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Transcript
Stoichiometry Mr Field Using this slide show The slide show is here to provide structure to the lessons, but not to limit them….go off-piste when you need to! Slide shows should be shared with students (preferable electronic to save paper) and they should add their own notes as they go along. A good tip for students to improve understanding of the calculations is to get them to highlight numbers in the question and through the maths in different colours so they can see where numbers are coming from and going to. The slide show is designed for my teaching style, and contains only the bare minimum of explanation, which I will elaborate on as I present it. Please adapt it to your teaching style, and add any notes that you feel necessary. Main Menu Menu: Lesson 1 – The Mole Lesson 2 – Moles and Molar Mass Lesson 3 - Solutions Lesson 4 - Formulas Lesson 5 - Equations Lesson 6 – Theoretical Yields Lesson 7 – Molar Volume of Gases Lesson 8 – Ideal Gases Lesson 9 - TEST Lesson 10 – Test Debrief Main Menu Lesson 1 The Mole and the Avogadro Constant Main Menu Intro to Unit Brainstorm everything you know about quantitative chemistry (equations, moles and all that stuff!) Main Menu Overview Copy this onto an A4 page. You should add to it as a regular review throughout the unit. Main Menu Assessment Test at the end of the unit (100%) Approx lesson 9 Main Menu We Are Here Main Menu Lesson 1: Meet the Mole Objectives: Understand the phrase ‘a mole’ as just a large number Define ‘a mole’ according to Avogadro’s number Complete an experiment to determine Avogadro’s number Calculate numbers of particles using Avogadro’s number Main Menu What do the following have in common? A dozen A grand A score A pair A trio Main Menu A Mole A mole is a large number: A mole is: 6.02 x 1023 602,000,000,000,000,000,000,000 Six hundred and two thousand quadrillion Given the symbol, L This number is called Avogadro’s number after the Italian scientist Amedeo Avogadro who first proposed it Picture – all the grains of sand, on all the beaches and in all the deserts in the world; that is about a tenth of a mole! Main Menu Some Calculations •We can use this equation to calculate a number of moles from a number of particles N n L •Where: •n = quantity in moles •N = number of particles •L = Avogadro’s Constant (6.02x1023) 1. Example 1: You have 3.01x1022 atoms of carbon. How many moles is this? •n = N / L •n(C) = 3.01x1022 / 6.02x1023 •n(C) = 0.0500 mol 2. Example 2: You have 6.02x1024 molecules of water. How many moles of hydrogen atoms are present? •n = N / L •n(H) = 2 x n(H2O) •n(H) = 2 x 6.02x1024 / 6.02x1023 •n(H) = 20.0 3. Example 3: How many atoms of hydrogen are there in 2.5 moles of methane (CH4)? •N(H) = 4 x N(CH4) •N(H) = 4 x n(CH4) x L •N(H) = 4 x 2.5 x 6.02x1023 •N(H) = 6.02x1024 Main Menu Some Calculations 1. How many moles of ethene (C2H4) are there in 1.5x1022 molecules? 2. How many moles of oxygen atoms in 1.20x1024 molecules of sulphuric acid (H2SO4)? How many moles of oxygen molecules could they make? 3. How many molecules in 3.0 moles of nitrogen gas? 4. How many chloride ions are released on dissolving 0.050 moles of calcium chloride (CaCl2)? BONUS QUESTION: How many moles of people are there on the planet? Main Menu How do you determine Avogadro’s number? There are a number of ways including: Electrolysis Measurement of electron mass X-ray crystal density Molecular monolayers You will be determining L by measuring the area of a molecular monolayer Not the most accurate but feasible in our lab! Main Menu Key Points N n L particles moles 23 6.02 x10 Main Menu Lesson 2 Moles and Molar Mass Main Menu Refresh How many molecules are present