Edexcel IGCSE Chemistry PDF Past Papers

Head to www.savemyexams.com for more awesome resources Edexcel IGCSE Chemistry Your notes Chemical Formulae, Equations, Calculations Contents Word & Chemical Equations Calculate Relative Mass Moles, Mass & RFM...

Head to www.savemyexams.com for more awesome resources Edexcel IGCSE Chemistry Your notes Chemical Formulae, Equations, Calculations Contents Word & Chemical Equations Calculate Relative Mass Moles, Mass & RFM Calculate Reacting Masses Calculate Percentage Yield Experiment: Finding Formulae of Compounds Practical: Determine the Formula of a Metal Oxide Empirical & Molecular Formulae Calculate Concentrations of Solutions Calculate Volumes of Gases Page 1 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Word & Chemical Equations Your notes Writing equations New substances are made during chemical reactions However, the same atoms are always present before and after reaction They have just joined up in different ways Atoms cannot be created or destroyed, so if they exist in the reactants then they absolutely must be in the products! Because of this the total mass of reactants is always equal to the total mass of products This idea is known as the Law of Conservation of Mass Conservation of mass The Law of Conservation of Mass enables us to balance chemical equations, since no atoms can be lost or created You should be able to: Write word equations for reactions outlined in these notes Write formulae and balanced chemical equations for the reactions in these note How to write word equations Word equations show the reactants and products of a chemical reaction using their full chemical names reactants → products The reactants are those substances on the left-hand side of the arrow They can be thought of as the chemical ingredients of the reaction They react with each other to form new substances, which are the products The products are on the right-hand side of the arrow The arrow (which is spoken as “to form” or “produces”) implies the conversion of reactants into products Reaction conditions or the name of a catalyst (a substance added to make a reaction go faster) can be written above the arrow Page 2 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources An example is the reaction of sodium hydroxide (a base) and hydrochloric acid to produce sodium chloride (common table salt) and water: sodium hydroxide + hydrochloric acid ⟶ sodium chloride + water Your notes Worked Example Word equations 1. Ammonia reacts with nitric acid to form the fertiliser ammonium nitrate. Write a word equation for the reaction taking place. 2. Iron(II) hydroxide and sodium sulfate are formed when iron(II) sulfate solution and sodium hydroxide react together. Write a word equation for the reaction taking place. 3. Carbon is the main element found in coal and burns in air to produce carbon dioxide. Write a word equation for the reaction taking place. Answers: 1. Ammonia + nitric acid → ammonium nitrate This question has all the information in the correct order Ammonia reacts with nitric acid This becomes ammonia + nitric acid to form This is the arrow in the equation to form the fertiliser ammonium nitrate This tells you that the product is ammonium nitrate 2. Iron(II) sulfate + sodium hydroxide → iron(II) hydroxide + sodium sulfate Careful: This question has all the required information but the products are written first Iron(II) hydroxide and sodium sulfate are formed This becomes → iron(II) hydroxide + sodium sulfate when iron(II) sulfate solution and sodium hydroxide react together This becomes Iron(II) sulfate + sodium hydroxide → 3. Carbon + oxygen → carbon dioxide Careful: Not all of the required information is given in the question You are expected to know that burning in air means that the chemical is reacting with oxygen Carbon......burns in air Page 3 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources This becomes carbon + oxygen to produce Your notes This is the arrow in the equation to produce carbon dioxide This tells you that the product is carbon dioxide How to write balanced equations A symbol equation uses the formulae of the reactants and products to show what happens in a chemical reaction When writing symbol equations, you should: Ensure reactants are on the left of the equation and products are on the right Write the following non-metals as molecules: H2, N2, O2, F2, Cl2, Br2 and I2 Include state symbols Solid = (s) Liquid = (l) Gas = (g) Aqueous = (aq) You need to be confident using the state symbols (s), (l), (g) and (aq) You will not need to include them in all equations unless you are specifically asked to However, it is good practice to include state symbols in your equations so that you don't miss any marks A symbol equation must be balanced to give the correct ratio of reactants and products: For example, the