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How do you know if it is balanced? • The number of atoms of each element must balance CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) • LHS: RHS: 1xC 1xC 4xH 2x2=4xH 2x2=4xO 2+2x1=4xO Balanced or unbalanced? Balanced or unbalanced? 2H2 + O2 2H2O Balanced or unbalanced? 2 Mg + O2 2MgO x Balanced or unbalanced? 2 K + Cl2 2KCl x • When NaCl is added to water, the ions are pulled off the structure, and they become free to move in the water. visible + invisible • The ions can move independently, so a solution of NaCl in water conducts electricity. Activity • mix copper (II) sulphate and barium chloride solutions • describe observations • did you get them all – – – – clear solutions colours of solutions colour of the precipitate the word ‘precipitate’ • Can you explain them Equations copper (II) sulfate + barium chloride copper (II) chloride + barium sulfate CuSO4(aq) + BaCl2(aq) CuCl2(aq) + BaSO4(s) Why is the balanced equation more useful? Why is the only chemical we can really see the barium sulfate? Cu2+(aq) + SO42-(aq) + Ba2+(aq) + 2 Cl-(aq) Cu2+(aq) + 2 Cl-(aq) + BaSO4(s) Is this too much writing? SO42-(aq) + Ba2+(aq) BaSO4(s) Learning objective: 01/05/2017 • To become familiar with the concept of the mole • Sponge: 350g unsalted butter 350g caster sugar ½ doz medium eggs, beaten 350g self-raising flour Filling: 6 tbsp strawberry jam, 1 l double cream • Give the ingredients for: twice as much and half as much sponge…comment on the units…especially for the eggs. • What assumption is it making about the eggs? Why is it useful? Today • Must: Understand the need for the mole concept • Should: Link Avogadro’s number to the mole • Could: Understand the basis of molar mass Masses and atoms • What is relative atomic mass? • What is relative isotopic mass? • Why are they relative? Some sums… • Radius carbon atom = 73 pm; density (diamond) = 3.5 gcm-3 • Calculate number of carbon atoms in 12 g • volume 12 g = mass/density … unit …convert to m3 • volume occupied by an atom …assumption cubic or spherical? • atoms in 12 g = vol 12g/vol atom • Approx 1x1024…should be 6.02x1023 mol-1 Definition • The Avogadro constant (symbols: L, NA), also called the Avogadro number, is the number of atoms in exactly 12 grams of 12C. • A mole is defined as this number of "entities" (usually, atoms or molecules) of any material. • NA = 6.02x1023 mol-1 • Molar mass – mass of one mole of entities…using 3 d.p. masses…why are they 3 d.p.? So how big is a mole…really… • • How many beaches like this would you need to hold one mole of sand grains? What assumptions will we make? – – – – Length Width Depth Volume of sand grain So how big is a mole…really… • What assumptions will we make? – – – – Length ~ 3 km Width ~ 200 m Depth ~ 2 m Volume of sand grain ~ 0.2x0.2x0.2 mm3 So how big is a mole…really… • A rough calculation – – – Volume of sand = 3000 x 200 x 2 = 1200000 m3 Volume of sand grain = 0.2 x 0.2 x 0.2 = 0.008 mm3 = 8 x 10-12 m3 Grains = 1.2 x106/8x10-12 – = 1.5x1017 For one mole sand: 4 million beaches needed! Part two… Must: Calculate amount of substance Should: Relate change in amount of substance to balanced equation Could: Comment on method Is this true? magnesium+ oxygen magnesium oxide 2 Mg + O2 2 MgO • How could we test this experimentally? Experiment • • • • Weigh a crucible and lid