in a drop of ethanol, C2H5OH, of mass 2.3 × 10–3 g? (L = 6.0 × 1023 mol–1) A. B. C. D. 3.0 × 10 19 3.0 × 1020 6.0 × 1020 6.0 × 1026 How many oxygen atoms are there in 0.20 mol of ethanoic acid, CH3COOH? A. 1.2 × 1023 B. 2.4 × 1023 C. 3.0 × 1024 D. 6.0 × 1024 Main Menu We Are Here Main Menu Lesson 2: Moles and Molar Mass Objectives: Calculate the relative mass and molar mass of substances Relate the mass of a substance to a quantity in moles Conduct an experiment to determine the number of moles of water of crystallisation Main Menu Atomic and Formula Mass Ar Mr is the relative atomic mass of an element found in the periodic table is the relative molecular mass of a compound To calculate Mr, you add the Ar of all the atoms in a compound 1. What does the term ‘relative’ mean? 2. What determines the mass of an atom and how can Ar be a decimal number? Main Menu Calculating Mr HCl Ar(H) = 1.01 Ar (Cl) = 35.45 Mr = 1.01 + 35.45 = 36.46 C 2 H4 Ar(C) = 12.01 Ar (H) = 1.01 H2SO4 Mr = 2x12.01 + 4x1.01 = 16.06 Mr = 2x1.01 + 32.06 + 4x16.00 = 98.08 Mg(OH)2 Ar(H) = 1.01 Ar (S) = 32.06 Ar(O) = 16.00 Main Menu Ar(Mg) = 24.31 Ar(O) = 16.00 Ar (H) = 1.01 Mr = 24.31 + 2x16.00 + 2x1.01 = 58.33 Calculate Mr for: Br2 (NH4)2SO4 C3H8 C6H12O6 Main Menu Molar Mass, Mm This is the mass of one mole of something. To calulate Mm, simply stick ‘g’ for grams on the end of Mr. For example: Mr(H2O) = 18.02 Mm(H2O) = 18.02 g Note: This is why the value of L was chosen to be what it was. Main Menu Relating ‘mass’ and ‘molar mass’ You need to be able to solve problems like: How many moles of Y is mass X? What is the mass of X moles of Y? X moles of Y has a mass of Z, what is it’s molar mass? Use this equation: Main Menu For example: How many moles of water are present in 27.03g? Calc Mm(H2O): Mm(H2O)= 2x1.01 + 16.00 = 18.02 Find n(H2O): n(H2O) = M / Mm = 27.03 / 18.02 = 1.50 mol What is the mass of 4.40 mol of iron (III) oxide (Fe2O3)? Calc Mm(Fe2O3): Mm(H2O)= 2x55.85 + 3x16.00 = 159.70 g Find M(Fe2O3): M(Fe2O3) = n x Mm = 4.40 / 159.70 = 703 g 1.30 moles of an unknown compound has a mass of 20.9g, what is it’s molar mass? Mm(unknown)= M / n = 20.9 / 1.30 = 16.1 g/mol Main Menu Questions 1. Calculate the mass of 0.10 mol of benzene (C6H6) 2. Calculate the mass of 0.75 mol of ammonium nitrate (NH4NO3) 3. 4. What quantity of iron (III) oxide is present in a 1.0 kg sample? What quantity of cobalt (II) chloride (CoCl2) is present in a 2.40 g sample? Main Menu 5. 8.8 moles of a compound has a mass of 1.41 kg. Calculate its molar mass. 6. 0.010 mol of an oxide of hydrogen has a mass 0.340 g. Deduce it’s formula. Moles of Water of Crystallisation Many compounds can incorporate water into their crystal structure, this is called the water of crystallisation. CoCl2 is blue The ‘.’ means the water is ‘associated’ with the CoCl2 CoCl2.6H2O is pink It is loosely bonded, but exactly how is unimportant In this experiment you will calculate the moles of water of crystallisation of a compound Main Menu Key Points mass moles molar mass Main Menu Lesson 3 Solutions Main Menu Refresh A student reacted some salicylic acid with excess ethanoic anhydride. Impure solid aspirin was obtained by filtering the reaction mixture. Pure aspirin was obtained by recrystallisation. The following data was recorded by the student. 