ࡱ> npklm'` 8bjbjDD .F&&jpppLLLLd&8888ӌ_Hʷh2<ip88z2աաա 8p8աաաrT0@p׬8t p|F^LOcz&]znn׬׬npw>,ա$5(&?hF?hF 1.3 Formulae, equations and amounts of substance 1.3a particles, substances and formulae An atom is the smallest particle of an element that has that elements properties. An atom contains protons, neutrons and electrons. The protons and neutrons are found in the nucleus of the atom while the electrons exist in the shells, energy levels or quantum shells around the nucleus. The existence of the nucleus was proved by Geiger and Marsden in their experiment when alpha particles were fired at gold foil. Deflection of the alpha particles could only be explained by the existence of a nucleus containing all of the atom's mass. particlerelative massrelative chargeelectron1/1840-1proton1+1neutron10 HYPERLINK "file:///C:\\Documents%20and%20Settings\\David%20Bateman\\My%20Documents\\Dave\\My%20Webs\\school%20science%20uk%20web\\asa2sums\\sum1.1\\story1.1a.htm" Story1.1a A molecule is a particle with two or more atoms chemically ``joined together. An ion is a particle of one or more atoms with an electric charge. An element is a pure substance which cannot be separated by chemical means. A compound is a substance with two or more different elements chemically joined. An empirical formula is the simplest formula for a substance showing the ratio of atoms that have been joined. A molecular formula shows the number of atoms of each type in a molecule of a substance. Task 1.3a.1 Classify the following: He, O2, Na+, bromine, sodium bromide, C2H4, CH2, H, H2O, CO32-, neon, lithium chloride, CH3, C2H6. Task 1.3a.2 Give examples of each defined word above. 1.3b Full and ionic equations When iron is added to copper (II) sulphate, copper and iron (II) sulphate is formed. The full equation is Fe(s) + CuSO4(aq) ---> Cu(s) + FeSO4(aq) In terms of ions this is: Fe(s) + Cu2+(aq) + SO42-(aq) ----> Cu(s) +Fe2+(aq) + SO42-(aq) Sulphate ions are spectator ions as they are unaffected by the reaction. Therefore they are left out. The ionic equation is: Fe(s) + Cu2+ (aq) ----> Fe2+ (aq) + Cu(s) Both the number of atoms and charges balance. Common ions Cations Sodium Na+, potassium K+, lithium Li+, magnesium Mg2+, calcium Ca2+, Aluminium Al3+, ammonium NH4+, hydrogen H+, Anions Oxygen O2-, fluoride F-, chloride Cl-, bromide Br- iodide I-, hydroxide OH-, carbonate CO3-, sulphate SO4-, nitrate NO3-, Task 1.3b Balancing equations Balance the following equations: Write ionic equations where appropriate 1) Ca + H2O Ca(OH)2 + H2 2) CO + O2 CO2 3) Ca + O2 CaO 4)* Fe2O3(s) + HCl(aq) FeCl3(aq) + H2O (l) 5)* NH3(aq) + H2SO4(aq) (NH4)2SO4 (aq) 6)* Al + H2SO4 Al2(SO4)3 + H2 7)* CaO + HCl CaCl2 + H2O 8) NH3 + O2 NO + H2O 9) Na2O + H2O NaOH 10)* Na2CO3 + HCl NaCl + CO2 + H2O 11)* Br2 + KI ( KBr + I2 12)* Ca(OH)2 + HNO3 Ca(NO3)2 + H2O 13) Fe + H2O Fe3O4 + H2 14)* Pb3O4 + HNO3 ( Pb(NO3)2 + PbO2 + H2O 15)* Cu + HNO3 ( Cu(NO3)2 + H2O + NO 16) HNO3 ( NO2 + H2O + O2 Answers 1) Ca + 2H2O Ca(OH)2 + H2 2) 2CO + O2 2CO2 3) 2Ca + O2 2CaO (s) 4) Fe2O3(s) + 6HCl(aq) 2FeCl3(aq) + 3H2O(l) Fe2O3(s) + 6H+ + 6Cl- 2Fe2+ + 6Cl- + 3H2O(l) Fe2O3(s) + 