in Group II.