1.Review on Mole Concept Mole ConceptAvogrado constantReview on Limiting and Excess & Past Year Exam questions Limiting ad ExcessTitrationEmpirical Formula

2. Mass reactants given, which is limiting and excess ? 1Balanced chemical eqn2Mole ratio/stoichiometry ratio31Pb(NO3)2(s) + 2NaI(aq) 1PbI2(s) + 2NaNO3 (aq)Mass of reactants added4Mole ratio1 : 2 1: 2 10.0g Pb(NO3)2 + 10.0g NaI addedMass = 10.0 RMM 331.2 = 0.0302 molMass MolesMass = 10.0 RMM 149.9 = 0.0667 mol1 mol Pb(NO3)2 react 2 mol NaI 10.0g Pb(NO3)2 + 10.0g NaI added0.0302 mol Pb(NO3)2 + 0.0667 mol NaIWhich is limiting and excess ?1st method 1 mol Pb(NO3)2 2 mol NaI 0.0302 mol Pb(NO3)2 2 x 0.0302 mol NaI = 0.0604 mol NaI needed = 0.0667 mol NaI add, (excess) = 0.0667 (added) > 0.0604 (needed)NaI is excess Pb(NO3)2 is limiting2nd method Reactants that produce least amt of product will be limitingAssume Pb(NO3)2 limiting 1 mol Pb(NO3)2 1 mol PbI2 0.0302 mol Pb(NO3)2 1x0.0302 mol PbI2 = 0.0302 mol PbI2 Pb(NO3)2 is limitingAssume NaI limiting 2 mol NaI 1 mol PbI2 0.0667 mol NaI 1 x 0.0667 2 = 0.0334 mol PbI2 3. Vol/Conc (solution) given, which is limiting and excess ?2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + H2O(l)1Balanced chemical eqn2Mole ratio/stoichiometry ratioMole ratio2 : 1 1: 1 Click here for animation34100ml, 0.2M, NaOH + 50.0ml, 0.5M H2SO4 addVol/Conc solution addedMole = M x V 1000 = 0.2 x 100 = 0.02 mol 1000Vol/Conc MolesMole = M x V 1000 = 0.5 x 50 = 0.025 mol 10002 mol NaOH react 1 mol H2SO4 100ml, 0.2M NaOH + 50ml, 0.5M H2SO40.02 mol NaOH + 0.025 mol H2SO4Which is limiting and excess ?1st method 2 mol NaOH 1 mol H2SO4 0.02 mol NaOH 1 x 0.02 mol H2SO4 2 = 0.01 mol H2SO4need = 0.025 mol H2SO4add, (excess) = 0.025 (add) > 0.01 (need)H2SO4 is excessNaOH is limiting2nd method Reactants produce least amt of product will be limiting Assume NaOH limiting 2 mol NaOH 1 mol H2O 0.02 mol NaOH 1 x 0.02 mol H2O 2 = 0.01 mol H2O NaOH is limitingAssume H2SO4 limiting 1 mol H2SO4 1 mol H2O 0.025 mol H2SO4 0.025 mol H2O = 0.025 mol H2O 4. Vol (gas) given, which is limiting and excess ? 122CO(g) + 1O2(g) 2CO2 (g)Balanced chemical eqnMole ratio/stoichiometry ratioMole ratio2: 1 2 Click here for animation3445.42L CO + 11.36L O2 addVol gas addedMole = Vol molar vol = 45.42 = 2.0 mol 22.4Vol MolesMole =Vol molar vol = 11.36 = 0.5 mol 22.42 mol CO react 1 mol O2 45.42L CO + 11.36L O22 mol CO+0.5 mol O2Which is limiting and excess ?2nd method1st method 2 mol CO 1 mol O2 2 mol CO 1 mol O2 = 1 mol O2 need = 0.5 mol O2 add, (limit) = 0.5 (add) < 1 (need)Reactants produce least amt of product will be limiting Assume CO limiting 2 mol CO 2 mol CO2 2 mol CO 2 mol CO2 = 2 mol CO2Assume O2 limiting 1 mol O2 2 mol CO2 0.5 mol O2 2 x 0.5 mol CO2 = 1 mol CO2O2 is limiting O2 is limiting 5. IB exam questions on limiting/excess. Q1StepsInvolve limiting excess.0.623g Mg react with 27.3cm3 of 1.25M HCI., H2 gas released Cal the vol of H2 released at T 273K, P 1 atm. Cal the vol occupied by H2 at T 22C, P 1.12 atm Cal % yield, if actual vol H2 collected is 382cm3 1Mole ratio/stoichiometry ratio3Mass/Vol reactants add4Involve Percentage yield.Mg(s) + 2HCI(aq) MgCI2(aq) + H2(g)Balanced chemical eqn2Involve ideal gas eqn.Mass Moles Solution MolesMole ratio1 : 2 1: 2 0.623g Mg + 1.25M, 27.3cm3 HCI Mass = 0.623 RMM 24.31 = 0.0256 molMole = M x V = 1.25 x 0.0273 = 0.0341 mol0.0256 mol Mg + 0.0341 mol HCIMg 5HCIWhich is limiting and excess ?1 mol Mg 2 mol HCI 0.0256 mol Mg 2 x 0.0256 mol HCI = 0.0512 mol HCI needed = 0.0341 mol HCI added (limiting)Mg is excessHCI is limiting continue next slide 6. IB exam questions on limiting/excess. Q1Find vol by H2 at T-273K, p- 1atmInvolve ideal gas eqn. Involve Percentage yield.Mg(s) + 2HCI(aq) MgCI2(aq) + H2(g)Balanced chemical eqn6Involve limiting excess.0.623g Mg react with 27.3cm3 of 1.25M HCI., H2 gas released Cal the vol of H2 released at T 273K, P 1 atm. Cal the vol occupied by H2 at T 22C, P 1.12 atm Cal % yield, if actual vol H2 collected is 382cm3Mass = 0.623 RMM 24.31 = 0.0256 molMole = M x V = 1.25 x 0.0273 = 0.0341 molAmt depend on limiting reagent0.0256 mol Mg + 0.0341 mol HCI Mg(s) + 2HCI(aq) MgCI2(aq) + H2(g) Mg Mole ratio 2 mol HCI 1 mol H2 0.0341 mol HCI 1 x 0.0341 mol H2 2 = 0.0171 mol H2Ideal gas eqn.Continue slidePV = n x R x T V=nxRxT P = 0.0171 x 8.31 x 273 101325 = 3.83 x 10-4 m3 = 383 cm37HCIFind vol by H2 at T-22C, p- 1.12atm Unit conversion T 22C 295K P- 1.12 atm 1.12 x 101325 Pa = 113500 PaPV = n x R x T V=nxRxT P = 0.0171 x 8.31 x 295 113500 = 3.69 x 10-4 m3 = 369 cm38Find % yield if actual vol H2 is 342cm3% Yield =Expt yield x 100% Theoretical yield = 342 x 100% 383 = 89.2% 7. Review on Empirical Formula & Past Year Exam questions 8. Empirical Formula Calculation Relationship bet Molecular Formula and Empirical FormulaCompoundEmpirical Formula (RMM)Molecular Formula (RMM)EtheneC1H2- 14C2H4 - 28Phosphorus(V) oxideP2O5- 142P4O10 - 284Hydrogen PeroxideH1O1- 17H2O2 - 34Ethanoic acidC1H2O1 30Empirical Formula CalculationEmpirical Formula Calculation Step 1: Write mass/ percentage of each element Step 2: Calculate number of moles of each element (dividing with molar mass/RAM) Step 3: Divide each by smallest number, obtain simplest ratioC2H4O2- 60Element M combines with O to form oxide, MO. Find the empirical formula for MO.ElementMOMass/g2.41.6RAM/RMM4816Step 2Number moles/mol2.4/48 = 0.051.6/16 = 0.1Step 3Simplest ratio0.05/0.05 =10.1/0.05 =2Step 1 Molecular formula = n x Empirical formula or RMM = n x formula mass of Empirical formulaEmpirical formulaRMM Molecular formula (C1H2O1 )n = 60 (30)n = 60 n=2(C1H2O1 )2 = 60 Molecular Formula = C2H4O2Empirical formula - M1O2. Video tutorial Molecular/empirical formula 9. Empirical Formula Calculation Q1Boric acid used to preserve food contains 4.8% hydrogen, 17.7% boron and rest is oxygen. Determine the empirical formula of boric acid.Empirical Formula Calculation Step 1: Step 2: Step 3:Write the mass/ percentage of each element Calculate the number of moles of each element (dividing with molar mass/RAM) Divide each by smallest number, obtain the simplest ratioElementHBOPercentage/ %4.8%17.7%77.5%RAM/RMM11116Step 2Number moles/mol4.8/1 = 4.817.7/11 = 1.677.5/16 = 4.84Step 3Simplest ratio4.8/1.6 =31.6/1.6 =14.84/1.6 =3Step 1Q2Answer Empirical formula - H3B1O3.2.5 g of X combined with 4 g of Y to form compound with formula XY 2. If the RAM of Y is 80, determine the relative atomic mass of X.ElementXYMass/g2.54RAM/RMMa80Step 2Number moles/mol2.5/a =?4/80 = 0.05Step 3Simplest ratio12Step 12.5 / a 1 0.05 2 2.5 0.05 a 2 2.5 x 2 a 100 0.05Answer Empirical formula given as X1Y2. a= 100 10. Challenging Empirical Formula Calculation Q3Gaseous hydrocarbon X contains 85.7% of carbon by weight. 4.2 g of gas X occupy volume of 3.36 dm 3 at stp. a) Determine the empirical formula of X b) Determine the RMM of X c) Determine the Molecular formula of X AnswerElementCHPercentage/ %85.714.3RAM/RMM121Step 2Number moles/mol85.7/12 = 7.1414.3/1 = 14.3Step 3Simplest ratio7.14/7.14 =114.3/7.14 =2Step 1Q4a) Empirical Formula = C1H2 b) Volume of 3.36dm3 at stp Mass, 4.2g Volume of 22.4dm3 at stp Mass, 4.2g x 22.4/3.36 = 28g RMM of X = 28 c) Assume molecular formula of X - (CH2)n RMM of X is (12+2)n = 28 n=2 Molecular formula of X is C2H4Compound X contain carbon, hydrogen and oxygen was analysed. 0.50g of compound on complete combustion, yielded 0.6875g of carbon dioxide and 0.5625g of water. Determine the empirical formula.ElementHOMass/g0.18750.06250.25RAM/RMMStep 1C12116Conservation of mass Mass carbon atoms before = Mass carbon atoms after Mass hydrogen atom before = Mass oxygen atoms afterCHO + O2 0.50gStep 2 Step 3Number moles/mol0.1875/1 2 = 0.015620.0625/1 = 0.06250.25/16 = 0.01562Simplest ratio0.01562 0.01562 =10.0625 0.01562 =40.01562 0.01562 =1AnswerEmpirical formula C1H4O10.75gMoles carbon atoms in CO2 = 0.6875 = 0.0156 mol 44 Mass carbon = moles x RAM C atoms = 0.015625 x 12 = 0.1875gCO2 + H2O 0.6875g0.5625gMoles hydrogen atoms in H2O = 0.5625 = 0.03125 x 2 mol 18 = 0.0625 mol Mass hydrogen = moles x RAM H atoms = 0.0625 x 1 = 0.0625gMass of oxygen atoms in CHO = (Mass CHO Mass C Mass O) = 0.5 (0.1875 + 0.0625) = 0.25g 11. IB exam questions on empirical formula. Q5Element X react with oxygen to form oxide, X2O3. If 2.199 g of oxide obtained from 1.239 g of X, find the RAM for X and identify the element X. Mass X = 1.239 gMass O = (2.199 1.239) = 0.960 gElementXOMass/g1.2390.960RAM/RMM?16Step 2Number moles/mol1.239/x = 0.050.960/16 = 0.1Step 3Simplest ratio23Step 1Q6Empirical formula = 2 : 3Mole ratio X : O = 2 : 3 1.239 : 0.960 = 2 : 3 X 16 1.239 = 0.12 X 3 X = 3 x 1.239 0.12 X = 30.97 (Phosphorus) Chemical eqn : 4 X + 3O2 2 X2O3Nitrogen forms oxide with oxygen. Oxide contains 25.9% nitrogen and 74.1% oxygen by mass. Cal the empirical formula of this oxide. ElementNO% by mass25.974.1RAM/RMM14.0116.00Step 2Number moles/mol25.9/14.01 = 1.8574.1/16.00 = 4.63Step 3Simplest ratio25Step 1Lowest ratio=2 : 5Empirical formula = N2O5 12. IB exam questions on empirical formula. Q7Crocetin consists of elements carbon, hydrogen and oxygen. Find empirical formula of crocetin, if 1.00g crocetin forms 2.68 g of carbon dioxide and 0.657 g water when combusted with oxygen. Find the molecular formula of crocetin given 0.300 mol has mass of 98.5 g. ElementStep 2 Step 3AnswerHOMass/g0.7310.07300.195RAMStep 1C12.011.0116.01Number moles/mol0.731/12.01 = 0.06090.0730/1.01 = 0.07300.195/16.01 = 0.0122Simplest ratio0.0609 0.0122 =50.0730 0.0122 =60.0122 0.0122 =1Conservation of mass Mass carbon atoms before = Mass carbon atoms after Mass hydrogen atom before = Mass oxygen atoms afterCHO + O2CO2 +Moles carbon atoms in CO2 = 2.68 = 0.0609 mol 44 Mass carbon = mole x RAM C atoms = 0.0609 x 12.01 = 0.731 gH2O2.68 g1.00g0.657 gMoles hydrogen atoms in H2O = 0.657 x 2 = 0.0729 18 Mass hydrogen = mole x RAM H atoms = 0.0729 x 1.01 = 0.0736 gEmpirical formula C5H6O1 Mass of oxygen atoms = (Mass CHO Mass C Mass H) atoms in CHO = 1.0 (0.731 + 0.0736) = 0.195gFind molecular formula- crocetin given 0.300 mol has mass of 98.5 g. a) Empirical Formula = C5H6O1 Mole Mass 0.300 mol 98.5 g 1 mol 98.5/0.300 = 328 gmol-1 c) Assume molecular formula - (C5H6O1)n = 328 RMM is ( (5 x 12.01) +(6 x 1.01) + ( 1 x 16.01) )n = 328 82.11 x n = 328 n=4 Molecular formula is (C5H6O1) 4 (C5H6O1) 4= C20H24O4Molecular formula = n x Empirical formula or RMM = n x formula mass of Empirical formulaEmpirical formulaRMM Molecular formula(C5H6O1 )n = 328 (82.11)n = 328 n=4 (C5H6O1 )4 = 328 Molecular Formula = C20H24O4 13. Review on Ideal Gas Law & Past Year Exam questions 14. Ideal Gas Equation P = Pressure Unit Nm-2/Pa/kPa/atmn = number of molesPV = nRT V = Volume gas Unit dm3 or m3T = Absolute Temperature in KR = universal gas constant Unit - 8.314Jmol-1K-1 or 0.0821 atm L mol-1 