6H+ 2Fe2+ + 3H2O(l) 5) 2NH3 + H2SO4 (NH4)2SO4 6) 2Al + 3H2SO4 Al2(SO4)3 + 3H2 7) CaO + HCl CaCl2 + H2O 8) NH3 + O2 NO + H2O 9) Na2O + H2O NaOH 10) Na2CO3 + HCl NaCl + CO2 + H2O 11) Br2 + KI ( KBr + I2 12) Ca(OH)2 + HNO3 Ca(NO3)2 + H2O 13) Fe + H2O Fe3O4 + H2 14) Pb3O4 + HNO3 ( Pb(NO3)2 + PbO2 + H2O 1.3c Chemical quantities The unit of mass is 1/12 of the mass of an atom of the isotope Carbon-12 (12C=12.0000 exactly) The Relative Atomic Mass (Ar)of an element is the ratio of the mass of an average atom of that element to 1/12 of the mass of an atom of the nuclide Carbon-12. Experiment 1.3c.1 Cut squares, to represent different atoms, from graph paper with areas of 4cm2, 8cm2 and 16cm2. Weigh each square on a two decimal place balance. Calculate the relative masses of your atoms. AtomMass/gRelative massA0.120.12g/0.12g = 1.00B0.230.07g/0.02g = 3.5C0.480.14/0.02 = 7(graph paper, scissors, 2dec place balances) 1.3c Amount of substance The mole is the amount of a substance that contains the same number of particles as there are atoms in 12.00g of carbon-12. This number of atoms is 6.02 x 1023 and is called the Avogadro constant.  INCLUDEPICTURE "http://www.drbateman.net/asa2sums/sum1.2/Cpile.gif" \* MERGEFORMATINET One mole of carbon atoms. Molar mass can be found. The number of g in one mole is the same number as the relative formula mass of a substance E.g. 23gmol-1 of Na atoms and 28g mol-1 of N2 molecules all contain the same number of particles, they are all one mole of atoms or molecules. Molar mass = mass of 1 mole amount of substance = mass of substance/molar mass of substanceTask 1.3c.1 What are molar masses of (a) K, (b) O2, (c) Br, (d) Br2, (e) H2O, (f) H2SO4 Task 1.3c.2 What amount is (a)12g C, (b)14g Li, (c)12g Mg, (d)18g H2O, (e)4.4g CO2? Task 1.3c.3 What mass is (a)1mol O, (b)2mol O2, (c)0.5mol N2, (d) 0.1mol CO, (e)0.2mol H2SO4? Task 1.3c.4 What is the molar mass of (a)X if 1mol X has mass of 5g, (b)2mol X has mass of 20g, (c)0.25 mol X has 25g. Answers: 1(a) 39gmol-1, (b) 32 gmol-1 (c) 80 gmol-1 (d) 160 gmol-1, (e) 18 gmol-1, (f) 98 gmol-1 2(a)1mol, (b)2mol, (c)0.5mol, (d)1mol, (e)0.1mol 3(a)16g, (b)64g, (c)14g, (d)2.8g, (e)19.6g 4(a)5gmol-1, (b)10gmol-1, (c)100gmol-1 1.3c CO - Carbon Monoxide USA Air Quality Standards These levels may not be exceeded more than once per year: 1-hour average concentration -- 35 ppm 8-hour average concentration -- 9 ppm NO2 - Nitrogen Dioxide Air Quality Standard Annual average concentration -- 0.053 ppm  HYPERLINK "http://www.airquality.co.uk/" http://www.airquality.co.uk/ Pollutants in air are measured in parts per million. Air that contains 4 cm3 of carbon dioxide in 1 million cm3 (1000dm3) has a carbon dioxide concentration of 4ppm. Pollutants in water are also measured as parts per million (ppm). A solution that contains 2 g of lead in 1 million grams of water (1,000 dm3) is a 2 ppm solution HYPERLINK "http://www.drbateman.net/asa2sums/sum1.2/task1.2d.htm"  Task 1.3c.5 (a) A car produces CO in the concentration 3ppm from its exhaust. What volume of CO is present in 2000dm3 of exhaust gas? (b) What is the concentration in ppm of nitrogen dioxide if 32 cm3 is found in 500dm3 of air? (c) What mass of water is there in 1 tonne (1000kg) of aviation fuel if the concentration is 1ppm? (d) What is the concentration of nitrate ion in water from a stream in which 3g of nitrate is found in 2000dm3 of water? 