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Mole Calculations

Mole Calculations. West Midlands Chemistry Teachers Centre Tuesday 23rd November 2010. Presenter: Dr Janice Perkins. 1 10 6 ‘formulae’. 3 10 6 molecules. 2 10 6 atoms. 3 10 6 molecules. 1 dozen ‘formulae’. 3 dozen molecules. 2 dozen atoms. 3 dozen molecules.

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Mole Calculations

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  1. Mole Calculations West Midlands Chemistry Teachers Centre Tuesday 23rd November 2010 Presenter: Dr Janice Perkins

  2. 1106 ‘formulae’ 3106 molecules 2106 atoms 3106 molecules 1dozen ‘formulae’ 3 dozen molecules 2 dozen atoms 3 dozen molecules Reacting ratio in equations Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g) 1 ‘formula’ 3 molecules 2 atoms 3 molecules 10 ‘formulae’ 30 molecules 20 atoms 30 molecules

  3. Funny numbers 12 Dozen = Gross = 12  12 = 144 Score = 20 Mole = 6.023  1023 That’s 602300000000000000000000

  4. 6.02  1023 Or The Avogadro Constant (L) 602300000000000000000000 It is just a number – no more special than a ton, a score or a dozen – its just a bit bigger!

  5. 6.021023 ‘formulae’ 18.06 1023 molecules 12.04 1023 atoms 18.06 1023 molecules Reacting Ratio Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g) 1 formula 3 molecules 2 atoms 3 molecules 10 ‘formulae’ 30 molecules 20 atoms 30 molecules 1106 ‘formulae’ 2106 atoms 3106 molecules 3106 molecules 2 moles 3 moles 1 mole 3 moles

  6. 3:2 1:2 1:3 3:3 Mole Ratio (Reacting Ratio) Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g) Fe2O3 : CO = 1:3 CO: CO2 = 1:1 Fe2O3 : Fe = 1:2 CO: Fe = 3:2

  7. Sufficient data to calculate moles?

  8. Mole Calculations Mass Mr Mass n = m Mr Moles of known substance Moles of unknown substance Volume of solution Mole ratio from equation n = v  c Volume of gas n = pV RT Moles = Mass Mr Moles = volume  conc Moles = P(in Pa)  V(in m3) R  T(in Kelvin) Ideal Gas Equation pV = nRT

  9. Mass Mass Moles Mr Moles Mr Rearranging the formula Moles = Mass Mr Mr = Mass Moles Mass = Moles  Mr Mass Moles Mr

  10. Mole Calculations Mass Mr m = n  Mr Mass n = m Mr Mr = m n Moles of known substance Moles of unknown substance Volume of solution Vol Conc Mole ratio from equation n = v  c Volume of gas n = pV RT

  11. Moles Moles Vol Conc Vol Conc Rearranging the formula Moles = Volume  Concentration Vol = Moles Conc Conc = Moles Vol Moles Vol Conc

  12. Mole Calculations Mass Mr m = n  Mr Mass n = m Mr Mr = m n Moles of known substance Moles of unknown substance v = n c Vol Conc Volume of solution Mole ratio from equation c = n v n = v  c V P T Volume of gas n = pV RT

  13. Rearranging the formula Moles = pV RT Volume = nRT p Units are vital: ‘V’ always in m3 ‘P’ always in Pa ‘T’ always in Kelvin Pressure = nRT V Temperature = pV nR

  14. Mole Calculations Mass Mr m = n  Mr Mass n = m Mr Mr = m n Moles of known substance Moles of unknown substance v = n c Vol Conc Volume of solution Mole ratio from equation c = n v n = v  c V P T V = nRT p Volume of gas p = nRT V n = pV RT T = pV nR

  15. Example 1: Calculating moles from masses Calculate the number of moles in 300g of CaCO3 Use the equation Mass = Mr x moles Mr = 100 Rearrange equation Moles = mass/Mr = 300 100 = 3 moles

  16. Example 2: Calculating concentration of solution Calculate the concentration, in mol dm-3, of the solution formed when 19.6 g of hydrogen chloride, HCl, are dissolved in water and the volume made up to 250 cm3. Calculate the concentration, in mol dm-3, of the solution formed when 19.6 g of hydrogen chloride, HCl, are dissolved in water and the volume made up to 250 cm3. First we need to calculate the number of moles of HCl. Moles = mass/Mr Moles HCl = 19.6 Moles HCl = 19.6 36.5 Moles HCl = 19.6 = 0.537 mol 36.5 Conc of HCl(aq) = moles = volume (in dm3) = Conc of HCl(aq) = moles = 0.537 volume (in dm3) = Conc of HCl(aq) = moles = 0.537 volume (in dm3) 250  10-3 = Conc of HCl(aq) = moles = 0.537 volume (in dm3) 250  10-3 = 2.15 mol dm-3