combustion of sulfur: S (s) + O2 (g)→ SO2 (g) This equation shows that one atom of sulfur, S, reacts with one molecule of oxygen, O2, to make one molecule of sulfur dioxide, SO2 Balancing equations When balancing equations, there must be the same number of atoms of each element on either side of the equation following the Law of Conservation of Mass Page 4 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources To balance an equation you work across the equation from left to right, checking one element after another Your notes If there is a group of atoms such as a nitrate group (NO3–) that has not changed from one side to the other, then count the whole group as one entity rather than counting the individual atoms Examples of balanced symbol / chemical equations include: Acid-base neutralisation reaction: NaOH (aq) + HCl (aq) ⟶ NaCl (aq) + H2O (l) Redox reaction: 2Fe2O3 (aq) + 3C (s) ⟶ 4Fe (s) + 3CO2 (g) In each equation, there are equal numbers of each atom on either side of the reaction arrow so the equations are balanced The best approach is to practice lot of examples of balancing equations This can be by trial and error - changing the coefficients (numbers) in front of the formulae one by one and checking the result on the other side Balance elements that appear on their own, last in the process Worked Example Aluminium reacts with copper(II) oxide to produce aluminium oxide and copper. Balance the symbol equation for the reaction taking place. Al (s) + CuO (s) ⟶ Al2O3 (s) + Cu (s) Answer: The balanced symbol equation is: 2Al (s) + 3CuO (s) ⟶ Al2O3 (s) + 3Cu (s) Step 1 - balancing aluminium atoms There are 2 aluminium atoms on the product side, so 2 aluminium atoms are needed on the reactant side 2Al + CuO ⟶ Al2O3 + _Cu Step 2 - balancing oxygen atoms There are 3 oxygen atoms on the product side, so 3 oxygen atoms are needed on the reactant side This means that 3 CuO will be needed as we cannot change the chemical formula Page 5 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 2Al + 3CuO ⟶ Al2O3 + Cu Step 3 - balancing copper atoms Your notes There are 3 copper atoms on the reactant side, so 3 copper atoms are needed on the product side 2Al + 3CuO ⟶ _Al2O3 + 3Cu The equation is now balanced Worked Example When magnesium oxide, MgO, reacts with nitric acid, HNO3, it forms magnesium nitrate, Mg(NO3)2, and water. Write a symbol equation for this reaction. Answer: The balanced symbol equation is: MgO (s) + 2HNO3 (aq) ⟶ Mg(NO3)2 (aq) + H2O (l) Step 1 - writing the unbalanced equation Magnesium oxide, MgO, reacts with nitric acid, HNO3, it forms magnesium nitrate, Mg(NO3)2, and water MgO + HNO3 ⟶ Mg(NO3)2 + H2O The Mg and O atoms (not including the O in the NO3 group appear to be balanced), so we should focus on the H atoms and NO3 groups Step 2 - balancing hydrogen atoms There are 2 hydrogen atoms on the product side, so 2 hydrogen atoms are needed on the reactant side This means that 2 HNO3 will be needed as we cannot change the chemical formula MgO + 2HNO3 ⟶ Mg(NO3)2 + H2O This also balances the nitrate, NO3, groups Step 3 - checking the equation The equation appears balanced so we need to check that it is Reactant side: 1 Mg atom 1 O atom - not including those in the NO3 group 2 H atoms 2 NO3 groups - remember to keep groups as a single entity if they are unchanged on both sides of the equation Product side: Page 6 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 1 Mg atom 2 NO3 groups - remember to keep groups as a single entity if they are unchanged on both sides of the equation Your notes 2 H atoms 1 O atom - not including those in the NO3 group The equation is now balanced Examiner Tips and Tricks Careful: A common mistake when balancing symbol equations is to add, change or remove small numbers in the chemical formula of a substance You cannot do this because it changes what the substance is For example, if a product was water, H2O, and you added a second oxygen to make it H2O2 then it is no longer water Page 7 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Calculate Relative Mass Your notes Relative formula (molecular) mass How to calculate relative formula mass The symbol for the relative atomic mass is Ar This is calculated from the mass number and relative abundances of all the isotopes of a particular element The symbol for the relative formula mass is Mr and it refers to the total mass of the molecule To calculate the Mr of a substance, you have to add up the relative atomic masses of all the atoms present in the formula Relative formula mass calculations table Substance Atoms present Calculation Mr Hydrogen 2xH (2 x 1) 2 H2 Water (2 x H) + (1 x O) (2 x 1) + (1 x 16) 18 H2 O Potassium carbonate (2 x K) + (1 x C) + (3 x O) (2 x 39) + (1 x 12) + (3 x 16) 138 K2CO3 Calcium hydroxide (1 x Ca) + (2 x O) + (2 x H) (1 x 40) + (2 x 16) + (2 x 1) 74 Ca(OH)2 