6cm3, (b) 64ppm, (c) 1g, (d) 1.5ppm. Answers: (a) 2000dm3 is 2000000cm3 3ppm 1000000cm3 exhaust contains 3cm3 of CO so 2000000cm3 exhaust contains 6cm3 CO 1.3d Solution concentration data The concentration of a solution can be stated as the mass of solute per cubic decimetre of solution (g/dm3) or the amount in moles of a solute present in 1dm3 of solution (mol/dm3). Concentration = mass of solute/volume Concentration (Molarity) = amount of solute /volume of solution1dm3 = 1000cm3 Task 1.3d Concentration and Molarity Problems 1. Calculate the mass of 1 mol of (a) NaOH, (b) NaCl, (c) H2SO4, (d) NH4OH, (e) HCl. 2. Calculate the amount present in (a) 10.6g Na2CO3, (b) 12.6 g of HNO3, (c) 21.3g Cl2, (d) 8.5g NH3, (e) 14.8g Ca(OH)2. Answers:Q1 40g, 58.5g, 98g, 35g, 36.5g Answers:Q2 0.100mol, 0.200mol, 0.300mol, 0.500mol, 0.200mol 3. Calculate, to three significant figures, the concentrations (molarity) in mol dm-3 of (a) 3.65g of hydrogen chloride, HCl, in 2.00dm3 of solution (b) 73.0g of hydrogen chloride, HCl, in 2.00dm3 of solution. (c) 14.8g Ca(OH)2 in 3.00dm3 of solution. (d) 4.25g NH3 in 1dm3 of solution. (e) 17.4g glucose, C6H12O6 in 0.5dm3 of blood. (f) 5.85g NaCl in 200cm3 aqueous solution Answers; 0.0500 mol dm-3, 1.00 mol dm-3, 0.0667 mol dm-3, 0.250 mol dm-3, 0.193 moldm-3 1.3e Using reacting masses and chemical equations What mass of sodium carbonate (molar mass = 106g/mol) can be made by heating 16.8g of sodium hydrogencarbonate (molar mass = 84g/mol)? Method 1 (recommended) 2NaHCO3(s) ----> Na2CO3(s) + CO2(g) + H2O(g) mass NaHCO3 =16.8g amount NaHCO3 = mass NaHCO3/molar mass NaHCO3 =16.8g/84gmol-1 = 0.20mol from equation: amount Na2CO3/amount NaHCO3 =1/2 amount Na2CO3 =1/2*0.2mol = 0.1 mol mass Na2CO3 = amount Na2CO3*molar mass Na2CO3 =0.1mol*106gmol =10.6g Method 2 formula mass 2NaHCO3 = 168 formula mass Na2CO3 = 106 so 168g of sodium hydrogen carbonate forms 106g of sodium carbonate so 1g of sodium hydrogen carbonate forms 106/168g of sodium carbonate so 16.8g of sodium hydrogen carbonate forms 16.8*106/168g of sodium carbonate so 16.8g of sodium hydrogen carbonate forms 10.6g of sodium carbonate Task 1.3e 1. What amounts of substance are the following masses? (a) 8g of S, (b) 8g of O, (c) 8g of O2, (d) 0.5g of H2, (e)25g of NaOH 2. What are the masses of the following amounts? (a) 2mol of Cl2, (b) 0.5 mol of CuO, (c) 0.1 mol of H2SO4, (d) 2.5mol of NaHCO3. 3. What mass of glucose can be fermented to give 5.00g of ethanol? C6H12O6(aq) -----> 2C2H5OH(aq) + 2CO2(g) 4. What mass of silver chloride can be precipitated from a solution which contains 1.00*10-3 mol of silver ions? Ag+(aq) + Cl-(aq) -----> AgCl(s) 5. The pollutant, sulphur dioxide, can be removed from the air by the reaction 2CaCO3(s) + 2SO2(g) +O2(g) -----> 2CaSO4(s) + 2CO2(g) What mass of calcium carbonate is needed to remove 10.0g of SO2? 