K-14 different variables P, V, n, TPV = nRT Boyles LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pP1V1 = P2V2Charless Law PV = nRT (n ,P fix) V = constant x T V = constant T VTV1 = V2 T1 T2Fix 2 variables change to different gas LawsPressure LawPV = nRT (n, V fix) P = constant x T PTP1 = P2 T1 T2Avogadros Law PV = nRT (P, T fix) V = constant x n VnV1 = V2 n1 n2 15. PV = nRT Boyles LawCharless LawV 1 PAvogadros LawV TPressure LawV nP T2 different variables+ +Combined Gas LawP1V 1 = P2V2 T1 T2 3 different variablesIdeal Gas Equation For 1 mole PV = RT For n mole PV = nRTPV = nRTPV = nRT4 different variablesPressure/P 101 325 Pa(Nm-2)n 1 mol Find R (Universal Gas Constant) at molar volume n = 1 mol T = 273K P = 101325Pa/Nm-2 V = 22.4 x 10-3m3 R=?Temp/T oC 273KR=P V n TVolume/V 22.4dm3 22.4 x 10-3 m3R = 101325 x 22.4 x 10-3 1 x 273R = 8.31 JK-1mol-1or NmK-1 When n = 1 mol Gas constant, R is 8.31 JK-1mol-1or NmK-1 16. Standard Temp/PressureStandard Molar VolumeP - 1 atm = 760 mmHg = 101 325Pa (Nm-2) = 101.325kPa 1 mole of any gas at fix STP (Std Temp/Pressure) occupies a volume of 22.4dm3/22400cm3/24LT 0C (273.15K) Unit conversion22.4L22.4L22.4L1 mole gas1 atm = 760 mmHg/Torr = 101 325Pa(Nm-2) =101.325kPa 1m3 = 103 dm3 = 106cm3 1dm3 = 1 litreAvogadros LawGasHeliumNitrogenOxygenMole/mol114.028.032.0Pressure/atm111Temp/K273273273Vol/L22.4L22.4L22.4LParticles6.02 x 10236.02 x 10236.02 x 1023 equal vol of gases at same temperature/pressure contain equal numbers of molecules1Mass/gmolar volume of all gases the same at given T and P 22.4Lhttp://leifchemistry.blogspot.kr/2011/01/molar-volume-at-stp.htmlVideo on Avogadros Law 17. IB Questions on Ideal Gas 1Find volume (dm3) of 2.00g CO at T 20C P 6250Nm-22Answer: (Ideal Gas Eqn) PV = nRT T (20 + 273) = 293K V = nRT n 2.00/28 = 0.0714 mol P = 0.0714 x 8.314 x 293 6250 =0.0278m3 = 27.8dm3Find volume (m3) of 1 mol of gas at T - 298K P - 101 325Pa3Answer: (Ideal gas eqn) PV = nRT V = nRT P V = 1 x 8.314 x 298 101325 = 0.0244m3What volume (dm3) of 1 mol gas at P - 101325Nm-2 T - 25C Answer: (Ideal gas eqn) pV = nRT V = nRT P V = 1 x 8.31 x (273 + 25) 101325 = 0.0244m3 = 24.4dm3Using PV = nRT (Ideal gas eqn) Need to convert to SI units 4 variables involved4A gas occupy at (constant P) V - 125cm3 T - 27C Calculate its volume at 35C Answer: (Charles Law) V1 = V2 (constant P) T1 T2 125 = V2 (27+273) (35 + 273) V2 = 128cm35Find final vol, V2, gas at (constant T) compressed to P2 = 250kPa V1 - 100cm3 P1 - 100kPa V2 - ? P2 250kPa Answer: (Boyle Law) p1V1 = p2V2 (constant T) 100 x 100 = 250 x V2 V2 = 40cm363.0 dm3 of SO2 reacted with 2.0 dm3 of O2 2SO2(g) + O2(g) 2SO3(g) Find volume of SO2 in dm3 at stp Answer: (Avogadro Law) PV = nRT (at constant P,T) Vn 2SO2(g) + 1 O2(g) 2SO3(g) 2 mol 1 mol 2 mol 2 vol 1 vol 2 vol 3dm3 2dm3 ?SO2 is limiting 2dm3 SO2 2dm3 SO3 3dm3 SO2 3dm3 SO3 Boyle, Charles, Avogadro Law no need to convert to SI units cancel off at both sides 2 variables involved 18. IB Questions on Ideal Gas 7Find empirical formula for composition by mass. S 23.7%, O 23.7%, CI 52.6% Density of its vapour at T- 70C and P- 98kNm-2 = 4.67g/dm3 What molecular formula? AnswerAnswerElementSOCIComposition23.723.752.6Moles23.7 32.1 = 0.73823.7 16.0 = 1.4852.6 35.5 = 1.480.738 = 1 0.738 11.48 = 2 0.738 21.48 = 2 0.738 2Mole ratioEmpirical formula - SO2CI2PV = nRT PV = m x R x T M M =mxRxT V P Density = m (mass) V (volume) M = x RT P Density = 4.67gdm-3 4.67 x 103 gm-3 P = 98kN-2 9.8 x 104 Nm-2 M = (4.67 x 103) x 8.31 x (273 +70) 9.8 x 104 = 135.8 135.8 = n [ 32 + (2x16)+(2 x 35.5) ] 135.8 = n [ 135.8] n=1 MF = SO2CI2 19. IB Questions on Ideal Gas 8Find molar mass gas by direct weighing, at T-23C and P- 97.7kPa Mass empty flask = 183.257g Mass flask + gas = 187.942g Mass flask + water = 987.560g Mass gas = (187.942 183.257) = 4.685g Vol gas = Vol water = Mass water = (987.560 183.257) = 804.303cm3 Vol gas = 804.303cm3 804.303 x 10-6m3 Pressure = 97.7kPa 97700Pa Temp = 23C (273 + 23) = 296K Assume density water = 1g/cm39Answer Mass gas = (26.017 25.385) = 0.632g Vol flask = (231.985 25.385) = 206.6 x 10-6 m3 P = 101kNm-2 101 x 103 Nm-2 M = m x RT PV = 0.632 x 8.314 x 305 101 x 103 x 206.6 x 10-6 = 76.8Answer PV = nRT PV = mass x R x T M M = mass x R x T PV = 4.685 x 8.314 x 296 97700 x 804.303 x 10-6 = 146.7g/mol 11 103.376g gas occupies 2.368dm3 at T- 17.6C, P - 96.73kPa. Find molar mass Unit conversion Answer PV = nRT T 17.6C (273 = 17.6 = 290.6) PV = mass x RT P 96.73kPa 96730Pa M M = mass x R x T PV = 3.376 x 8.314 x 290.6 96730 x 2.368 x 10-3 = 35.61g/molCalculate RMM of gas Mass empty flask 25.385g Mass flask filled gas = 26.017 Mass flask filled water = 231.985g Temp = 32C Pressure = 101kPa6.32 g gas occupy 2200cm3, T- 100C P -101kPa. Calculate RMM of gas Answer PV = nRT n = PV RT n = (101 x 103) (2200 x 10-6) 8.31 x (100+273) n = 7.17 x 10-2 mol n = mass M RMM = mass n RMM = 6.32 7.17 x 10-2 = 88.15 20. Review on Mole Concept, RMM and Molar mass & Past Year Exam questions 21. Molar Mass Molar Mass Mass for 1 mole of any substance Symbol M, Unit - g /mole Molar mass contain 6.02 x 1023 particles 1 mole atom relative atomic mass in gram ( 1 mole carbon 12g) 1 mole molecule - relative molecular mass in gram ( 1 mole water 18g) Number moles = mass or mass RAM/RMM Molar Mass1 MoleMolecule CO2Ionic Compound NaCIIonic Compound MgCI21 mole of Fe contains 6.02 x 1023 Fe atoms1 mole of iron element (Fe) weigh RAM = 55.85 Molar Mass = 55.85g1 MoleElement FeMolar MassMolar Mass1 mole of CO2 contains 6.02 x 1023 CO2 molecules1 mole of CO2 molecule weighs RMM = 12.01 + 16.00 + 16.00 = 44 Molar Mass = 44gMolar Mass1 Mole1 mole of NaCI molecule weighs RMM = 22.99 + 35.45 = 55.45 Molar Mass = 55.45g1 mole of NaCI contains 6.02 x 1023 NaCI particlesMolar Mass1 Mole1 mole of MgCI2 molecule weighs 1 mole of MgCI2 contains 6.02 x 1023 MgCI2 particles RMM = 24..31 + 35.45 + 35.45 = 95.21 Molar Mass = 95.21gTED Video on Mole 22. Conversion from Volume to Number of particlesConversion from Mass to Number of particlesMassVolumeNumber particles 1Calculate the number of particles in a) 12.8g of copper atoms, Cu b) 8.5g of ammonia NH3 moleculesMassMolesNumber particles Answer: a) 64g Cu atoms 1 mole 12.8g Cu atoms 1 x 12.8 = 0.2 mol 64 1 mol Cu 6 x 1021 Cu atoms 0.2 mol Cu 6 x 1021 x 0.2 = 1.2 x 1021 Cu atoms b) 17g NH3 molecules 1 mole 8.5g NH3 molecules 1 x 8.5 = 0.5 mol 17 1 mole NH3 6 x 1023 NH3 molecules 0.5 mole NH3 0.5 x 6 x 1023 = 3 x 1023 NH3 molecules2Calculate number of particles at stp a) 0.28dm3 of N2 gasVolumeMolesNumber particles Answer: a) 22.4dm3 N2 1 mole 0.28dm3 N2 1 x 0.28 22.4 = 0.0125mol 1 mol N2 6.02 x 1023 particles 0.0125 mol N2 6.02 x 1023 x 0.0125 = 7.5 x 1021 particles 23. Conversion from Number of particles to VolumeConversion from Number of particles to Mass Number particlesMass3VolumeCalculate volume of 1.5 x 1023 molecules of N2 at stp ( 1 mole occupies 22.4dm3 at stp)Calculate the mass in a) 1.2 x 1022 zinc atoms b) 3 x 1023 ethanol, C2H5OH molecules4MassVolumeMolesNumber particles Answer: a) 6 x 1023 Zn atoms 1 mole 1.2 x 1022 Zn atoms 1.2 x 1022 = 0.02 mol 6 x 1023 1 mol Zn atoms 65g o.o2 mol Zn atoms 0.02 x 65g = 1.3gb) 6 x 1023 C2H5OH molecules 1 mole 3 x 1023 C2H5OH molecules 3 x 1023 = 0.5mol 6 x 1023 1 mole C2H5OH 46g 0.5 mole C2H5OH 46 x 0.5 = 23gMolesNumber particles Answer: a) 6.02 x 1023 N2 molecules 1 mole 1.5 x 1023 N2 molecules 1 x 1.5 x 1023 6.02 x 1023 = 0.25mol1 mol N2 22.4dm3 0.25 mol N2 22.4 x 0.25 = 5.6dm3 24. Conversion from Mass to MolesConversion from Mass to Volume Mass5Calculate the moles for a) 23.5g of copper(II)nitrate, Cu(NO3)2 b) 0.97g of caffeine C8H10N4O2 moleculesMolesVolumeAns a) 1 mole copper(II)nitrate, Cu(NO ) 64 + 2 *14 + 3(16)+ 3 23.4g NH3 1 x 3.4 = 0.2 mol 171 mol NH3 22.4dm3 0.2 mol NH3 22.4 x 0.2 dm3 = 4.48 dm3Conversion from Moles to MassAnsa) 1 mole Al atoms 27g 2/3 mole AI atoms 2/3 x 27 g 18g b) 1 mole, C6H8O6 6(12) + 8(1) + 6(16) = 176g 0.08 mole C6H8O6 0.08 x 176 g = 14.08g c) 1 mole Mg(OH)2 24 + 2( 16 + 1) = 58g 0.125 mole Mg(OH)2 0.125 x 58g = 7.25gCalculate the volume of 3.4g of NH3 at stp ( 1 mole occupies 22.4dm3 at stp)Ans a) 17g NH 1 mole 3= 188g 188g Cu(NO3)2 1 mole 23.5g Cu(NO3)2 1 x 23.5 = 0.125 mol 188 b) 1 mole, C8H10N4O2 8(12) + 10 + 4(14) + 2(16) = 194g 194g C8H10N4O2 1 mole 0.97g C8H10N4O2 1 x 0.97 = 0.005 mol 194Calculate the mass for a) 2/3 mole of aluminium atoms b) 0.08 mole of C6H8O6 molecules c) 0.125 mole Mg(OH)26Conversion from Moles to VolumeMolesCalculate the volume of 0.75 mole of nitrogen at stp ( 1 mole occupies 22.4dm3 at stp)MassVolume Ans1 mole occupies 22.4dm3 0.75 mole 0.75 x 22.4dm3 = 16.8dm3 25. IB Exam Question on Stoichiometry Q112XO43- ions form precipitate with silver ions, Ag+. When 41.18cm3 of Ag+ ions added to solution of XO43- ions, 1.172g of precipitate formed. Cal amt (mol) of Ag+ ions, molar mass of ppt, RAM of X and identify element X.Balanced chemical eqn3Ag+ + XO43- Ag3XO4 Mole ratioMole ratio/stoichiometry ratio3:1 3 3Vol/Conc solution added4Vol/Conc Moles41.18ml, 0.2040M Ag+ + excess XO43- added Mole = M x V = 0.2040 x 0.04118 = 8.401 x 10-3 mol Ag+ 8.401 x 10-3 mol Ag+ExcessExcess XO43-Cal amt (mol) of ppt Ag3XO4 and Ag+ ions1.172g precipitate Ag3XO43 mol Ag+ 1 mol Ag3XO4 8.401 x 10-3 mol Ag+ 8.401 x 10-3 /3 mol Ag3XO4 = 2.80 x 10-3 mol Ag3XO4Mole Mass 2.8 x 10-3 Ag3XO4 1.172g 1 Ag3XO4 1.172g 2.80 x 10-3 Molar mass = 418.6gmol-1RMM Ag3XO4 = 418.6 (108) x 3 + X + (16 x 4) = 418.6 324 + X + 64 = 418.6 X = 30.99 (PHOSPORUS)Find RAM X Identify X Molar mass Ag3XO4 26. IB Exam Question on Stoichiometry Q2120.502 g of alkali metal sulfate, M2SO4 dissolved in water and excess BaCI2 added to precipitate all sulfate ions as barium sulfate, BaSO4(s). Ppt filtered and weighs 0.672 g. Cal the amt of BaSO4 , alkali metal sulfate, molar mass of alkali metal sulfate, state its units and identity of alkali metal.SO42- + Ba2+ BaSO4Balanced ionic eqnMole ratioMole ratio/stoichiometry ratio1:1 1 3Mass reactants added4Ions produced0.502 g MSO4 + excess Ba2+ added M2SO4 M+ + SO42-Added SO42-BaCI2 Ba2+ + 2CI-Excess Ba2+ BaCI20.672 g precipitate BaSO4MM BaSO4 = 233.40 gmol-1 Mole BaSO4 = Mass = 0.672 M 233.40 = 2.88 x 10-3 molFind amt ppt BaSO4 and M2SO4Salt M2SO4/BaCI2 dissolve produce ion M2SO4 2M+ + SO42- / BaCI2Ba2+ + 2CIStoichiometry ratio 1 SO42- + 1Ba2+ 1BaSO4 1 mol 1 mol 1 mol -3 SO 2.88 x 10 2.88 x 10-3 BaSO4 4 Moles of M2SO4 = 2.88 x 10-3Mole Mass 2.88 x 10-3 M2SO4 0.502 g 1 M2SO4 0.502 2.88 x 10-3 Molar mass = 174.31 gmol-1RMM M2SO4 = 174.31 2M+ 32+(16 x 4) = 174.31 M = 39 ( Potassium) M2SO4 = K2SO4Find Molar mass M2SO4 RAM and identify M 27. IB Exam Question on Stoichiometry Q3120.982 g of Fe(NH4)2(SO4)2 is dissolved in water and excess BaCI2 was added. Precipitate of BaSO4 separated, dried and weigh 1.17 g. Cal amt of