Ammonium sulfate (2 x N) + (8 x H) + (1 x S) + (4 x O) (2 x 14) + (8 x 1) + (1 x 32) + (4 x 16) 132 (NH4)2SO4 In accordance with the Law of Conservation of Mass, the sum of the relative formula masses of the reactants will be the same as the sum of the relative formula masses of the products Worked Example Page 8 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Calculate the relative formula mass of: 1. Sodium chloride, NaCl Your notes 2. Copper oxide, CuO 3. Magnesium nitrate, Mg(NO3)2 Answers: 1. Sodium chloride NaCl = 23 + 35.5 = 58.5 2. Copper oxide CuO = 63.5 + 16 = 79.5 3. Magnesium nitrate Mg(NO3)2 = 24 + (14 x 1 x 2) + (16 x 3 x 2) = 148 Examiner Tips and Tricks The relative atomic mass of every element is given on the Periodic Table. It is the larger of the two numbers. Page 9 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Moles, Mass & RFM Your notes The mole Chemical amounts are measured in moles The symbol for the unit mole is mol One mole of a substance contains the same number of the stated particles, atoms, molecules, or ions as one mole of any other substance The number of atoms, molecules or ions in a mole (1 mol) of a given substance is the Avogadro constant. The value of the Avogadro constant is 6.02 x 1023 per mole For example: One mole of sodium (Na) contains 6.02 x 1023 atoms of sodium One mole of hydrogen (H2) contains 6.02 x 1023 molecules of hydrogen One mole of sodium chloride (NaCl) contains 6.02 x 1023 formula units of sodium chloride The mass of 1 mole of a substance is known as the molar mass For an element, it is the same as the relative atomic mass written in grams For a compound, it is the same as the relative molecular or formula mass in grams If you had 6.02 x 1023 atoms of carbon in your hand, that number of carbon atoms would have a mass of 12 g (because the Ar of carbon is 12) So one mole of helium atoms would have a mass of 4 g (Ar of He is 4), one mole of lithium would have a mass of 7 g (Ar of Li is 7) and so on To find the mass of one mole of a compound, we add up the relative atomic masses So one mole of water would have a mass of (2 x 1) + 16 = 18 g So one carbon atom has the same mass as 12 hydrogen atoms Examiner Tips and Tricks You need to appreciate that the measurement of amounts in moles can apply to atoms, molecules, ions, electrons, formulae and equations. E.g. in one mole of carbon (C) the number of atoms is the Page 10 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources same as the number of molecules in one mole of carbon dioxide (CO2). Your notes Calculating moles and masses Although elements and chemicals react with each other in molar ratios, in the laboratory we use digital balances and grams to measure quantities of chemicals as it is impractical to try and measure out moles Therefore we have to be able to convert between moles and grams We can use the following formula to convert between moles, mass in grams and the molar mass: Formula triangle for moles, mass and molar mass Worked Example What is the mass of 0.250 moles of zinc? Answer: From the Periodic Table the relative atomic mass of Zn is 65.38 So, the molar mass is 65.38 g mol-1 The mass is calculated by moles x molar mass Page 11 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources This comes to 0.250 mol x 65.38 g mol-1 = 16.3 g Your notes Worked Example How many moles are in 2.64 g of sucrose, C12H22O11 (Mr = 342.3)? Answer: The molar mass of sucrose is 342.3 g mol-1 The number of moles is found by mass ÷ molar mass This comes to 2.64 g ÷ 342.3 g mol-1 = 7.71 x 10-3 mol Examiner Tips and Tricks Always show your workings in calculations as its easier to check for errors and you may pick up credit if you get the final answer wrong. Page 12 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Calculate Reacting Masses Your notes Reacting mass calculations Chemical / symbol equations can be used to calculate: The moles of reactants and products The mass of reactants and products To do this: Information from the question is used to find the amount in moles of the substances being considered Then, the ratio between the substances is identified using the balanced chemical equation Once the moles have been determined they can then be converted into grams using the relative atomic or relative formula masses Worked Example Magnesium undergoes combustion to produce magnesium oxide. The overall reaction that is taking place is shown in the equation below. 2Mg (s) + O2 (g) ⟶ 2 MgO (s) Calculate the mass of magnesium oxide that can be made by completely burning 6.0 g of magnesium in oxygen in the following reaction: Ar(O) = 16 Ar(Mg) = 24 Answer: Step 1 - calculate the moles of magnesium ⎛ mass ⎞⎟ ⎛⎜ 6 ⎞⎟ Moles = ⎜⎜⎜ ⎟⎟ = ⎜ ⎟ = 0.25 ⎝ Mr ⎠ ⎝ 24 ⎠ Step 2 - use the molar ratio from the balanced symbol equation 2 moles of magnesium produce 2 moles of magnesium oxide The ratio is 1 : 1 Therefore, 0.25 moles of magnesium oxide is produced Step 3 - calculate the mass of magnesium oxide Page 13 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Mass = moles x Mr = 0.25 moles x (24 + 16) = 10 g Your notes Worked Example In theory, aluminium could decompose as shown in the equation below. 