6. What mass of sodium carbonate can be made by heating 100g of sodium hydrogencarbonate? 2NaHCO3(s) -----> Na2CO3(s) + CO2(g) H2O(g) 7. What mass of KOH is fromed from 5 g of K when it reacts with water 8. What mass of HCl forms 4.4g carbon dioxide in a reaction with calcium carbonate. Answers: 1. (a) 8g/32gmol-1 = 0.25mol, (b) 0.5mol (c)0.25mol (d) 0.25mol (e) 0.625mol 2. (a) 2mol*71gmol-1 = 142g, (b) 40g, (c)9.8g, (d) 210g. 3. 9.81g 4. 0.144g 5. 15.6g 6. 63.1g 7. amount K = 5/39 =0.128mol Amount of KOH/amount of K =2/2 So amount of KOH = 0.128mol Mass of KOH = amount KOH * molar mass KOH = 0.128mol*56gmol-1 = 7.18g 8. amount CO2 = 4.4/44 = 0.1mol Amount HCl/amount CO2 =2/1 So amount HCl = 2* amount CO2 =2*0.1 = 0.2mol Mass HCl = amount * molar mass = 0.2 * 36.5= 7.3g Calculating percentage yield Mass of lead ethanoate = 3.78g Mass of lead sulfate precipitated from solution = 2.80g (CH3COO)2Pb.3H2O(aq) + H2SO4(aq) ( PbSO4(s) + 2CH3COOH + 3H2O(l) amount of (CH3COO)2Pb.3H2O = mass (CH3COO)2Pb.3H2O / molar mass of (CH3COO)2Pb.3H2O amount of (CH3COO)2Pb.3H2O = 3.78g / 378 g/mol = 0.0100mol Chemical equation shows that 1 mol (CH3COO)2Pb.3H2O forms 1 mol PbSO4 so 0.0100 mol of (CH3COO)2Pb.3H2O forms 0.0100 mol PbSO4 So theoretical mass of PbSO4 = amount PbSO4 * molar mass PbSO4 = 0.0100 mol * 302 gmol-1 = 3.02g % yield of PbSO4 = actual mass of PbSO4 obtained*100 / theoretical mass of PbSO4 = 2.80g *100 / 3.02g = 92.7% Predicted mass Mass empty tube = 44.0g Mass tube + NaHCO3 = 45.90g Mass tube + solid product = 45.40g Mass of NaHCO3 = 1.90g Mass of solid product = 1.40g 2NaHCO3 ( Na2CO3 +H2O + CO2 Amount NaHCO3 = mass NaHCO3/molar mass NaHCO3 = 1.90g/84gmol-1 = 0.226mol From equation: expected amount of Na2CO3 / Amount NaHCO3 = 1/2 expected amount of Na2CO3 = amount NaHCO3 * = 0.226/2mol = 0.113mol Expected mass of Na2CO3 = amount NaCO3 * molar mass Na2CO3 = 0.113mol * 106gmol-1 = 1.20g Above the expected mass is less than the actual mass so perhaps decomposition was not complete. Calculation for Standard solution Mass of sodium carbonate = 1.30g Na2CO3 Volume of solution = 250cm3 Amount of Na2CO3 = mass Na2CO3/ molar mass Na2CO3 = 1.30g/106gmol-1 = 0.0123mol Concentration of Na2CO3 = amount Na2CO3/ vol Na2CO3 solution = 0.0123mol/(250/1000)dm3 = 0.0491moldm-3 Titration of HCl v sodium carbonate Vol of sodium carbonate = 25.0cm3 Conc of sodium carbonate = 0.0500moldm-3 Volume of HCl used = 9.40cm3 Amount of Na2CO3 = conc of Na2CO3 solution * vol Na2CO3 solution = 0.0500 moldm-3* (25.0/1000)dm3 = 0.00125mol 2HCl(aq) + Na2CO3 ( 2NaCl(aq) +H2O(l) + CO2(g) From equation amount HCl/amount Na2CO3 = 2/1 so amount HCl =(2/1)* amount Na2CO3 = 2*0.00125mol = 0.00250mol Conc of HCl = amount HCl/vol HCl solution = 0.0025mol/(9.40/1000)dm3 = 0.266moldm-3 Conc of NaOH by titration against HCl Conc of acid = 0.100moldm-3 Vol of acid = 24.90cm3 Vol of alkali = 25.0cm3 Amount of HCl = conc HCl * vol HCl = 0.100moldm-3 * (24.90/1000) dm3 = 0.00249mol HCl + NaOH ( NaCl +H2O From equation amount of alkali/amount of HCl = 1/1 So amount of NaOH = amount HCl = 0.00249mol Conc NaOH = amount NaOH/vol