1. Mass of salicylic acid used Mass of pure aspirin obtained 3.15 ± 0.02 g 2.50 ± 0.02 g Determine the amount, in mol, of salicylic acid, C6H4(OH)COOH, used. Main Menu We Are Here Main Menu Lesson 3: Solutions Objectives: Understand the relationship between concentration, volume and moles Prepare a standard solution of silver nitrate. Pose and solve problems involving solutions (of the chemical kind not the answers kind) Main Menu Solutions Basics Aqueous copper sulfate solution: + SOLUTE SOLVENT Main Menu SOLUTION Concentration This is the strength of a solution. Most Concentrated Least Concentrated Main Menu Molarity The number of moles of a substance dissolved in one litre of a solution. moles concentration volume Units: mol dm-3 Pronounced: moles per decimetre cubed Units often abbreviated to ‘M’ (do not do this in an exam!) Volume must be in litres (dm3) not ml or cm3 This is the most useful measure of concentration but there are others such as % by weight, % by volume and molality. Main Menu Example 1: 25.0 cm3 of a solution of hydrochloric acid contains 0.100 mol HCl. What is it’s concentration? Answer: Concentration = moles / volume = 0.100 / 0.0250 = 4.00 mol dm-3 Note: the volume was first divided by 1000 to convert to dm3 Main Menu Example 2: Water is added to 4.00 g NaOH to produce a 2.00 mol dm-3 solution. What volume should the solution be in cm3? Calculate quantity of NaOH: n(NaoH) = mass / molar mass = 4.00/40.0 = 0.100 Calculate volume of solution: Volume = moles / concentration = 0.100 / 2.00 = 0.0500 dm3 = 50.0 cm3 Main Menu Example 3: It is found by titration that 25.0 cm3 of an unknown solution of sulfuric acid is just neutralised by adding 11.3 cm3 of1.00 mol dm-3 sodium hydroxide. What is the concentration of sulfuric acid in the sample. H2SO4 + 2 NaOH Na2SO4 + 2 H2O Use: n 1 C1 V 1 = n 2 C2 V 2 1 x C1 x 25.0 = 2 x 1.00 x 11.3 Where: n = coefficient C = concentration V = volume ‘1’ refers to H2SO4 ‘2’ refers to NaOH C1 = (2 x 1.00 x 11.3) / (1 x 25.0) = 0.904 mol dm-3 Main Menu Questions 1. You have 75.0 cm3 of a 0.150 mol dm-3 solution of zinc sulphate (ZnSO4). What mass of zinc sulphate crystals will be left behind on evaporation of the water? 2. What mass of copper (II) chloride (CuCl2) should be added to 240 cm3 water to form a 0.100 mol dm-3 solution? 3. A 10.0 cm3 sample is removed from a vessel containing 1.50 dm3 of a reaction mixture. By titration, the sample is found to contain 0.0053 mol H+. What is the concentration of H+ in the main reaction vessel? 4. The titration of 50.0 cm3 of an unknown solution of barium hydroxide was fully neutralised by the addition of 12 cm3 of 0.200 mol dm-3 hydrochloric acid solution. What concentration is the barium hydroxide solution? Ba(OH)2 + 2 HCl BaCl2 + 2 H2O Main Menu Preparing a Standard Solution Prepare standard solutions silver nitrate with concentrations of 0.25-0.75 mol dm-3 You will use these in a future lesson so make them well! Main Menu Problem Time Write three stoichiometry problems using any of the ideas from the unit so far (make sure you know the answer). In ten minutes time you will need to give your problems to a classmate to solve. Main Menu Key Points moles concentration volume Main Menu Remember! Alcohol is not the answer but it is a solution*! *Of ethanol, water and various other bits and pieces Main Menu Lesson 4 Formulas Main Menu Refresh Which statement about solutions is correct? A. B. C. D. When vitamin D dissolves in fat, vitamin D is the solvent and fat is the solute. In a solution of NaCl in water, NaCl is the solute and water is the solvent. An aqueous solution consists of water dissolved in a solute. The concentration of a solution is the amount of solvent dissolved in 1 dm3 of solution A student added 27.20 cm3 of 0.200 mol dm–3 HCl to 0.188 g of eggshell. Calculate the amount, in mol, of HCl added. Main Menu We Are Here Main Menu Lesson 4: Formulas Objectives: Understand the difference between empirical and molecular formulas Experimentally determine the empirical formula of a compound. Solve problems involving empirical and molecular formulas Main Menu Formulas You have one minute to write down as