BaSO4 , amt of sulfate and iron in 0.982g of Fe(NH4)2(SO4)2. Cal mass of Fe, NH4 and SO4 in 0.982 g of Fe(NH4)2(SO4)2.SO42- + Ba2+ BaSO4Balanced ionic eqnMole ratio/stoichiometry ratioMole ratio1:1 1 3Mass reactants added4Ions produced0.982 g Fe(NH4)2(SO4)2. + excess Ba2+ add Fe(NH4)2(SO4)2 SO42Added SO42-BaCI2 Ba2+ + 2CIExcess Ba2+ BaCI21.17 g precipitate BaSO4MM BaSO4 = 233.40 Mole BaSO4 = Mass M = 1.17 233.40 = 5.01 x 10-3 molFe(NH4)2(SO4)2 / BaCI2 dissolve produce ion Fe(NH4)2(SO4)2 2SO42- / BaCI2Ba2+ + 2CIStoichiometry ratio 1 SO42- + 1Ba2+ 1BaSO4 1 mol 1 mol 1 mol 5.01 x 10-3 SO4 5.01 x 10-3 BaSO4 Moles SO4 = 5.01 x 10-3 from Fe(NH4)2(SO4)2Find amt ppt BaSO4 SO4 in Fe(NH4)2(SO4)2Fe(NH4)2(SO4)2 1 mol Mole - 2.5 x 10-3Fe2+ + 2NH4+ + 1 mol 2 mol 2.5 x 10-3 5.01 x 10-3 X RMM X 55.85Mass -0.140g2SO422 mol 5.01 x 10-3X RMM X 18.05X RMM X 96.060.0904g0.481gFind amt and mass of Fe, NH4, SO4 28. Review on Titration Calculation & Past Year Exam questions 29. TitrationRedox TitrationAcid Base TitrationComplexometric titrationNeutralizationCondition for Acid/Alkali Titration -One reactant must be standard (known conc) or capable being standardised - Equivalent point equal amt neutralize each other - End point measurable/detectable by colour change (indicator), pH change /conductivityAccurate known concPrimary standard acids - Potassium hydrogen phthalate 20.4 g in 1L20.4 g KHPBurette Volumetric flaskStandard 0.1M KHPAcid/Base used as primary standard -Stable/solid -Soluble in water -Does not decompose over timeUnknown Conc NaOHAccurate known concPrimary standard bases - Anhydrous sodium carbonate 10.6 gNa2CO310.6g in 1 LUnknown Conc HCIVolumetric flask??Unable to prepare accurate conc of NaOH/HCI due to Hygroscopic nature NaOH absorb water vapour HCI is in vapour state difficult to measure amtStandard 0.1M Na2CO3Burette 30. Sample Titration Calculation Standardization of (BASE) with standard (ACID)Standardization of (ACID) with standard (BASE)KHP M = 0.100M V = 26.4 mlNa2CO3 M = 0.100M V = 26.4 ml HCI M=? V = 25.0mlNaOH M=? V = 25.0mlCalculation KHP+M = 0.100M V = 26.40mlNaOHCalculation NaKPM= ? V = 25.0mlUsing mole ratioMoles of Acid = MV = (0.100 x 0.0264) = 2.64 x 10-3 Mole ratio (1 : 1) 1 mole acid neutralize 1 mole base 2.64 x 10-3 acid neutralize 2.64 x 10-3 base Moles of Base = M x V = M x 0.025 M x 0.025 = 2.64 x 10-3 M = 0.106M+ H2ONa2CO3+ 2HCIM = 0.100M V = 26.4mlMole ratio 1: 1 2NaCI + H2O + CO2M=? V = 25.0mlUsing mole ratioUsing formulaM aVa = 1 Mb Vb 1 0.1 x 26.40 = 1 M x 25.0 1 M b = 0.106MMoles of Base = MV = (0.100 x 0.0264) = 2.64 x 10-3 Mole ratio (1 : 2) 1 mole base neutralize 2 mole acid 2.64 x 10-3 base neutralize 5.28 x 10-3 acid Moles of Acid = M x V = M x 0.025 M x 0.025 = 5.28 x 10-3 M = 0.211MMole ratio 1: 2Using formulaM bVb = 1 Ma Va 2 0.1 x 26.4 = 1 Ma x 25.0 2 Ma = 0.211MVideo on TitrationClick here Acid/Base calculationClick here Titration calculationClick here cal Na2CO3/HCIClick here cal NaOH/H2SO4 31. Acid/Base Titration Calculation 125.0 cm3 of NaOH of unknown conc require 26.5cm3 of 1.0M sulphuric acid for complete neutralization. Find its molarity of NaOH.2Find vol of 1.5M aq NH3 required to completely neutralize 30.0cm3 of 0.5M sulphuric acid NH3 / NH4OH M = 1.5M V = ? mlH2SO4 M = 1.00M V = 26.5cm3H2SO4 M = 0.5M V = 30.0mlNaOH M=? V = 25.0mlCalculation 2NaOH M=? V = 25.0ml+ H2SO4 M = 1.00M V = 26.5mlCalculationNa2SO4 + 2H2O Mole ratio 2: 1Using mole ratioMoles of Acid = MV = (1.00 x 0.0265) = 2.65 x 10-2 Mole ratio (1 : 2) 1 mole acid neutralize 2 mole base 2.65 x 10-2 acid neutralize 5.30 x 10-2 base Moles of Base = M x V = M x 0.025 M x 0.025 = 5.30 x 10-2 M = 2.12MUsing formulaMb Vb = 2 Ma Va 1 M x 25.0 = 2 1.0 x 26.5 1 Mb = 2.12M2NH4OH +H2SO4 M = 1.5M V = ? mlM = 0.5M V = 30.0ml(NH4)2SO4 + 2H2OUsing mole ratioMoles of Acid = MV = (0.5 x 0.030) = 1.5o x 10-2 Mole ratio (2 : 1) 1 mole acid neutralize 2 mole base 1.50 x 10-2 acid neutralize 3.00 x 10-2 base Moles of Base = M x V = 1.5 x V 1.5 x V = 3.00 x 10-2 Vb = 0.02 dm3 = 20mlMole ratio 2: 1Using formulaM bVb = 2 Ma Va 1 1.5 x Vb = 2 0.5 x 30.0 1 Vb = 20ml 32. Acid/Base Titration Calculation 3Find vol of 2.0M HCI needed to neutralize 2.65g of sodium carbonate (Na2CO3) in 50ml water.410.0cm3 of 0.200M Na2CO3 needed 25.0cm3 of HCI for neutralization. Find molarity of HCI.HCI M = 2.0M V = ? mlNa2CO3 M = 0.200M V = 10.0mlNa2CO3 2.65g V = 50mlMoles = Mass/M = 2.65 106 = 0.025 molHCI M=? V = 25.0mlCalculationCalculationNa2CO3 +2HCI 2NaCI + CO2 + H2OM = 0.5M V = 50mlM = 2.0M V = ? mlUsing mole ratioMoles of Base = Mass/M = (2.65 106) = 2.5 x 10-2 Mole ratio (1 : 2) 1 mole base neutralize 2 mole acid 2.5 x 10-2 base neutralize 5.0 x 10-2 acid Moles of Acid = M x V = 2.0 x V 2.0 x V = 5 x 10-2 V = 0.25 dm3 = 25cm3Na2CO3Mole ratio 1: 2Using formulaMbVb = 1 Ma Va 2 0.5 x 50.0 = 1 2.0 x V 2 Va = 25cm3+M = 0.200M V = 10.0ml2HCI 2NaCI + H2O + CO2M=? V = 25.0mlUsing mole ratioMoles of Base = MV = (0.200 x 0.010) = 2.00 x 10-3 Mole ratio (1 : 2) 1 mole base neutralize 2 mole acid 2.00 x 10-3 base neutralize 4.00 x 10-3 acid Moles of Acid = M x V = M x 0.025 M x 0.025 = 4.00 x 10-3 M = 0.160MMole ratio 1: 2Using formulaMbVb = 1 Ma Va 2 0.2 x 10.0 = 1 Ma x 25.0 2 Ma = 0.160M 33. Acid/Base Titration Calculation 5Vol NaOH need to neutralize 1.325 g ethanedioic acid (C2H2O4.2H2O) is 27.52cm3, calculate molarity of NaOH.633.7cm3 of HCI neutralizes 20cm3 of