2Al2O3 ⟶ 4Al + 3O2 Calculate the maximum possible mass of aluminium, in tonnes, that can be produced from 51 tonnes of aluminium oxide. Ar(O) = 16 Ar(Al) = 27 Answer: Step 1 - calculate the moles of aluminium oxide Mass = 51 tonnes x 106 = 51 000 000 g ⎛ mass ⎞⎟ ⎛⎜ 51000000 ⎞⎟ Moles = ⎜⎜⎜ ⎟=⎜ ⎟ = 500 000 ⎝ Mr ⎟⎠ ⎝ 102 ⎠ Step 2 - use the molar ratio from the balanced symbol equation 2 moles of aluminium oxide produces 4 moles of aluminium The ratio is 1 : 2 Therefore, 2 x 500 000 = 1 000 000 moles of aluminium is produced Step 3 - calculate the mass of aluminium Mass = moles x Mr = 1 000 000 moles x 27 = 27 000 000 g ⎛ 27000000 ⎞⎟ Mass in tonnes = ⎜⎜ ⎟ = 27 tonnes ⎝ 10 6 ⎠ Examiner Tips and Tricks When you see any of the following phrases in a question, make sure you work methodically through the process of calculating moles, using molar ratios and calculating mass Calculate the mass... Calculate the minimum mass... Calculate the maximum mass... Page 14 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources As long as you are consistent, it doesn't matter whether you work in grams or tonnes or any other mass unit as the reacting masses will always be in proportion to the balanced equation Your notes Balancing Equations using Reacting Masses If the masses of reactants and products of a reaction are known then we can use them to write a balanced equation for that reaction This is done by converting the masses to moles and simplifying to find the molar ratios Worked Example A student reacts 1.2 g of carbon with 16.2 g of zinc oxide. The resulting products are 4.4 g of carbon dioxide and 13 g of zinc. Determine the balanced equation for the reaction. Answer: Step 1: Write the unbalanced chemical equation: C + ZnO → CO2 + Zn Step 2: Write down the masses of each substance: C = 12 ZnO = 65 + 16 = 81 CO2 = 12 + (2 x 16) = 44 Zn = 65 Step 3: Calculate the moles of each substance: C = 1.2 / 12 = 0.1 ZnO = 16.2 / 81 = 0.2 CO2 = 4.4 / 44 = 0.1 Zn = 13 / 65 = 0.2 Step 4: Deduce the whole number ratio of substances: C : ZnO : CO2 : Zn 0.1 : 0.2 : 0.1 : 0.2 1:2:1:2 Step 5: Apply the whole number ratio to the unbalanced equation: 1C + 2ZnO → 1CO2 + 2Zn C + 2ZnO → CO2 + 2Zn Page 15 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Examiner Tips and Tricks Your notes These questions look hard but they are actually quite easy to do, as long as you follow the steps and organise your work neatly. Remember the molar ratio of a balanced equation gives you the ratio of the amounts of each substance in the reaction. Page 16 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Calculate Percentage Yield Your notes How to calculate percentage yield Yield is the term used to describe the amount of product you get from a reaction In practice, you never get 100% yield in a chemical process for several reasons These include: Some reactants may be left behind in the equipment The reaction may be reversible and in these reactions a high yield is never possible as the products are continually turning back into the reactants Some products may also be lost during separation and purification stages such as filtration or distillation There may be side reactions occurring where a substance reacts with a gas in the air or an impurity in one of the reactants Products can also be lost during transfer from one container to another Actual and theoretical yield The actual yield is the recorded amount of product obtained The theoretical yield is the amount of product that would be obtained under perfect practical and chemical conditions It is calculated from the balanced equation and the reacting masses The percentage yield compares the actual yield to the theoretical yield For economic reasons, the objective of every chemical producing company is to have as high a percentage yield as possible to increase profits and reduce costs and waste Calculating percentage yield The percentage yield is a good way of measuring how successful a chemical process is There are often several methods of creating a compound and each method is called a reaction pathway Reaction pathways consist of a sequence of reactions which must occur to produce the required product Page 17 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Companies often investigate and try out different reaction pathways and these are then compared and evaluated so that a manufacturing process can be chosen Your notes The percentage yield of each pathway is a significant factor in this decision making process The equation to calculate