Add between 1 and 5 cm strip of magnesium ribbon Weigh the crucible, lid and magnesium Place on a pipeclay triangle resting on a tripod. Leave a small gap between the lid and the crucible and heat strongly for five minutes (CARE: very hot) • Allow the apparatus to cool • Reweigh • If time available: Repeat the heating/cooling/weighing sequence until constant mass is achieved 12T slap NaCl NaCl Cu Ar SiO2 H2 A metal An acidic gas An element and a A molecule monatomic element A covalently An ionic solid bonded lattice Cu Ar SiO2 H2 HCl HCl Calculation • (By definition) one mole of a substance has a mass equal to its relative mass in grams • Amount (moles) = mass (g)/molar mass(gmol-1) • For molecules and compounds: Relative formula mass = sum of atomic masses in the formula Calculation • Amount of solid (moles) = mass (g)/relative mass(gmol-1) • • • • • Calculate the mass of Mg in the crucible Calculate the final mass of MgO in the crucible Calculate the mass of O added Calculate the amount of Mg in the crucible Calculate the final amount of O in the crucible after the reaction • What is the ratio of Mg:O Some sums… • Calculate the mass of Mg in the crucible • Calculate the final mass of MgO in the crucible • Calculate the amount of Mg in the crucible • Calculate the final amount of MgOin the crucible Does the data fit? What should the relationship be? • Estimating error: If X = A + B or X = A – B , then: DX = √{(DA)2 + (DB)2} • • • • Find the % error with the masses Use this to estimate the errors in the amounts How confident are you in your conclusion? What other sources of error are present? Formula Mass Formula mass (molar mass) = sum of atomic masses in formula KCl 1 x 39.1 + 1 x 35.5 = 74.6 MgBr2 1 x 24.3 + 2 x 79.9 = 184.1 CuSO4 1 x 63.5 + 1 x 32.1 + 4 x 16.0 = 159.6 Mg(NO3)2 1 x 24.3 + 2 x 14.0 + 6 x 16.0 = 148.3 Empirical formula • The simplest whole number ratio of the elements in a compound A hydrocarbon contains 2.625g carbon and 0.875g hydrogen. 1. Calculate amount of each element amount (mol) = mass/molar mass amount C = 2.625/12.0 = 0.219mol amount H = 0.875/1.0 = 0.875mol 2. Calculate ratio by dividing through by smallest number C:H = 0.219/0.219:0.875/0.219 C:H = 1.000:3.995 i.e. 1:4 3. Write formula CH4 Molecular formula • The formula of a molecule A hydrocarbon contains 5.00g carbon and 1.25g hydrogen. One mole of the hydrocarbon has a mass of 30.00g 1. Calculate amount of each element amount (mol) = mass/molar mass amount C = 5.00/12.0 = 0.417mol amount H = 1.25/1.0 = 1.25mol 2. Calculate ratio by dividing through by smallest number C:H = 0.417/0.417:1.25/0.417 C:H = 1.00:3.00 i.e. 1:3 3. Write formula CH3 Molecular formula • The formula of a molecule A hydrocarbon contains 5.00g carbon and 1.25g hydrogen. One mole of the hydrocarbon has a mass of 30.00g 4. Calculate molar mass of formula = 12.0 + 3 x 1.0 = 15.0 5. Calculate ratio of empirical formula mass to molar mass of molecule = 30.0/15.0 =2 6. Molecule contains two times the empirical formula C2H6 Empirical formula from % by mass • The simplest whole number ratio of the elements in a compound A compound contains the following percentages by mass: Mg – 28.8%; C – 14.2%; O – 56.9% 1. Calculate amount of each element – assume % is mass in grams amount (mol) = mass/molar mass amount Mg = 28.8/24.3 = 1.19mol amount C = 14.2/12.0 = 1.18mol amount O = 