Na2CO3 of 1.37M. Find molarity of acid. HCI M=? V = 33.7mlNaOH M=?M V = 27.52 ml Moles = Mass/M = 1.325 126.08 = 0.0105 molAcid Mass = 1.325gNa2CO3 M = 1.37M V = 20.0mlCalculation C2H2O4+Mass - 1.325g Mole 0.0105Calculation2NaOH Na2C2O4 + 2H2O M= ? V = 27.52mlUsing mole ratioMoles of Acid = Mass/M = (1.325 126.08) = 1.05 x 10-2 Mole ratio (1 : 1) 1 mole acid neutralize 2 mole base 1.05 x 10-2 acid neutralize 2.10 x 10-2 base Mole of Base = M x V = M x 0.02752 M x 0.02752 = 2.10 x 10-2 M = 0.764MMole ratio 1: 2Using formulaM aVa = 1 Mb Vb 2 0.0105 = 1 M x 0.02752 2 Mb = 0.764MNa2CO3+ 2HCIM = 1.37M V = 20.0ml 2NaCI + H2O + CO2M=? V = 33.7mlMole ratio 1: 2Using mole ratioUsing formulaMoles of Base = MV = (1.37 x 0.020) = 2.74 x 10-2 Mole ratio (1 : 2) 1 mole base neutralize 2 mole acid 2.74 x 10-2 base neutralize 5.48 x 10-2 acid Moles of Acid = M x V = M x 0.0337 M x 0.0337 = 5.48 x 10-2 M = 1.63MM bVb = 1 Ma Va 2 1.37 x 20.0 = 1 Ma x 33.7 2 Ma = 1.63M 34. Acid/Base Titration Calculation Ethanoic acid determination in vinegar 725.0ml of conc vinegar (ethanoic acid) was diluted to a total vol of 250 in a volumetric flask. Diluted vinegar was transfer to a burette. 25.0ml, 0.1M NaOH is pipette into a conical flask, with indicator added. End point reached when average 29.2 ml of diluted vinegar added. Find its molarity. Diluted 25xCH3COOH M=? V = 29.2mlNaOH M = 0.1M V = 25.0ml V = 25 ml M=?V = 250ml M = 0.08561NaOH + CH3COOH CH3COONa + H2O M = 0.1M V = 25.0mlM=? V = 29.2mlMole ratio 1: 1Using formulaUsing mole ratioMoles bef dilution = Moles aft dilution M1 V1 = M2V2 M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol3M1 x V1 = M2 x V2 M1 x 25 = 0.0856 x 250 M1 = 0.0856 x 250 25 M1 = 0.856M2Mole NaOH = MV = (0.1 x 0.025) = 2.5 x 10-3 Mole ratio (1 : 1) 1 mole NaOH react 1 mole acid 2.5 x 10-3 mole NaOH react 2.5 x 10-3 acid Mole acid = M V = M x 0.0292 M x 0.0292 = 2.5 x 10-3 qcid M = 2.5 x 10-3 0.0292 M = 0.0856MM aVa = 1 Mb Vb 1 0.1 x 0.025 = 1 M x 0.0292 1 M = 0.0856MVideo on determination of ethanoic acid in vinegar 35. Acid/Base Titration Calculation - Empirical formula determination for Na2CO3. x H2O 827.82g of hydrated (Na2CO3 . x H2O) dissolved in water, making up to 1L. 25.0ml of solution was neutralized by 48.8ml of 0.100M HCI. Cal molarity and mass of Na2CO3 present in 1L of solution. Find x . HCI M = 0.100 M V = 48.8mlDiuted to 1LNa2CO3 M=?M V = 25.0ml25 ml transfer27.82g Na2CO3. xH2O1V = 1L M=?2HCI + Na2CO3 2NaCI +CO2 + H2O M = 0.1M V = 48.8mlM=? V = 25.0mlMole ratio 2: 1Using mole ratio 24Mass Na2CO3 . x H2O = 27.82 g Mass Na2CO3 = 10.36 g Mass of water = (27.82 10.36) g = 17.46g Empirical formula5Mole HCI = MV = (0.1 x 0.0488) = 4.88 x 10-3 Mole ratio (2 : 1) 2 mole HCI react 1 mole base 4.88 x 10-3 mole HCI react 2.44 x 10-3 base Mole base = M V = M x 0.0250 M x 0.0250 = 2.44 x 10-3 base M = 2.44 x 10-3 0.0250 M = 0.0976MNa2CO317.463RMM10618.02Mole10.34/106 = 0.0977317.46/18.02 = 0.9689mol/dm3 g/dm3 mol/dm3 x M g/dm3 0.0976 x 106 = 10.36g/dm30.09773/0.09733 10.9689/0.09733 10M aVa = 2 Mb Vb 1 0.1 x 0.0488 = 2 M x 0.0250 1 M = 0.0976MH2O10.36Using formulaMass/gLowest ratioEmpirical formula Na2CO3 . 10 H2OConvert moles to mass in 1LVideo on Na2CO3 calculation 36. Redox Titration Calculation- % Iron in iron tablet 9Iron tablet contain hydrated iron (II) sulphate (FeSO4.7H2O). One tablet weighing 1.863g crushed, dissolved in water and solution made up to total vol of 250ml. 10ml of this solution added to 20ml of H2SO4 and titrated with 0.002M KMnO4. Average 24.5ml need to reach end point. Cal % iron(II) sulphate in iron tablet.Video on % Iron in iron tabletKMnO4 M = 0.002M V = 24.5 ml1.863 g 250mlMnO4M = 0.002M V = 24.5mlFe2+ M=? V = 30ml+ 5Fe2+ + 8H+ Mn2+ + 5Fe2+ + 4H2O M= ?Mole ratio 1: 5Video on Fe2+/KMnO4 titration calculation Using mole ratioUsing formulaMole KMO4- = MV = (0.002 x 0.0245) = 4.90 x 10-5 Mole ratio (1 : 5) 1 mole KMO4- react 5 mole Fe2+ 4.90 x 10-5 KMO4-react 2.45 x 10-4 Fe2+12M aVa = 1 Mb Vb 5 0.002 x 0.0245 = 1 Moles Fe2+ 5 Moles = 2.45 x 10-4 Fe2+10ml sol contain - 2.45 x 10-4 Fe2+ 250ml sol contain - 250 x 2.45 x 10-4 Fe2+ 10 = 6.125 x 10-3 mole Fe2+ FeSO4.7H2O FeSO4 + 7H2O 1 mol 1 mol + 7 mol FeSO4 Fe2+ + SO4 21 mol 1mol + 1mol 6.125 x 10-3 mol 6.125 x 10-3 mole Fe2+4 3Mole Mass Mole x RMM = Mass FeSO4 6.125 x 10-3 x 278.05 = 1.703g FeSO4Mass of (expt yield) = 1.703g Mass of (Actual tablet) = 1.863g % Fe in iron tablet = 1.703 x 100% 1.863 = 91.4% 37. Redox Titration Calculation CIO- in Bleach 1010.0ml of bleach (CIO-) diluted to a total vol of 250ml. 20.0ml is added to 1g of KI (excess) and iodine produced is titrated with 0.0206M Na2S2O3.Using starch indicator, end point was 17.3ml. Cal molarity of CIO in bleach. Diuted 25xVideo on CIO- in bleachWater added till 250ml10.0ml CIOtransferNa2S2O3 M = 0.0206M V = 17.3ml1g KI excess added20ml transferI2 M=?titrated V = 10 M=?V = 250ml M = 1.78 x 10-2 M 2S2O32-1 6M = 0.0206 V = 17.3mlMole bef dilution = Mole aft dilution M1 V1 = M2V2 M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol M1 V1 = M2 V2 M1 x 10 = 1.78 x 10-2 x 250 M1 = 1.78 x 10-2 x 250 10 M1 = 0.445M+I2 Mole = ? V = 0.022Mole ratio 2: 1Mole S2O32- = MV = (0.0206 x 0.0173) = 3.56 x 10-4 Mole ratio (2 : 1) 2 mole S2O32- react 1 mole I2 3.56 x 10-4 S2O32--react 1.78 x 10-4 I22CIO- + 2I- + 2H+ I2 + 2CI- + H2O 2CIOI2 Mole ratio 2: 1Mole = ? Moles of CIO- = M x V MxV = 3.56 x 10-4 M x 0.02 = 3.56 x 10-4 M = 3.56 x 10-4 002 M = 1.78 x 10-2 M diluted 25xS4O62- + 2IUsing direct formula354Mole ratio (2 : 1) 2 mole CIO 3.56 x 10-4 CIO-2 mol 2CIO- +I2 + 1 mol2I- +1 mol 2H+ I2 + 2CI- + H2O2S2O322 mol S4O62- + 2I-Mole ratio ( 1 : 1) 2 mole CIO- : 1 mole I2 : 2 mole S2O322 mole CIO2 mole S2O32-Mole = 1.78 x 10-41 mole I2 1.78 x 10-4 I2M V (CIO+) = 2 = 1 M V (S2032-) 2 1 Moles of CIO+ = 1 0.0206 x 0.0173 1 Moles of CIO- = 3.56 x 10-4 38. Redox Titration Calculation Vitamin C quantification 11Iodometric titration was performed on Vit C, (C6H8O6). 