the percentage yield is: actual yield percentage yield = × 100 theoretical yield Worked Example Copper(II) sulfate may be prepared by the reaction of dilute sulfuric acid on copper(II) oxide. A student prepared 1.6 g of dry copper(II) sulfate crystals. Calculate the percentage yield if the theoretical yield is 2.0 g. Answer: Actual yield of copper(II) sulfate = 1.6 g Percentage yield of copper(II) sulfate = (1.6 / 2.0) x 100 Percentage yield = 80% Examiner Tips and Tricks The actual yield and the percentage yield can be determined by experiment only The theoretical yield can be calculated assuming there is 100% conversion of reactants to products. You are expected to remember the equation for percentage yield If you remember it incorrectly and get a percentage yield greater than 100%, then you have made an error! The most common error is to divide the theoretical yield by the actual yield In this case, you just need to swap the numbers around in your calculation Page 18 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Experiment: Finding Formulae of Compounds Your notes Formulae of a simple compound by experiment The formulae of simple compounds can be found by careful experimentation and accurate measurements of mass changes The principle is to use mass measurements before and after a reaction and then convert masses into moles Using the moles of reactants and products it is possible to deduce molar ratios and hence an empirical formula Experiments which are easier to do using this process involve gases being lost or gained In this example a hydrated salt is heated to drive off the water as water vapour The formula of a hydrated salt Aim: To determine the formula of hydrated copper sulfate, CuSO4. xH2O Diagram: Heating a hydrated salt to remove the water of crystallisation Method Page 19 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 1. Measure the mass of evaporating dish 2. Add a known mass of hydrated salt Your notes 3. Heat over a Bunsen burner, gently stirring, until the blue salt turns completely white, indicating that all the water has been lost 4. Record the mass of the evaporating dish and its contents Practical tip: Avoid overheating the salt as it could decompose and give you a larger mass change Results: Mass of the white anhydrous salt Measure the mass of white anhydrous salt remaining Mass of water Subtract the mass of the white anhydrous salt remaining from the mass of known hydrated salt Step 1 – Divide the mass of the copper sulfate and the water by their respective molar masses Step 2 – Simplify the ratio of water to copper sulfate: anhydrous salt water Mass a b Moles (Step 1) a / Mr b / Mr =y =x Ratio (Step 2) 1: x Step 3 – Represent the ratio in the form ‘salt.xH2O’ Examiner Tips and Tricks It is unlikely that you will get a whole number for the number of moles of water in the ratio, so you will need to round up or down to the nearest whole number. Page 20 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Practical: Determine the Formula of a Metal Oxide Your notes Practical: Determine the formula of magnesium oxide Aim: To determine the empirical formula of magnesium oxide by combustion of magnesium Diagram: Method: 1. Measure the mass of the crucible with the lid 2. Add a sample of magnesium into the crucible and measure the mass with the lid (calculate the mass of the metal by subtracting the mass of the empty crucible) 3. Strongly heat the crucible over a Bunsen burner for several minutes 4. Lift the lid frequently to allow sufficient air into the crucible for the magnesium to fully oxidise without letting magnesium oxide smoke escape 5. Continue heating until the mass of the crucible remains constant (maximum mass), indicating that the reaction is complete Page 21 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 6. Measure the mass of the crucible and its contents (calculate the mass of metal oxide by subtracting the mass of the empty crucible) Your notes Results Mass of metal: Subtract the mass of the crucible from magnesium and the mass of the empty crucible Mass of oxygen: Subtract the mass of the magnesium used from the mass of magnesium oxid Step 1 – Divide each of the two masses by the relative atomic masses of the element Step 2 – Simplify the ratio magnesium oxygen mass a b moles a / Ar a / Ar x y Ratio = x : y Step 3 – Represent the ratio in the form ‘MxOy‘ E.g, MgO Practical: Determine the formula of copper(II) oxide Aim: To determine the formula of copper(II)oxide by reduction with methane Diagram: Page 22 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Method: 1. Measure mass of the empty boiling tube 2. Place metal oxide into a horizontal boiling tube and measure the mass again 3. Support the tube in a horizontal position held by a clamp 4. A steady stream of natural gas(methane) is passed over the copper(II)oxide and the