  17. Example 3: Calculating moles of solution and solid, then Mr andAr (Jan 09 chem1) • A metal carbonate MCO3 reacts with HCl as in the following equation. • MCO3 + 2HCl  MCl2 + H2O + CO2 • A 0.548 g sample of MCO3reacted completely with 30.7 cm3 of 0.424 mol dm-3 HCl. • Calculate the amount, in moles, of HCl which reacted with the 0.548 g of MCO3 • Moles = vol x conc (in dm3) • = 30.7/1000 x 0.424 = 0.0130 mol

  18. (b) Calculate the amount, in moles, of MCO3 in 0.548 g Look at equation again MCO3 + 2HCl  MCl2 + H2O + CO2 There is a 2:1 ratio of reactants. We have calculated that there are 0.0130 mol of HCl so there must be half that amount of MCO3 = 0.0130/2 = 0.0065 mol = 6.50 x 10-3 mol

  19. (c) Calculate the Mr of MCO3 A mass of 0.548g of MCO3 contains 0.0065 mol. Use the equation Mr = mass/moles Mr = 0.548/0.0065 = 84.3 (d) Use your answer to deduce the Ar of M We have just calculated the Mrof MCO3 to be 84.3 Since there is 1xC and 3xO this makes 12 + 48 =60. So this means M must have an Ar of 84.3 – 60= 24.3 This isMagnesium.

  20. Example 4: Identity unknown – less structured calculation The carbonate of metal M has the formula M2CO3. The equation for the reaction of this carbonate with hydrochloric acid is given below. M2CO3 + 2HCl  2MCl + CO2 + H2O A sample of M2CO3, of mass 0.394 g, required the addition of 21.7 cm3 of 0.263 mol dm-3 solution of hydrochloric acid for complete reaction. Calculate the Ar of metal M and deduce its identity. The carbonate of metal M has the formula M2CO3. The equation for the reaction of this carbonate with hydrochloric acid is given below. M2CO3 + 2HCl  2MCl + CO2 + H2O A sample of M2CO3, of mass 0.394 g, required the addition of 21.7 cm3 of 0.263 mol dm-3 solution of hydrochloric acid for complete reaction. Calculate the Ar of metal M and deduce its identity. The carbonate of metal M has the formula M2CO3. The equation for the reaction of this carbonate with hydrochloric acid is given below. M2CO3 + 2HCl  2MCl + CO2 + H2O A sample of M2CO3, of mass 0.394 g, required the addition of 21.7 cm3 of 0.263 mol dm-3 solution of hydrochloric acid for complete reaction. Calculate the Ar of metal M and deduce its identity. (a)Find moles of known substance - in this case HCl Moles HCl(aq) = volume  concentration = = Moles HCl(aq) = volume  concentration = 21.7 = Moles HCl(aq) = volume  concentration = 21.7  10-3 = Moles HCl(aq) = volume  concentration = 21.7  10-3  0.263 = Moles HCl(aq) = volume  concentration = 21.7  10-3  0.263 = 5.71  10-3 mol

  21. M2CO3 + 2HCl  2MCl + CO2 + H2O (b) Calculate the moles of the other substance- in this case M2CO3 The mole ratio is 2:1 We have calculated the moles HCl = 5.71  10-3 mol So the moles M2CO3 = 5.71  10-3 /2 = 2.85 x 10-3 mol (c) Now find Mr of M2CO3 Mr of M2CO3 = mass = moles = Mr of M2CO3 = mass = 0.394 moles = Mr of M2CO3 = mass = 0.394 moles 2.85  10-3 = Mr of M2CO3 = mass = 0.394 moles 2.85  10-3 = 138 (d) find Ar of metal M and hence deduce its identity 2  Ar(M) = Mr(M2CO3) - Mr(‘CO3’) = 138 – 60 = 78 Ar(M) = 78/2 = 39 M = Potassium