25ml Vit C is titrated with 0.002M KIO3 from burette, using excess KI and starch. Average vol KIO3 is 25.5ml. Cal molarity of Vit C. KIO3 M = 0.002M V = 25.5ml1g KI excess + starch added25ml transferVitamin CVit C M=? V = 25mltitrated V = 25ml M= ? 1KIO3 + 5KI + 6H+ M = 0.002M V = 25.5ml3I2 +Mole = ?3H2O + 6K= Mole ratio 1: 3Using mole ratio 23Mole KIO3 = MV = (0.002 x 0.0255) = 5.10 x 10-5 Mole ratio (1 : 3) 1 mole KIO3 produce 3 mole I2 5.10 x 10-5 KIO3 produce 1.53 x 10-4 I23C6H8O6 Mole = ?+ 3I2 3C6H6O6 + 6I- + 6H+ 1.53 x 10-4 Mole ratio 3: 3Using formula1 mol 3 mol KIO3 + 5KI + 6H+ 3I2 + 6K+ + 3H2O 3C6H8O6 + 3I2 3C6H6O6 + 6I- + 6H+ 3 mol 3 molMole ratio (1 :3) 1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6 1 mol KIO3 3 mol C6H8O6 Using formula4Mole ratio (1 : 3) 1 mol KIO3 react 3 mol C6H8O6 5.10 x 10-5 KIO3 react 1.53 x 10-4 C6H8O6 Mole C6H8O6 = M x V MxV = 1.53 x 10-4 M x 0.025 = 1.53 x 10-4 M = 1.53 x 10-4 0025 M = 6.12 x 10-3 MM aVa (KIO3) = 1 Mb Vb (C6H8O6) 3 0.002 x 0.0255 = 1 Mole C6H8O6 3 Mole C6H8O6 = 1.53 x 10-4 Mole C6H8O6 = M x V M x V = 1.53 x 10-4 M x 0.025 = 1.53 x 10-4 M = 1.53 x 10-4 0025 M = 6.12 x 10-3 M 39. Redox Titration Calculation - % Cu in Brass 122.5g brass react with 10ml conc HNO3 producing Cu2+ ions. Solution made up to 250ml using water in a volumetric flask. Pipette 25ml of solution into conical flask. Na2CO3 added to neutralize excess acid. 1g KI (excess) and few drops of starch added to flask. Titrate with 0.1M S2O32- and end point, reached when 28.2ml added. Find molarity copper ions and % copper found in brass. Water added till 250ml 1g KI excess + starch addedPour into Volumetric flask25ml transfer 2.5g brass10 ml HNO3Na2S2O3 M = 0.1M V = 28.2mlV = 250ml M=?I2 M=?titrated2S2O32-1M = 0.1M V = 28.2ml+I2Mole = ?S4O62- + 2IMole ratio 2: 1Using mole ratio 2Using formulaMole S2O32- = MV = (0.1 x 0.0282) = 2.82 x 10-3 Mole ratio (2 : 1) 2 mole S2O32- react 1 mole I2 2.82 x 10-3 S2O32-- react 1.41 x 10-3 I22Cu2+ + 4I- I2 + 2CuI Mole = ? 1.41 x 10-3 I23Mole ratio 2: 12 mol 2Cu2+ +I2 + 1 mol4I-2S2O32- 2 mol1 mol I2 + 2CuIS4O62- + 2I-Mole ratio ( 1 : 1) 2 mole Cu2+ : 1 mole I2 : 2 mole S2O322 mole Cu2+ 2 mole S2O32Using formula5Mole of Cu2+ = M x V MxV = 2.82 x 10-3 M x 0.025 = 2.82 x 10-3 M = 2.82 x 10-3 0.025 M = 1.13 x 10-1 M Mass Cu = Molarity Cu x M Mass Cu = (1.13 x 63.5)g Cu in 1000ml = 7.18g Cu in 1000ml = 1.79g Cu in 250ml46Mole ratio (2 : 1) 2 mole Cu2+ 2.82 x 10-3 Cu2+1 mole I2 1.41 x 10-3 I2% Cu in brass = mass Cu x 100% mass brass = 1.79 x 100% 2.5 = 71.8%M V (Cu2+) = 2 = 1 M V (S2032-) 2 1 Moles of Cu2+ = 1 0.1 x 0.0282 1 Moles of Cu2+ = 2.82 x 10-3 40. Redox Titration Calculation - % Cu in Brass 130.456 g brass react with 25ml conc HNO3 producing Cu2+ ions. Solution produced was titrate with 0.1M S2O32- and end point, reached when 28.5ml added. Cal moles, mass, molarity of copper ions and % copper found in brass. Reaction mechanism/processStep 1 Cu + 2HNO3 + 2H+ Cu2+ + 2NO2 + 2H2O Step 2 4I- + 2Cu2+ 2 CuI + I2 Step 3 I2 + 2S2O32- 2I- + S4O62-1g KI excess + starch addedtransfer0.456g brassNa2S2O3 M = 0.1M V = 28.5mlI2 M=?titrated 25 ml HNO3 2S2O32- +1M = 0.1M V = 28.5mlI2Mole = ?S4O62- + 2IMole ratio 2: 1Using mole ratio 2Using formulaMole S2O32- = MV = (0.1 x 0.0285) = 2.85 x 10-3 Mole ratio (2 : 1) 2 mole S2O32- react 1 mole I2 2.85 x 10-3 S2O32-- react 1.425 x 10-3 I22Cu2+ + 4I- I2 + 2CuI Mole = ? 1.425 x 10-3 I23Mole ratio 2: 12 mol 2Cu2+ +I2 + 1 mol4I-2S2O32- 2 mol1 mol I2 + 2CuIS4O62- + 2I-Mole ratio ( 1 : 1) 2 mole Cu2+ : 1 mole I2 : 2 mole S2O322 mole Cu2+ 2 mole S2O32Using formula5Mole of Cu2+ = M x V MxV = 2.85 x 10-3 M x 0.025 = 2.85 x 10-3 M = 2.85 x 10-3 0.025 M = 1.14 x 10-1 M Mass Cu = Moles x RAM = 2.85 x 10-3 x 63.55 = 0.18g46Mole ratio (2 : 1) 2 mole Cu2+ 2.85 x 10-3 Cu2+1 mole I2 1.41 x 10-3 I2% Cu in brass = mass Cu x 100% mass brass = 0.18 x 100% 0.456 = 39.7%M V (Cu2+) = 2 = 1 M V (S2032-) 2 1 Moles of Cu2+ = 1 0.1 x 0.0285 1 Moles of Cu2+ = 2.85 x 10-3 41. Review on Back Titration Calculation & Past Year Exam questions 42. TitrationDirect TitrationBack Titration ConditionCondition Both titrant and analyte soluble Reaction is fast Sparingly soluble acid/base. Rxn too slow, unable to dissolve/react Calcium carbonate (egg shell) and impurities (base) from antacid tablet/impure limestone. Salicyclic acid in aspirin tablet.Titrant - solubleAnalyte - solubleaddedRedox TitrationAcid Base TitrationComplexometric TitrationImpure limestoneCaCO3 in egg shell Impure antacid Known conc /vol of acid used Left overnight in acidH2SO4 HCI HNO3Soluble acidNaOH NH4OH Soluble base KOH (Alkali) Ba(OH)2 LiOHAmt of known acid added Titrated with known conc/vol alkali known unknownAmt of excess acid left Transfer to flaskAmt of base (solid) react with acid Video on back titration calculationAmt base(solid) = Amt Known Amt excess acid added acid leftExcess acid left 43. IB exam questions on empirical formula. Back Titration% Calcium carbonate in egg shell - Back Titration Calculation Q125.0g of egg shell (CaCO3) dissolved in 250.0ml, 2.0M HNO3. Solution was titrated with NaOH. 17.0ml, 1.00M NaOH needed to neutralize excess acid. Cal percentage CaCO3 by mass in egg shell. 