excess gas is burned off 5. The copper(II)oxide is heated strongly using a Bunsen burner 6. Heat until metal oxide completely changes colour, meaning that all the oxygen has been removed 7. Measure mass of the tube remaining metal powder and subtract the mass of the tube Results: Working out empirical formula: Mass of Metal: Measure mass of the remaining metal powder Page 23 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Mass of Oxygen: Subtract mass of the remaining metal powder from the mass of metal oxide Your notes Step 1 – Divide each of the two masses by the relative atomic masses of elements Step 2 – Simplify the ratio: metal oxygen mass a b moles a / Mr b / Mr Ratio x y Step 3 – Represent the ratio in the form ‘MxOy‘ E.g, CuO Page 24 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Empirical & Molecular Formulae Your notes Empirical & Molecular Formulae The molecular formula is the formula that shows the number and type of each atom in a molecule E.g. the molecular formula of ethanoic acid is C2H4O2 The empirical formula is the simplest whole number ratio of the atoms of each element present in one molecule or formula unit of the compound E.g. the empirical formula of ethanoic acid is CH2O Organic molecules often have different empirical and molecular formulae The formula of an ionic compound is always an empirical formula Calculating empirical and molecular formula How to calculate empirical formulae Empirical formula calculations are very methodical Use a table and the following steps to complete an empirical formula calculation: 1. Write the element 2. Write the value given for each element This may be given as a mass, in g, or as a percentage There are exam questions where you are required to calculate the value of one of the elements 3. Write the relative atomic mass of each element 4. Calculate the moles of each element mass Moles = Ar 5. Calculate the ratio of elements Divide all the moles by the smallest number of moles If you get a ratio that does not have whole numbers, you multiply by an appropriate number to make all the values into whole numbers 6. Write the final empirical formula Page 25 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Worked Example A sample of a compound was found to contain 10 g of hydrogen and 80 g of oxygen. Calculate the empirical formula of this compound. Ar (H) = 1 Ar (O) = 16 Answer: 1. Element H O 2. Value 10 80 3. Relative atomic mass 1 16 mass 10 80 4. Moles = = 10 =5 Ar 1 16 5. Ratio (divide by smallest) 10 5 =2 =1 5 5 6. Answer The empirical formula is H2O Worked Example Carbohydrate X was analysed and found to contain 31.58% carbon and 5.26% hydrogen by mass. Find the empirical formula of carbohydrate X. Ar (H) = 1 Ar (C) = 12 Ar (O) = 16 Answer: A carbohydrate contains carbon, hydrogen and oxygen The percentages do not add up to 100%, which means that you need to calculate the percentage of oxygen needs to be calculated Percentage of oxygen = 100 - 31.58 - 5.26 = 63.16% 1. Element C H O Page 26 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 2. Value 31.58 5.26 63.16 3. Relative atomic mass 12 1 16 Your notes mass 31. 58 5. 26 63. 16 4. Moles = = 2.63 = 5.26 = 3.95 Ar 12 1 16 5. Ratio (divide by smallest) 2. 63 5. 26 3. 95 =1 =2 = 1.5 2. 63 2. 63 2. 63 5. Whole number ratio 1x2=2 2x2=4 1.5 x 2 = 3 6. Answer The empirical formula is C2H4O3 Examiner Tips and Tricks The molar ratio must be a whole number. If you don't get a whole number when calculating the ratio of atoms in an empirical formula, such as 1.5, multiply that and the other ratios to achieve whole numbers. How to calculate molecular formula Molecular formula gives the actual numbers of atoms of each element present in the formula of the compound Table showing the relationship between empirical and molecular formulae Compound Empirical formula Molecular formula Methane CH4 CH4 Ethane CH3 C2 H 6 Ethene CH2 C2 H 4 Benzene CH C 6H 6 To calculate the molecular formula: Page 27 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 1. Find the relative formula mass of the empirical formula Add the relative atomic masses of all the atoms in the empirical formula Your notes 2. Use the following equation: relative formula mass of molecular formula relative formula mass of empirical formula 3. Multiply the number of each element present in the empirical formula by the number from step 2 to find the molecular formula Worked Example The empirical formula of X is C4H10S1 The relative formula mass (Mr ) of X is 180. Calculate the molecular formula of X. Ar (C) = 12 Ar (H) = 1 Ar (S) = 32 Answer: 1. Calculate the relative formula mass of the empirical formula: Mr = (12 x 4) + (1 x 10) + (32 x 1) = 90 2. Divide the relative formula mass of X by the relative formula mass of empirical formula: 180 / 90 = 2 3. Multiply for the molecular formula: The number of atoms