many different chemical formulas as you can. Go! No!!!! What is the job of a formula? Main Menu Types of Formula Empirical: Molecular: The ratio of the atoms of each element in a compound in its lowest terms The number of atoms of each element in a molecule You will meet other types in the organic chemistry unit including structural, displayed and skeletal. Main Menu Examples Oxygen Molecular: O2 Empirical: O Water Molecular: C2H6 Empirical: CH3 Glucose Molecular: H2O Empirical: H2O Ethane Sodium Chloride Molecular: C6H12O6 Empirical: CH2O Molecular: n/a* Empirical: NaCl Copper Sulphate Molecular: n/a* Empirical: CuSO4 *Why do these two not have an empirical formula? Main Menu Your turn… Write empirical and molecular formulas for each of the following: Chlorine gas Sulphuric acid Propane Ethanol Main Menu Determining empirical formulas from % by mass data* For example: a sample of a compound contains 20% hydrogen and 80% carbon by mass. C: 80% 6.67 1 H 20% 20 3 write out the elements as a ratio write the % composition below divide each % by the Ar of each element divide each number by the smallest of the two numbers Empirical formula = CH3 *This may seem like an odd thing to do but this data is easily determined in the lab and so is very useful Main Menu Determining Molecular Formulas From Empirical Formulas You need to know the molecular mass (can be found from mass spectrometry (see Atomic Structure unit). Following on from the previous question. Your unknown molecule is found to have Mr = 30.06 Determine empirical formula mass: Divide molecular mass by empirical formula mass to tell you the multiplier for the empirical formula: Mr(CH3) = 12.00 + (3 x 1.01) = 15.03 Multiplier = 30.06 / 15.03 = 2 Multiply the empirical formula by the multiplier: Molecular formula = CH3 x 2 = C2H6 Main Menu Determining an empirical formula You will be determining the empirical formula of an unknown compound of potassium, chlorine and oxygen. Follow the instructions here Main Menu Practice Questions Begin the practice questions found here Complete for homework. Main Menu Key points: Empirical formula gives the lowest ratio of atoms The molecular formula is a whole multiple of the empirical formula When using % by mass data, start by dividing each % by the Ar of that element. Main Menu Lesson 5 Equations Main Menu Refresh A toxic gas, A, consists of 53.8 % nitrogen and 46.2 % carbon by mass and has a molar mass of 50.81 g/mol. Determine the empirical formula of A. Determine the molecular formula of A Main Menu We Are Here Main Menu Lesson 5: Equations Objectives: Know how to construct and balance symbol equations Apply the concept of the mole ratio to determine the amounts of species involved in chemical reactions Meet the idea of the ‘limiting reagent’ Main Menu Word Equations Hydrogen + Oxygen REACTANTS Water PRODUCTS The arrow means: ‘makes’ or ‘becomes’ not ‘equals’ Main Menu Symbol Equations 2 H2 + O2 2 H2O 4H 4H 2O 2O The bold numbers are called coefficients and tell you the number of each molecule involved in the reaction vs H2 + O2 H2 O 2H 2H 2O 1O Required to balance the equation Without them the equation does not balance – each side of the reaction would have different numbers of each atom – which would break physics You can’t change the little numbers in the formulas as this changes the chemical If there is no coefficient, it is ‘1’ Main Menu Tips for balancing equations Only change the coefficients Only make one change at a time Re-count the numbers of each atom after each change Try keeping a tally-chart of the numbers of atoms on each side of the equation If you get to a point where you just need a fractional amount of a compound, write it in and then multiply everything by the denominator