NaOH solution = 0.00249mol/(25.0/1000)dm3 = 0.0996moldm-3 1.3f The molar volume of a gas Avogadro's Law states that equal volumes of all gases under the same conditions (temperature and pressure) contain the same number of particles. One mole of any ideal gas occupies 22.4dm3 at stp (Standard temperature and pressure) and 24dm3 at rtp (Room temperature and pressure). This is the molar volume of a gas. amount of gas = volume of gas/molar volume H2(g) + Cl2(g) ----> 2HCl(g) (At stp) 22.4dm3 of hydrogen reacts with 22.4dm3 of chlorine forms 44.8dm3 of hydrogen chloride 1 volume of hydrogen reacts with 1 volume of chlorine to form 2 volumes of hydrogen chloride 1 mole of hydrogen molecules reacts with 1 mole of chlorine molecules to form 2 moles of hydrogen chloride molecules.  HYPERLINK "file:///C:\\Documents%20and%20Settings\\David%20Bateman\\My%20Documents\\Dave\\My%20Webs\\school%20science%20uk%20web\\asa2sums\\sum1.2\\task1.2g.htm" task 1.3f.1 Equations can be used to deduce reacting gas volumes: What volume of nitrogen dioxide can be formed when 16.4g of Calcium nitrate (molar mass 164g/mol) decomposes on heating? 2Ca(NO3)2(s)----> 2CaO(s) + 4NO2(g) + O2(g) amount of calcium nitrate = mass/molar mass = 16.4/164 = 0.100mol from equation amount of nitrogen dioxide/amount of calcium nitrate = 4/2 so amount of nitrogen dioxide = 0.200mol volume of nitrogen dioxide = amount *molar volume volume of nitrogen dioxide = 0.200mol*22.4dm3 mol-1 = 4.48dm3 .  HYPERLINK "file:///C:\\Documents%20and%20Settings\\David%20Bateman\\My%20Documents\\Dave\\My%20Webs\\school%20science%20uk%20web\\asa2sums\\sum1.2\\task1.2h.htm" Task 1.3f.2 Answers 1.(a) O2 20cm3, CO2 10cm3, H2O 20cm3. (b) O2 25cm3, CO2 20cm3, H2O 10cm3. (c) H2 30cm3, NH3 20cm3. 1.3g Percentage yields and atom economies Many organic reactions produce side products and during purification some of the main product is normally lost. The result is that the amount of product obtained, the yield, is less than the theoretical maximum yield. The theoretical yield is calculated using the balanced chemical equation. E.g. C2H5OH + HBr ----> C2H5Br + H2O 1 mol of ethanol forms 1 mol of bromoethane so 46.0g of ethanol forms 109g of bromoethane so if we start with 2.30g of ethanol 1g of ethanol forms 109/46.0g of bromoethane 2.30g of ethanol forms 2.30*109/46.0g = 5.45g of bromoethane so the theoretical yield of bromoethane is 5.45g if we start with 2.30 g of ethanol. If only 4.00g are actually formed (actual yield = 4.00g) in the preparation then percentage yield = 100*actual yield/theoretical yield percentage yield = 100*4.00/5.45 = 73.4% Task 1.3g.1 (b) What is the % yield if 20.2g ethanol is oxidised to 15.0g ethanoic acid. C2H5OH +[O] ( CH3COOH amount of C2H5OH =mass /molar mass = 20.2/46 =0.439mol Amount of CH3COOH = mass/molar mass = 35.0/60 = 0.583mol So only 0.439mol of ethanol reacts % atom economy =  INCLUDEPICTURE "http://upload.wikimedia.org/math/0/7/9/079371dbc8ff9ef3dbd539a6ff6650a2.png" \* MERGEFORMATINET  100% atom economy = all atoms in reactants turn into products. The atom economy can be low even if percentage yield is