56.9/16.0 = 3.56mol 2. Calculate ratio by dividing through by smallest number Mg:C:O = 1.19/1.18:1.18/1.18:3.56/1.18 Mg:C:O = 1.01:1.00:3.02 i.e. 1:1:3 3. Write formula MgCO3 Empirical formula from % by mass • The simplest whole number ratio of the elements in a compound A compound contains the following percentages by mass: Al – 52.9%; O – 47.1% 1. Calculate amount of each element – assume % is mass in grams amount (mol) = mass/molar mass amount Al = 52.9/27.0 = 1.96mol amount O = 47.1/16.0 = 2.94mol 2. Calculate ratio by dividing through by smallest number Al:O = 1.96/1.96:2.94/1.96 Al:O = 1.00:1.50 Al:O = 2:3 (If you get around 0.5 in the ratio double; around 0.33 or 0.67 triple) 3. Write formula Al2O3 Amounts and balanced equations 1. 2. 3. 1. 2. This is easier than you are prepared to believe! Find the ratio of substances in the equation (don’t forget that no number = 1) Use the ratio to convert the amount of known substance to the amount of unknown There isn’t a 3 CuO + H2 Cu + H2O How many moles of copper from 0.1 mole copper oxide? Ratio 1:1 0.1 gives 0.1 0.1 moles copper from 0.1 moles copper oxide Amounts and balanced equations C6H12O6 + 6 O2 6 CO2 + 6 H2O How many moles of oxygen are required to aerobically respire 0.2 moles glucose? Ratio C6H12O6:O2 = 1:6 0.2 moles requires 6 x 0.2 = 1.2 moles O2 Mg + 2 HCl MgCl2 + H2 How many moles of hydrogen can be made from 0.1 moles of hydrochloric acid reacted with excess magnesium Ratio HCl:H2 = 2:1 0.1 moles makes ½ x 0.1 = 0.05 moles H2 By the end of this lesson you: Must be able to calculate amounts (moles) Activity: 01/05/2017 To use balanced equations to perform mole calculations Should be able to use balanced equations to work out number of moles Could mass (g) amount (moles) formula mass (gmole-1) relate gas volumes to number of moles How many moles of oxygen can be made from 0.5 moles hydrogen peroxide 2 H2O2 2 H2O + O2 Experiment • Weigh 5 1 cm strips of magnesium ribbon • Place 5 pieces of magnesium ribbon in the ignition tube • EITHER: Fill a 100 cm3 measuring cylinder with water and place upside down in a plastic bowl 2/3rds full of water. Arrange an upward delivery tube to collect the hydrogen gas • OR: Attach a gas syringe • Add 40+ cm3 hydrochloric acid to the conical flask • Push the bung into the boiling tube • Tip the ignition tube so acid and magnesium can react • Measure the volume of hydrogen gas collected • Enter data on the spread sheet Results Table Mg (s) + 2 HCl (aq) MgCl2 (aq) + H2 (g) Experiment • Weigh accurately between 0.9 and 1.0g sodium carbonate decahydrate (Na2CO3.10H2O) and transfer to ignition tube • Place solid in an ignition tube • Put about 40 cm3 in the flask • Set up a gas syringe • Push the bung into the flask • Attach the gas syringe • Topple ignition tube • Measure the volume of carbon dioxide gas collected Na2CO3 (s) + 2 HCl (aq) 2 NaCl (aq) + H2O (l) + CO2 (g) Na2CO3.10H2O (s) + 2 HCl (aq) 2 NaCl (aq) + 11 H2O (l) + CO2 (g) Gas Molar Volume 1 mole of any gas occupies 24000 cm3 (at room temperature and pressure) 1. Volume occupied by: 1 mole CO2 (g) 0.1 mole Cl2 (g) 1 x 24000 = 24000 cm3 0.1 x 24000 = 2400 cm3 volume (cm3) 2. Amount (moles) in: 12000 cm3 SO2 (g) 240 cm3 NH3 (g) Gas molar 12000/24000 = 0.5 mole amount (moles) 240/24000 = 0.01 mole volume (24 000cm3) Calculation CaCO3 (s) –(heat) CaO (s) + CO2 (g) Calculate