1added25.0g impure CaCO3 in egg shell 2250.0ml, 2.0M HNO3Titrated with known conc/vol NaOH M = 1.0 V = 17.0mlLeft overnight in acidAmt of HNO3 addedHNO3 left346 NaNO3 + H2Omoles = ?Mole ratio 1: 1Using formulaMole NaOH = MV = (1.00 x 0.017) = 1.7 x 10-2 Mole ratio (1 : 1) 1 mole NaOH react 1 mole HNO3 1.7 x 10-2 mole NaOH react 1.7 x 10-2 HNO3 HNO3 left = 1.7 x 10-2 mol3Mole ?Amt HNO3 react with base7Mass = Mole CaCO3 x RMM CaCO3 = 0.242 x 100 = 24.2g8% by mass = mass CaCO3 x 100% CaCO3 mass impure = (24.2/25.0) x 100% = 96.8%Mole ratio 2: 1Mole ratio (2 : 1) 2 mol HNO3 react 1 mol CaCO3 0.483 mol HNO3 react o.242 mol CaCO3MbVb = 1 Ma Va 1 1.00 x 0.017 = 1 moles 1 mole HNO3 = 1.7 x 10-2 Amt HNO3 leftAmt HNO3 react = Amt HNO3 add Amt HNO3 left with NaOH = 0.50 1.7 x 10-2 = 0.483 mol2HNO3 + CaCO3 (CaNO3)2 + H2O + 2CO2 Mole 0.483Amt HNO3 react = Amt HNO3 Amt HNO3 with base add leftAmt HNO3 addUsing mole ratioAmt of base (egg)5NaOH + HNO3 M = 1.00M V = 17.0mlTransfer to flaskAmt of HNO3 leftAmt HNO3 add = M V = 2.0 x 0.250 = 0.50 mol 44. IB exam questions on empirical formula. Back Titration% Calcium carbonate in egg shell - Back Titration Calculation Q20.188g of egg shell (CaCO3) dissolved in 27.20ml, 0.200M HCI. Solution was titrated with NaOH. 23.8ml, 0.10M NaOH needed to neutralize excess acid. Cal the number of moles, mass and percentage of CaCO3 by mass in egg shell. 1added0.188g impure CaCO3 in egg shell 27.20ml, 0.200M HCI Left overnight in acidAmt of HCI added2Titrated with known conc/vol NaOH M = 0.10 V = 23.8mlAmt HCI add NaCI + H2Omoles = ?Mole ratio 1: 1Using formulaUsing mole ratio 3Mole NaOH = MV = (0.10 x 0.238) = 2.38 x 10-3 Mole ratio (1 : 1) 1 mole NaOH react 1 mole HCI 2.38 x 10-3 mole NaOH react 2.38 x 10-3 HCI HCI left = 2.38 x 10-3 mol3MbVb = 1 Ma Va 1 0.10 x 0.238 = 1 moles 1 mole HCI = 2.38 x 10-3 Amt HCI leftHCI left 4Amt of base (egg)56Amt HCI react = Amt HCI add Amt HCI left with NaOH = 0.0544 2.38 x 10-3 = 0.00306 mol2HCI + CaCO3 CaCI2 + H2O + CO2 Mole 0.00306Amt HCI react = Amt HCI Amt HCI with base add leftNaOH + HCI M = 0.10M V = 23.8mlTransfer to flaskAmt of HCI leftAmt HCI add = M V = 0.200 x 0.270 = 0.0544 molMole ?Amt HCI react with base 7Mass = Mole CaCO3 x RMM CaCO3 = 0.00153 x 100 = 0.153g8% by mass = mass CaCO3 x 100% CaCO3 mass impure = (0.153/0.188) x 100% = 81.4%Mole ratio 2: 1Mole ratio (2 : 1) 2 mol HCI react 1 mol CaCO3 0.00306 mol HCI react o.00153 mol CaCO3 45. IB exam questions on empirical formula. Back Titration Q3% Calcium hydroxide in an impure antacid tablet - Back Titration Calculation0.5214g of impure Ca(OH)2 from antacid was dissolved in 50.0ml, 0.250M HCI. 33.64ml, 0.1108M NaOH needed to neutralize excess acid. Cal percentage sample containing Ca(OH)2 in a tablet. 1added0.5214g impure Ca(OH)2 250.0ml, 0.250M HCI Left overnight in acidTitrated with known conc/vol NaOH M = 0.1108 V = 33.64mlAmt of HCI addedHCImoles = ?Amt HCI AddNaCI + H2O Mole ratio 1: 1Using formulaUsing mole ratio 3Mole NaOH = MV = (0.1108 x 0.03364) = 3.727 x 10-3 Mole ratio (1 : 1) 1 mole NaOH react 1 mole HCI 3.727 x 10-3 mole NaOH react 3.727 x 10-3 HCI HCI left = 3.727 x 10-3 mol3MbVb = 1 Ma Va 1 0.1108 x 0.03364 = 1 moles 1 mole HCI = 3.727 x 10-3 Amt HCI LeftHCI left 4Amt of base (solid)56Amt HCI react = Amt HCI add Amt HCI left with NaOH = 0.01250 3.727 x 10-3 = 0.008773 mol2HCI + Ca(OH)2 Mole 0.008773Amt HCI react = Amt HCI Amt HCI with NaOH add leftNaOH + M = 0.1108M V = 33.64mlTransfer to flaskAmt HCI leftAmt HCI add = M V = 0.250 x 0.050 = 0.01250 molMole ? CaCI2 + 2H2O7Mass = Mole Ca(OH)2 x RMM Ca(OH)2= 0.004386 x 74.1 = 0.3250g8% by mass = mass Ca(OH)2 x 100% Ca(OH)2 mass impure = (0.3250/0.5214) x 100% = 62.3%Mole ratio 2: 1Mole ratio (2 : 1) 2 mol HCI react 1 mol Ca(OH)2 0.008773 mol HCI react o.oo4386 mol Ca(OH)2Amt HCI react with base 46. IB exam questions on empirical formula. Back Titration Q4Molar mass of insoluble acid in a tablet -Back Titration Calculation2.04g insoluble dibasic acid dissolve in 20.0ml, 2.00M NaOH. Excess NaOH require 17.6ml, 0.5M HCI to neutralize it. Find molar mass acid 1added2.04g impure dibasic acid H2A 220.0ml, 2.00M NaOH Left overnightTitrated with known conc/vol HCI M = 0.50 V = 17.6mlAmt NaOH addedHCI+M = 0.50M V = 17.6mlNaOH moles = ?Amt NaOH addNaCI + H2O Mole ratio 1: 1Using formulaUsing mole ratio 3Transfer to flaskAmt of NaOH leftAmt NaOH add = M V = 2.00 x 0.02 = 0.040 molMole HCI = MV = (0.50 x 0.0176) = 8.8 x 10-3 Mole ratio (1 : 1) 1 mol HCI react 1 mol NaOH 8.8 x 10-3 mol HCI react 8.8 x 10-3 NaOH NaOH left = 8.8 x 10-3 mol3MaVa = 1 Mb Vb 1 0.50 x 0.0176 = 1 moles 1 mole HCI = 8.8 x 10-3 Amt NaOH LeftNaOH left 4Amt of acid (solid)5Amt NaOH = Amt NaOH Amt NaOH react with acid add left = 0.040 8.8 x 10-3 = 0.0312 mol 2NaOH + H2A Na2 A + 2H2O Mole 0.00312Amt NaOH react = Amt NaOH Amt NaOH with acid add left 6Amt NaOH react with acidMole ?Mole ratio 2: 1Mole ratio (2 : 1) 2 mol NaOH react 1 mol H2 A 0.0312 mol NaOH react o.o156 mol H2A7Molar Mass 0.0156 mol H2A - 2.04g 1 mol H2A - 2.04 0.0156 = 131