high. The atom economy idea is popular as high raw material costs and environmental concerns make waste an increasing problem. Task 1.3g.2 Calculate the percentage atom economy for each of the following organic products: (a) In the reaction shown below, 4.60g of ethanol gave 8.50g of bromoethane. Find the % yield (78.0%) and % atom economy(41.4%) for the organic product. C2H5OH + KBr + H2SO4 ---> C2H5Br + KHSO4 + H2O (b) In a reaction to make ethyl ethanoate by the reaction C2H5OH + CH3COOH ---> CH3COOC2H5 + H2O 20.2g of ethanol was heated under reflux with 35.0g of ethanoic acid. 15.0g of ethyl ethanoate was obtained. Find the theoretical and percentage yield of ethyl ethanoate(83.0%) . Find the atom economy of the organic product. (c) What is the theoretical yield of cyclohexene when dehydrating 10g of cyclohexanol by heating it with phosphoric acid? What is the percentage yield is the actual yield of cyclohexene is 7.1g? C6H11OH --> C6H10 + H2O (d) A synthesis of 1-bromobutane produced 6.5g of product from 6.0g of butan-1-ol using excess sodium bromide and conc. sulfuric acid. Calculate the theoretical and % yields of the product. 1.3h Avogadro Constant Experiment 1.3c.2 Describe the amounts of substance in each of 4 bottles. Consider the masses and the number of atoms. (bottles containing 12g of carbon, 32g sulphur, 20 black model atoms, 20 yellow model atoms) amount of substance = number of particles/Avogadro constantChemist cannot find numbers of particles easily so mass is used. The mass of a substance which contains the Avogadro number of particles is called the molar mass. The Avogadro number is about 6*1023mol-1 1. Calculate the number of atoms in (a) 1mol of carbon, C. (b) 0.5mol of copper, Cu. (c) 7.20g of sulfur, S 2. Calculate the number of molecules in (a) 1.00g of ammonia, NH3. (b) 3.28g of sulfur dioxide, SO2. (c) 7.83g of hydrogen chloride, HCl. 3. Calculate the number of ions in (a) 0.500mol of sodium chloride, NaCl [Na+,Cl-] (b) 14.6g of sodium chloride, NaCl. (c) 18.5g of calcium chloride, CaCl2. 4. Calculate the amount of hydrogen molecules, H2 in 3*1023 molecules if the Avogadro number is 6*1023. 5. Calculate the Avogadro number if 1.5*1023 mol of helium atoms He has a mass of 1g. Answers 1(a) 9.03*1022, (b) 1.71*1022, (c) 1.35*1023. 2(a) 3.54*1022 , (b) 3.08*1022 (c) 1.29*1023 . 3(a)6.02*1023. (b)3.01*1023 , (c) 3.01*1023 . 4. 0.5mol 5. 6*1023 1.3i Formulae and equations by experiment elements reactingmagnesiumchlorinesymbols of elementsMgClmasses reacting (from experiment)2.4g7.1gmolar mass (look up relative atomic mass in periodic table)24gmol-135.55gmol-1amounts (amount = mass/molar mass2.4g/24gmol-1 =0.1mol7.1g/35.5gmol-1 0.2molratio of atoms (divide by smallest)12 formulaMgCl2 Task 1.3i.1  HYPERLINK "http://www.drbateman.net/gcse2003/gcsesums/modsum10/mgburnpic.JPG" mgburnpic.JPG  HYPERLINK "http://www.drbateman.net/gcse2003/gcsesums/modsum10/mgburnvid.3gp" mgburnvid.3gp 1.What is the formula of magnesium oxide given experimental results below Mass an empty crucible = 10.00g Mass of crucible + magnesium = 11.20g Mass of heated crucible + magnesium oxide = 12.00g Mass of magnesium = Mass of magnesium oxide = Mass of oxygen reacting with magnesium = 2.Work out formulae of compounds formed when the following react: (a) 56g of iron and 32g of sulphur (Fe =56, S =32) (b) 2g of hydrogen and 16g of oxygen (H=1, O=16) (c) 14g of lithium and 16g of oxygen (Li=7) (d) 32g of copper and 8g of oxygen (Cu=64) (e) 6.4g of copper and 0.8g of oxygen. Lithium and water react to form lithium hydroxide and hydrogen. The expected reaction might be: 2Li + 2H20 ( 2LiOH + H2 This can be checked by experiment. Draw apparatus to mix lithium and water and measure the volume of hydrogen gas formed. If the molar mass of Li is 7.0gmol-1 and the 0.35g Li reacts with water to form 600cm3 of hydrogen at 25oC is the equation correct? Assume molar vol =24000cm3 at 25oC. Amount of Li = mass Li/molar mass Li =0.35g/7.0gmol-1 =0.05mol Amount of H2 = vol H2/molar vol = 600cm3/24000cm3mol-1 = 0.025mol So by experiment 0.05 mol Li forms 0.025 mol H2 So by experiment 1 mol Li forms 0.5 mol H2 So by experiment 2 mol Li forms 1 mol H2 The equations shows 2 mol Li forms 1 mol H2 So the equation must be correct. Titanium can be made from titanium chloride using sodium. Experiment shows that 1.92 g of Na is needed to make 1g of Ti. Which of the following equations must be correct? TiCl4 + 4Na ( Ti + 4NaCl TiCl3 + 3Na ( Ti + 3NaCl 1.3j Making salts and percentage yields Methods include: metal + acid ( salt + hydrogen metal oxide +acid ( salt + water metal hydroxide + acid ( salt + water metal carbonate + acid ( salt +water + carbon dioxide soluble salt1 + soluble salt2 ( insoluble salt + soluble salt3 Exp 1.3j making sodium chloride using method 4 1dm3 1.00M HCl, anhydrous sodium carbonate, balances, matches Add solid anhydrous sodium carbonate, with stirring, to 25.0cm3 of 1.00M HCl until the effervescence ends. Weigh an evaporating basin. Filter the mixture and collect the filtrate in the evaporating basin. Evaporate the solution to dryness. Reweigh the evaporating basin. Results: Mass of basin = Mass of basin + salt = Analysis and conclusion Actual mass of sodium chloride formed (in basin) = Amount of HCl used in experiment = vol HCl * concentration HCl = Na2CO3 +2HCl ( 2NaCl + H2O + CO2 Amount NaCl formed theoretically/amount HCl = / Amount of NaCl formed theoretically = Mass of NaCl formed theoretically = amount of NaCl * molar mass NaCl Percentage yield NaCl = actual mass of NaCl formed *100/theoretical mass NaCl = Sample results: mass of basin =50.00g, mass of basin + salt = 51.20g. 1.3k Simple test tube reactions Method: Pour about 5cm3 of dilute sulphuric acid into a test tube and add a small piece of magnesium ribbon. Use a lighted splint to identify the gas given off. Observations: Effervescence, the gas burns with a squeaky pop.Inferences: The gas evolved is hydrogen. Magnesium + sulphuric acid ( magnesium sulphate + hydrogen Mg(s) + H2SO4(aq) ( MgSO4(aq) + H2(g) Mg(s) + 2H+(aq) ( Mg2+(aq) + H2(g) Method: Pour about 5cm3 of dilute hydrochloric acid into a test tube and add an approximately equal volume of dil. sodium hydroxide.Observations: No observable