the volume of carbon dioxide formed by the thermal decomposition of 10 g calcium carbonate 1. Calculate amount (moles) CaCO3 amount = mass/formula mass [40Ca; 12C; 16O] = 10/(40+12+3x16) = 0.1 moles 2. Use balanced equation ratio to state amount CO2 (g) 1 CaCO3 gives 1 CO2 so 0.1 moles CaCO3 gives 0.1 mole CO2 3. Calculate volume of gas gas volume = amount x 24000 = 0.1 x 24000 = 2400 cm3 Bingo Moles CO2 in 22g NaCl (aq) Formula mass NH3 Formula mass H2O H2O, CO2 but not NaCl % Mg (by mass) in MgO Na+ Volume CO2 in 22g (at RT) Moles BaSO4 in 466.8 g CuO 6x1023 Formula mass Ca(OH)2 Moles SO2 in 36 dm3 at RT Moles He in 1.2 dm3 at RT Cl- Cu2O Learning objective: • To 01/05/2017 • Which beaker contains most solute? • Concentration (moldm-3) = amount (mol)/volume (dm3) • The unit (moles per decimetre cubed) gives you the formula: amount per (divided by) volume What is a standard solution? • A standard solution has an accurately known concentration • HCl (aq) and NaOH (aq) cannot be used as standard solutions… …why not? • Anhydrous sodium carbonate (Na2CO3) is a good primary standard • You are going to make a solution of accurately known concentration around 0.05 moldm-3 Preparing a standard solution 1. Calculate the mass of sodium carbonate required to produce 250 cm3 of 0.05 moldm-3 Na2CO3: – – Amount (mol) = volume (dm3) x concentration (moldm-3) 1 dm3 = 1000 cm3 and 1 cm3 = 1 x10-3 dm3 2. Calculate the mass of this amount of Na2CO3 3. Rinse a 250 ml beaker and a 250 cm3 volumetric flask with distilled water Why is this necessary? Why have I used ml & cm3? 4. Weigh a clean dry weighing boat (do not zero balance) 5. Add the mass Na2CO3 calculated in 2 (+/- 0.1 g ) 6. Tip into the beaker and reweigh the weighing boat Preparing a standard solution 1. Dissolve in about 100 ml distilled water – use a stirring rod to crush lumps 2. When it has all dissolved transfer to the rinsed volumetric flask using a rinsed funnel 3. Rinse the beaker, rod and funnel into the volumetric flask 4. Make up the solution in the flask until the bottom of the meniscus touched the line 5. Mix by inversion 6. Check that the volume is the same after inversion Preparing a standard solution 1. Calculate the actual [Na2CO3 (aq)] – Concentration (moldm-3) = amount (mol)/volume (dm3) 2. Label the volumetric flask with: your intitials; [Na2CO3 (aq)] 0.0XXX moldm-3 3. We will use the solution to standardise some HCl (aq) next week 4. The HCl (aq) will then be used to test the concentration of the bench NaOH (aq) that is approximately 1 moldm-3 Homework • • • • • • Show working Look up symbols/formula e.g. type barium sulphate into google and the first link will give you the correct formula Ask for help before the work is due in – in person or via email Attempt all of the questions %Mg in Mg3N2 Mr (Mg3N2) = 24 x 3 + 14 x 2 = 100; Mass Mg = 3 x 24 = 72 % Mg = 72 x 100/100 = 72 % Formula of: 29.1% sodium, 40.5% sulfur & 30.4% oxygen Amount Na = 29.1/23 = 1.27; Amount S = 40.5/32 = 1.27; Amount O = 30.4/16 = 1.9; Ratio (Na:S:O) = 1.9/1.27 = 1: 1: 1.5; Formula = Na2S2O3 Homework • • Formula chloride of calcium. Calcium chloride: 9.435g of which calcium 3.400g Amount Ca = 3.4/40 = 0.085; Amount Cl = (9.435-3.4);/35.5 = 0.17; Ratio (Cl:Ca) = 0.17/0.085 = 2:1; Formula = CaCl2 Mass toluene and nitric acid required to make 10 tonnes TNT Amount TNT = 10 x 106/(7 x 12 + 5 x 1 + 3 x 14 + 6 x 16); = 44.1 x 103 mol; mass toluene = 44.1 x 103 x (7 x 12 + 8 x 1); = 4.06 tonnes; mass nitric acid = 44.1 x 103 x 3 x (1 + 14 + 3 x 16); = 8.33 tonnes Homework • Volume chlorine gas made by electrolysing 60g NaCl Amount NaCl = 60/(23 + 35.5); = 1.03 mol; Amount Cl2 = 1.03/2; = 0.51 mol; Volume = 0.51 x 22.4; = 11.5 dm3 Titrations Standardising HCl (aq) 1. Rinse burette with distilled water then your solution of sodium carbonate. – Why is this necessary? 