  22. Example 5: Gases – calculating the volume A sample ofethanol vapour, C2H5OH (Mr = 46), was maintained at a pressure of 100 kPa and at a temperature of 366 K. Use the ideal gas equation to calculate the volume, in cm3, that 1.36 g of ethanol vapour would occupy under these conditions. (The gas constant R = 8.31 J K-1 mol-1) A sample ofethanol vapour, C2H5OH (Mr = 46), was maintained at a pressure of 100 kPa and at a temperature of 366 K. Use the ideal gas equation to calculate the volume, in cm3, that 1.36 g of ethanol vapour would occupy under these conditions. (The gas constant R = 8.31 J K-1 mol-1) A sample ofethanol vapour, C2H5OH (Mr = 46), was maintained at a pressure of 100 kPa and at a temperature of 366 K. Use the ideal gas equation to calculate the volume, in cm3, that 1.36 g of ethanol vapour would occupy under these conditions. (The gas constant R = 8.31 J K-1 mol-1) Moles ethanol = 1.36 Moles ethanol = 1.36 46 Moles ethanol = 1.36 = n = 0.0296 mol 46 Use pV = nRT Use pV = nRT V = nRT = p = = Use pV = nRT V = nRT = 0.0296  8.31  366 p 100000 = = Use pV = nRT V = nRT = 0.0296  8.31  366 p 100000 = 8.992  10-4 = Use pV = nRT V = nRT = 0.0296  8.31  366 p 100000 = 8.992  10-4m3 = Use pV = nRT V = nRT = 0.0296  8.31  366 p 100000 = 8.992  10-4m3 = 8.992  10-4 106 Use pV = nRT V = nRT = 0.0296  8.31  366 p 100000 = 8.992  10-4m3 = 8.992  10-4 106= 899 cm3

  23. Example 6: Empirical formula Jun 09 An oxide of nitrogen contains 30.4% by mass of nitrogen, Calculate the empirical formula. First calculate the % of O 100 – 30.4 = 69.6% Now put in columns N O mass30.469.6 Ar 14 16 = moles 2.17 4.35 Ratio 2.17/2.17 = 1 4.35/2.17= 2 Empirical formula = NO2

  24. Example 7: Water of crystallisation The Mr of a hydrated sodium carbonate was found to be 250. Use this value to calculate the number of molecules of water of crystallisation, x, in this hydrated sodium carbonate, Na2CO3•xH2O. The Mr of a hydrated sodium carbonate was found to be 250. Use this value to calculate the number of molecules of water of crystallisation, x, in this hydrated sodium carbonate, Na2CO3•xH2O. Mr(Na2CO3•xH2O) = 250 Mr(Na2CO3) = 106 Mr(Na2CO3•xH2O) = 250 Mr(Na2CO3) = 106 x  Mr(H2O) = Mr(Na2CO3•xH2O) - Mr(Na2CO3) = 250 – 106 Mr(Na2CO3•xH2O) = 250 Mr(Na2CO3) = 106 x  Mr(H2O) = Mr(Na2CO3•xH2O) - Mr(Na2CO3) = 250 – 106 = 144 Mr(Na2CO3•xH2O) = 250 Mr(Na2CO3) = 106 x  Mr(H2O) = Mr(Na2CO3•xH2O) - Mr(Na2CO3) = 250 – 106 = 144 x  18 = 144 Mr(Na2CO3•xH2O) = 250 Mr(Na2CO3) = 106 x  Mr(H2O) = Mr(Na2CO3•xH2O) - Mr(Na2CO3) = 250 – 106 = 144 x  18 = 144 x = 144/18 = 8 Mr(Na2CO3•xH2O) = 250 Mr(Na2CO3•xH2O) = 250 Mr(Na2CO3) = 106 x  Mr(H2O) = Mr(Na2CO3•xH2O) Mr(Na2CO3•xH2O) = 250 Mr(Na2CO3) = 106 x  Mr(H2O) = Mr(Na2CO3•xH2O)- Mr(Na2CO3) Mr(Na2CO3•xH2O) = 250 Mr(Na2CO3) = 106 x  Mr(H2O) = Mr(Na2CO3•xH2O) - Mr(Na2CO3) = 250 – 106 = 144 x  18 = 144 x = 144/18 Mr(Na2CO3•xH2O) = 250 Mr(Na2CO3) = 106 x  Mr(H2O) = Mr(Na2CO3•xH2O) - Mr(Na2CO3) = 250 – 106 = 144 x  18 = 144 x = 144/18 = 8