change. Test tube warms slightly.Inferences: Reaction exothermic. Acid base reaction. Hydrochloric acid + sodium hydroxide ( sodium chloride + water HCl(aq) + NaOH(aq) ( NaCl(aq) + H2O(l) H+(aq) + OH-(aq) ( H2O(l) Method: Put half a spatula measure of copper (II) oxide into a boiling tube and add dilute sulphuric acid so that the tube is about 1/3 full. Carefully warm the mixture.Observations: The copper (II) oxide dissolves on heating to form a blue solution.Inferences: Acid base reaction. Neutralisation. Blue solution is aqueous copper sufate. Copper (II) oxide + sulfuric acid ( copper (II) sulfate + water CuO(s) + H2SO4(aq) ( CuSO4(aq) + H2O(l) CuO(s) + 2H+(aq) (Cu2+(aq) + H2O(l) Method: Mix equal volumes of lead (II) ethanoate solution and potassium iodide solution in a boiling tube. Filter the mixture.Observations: A yellow precipitate forms. The filtrate is a colourless solution.Inferences: A precipitation reaction. The precipitate is insoluble lead iodide. The filtrate contains aqueous potassium iodide. Lead(II) ethanoate + potassium iodide ( lead iodide + potassium ethanoate (CH3COO)2 Pb (aq) + 2KI(aq) ( PbI2(s) +2CH3COOK(aq) Pb2+(aq) + 2I-(aq) ( PbI2(s) Method: To about 5 cm3 of dilute hydrogen peroxide in a boiling tube add approximately half a spatula of manganese (IV) peroxide. This black solid acts as a catalyst in this reaction. Identify the gas given off using a glowing splint.Observations: Vigorous effervescence. Gas relights glowing splint.Inferences: Gas is oxygen. Hydrogen peroxide ( water + oxygen 2H2O2(aq) ( 2H2O(l) + O2(g) Method: Pour about 5cm3 of copper sulphate solution into a boiling tube. Add a spatula of zinc powder, stir thoroughly and then leave to stand. Note any colour changes in both the solution and solid.Observations: Blue colour of solution changes to colourless. Red solid appears on surface of zinc.Inferences: Displacement reaction. Copper is the red solid displaced from solution. Colourless zinc sulfate replaces blue copper sulfate in solution. Zinc + copper sulfate ( copper + zinc sulfate Zn(s) + CuSO4(aq) ( Cu(s) + ZnSO4(aq) Zn(s) + Cu2+(aq) ( Cu(s) + Zn2+ (aq) Revision Expressions (write as is or rearranged) Amount = mass/molar mass Amount = number of particles/Avogadro number Concentration of solution = amount/volume of solution Amount = volume of gas /Molar volume of gas Atom economy = 100*molar mass of product/sum of molar masses of reactants Yield = 100*actual mass of product/maximum theoretical mass of product Empirical formulae 1. A compound is found to contain the following % by mass: 69.58% Ba, 6.090% C, 24.32% O. What is the simplest (i.e. empirical) formula? 2. A compound consists of 29.1%Na, 40.5% S, and30.4%O. Determine the simplest formula. 3. Determine the simplest formula if a compound is composed of 7.20 g carbon, 1.20 g hydrogen, and 9.60 g oxygen. 4. Combustion analysis gives the following:26.7% C, 2.2% hydrogen, 71.1% oxygen. If the molecular mass of the compound is 90 g/mol, determine its molecular formula. 5. A compounds empirical formula is CH, and it weighs 104g/mol. 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