of the fraction. BE PATIENT!!!! Main Menu Construct equations and then balance each of the following: Magnesium (Mg) reacts with hydrochloric acid (HCl) to make magnesium chloride (MgCl2) and hydrogen gas Ethane (C2H6) reacts with oxygen gas to make carbon dioxide and water Lead nitrate (Pb(NO3)2) reacts with aluminium chloride (AlCl3) to make aluminium nitrate (Al(NO3)3) and lead chloride (PbCl2) Barium nitride (Ba3N2) reacts with water to make barium hydroxide (Ba(OH)2) and ammonia (NH3) Extension: Write a flow chart that enables you to balance equations Main Menu Mole Ratios This is the ratio of one compound to another in a balanced equation. For example, in the previous equation 2 H2 + O2 2 H2O Hydrogen, oxygen and water are present in 2:1:2 ratio. This ratio is fixed and means for example: 0.2 mol of H2 reacts with 0.1 mol of O2 to make 0.2 mol H2O 5 mol of H2 reacts with 2.5 mol of O2 to make 5 mol of H2O To make 4 mol of H2O you need 4 mol of H2 and 2 mol of O2 Main Menu Mole Ratios cont…. 2 Al(OH)3 + 3 H2SO4 Al2(SO4)3 + 3 H2O In the above equation, what quantity of Al(OH)3 in moles is required to produce 5.00 mol of H2O? Divide the given moles by the given coefficient and multiply by the target coefficient: n(Al(OH)3) = 5.00/3 x 2 = 3.33 mol* *Dividing by ‘3’ and multiplying by ‘2’ uses the H2O:Al(OH)3 mole ratio of 3:2 to adjust the number of moles. Main Menu The Limiting Reagent In a reaction, there is we can describe reactants as being ‘limiting’ or in ‘excess’ Limiting – this is the reactant that runs out Excess – the reaction will not run out of this 2 H2 + O2 2 H2O For example, if you have 2.0 mol H2 and 2.0 mol O2 To determine this, divide the quantity of each reactant by its coefficient in the equation. The smallest number is the limiting reactant: H2 is the limiting reactant – it will run out O2 is present in excess – there is more than enough H2: 2.0 / 2 = 1.0 – smallest therefore limiting O2: 2.0 / 1 = 2.0 You should use the limiting reactant when doing all further calculations including: Determining amounts of products formed Determining amounts of other reactants used Main Menu Use the idea of limiting reactants to identify the limiting reactant and solve the problem. Mg + 2 HCl MgCl2 + H2 2 C2H6 + 10 O2 4 CO2 + 6 H2O 0.55 mol C2H6 reacting with 3.50 mol O2. How many moles of CO2 is formed? 3 Pb(NO3)2 + 2 AlCl3 2 Al(NO3)3 + 3 PbCl2 0.15 mol Mg reacting with 0.25 mol HCl. How many moles of H2 is formed? 2.60 mol Pb(NO3)2 reacting with 3.00 mol AlCl3. Which element is in excess and how much remains after the reaction? Ba3N2 + 6 H2O 3 Ba(OH)2 + 2 NH3 1.45 mol Ba3N2 reacting with 10.0 mol H2O. Which reactant is present in excess and how much remains after the reaction? Main Menu Key Points Balance equations by playing with coefficients Use mole ratios to work quantities of chemicals involved in reactions Divide the quantity of each reactant by its coefficient to determine the limiting reactant Main Menu Lesson 6 Theoretical Yields Main Menu Refresh Equal masses of the metals Na, Mg, Ca and Ag are added to separate samples of excess HCl(aq). Which metal produces the greatest total volume of H2(g)? A. Na B. Mg C. Ca D. Ag Chloroethene, C2H3Cl, reacts with oxygen according to the equation below. 