of each elements should be multiplied by 2 (C4 x 2) + (H10 x 2) + (S1 x 2) Molecular formula of X = C8H20S2 Deducing formulae of hydrated salts A hydrated salt is a crystallised salt that contains water molecules as part of its structure The formula of a hydrated salt shows the water molecules, e.g. CuSO4 2H2O The symbol shows that the water present is water of crystallisation The formula of hydrated salts can be determined experimentally by: Weighing a sample of the hydrated salt Page 28 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Heating it until the water of crystallisation has been driven off This is achieved by heating until a constant mass Your notes Re-weighing the anhydrous salt From the results, you can determine the mass of anhydrous salt and the mass of the water of crystallisation Applying a similar approach to deducing empirical formulae, the formula of the hydrated salt can be calculated How to calculate water of crystallisation The steps for empirical formula can be adapted for hydrated salt / water of crystallisation calculations Instead of writing elements, write the two components of a hydrated salt The salt Water Instead of writing relative atomic mass, write the relative molecular / formula mass of the salt and water Use a table and the following steps to complete the calculation: 1. Write the salt and water 2. Write the value given for the salt and water There are exam questions where you are required to calculate one of these values 3. Write the relative molecular / formula mass of the salt and water 4. Calculate the moles of the salt and water mass Moles = Mr 5. Calculate the ratio salt : water Divide all the moles by the smallest number of moles The calculation should give a ratio of 1 salt : x water 6. Write the final hydrated salt formula Worked Example Page 29 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 11.25 g of hydrated copper sulfate, CuSO4.xH2O, is heated until it loses all of its water of crystallisation. Your notes It is re-weighed and its mass is 7.19 g. Calculate the formula of the hydrated copper(II) sulfate. Ar (Cu) = 63.5 Ar (S) = 3. Ar (O) = 16 Ar (H) = 1 Answer: 1. Salt and water CuSO4 H 2O 2. Value 7.19 11.25 - 7.19 = 4.06 3. Mr 63.5 + 32 + (16 x 4) (1 x 2) + 16 = 159.5 = 18 mass 7. 19 4. 06 4. Moles = = 0.045 = 0.226 Mr 159. 5 18 5. Salt : water ratio 0. 045 0. 226 =1 =5 0. 045 0. 045 6. Formula of hydrated salt The formula is CuSO4 5H2O Examiner Tips and Tricks The specification is not clear about whether deducing the formula of hydrated salts is required. However, it is an application of deducing empirical formulae so it is worth knowing how to do this. Page 30 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Calculate Concentrations of Solutions Your notes Calculate Concentrations of Solutions A solute is a solid substance that dissolves into a liquid The amount of solute can be expressed in grams (g) or moles (mol) A solvent is the liquid that a solute dissolves in The amount / volume of a solvent is measured in cm3 or dm3 Most chemical reactions occur between solutes which are dissolved in solvents, such as water or an organic solvent A solution is the mixture formed when a solute dissolves in a solvent The amount / volume of a solution measured in cm3 or dm3 Concentration refers to the amount of solute there is in a specific volume of the solvent The greater the amount of solute in a given volume, the greater the concentration Concentration is sometimes commonly referred to as strength For example, dissolving more coffee in hot water results in a stronger coffee Typically, concentration is expressed in terms of the amount of substance per dm3 Therefore, the units of concentration are: g / dm3 mol / dm3 It is more useful to a chemist to express concentration in terms of moles per unit volume rather than mass per unit volume To calculate concentration in mol / dm3 we use the following equation: number of moles of solute (mol) concentration (mol/dm3 ) = volume of solution (dm3 ) Page 31 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes The concentration-moles formula triangle Volumes are often expressed in cm3, but dm3 must be used when calculating concentration To convert cm3 to dm3, divide by 1000 To convert dm3 to cm3, multiply by 1000 Converting between cm3 and dm3 Worked Example Page 32 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Calculate the amount of solute, in moles, present in 2.5 dm3 of a solution whose concentration is 0.2 mol / dm3. Your notes Answer: Write down the information you are given in the question: Concentration of solution: 0.2 mol / dm3 Volume of solution: 2.5 dm3 Calculate the number of moles: Moles = concentration x volume Moles = 0.2 x 2.5 = 0.5 mol Worked Example Calculate the concentration of a solution of sodium hydroxide, NaOH, in mol / dm3, when 80 g is dissolved in 500 cm3 of water. Relative atomic masses, Ar: Na = 23; H = 1; O = 16 Answer: Calculate the Mr of NaOH: 23 + 16 + 1 = 40 Determine the number of moles of NaOH: 40 g = 1 mole So, 80 g = 2 moles Convert cm3 to dm3: 500 = 0.5 dm3 1000 Calculate the concentration: moles Concentration = volume 2 Concentration = = 4 mol / dm3 0. 