  25. Example 8: Water of crystallisation – less structured Sodium carbonate forms a number of hydrates of general formula Na2CO3•xH2O. A 3.01 g sample of one of these hydrates was dissolved in water and the solution made up to 250 cm3. In titration, a 25 cm3 portion of this solution required 24.3 cm3 of 0.200 mol dm-3 hydrochloric acid for complete reaction Na2CO3 + 2HCl  2NaCl + H2O + CO2 Calculate the Mr of Na2CO3 • xH2O Sodium carbonate forms a number of hydrates of general formula Na2CO3•xH2O. A 3.01 g sample of one of these hydrates was dissolved in water and the solution made up to 250 cm3. In titration, a 25 cm3 portion of this solution required 24.3 cm3of 0.200 mol dm-3 hydrochloric acid for complete reaction Na2CO3 + 2HCl  2NaCl + H2O + CO2 Calculate the Mr of Na2CO3 • xH2O • Calculate moles HCl since you can do this • Moles HCl = volume  concentration • = 24.3  10-3  0.200 • = 4.86  10-3 mol

  26. (b) Now calculate the moles of Na2CO3 in the 25 cm3 sample of solution. Na2CO3 + 2HCl  2NaCl + H2O + CO2 Moles Na2CO3 = Mole Ratio  Moles HCl Moles Na2CO3 = Mole Ratio  Moles HCl = 1/2  4.86  10-3 Moles Na2CO3 = Mole Ratio  Moles HCl = ½ 4.86  10-3 = 2.43  10-3 (c) deduce the moles of Na2CO3 in 250 cm3 of solution Moles Na2CO3 (250) = Moles Na2CO3 (25)  Moles Na2CO3 (250) = Moles Na2CO3 (25)  250 25 Moles Na2CO3 (250) = Moles Na2CO3 (25)  250 25 = 2.43  10-2 (d) Calculate the Mr of hydrated Na2CO3 [Mass = 3.01 g] Mr = Mass / Moles Mr = 3.01/2.43  10-2= 124

  27. Example 9: Percentage yield Jan 09 In an experiment 165 g TiCl4 were added to an excess of water. The equation for the reaction is as follows TiCl4 + 2H2O  TiO2 + 4HCl (a) Calculate the amount, in moles, of TiCl4 in 165 g Moles = mass = 165 Mr 189.9 = 0.869 (b) Calculate the maximum amount, in moles, of TiO2 which can be formed in this experiment. Look at the equation. 1 mole of TiCl4 gives 1 mole ofTiO2 So 0.869 moles ofTiCl4 gives 0.869moles ofTiO2

  28. Calculate the maximum mass of TiO2 formed in this experiment. • mass = Mr x moles • = 79.9 x 0.869 • = 69.4 g (d) In this experiment only 63.0 g of TiO2 were produced. Calculate the percentage yield. Percentage yield = actual (experimental) mass of product theoretical ( calculated) mass of product = 63.0 69.4 = 90.8%

  29. Example 10 : % atom economy You should learn this formula and be able to use it correctly. % atom economy = mass of desired product x 100 total mass of reactants

  30. Calculate the % atom economy for the formation of CH2Cl2 in this reaction. CH4 + 2Cl2  CH2Cl2 +2HCl There are no numbers given You only need the Mr values Desired product = CH2Cl2Mr =(12+ 2 +71) = 85 Reactants =CH4+2Cl2Mr = (12+4)+(2 x71)= 158 %atom economy = 85 x 100 = 53.8% 158