2C2H3Cl(g) + 5O2(g) → 4CO2(g) + 2H2O(g) + 2HCl(g) What is the amount, in mol, of H2O produced when 10.0 mol of C2H3Cl and 10.0 mol of O2 are mixed together, and the above reaction goes to completion? A. 4.00 B. 8.00 C. 10.0 D. 20.0 Main Menu We Are Here Main Menu Lesson 6: Theoretical Yields Objectives: Use the idea of limiting reactant to determine the proportions of an unknown mixture Use stoichiometry to calculate theoretical yields Main Menu Preparation of Silver Chromate In this experiment you will prepare silver chromate The challenge is that you will start with an unknown mixture of the reactants and will have to use your knowledge of stoichiometry to work out its composition Main Menu Determining Theoretical Yield The theoretical yield is the amount of product you expect to make if all of your limiting reactant fully reacts. It can be calculated by following these steps: Calculate moles of all reactants Determine limiting reactant Use mole ratios to calculate moles of product expected Convert moles of product to mass/volume/solution etc. All this does is string together all the calculations you have already met. Main Menu Some problems Have a go at the problems found here. Main Menu Key Points Calculate moles of all reactants Determine limiting reactant Use mole ratios to calculate moles of product expected Convert moles of product to mass/volume/solution etc. Main Menu Lesson 7 Molar Volumes of Gases Main Menu Refresh 0.600 mol of aluminium hydroxide is mixed with 0.600 mol of sulfuric acid, and the following reaction occurs: 2 Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l) a. Determine the limiting reactant. b. Calculate the mass of Al (SO ) produced. 2 4 3 Main Menu We Are Here Main Menu Lesson 7: Equations Objectives: Understand that a fixed quantity (in moles) of gas, always occupies the same volume at room temperature Perform calculations using the molar volume of a perfect gas Apply the above concept to design the best possible bottle rocket. Main Menu The Molar Volume of an Ideal Gas At standard temperature and pressure (STP): = 2.24x10-2 m3 mol-1 = 22.4 dm3 mol-1 T = 273K, P = 1.01x105 Pa At room temperature and pressure (RTP): Molar Volume of Gas Molar Volume of Gas = 2.45x10-2 m3 mol-1 = 24.5 dm3 mol-1 This is true (ish) no matter what the gas!...we will look at why next lesson Main Menu In Calculations…. What volume of H2 gas is produced when 0.050 mol Mg reacts with excess acid at S.T.P.? 2 Li(s) + 2 HCl(aq) 2 LiCl(aq) + H2(g) n(H2) = n(Li) /2 x 1 = 0.050/1 x 1 = 0.025 V(H2) = 0.025 x 22.4 = 0.56 dm3 Main Menu More calculations At RTP, 30 cm3 ethane reacts with 60 cm3 oxygen. Which reactant is present in excess, how much remains after the reaction and what volume of CO2 is produced? 2 C2H6(g) + 10 O2(g) 6 H2O(l) + 4 CO2(g) Limiting reactant: C2H6: 30/2 = 15 O2: 60/10 = 6 C2H6 remaining: therefore O2 limiting, ‘6’ will be the number used in all further calculations as there is enough O2 for ‘6’ of the reaction V(C2H6 used) = 6 x 2 = 12 cm3 V(C2H6 remaining) = 30 – 12 = 18 cm3 Volume CO2 produced: V(CO2) = 6 x 4 = 24 cm3 Note: there is no need to convert to moles as they are all in a ratio of moles as you would divide by 24.0 to get to moles, do your sums and then multiply by 24 .0 to get back to volumes…so why bother? Main Menu Time to practice What is the minimum volume of H2 gas required to fully reduce 10.0 g copper (II) oxide to copper (assume RTP…poor assumption but will do for the time being!)? CuO(s) + H2(g) Cu(s) + H2O(l) In a car airbag, sodium azide (NaN3) decomposes explosively to make N2 gas. What is the minimum mass of sodium azide required to fully inflate a 60.0 dm3 airbag, assuming RTP? 2 NaN3(s) 2 Na(s) + 3 N2(g) 500 cm3 carbon monoxide reacts with 300 cm3 oxygen to produce carbon dioxide. What are the final volumes of each of the three gases on completion of the reaction? CO(g) + O2(g) CO2(g) Main Menu Bottle Rockets Hydrogen reacts explosively with oxygen Bad if you hate fun! Excellent if you like to make stuff go bang or whoosh! You will need to design and build rockets (from standard 500 ml drinks bottles). The best will win an awesome prize.You need to consider: A suitable reaction to generate