2. Fill the burette with sodium carbonate solution using a funnel – remove the funnel 3. Rinse a pipette with distilled water and dilute hydrochloric acid (approx. 0.1 moldm-3) 4. Rinse a wide necked conical flask with distilled water only. 5. Use a safety filler to transfer exactly 25.0 cm3 of acid into the conical flask and add 3-4 drops of methyl orange indicator Titrations 1. Place the conical flask on a white tile and add sodium carbonate, with swirling, until the indicator changes colour 2. Record the volume to the nearest 0.05 cm3 3. Record to 2 d.p. even if you get 24.00 cm3 4. Rinse the flask with distilled water and repeat the process until you have two concordant titres (two results within 0.20 cm3). 5. Calculate the mean of the two concordant titres Calculation Na2CO3 (aq) + 2 HCl (aq) 2 NaCl (aq) + H2O (l) + CO2 (g) 1. Calculate the amount of Na2CO3 in your mean titre Amount (mol) = volume (dm3) x concentration (moldm-3) 2. Use the balanced equation to calculate the amount of HCl in the conical flask – this will be a simple sum – For example: the same, half as much or twice as much as your answer to 1 3. Calculate the concentration of the hydrochloric acid Concentration (moldm-3) = amount (mol)/volume (dm3) Titrations Standardising bench NaOH (aq) 1. Rinse a pipette with distilled water and bench sodium hydroxide (approx.1 moldm-3) 2. Rinse a volumetric flask with distilled water only. 3. Use a safety filler to transfer exactly 25.0 cm3 of hydroxide into the volumetric flask and make up to the line – mix by inversion 4. Rinse a burette with distilled water and then your dilute sodium hydroxide solution 5. Fill the burette using a funnel BELOW EYE LEVEL Titrations 1. Rinse the pipette with distilled water and then your standardised hydrochloric acid 2. Pipette 25.0 cm3 hydrochloric acid into a wide necked conical flask and add 2-3 drops phenolphthalein indicator 3. Place the conical flask on a white tile and add sodium hydroxide, with swirling, until the indicator changes colour permanently 4. Record the volume to the nearest 0.05 cm3 5. Record to 2 d.p. even if you get 24.00 cm3 6. Rinse the flask with distilled water and repeat the process until you have two concordant titres (two results within 0.20 cm3). 7. Calculate the mean of the two concordant titres Calculation NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l) 1. Calculate the amount of HCl in your conical flask Amount (mol) = volume (dm3) x concentration (moldm-3) 2. Use the balanced equation to calculate the amount of NaOH in the mean titre – this will be a very simple sum 3. Calculate the concentration of the sodium hydroxid Concentration (moldm-3) = amount (mol)/volume (dm3) 4. Calculate the concentration of bench sodium hydroxide – How much did you dilute the original solution by? Homework • Bestchoice section 1.3 Balancing equations to concentrations • AS labskills – work through titration exercise as far as acid/base titration exercises Amount (moles) – concentrations For solutions: amount (moles) = volume (dm3) x concentration (moldm-3) concentration (moldm-3) = amount (moles) / volume (dm3) volume (dm3) = amount (moles) / concentration (moldm-3) Find the amount of solute in: 25.0 cm3 0.1 moldm-3 KCl 25.0 x 10-3 x 0.1 = 2.50 x 10-3 mol 12.5 cm3 0.5 moldm-3 MgBr2 12.5 x 10-3 x 0.5 = 6.25 x 10-3 mol 50.0 cm3 0.01 moldm-3 CuSO4 50.0 x 10-3 x 0.01 = 5.0 x 10-4 mol Amount (moles) – concentrations For solutions: amount (moles) = volume (dm3) x concentration (moldm-3) concentration (moldm-3) = amount (moles) / volume (dm3) volume (dm3) = amount (moles) / concentration (moldm-3) Find the concentration of solute in: 2.50 x 10-3 moles in 25.0 cm3 KCl 2.50 x 10-3 / 25.0 x 10-3 = 0.1 moldm-3 2.50 x 10-3 moles in 50.0 cm3 MgBr2 2.50 x 10-3 / 50.0 x 10-3 = 0.05 moldm-3 1.0 x 10-3 moles in 100.0 cm3 CuSO4 1.0 x 10-3 / 100.0 x 10-3 = 0.01 moldm-3 Amount (moles) – concentrations For solutions: amount (moles) = volume (dm3) x concentration (moldm-3) concentration (moldm-3) = amount (moles) / volume (dm3) volume (dm3) = amount (moles) / concentration (moldm-3) Find the volume of solution containing: 0.01 mol of 0.2 moldm-3 KCl 0.01/0.2 = 0.05 dm3 2.50 x 10-3 mol of 0.1 moldm-3 MgBr2 2.50 x 10-3/0.1 = 25.0 x 10-3 dm3 = 25.0 cm3 1.25 x 10-3 mol of 0.5 moldm-3 CuSO4 1.25 x 10-3/0.5 = 2.50 x 10-3 dm3 = 2.50 cm3 Combining ideas • • • • lithium + water lithium hydroxide + hydrogen 2 Li (s) + 2 H2O (l) 2 LiOH (aq) + H2 (g) How could we demonstrate this is true? What can we measure? How would we measure it? Which reactant should be in excess? Instructions 1. Pour about 50 cm3 distilled water into a rinsed narrow mouth conical flask 2. Set up a gas jar (capacity c. 400 ml) to collect gas over water 3. Cut and clean a small piece of lithium (GLOVES and GOGGLES – corrosive and flammable) 4. You are aiming for a mass of about 0.13g – measured accurately 5. Add the lithium to the water and connect the delivery tube 6. When the reaction has finished mark the volume of gas collected on the gas jar …and then 1. Use water and a 100 ml measuring cylinder to estimate the volume of gas collected 2. Transfer the lithium hydroxide solution into a rinsed 250 cm3 volumetric flask, rinse the conical flask and add to the volumetric flask and then make up to the mark with distilled water. 3. Titrate against standardised hydrochloric acid using phenolphthalein indicator 4. Repeat until you have recorded two concordant titres Calculation 1. Calculate the amount of lithium you started with – Amount (mol) = mass (g)/formula mass (gmol-1) 2. Calculate the amount of hydrogen gas you collected – Amount (mol) = volume (dm3)/gas molar volume (dm3mol-1) – assume it is 24.0 dm3mol-1 3. Calculate the concentration of lithium hydroxide in the volumetric flask – Concentration (moldm-3) = amount (mol)/volume (dm3) 4. Calculate the amount of lithium hydroxide made in the reaction – Amount (mol) = concentration (moldm-3) x volume (dm3) Combining ideas • • • • • lithium + water lithium hydroxide + hydrogen 2 Li (s) + 2 H2O (l) 2 LiOH (aq) + H2 (g) Do your results match the equation? Comment on the accuracy of the data Which do you think was the most accurate value? Explain your choice I’d expect it to be the titration – explain my choice!