  31. Example 11: Moles HCl, moles & identity ‘Z’ The chloride of an element Z reacts with water according to the following equation. ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq) A 1.304g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250cm3 in a volumetric flask. 25.0cm3 of this solution was titrated against 0.112moldm-3 NaOH(aq), and 21.7cm3 were needed to reach the end point. Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample. Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq) A 1.304g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250cm3 in a volumetric flask. 25.0cm3 of this solution was titrated against 0.112moldm-3NaOH(aq),and 21.7cm3 were needed to reach the end point. Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample. Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq) A 1.304g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250cm3 in a volumetric flask. 25.0cm3 of this solution was titrated against 0.112moldm-3NaOH(aq),and 21.7cm3 were needed to reach the end point. Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample. Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq) A 1.304g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250cm3 in a volumetric flask. 25.0cm3 of this solution was titrated against 0.112moldm-3NaOH(aq),and 21.7cm3 were needed to reach the end point. Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample. Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq) A 1.304g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250cm3 in a volumetric flask. 25.0cm3 of this solution was titrated against 0.112moldm-3NaOH(aq),and 21.7cm3 were needed to reach the end point. Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample. Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq) A 1.304g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250cm3 in a volumetric flask. 25.0cm3 of this solution was titrated against 0.112moldm-3NaOH(aq),and 21.7cm3 were needed to reach the end point. Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample. Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq) A 1.304g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250cm3 in a volumetric flask. 25.0cm3 of this solution was titrated against 0.112moldm-3NaOH(aq),and 21.7cm3 were needed to reach the end point. Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample. Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq) A 1.304g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250cm3 in a volumetric flask. 25.0cm3 of this solution was titrated against 0.112moldm-3NaOH(aq),and 21.7cm3 were needed to reach the end point. Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample. Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq) A 1.304g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250cm3 in a volumetric flask. 25.0cm3 of this solution was titrated against 0.112moldm-3NaOH(aq),and 21.7cm3 were needed to reach the end point. Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample. Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq) A 1.304g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250cm3 in a volumetric flask. 25.0cm3 of this solution was titrated against 0.112moldm-3NaOH(aq),and 21.7cm3 were needed to reach the end point. Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample. Calculate the Mr of ZCl4. From your answer deduce the Ar of element Z and hence its identity.

  32. The only complete data we have is for NaOH(aq) • From this we can calculate the moles of HCl in 25.0 cm3 • We scale this to give the number of moles of HCl, in 250 cm3, formed in the reaction Moles NaOH(aq) = volume  concentration Moles NaOH(aq) = volume  concentration = 21.7 Moles NaOH(aq) = volume  concentration = 21.7  10-3 Moles NaOH(aq) = volume  concentration = 21.7  10-3  0.112 Moles NaOH(aq) = volume  concentration = 21.7  10-3  0.112 = 2.43  10-3 mol Moles HCl(25) = mole ratio  moles of NaOH Moles HCl(25) = mole ratio  moles of NaOH = 2.43  10-3 Moles HCl(25) = mole ratio  moles of NaOH = 1  2.43  10-3 Moles HCl(25) = mole ratio  moles of NaOH = 1  2.43  10-3 = 2.43  10-3 Moles HCl(25) = mole ratio  moles of NaOH = 1  2.43  10-3 = 2.43  10-3 Moles HCl(250) = 2.43  10-3 250/25 Moles HCl(25) = mole ratio  moles of NaOH = 1  2.43  10-3 = 2.43  10-3 Moles HCl(250) = 2.43  10-3  250/25= 0.0243 mol

  33. ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq) ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq) Find HCl : ZCl4mole ratio and hence find moles ZCl4 Moles ZCl4 = Mole Ratio  Moles HCl = 1/4 0.0243 Moles ZCl4 = Mole Ratio  Moles HCl = 1/4 or 4/1 0.0243 Moles ZCl4 = Mole Ratio  Moles HCl =  0.0243 Moles ZCl4 = Mole Ratio  Moles HCl = 1/4  0.0243 = 6.076  10-3 Moles ZCl4 = Mole Ratio  Moles HCl Find Mr of ZCl4– then Arof Z and henceidentifyZ Mr of ZCl4 = mass = 1.304(given in q)= 214.6 moles 6.076  10-3 Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = 214.6 - 142 = 72.6 Element Z= GeSo, ZCl4 = GeCl4 Mr of ZCl4 = mass = 1.304= 214.6 moles 6.076  10-3 Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = 214.6 - 142 = 72.6 Element Z= Ge Mr of ZCl4 = mass = 1.304 moles Mr of ZCl4 = mass = 1.304 moles 6.076  10-3 Mr of ZCl4 = mass = 1.304=214.6 moles 6.076  10-3 Mr of ZCl4 = mass = 1.304= 214.6 moles 6.076  10-3 Ar of Z = Mr (ZCl4) Mr of ZCl4 = mass = 1.304= 214.6 moles 6.076  10-3 Ar of Z = Mr (ZCl4) – 4 Ar (Cl) Mr of ZCl4 = mass = 1.304= 214.6 moles 6.076  10-3 Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = 214.6 - 142 Mr of ZCl4 = mass = 1.304= 214.6 moles 6.076  10-3 Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = 214.6 - 142 = 72.6 Mr of ZCl4 = mass = moles

  34. The End

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