hydrogen The stoichiometry of the reaction The oxygen content of the air Suitable ways to decide the winner Making it look cool (if you finish early!) Main Menu Key Points The volume of gas depends on the temperature, pressure and number of moles, NOT THE TYPE OF GAS Molar Volume at STP = 22.4 dm3 Molar Volume at RTP = 24.5 dm3 Main Menu Lesson 8 Ideal Gases Main Menu Refresh What volume of sulfur trioxide, in cm3, can be prepared using 40 cm3 sulfur dioxide and 20 cm3 oxygen gas by the following reaction? Assume all volumes are measured at the same temperature and pressure. 2SO2(g) + O2(g) → 2SO3(g) A. B. C. D. 20 40 60 80 Main Menu We Are Here Main Menu Lesson 8: Ideal Gas Equation Objectives: Understand the ideal gas equation Complete a circus of short experiments to explore the ideal gas equation Perform calculations using the ideal gas equation Main Menu Molar Volume of a Perfect Gas We learnt about the molar volume of gases last lesson….how can they be the same? The distance between particles is much bigger than the size of the particles….so particle size makes very little difference: 10 units 10 units In reality, the relative distance between the molecules is much much greater than this. Main Menu The Ideal Gas Equation The volume a gas takes up is determined by: Pressure Temperature Moles of gas This combines to form the ideal gas equation PV = nRT Where: P = pressure in Pa V = volume in m3 n = moles of gas R = gas constant, 8.31 J K-1 mol-1, this appears in many places in chem T = temperature in K Main Menu Ideal Gas Assumptions Particles occupy no volume Particles have zero intermolecular forces These are not always valid, particularly at: Low temperature High pressure Main Menu Study the equation to make some predictions: How would does temperature change if you decrease pressure at fixed volume? How would volume change if you heat something at fixed pressure? How would pressure change if you decrease the volume at fixed temperature? Main Menu Exploring Ideal Gases The best way to explore the behaviour of gases is to have a little play. Complete these four short experiments Main Menu Some calculations: 1.048 g of unknown gas A, occupies 846 cm3. at 500K What is it’s molecular mass? n( A) PV RT 846 1000 1000 0.02056 8.31 500 101,000 Mm(A) = mass / moles = 1.048 / 0.02056 = 60.0 g/mol Main Menu More calculations The volume of an ideal gas at 54.0 °C is increased from 3.00 dm3 to 6.00 dm3. At what temperature, in °C, will the gas have the original pressure? Use a modified version of the ideal gas equation: PV T 1 1 1 PV T 2 2 2 Since original and final pressure should be the same, we can remove this from the equation as they cancel out: 54.0 converted to Kelvin by adding 273 V V T T 1 2 1 2 3.00 / 327.0 = 6.00 / T2 T2 = (6.00 x 327.0 / 3.00) = 653 K Main Menu Time to practice Complete the questions found here Main Menu Key Points The Ideal Gas equation: PV = nRT Also: PV T 1 1 1 PV T 2 2 2 Provided that: Molecules have zero volume Molecules experience no attraction to each other Main Menu Lesson 9 Test Main Menu Good Luck You have 80 minutes! Main Menu Lesson 10 Test Debrief Main Menu Personal Reflection Spend 15 minutes looking through your test: Make a list of the things you did well Use your notes and text book to make corrections to anything you struggled with. Main Menu Group Reflection Spend 10 minutes working with your classmates: Help classmates them with corrections they were unable to do alone Ask classmates for support on questions you were unable to correct Main Menu Go Through The Paper Stop me when I reach a question you still have difficulty with. Main Menu Targeted Lesson PREPARE AFTER MARKING THE TEST SHORT LESSON ON SPECIFIC AREAS OF DIFFICULTY Main Menu