5 Worked Example Page 33 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 25.0 cm3 of 0.050 mol / dm3 sodium carbonate was completely neutralised by 20.00 cm3 of dilute hydrochloric acid. Calculate the concentration in mol / dm3 of the hydrochloric acid. Your notes Na2CO3 + 2HCl → 2NaCl + H2O + CO2 Answer: Calculate the moles of sodium carbonate: Moles of Na2CO3 = concentration x volume Remember: The volume needs to be in dm3 25. 0 Moles of Na2CO3 = 0.05 x = 0.00125 1000 Calculate the moles of hydrochloric acid: The balanced symbol equation shows that 1 mole of Na2CO3 reacts with 2 moles of HCl So, 0.00125 moles of Na2CO3 reacts with 0.00250 moles of HCl Calculate the concentration of hydrochloric acid: moles Concentration = volume Remember: The volume needs to be in dm3 20 cm3 ÷ 1000 = 0.02 dm3 0. 00250 Concentration = = 0.125 mol / dm3 0. 02 Examiner Tips and Tricks Don't forget your unit conversions: To go from cm3 to dm3 : divide by 1000 To go from dm3 to cm3 : multiply by 1000 Page 34 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Calculate Volumes of Gases Your notes Calculate Volumes of Gases Avogadro's Law Avogadro’s Law states that at the same conditions of temperature and pressure, equal amounts of gases occupy the same volume of space At room temperature and pressure, the volume occupied by one mole of any gas was found to be 24 dm3 or 24,000 cm3 This is known as the molar gas volume at RTP RTP stands for “room temperature and pressure” and the conditions are 20 ºC and 1 atmosphere (atm) From the molar gas volume the following formula triangle can be derived: Page 35 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Formula triangle showing the relationship between moles of gas, volume in dm3 and the molar volume If the volume is given in cm3 instead of dm3, then divide by 24,000 instead of 24: Your notes Formula triangle showing the relationship between moles of gas, volume in cm3 and the molar volume The formula can be used to calculate the number of moles of gases from a given volume or vice versa Simply cover the one you want and the triangle tells you what to do To find the volume Volume = Moles x Molar Volume Examples of Converting Moles into Volumes Table Name of Gas Amount of Gas Volume of Gas Page 36 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Hydrogen 3 mol (3 x 24) = 72 dm3 Your notes Carbon Dioxide 0.25 mol (0.25 x 24) = 6 dm3 Oxygen 5.4 mol (5.4 x 24,000) = 129,600 cm3 Ammonia 0.02 mol (0.02 x 24) = 0.48 dm3 To find the moles Moles = Volume ÷ Molar Volume Examples of Converting Volumes into Moles Table Name of Gas Volume of Gas Amount of Gas Methane 225.6 dm3 (225.6 ÷ 24) = 9.4 mol Carbon Monoxide 7.2 dm3 (7.2 ÷ 24) = 0.3 mol Sulfur Dioxide 960 dm3 (960 ÷ 24) = 40 mol Oxygen 1200 cm3 (1200 ÷ 24,000) = 0.05 mol Using mass to calculate the volume of a gas You may be asked to calculate the volume of a gas from a given amount stated in grams instead of moles To answer these type of questions you must first convert grams to moles and then calculate the volume. Worked Example What is the volume of 154 g of nitrogen gas at RTP? Answer: Step 1: Calculate the moles of nitrogen: Page 37 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Mr (N2) = 2 x 14 = 28 154 Moles of N2 = = 5.5 mol Your notes 28 Step 2: Calculate the molar volume of nitrogen: Volume = moles x 24 Volume = 5.5 x 24 = 132 dm3 A second style of gas calculation involves calculating the volumes of gaseous reactants and products from a balanced equation and a given volume of a gaseous reactant or product These problems are straightforward as you are applying Avogadro's Law, so the moles ( and coefficients) in equations are in the same ratio as the gas volumes Worked Example The complete combustion of propane gives carbon dioxide and water vapour as the products. C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g) Determine the volume of oxygen needed to react with 150 cm3 of propane and the total volume of the gaseous products Answer The balanced equation shows that 5 moles of oxygen are needed to completely react with 1 mole of propane Therefore, the volume of oxygen needed would be = 5 moles x 150 cm3 = 750 cm3 The total number of moles of gaseous products is = 3 + 4 = 7 moles The total volume of gaseous products would be = 7 moles x 150 cm3 = 1050 cm3 Examiner Tips and Tricks Make sure you use the correct units as asked by the question when working through reacting gas volume questions. Page 38 of 38 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers