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West Midlands Chemistry Teachers Centre Tuesday 11th November 2008
Mole Calculations Presenter: Dr Janice Perkins
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Extracting Iron – The Blast Furnace
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Chemistry of the Blast Furnace Process
At 2000 C C(s) + O2(g) CO2(g) 1 atom 1 molecule 1 molecule At 1500 C C(s) + CO2(g) 2CO(g) 1 atom 1 molecule 2 molecules At 1000 C Fe2O3(s) + 3CO(g) 2Fe + 3CO2(g) 1 ‘formula’ 3 molecules 2 atoms 3 molecules
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Fe2O3(s) + 3CO(g) 2Fe + 3CO2(g)
Reacting Ratio Fe2O3(s) + 3CO(g) 2Fe + 3CO2(g) 1 ‘formula’ 3 molecules 2 atoms 3 molecules 10 ‘formulae’ 30 molecules 20 atoms 30 molecules 1106 ‘formulae’ 3106 molecules 2106 atoms 1dozen ‘formulae’ 3 dozen molecules 2 dozen atoms
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Funny numbers Dozen = 12 Gross = 12 12 = 144 Ton = 100 Monkey = £500 Score = 20 Mole = 1023 That’s
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The Avogadro Constant (L)
6.02 1023 Or It is just a number – no more special than a ton, a score or a dozen – its just a bit bigger!
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Fe2O3(s) + 3CO(g) 2Fe + 3CO2(g)
Reacting Ratio Fe2O3(s) + 3CO(g) 2Fe + 3CO2(g) 1 formula 3 molecules 2 atoms 3 molecules 10 ‘formulae’ 30 molecules 20 atoms 30 molecules 1106 ‘formulae’ 3106 molecules 2106 atoms 3106 molecules 6.021023 ‘formulae’ 18.06 1023 molecules 12.04 1023 atoms 18.06 1023 molecules 1 mole 3 moles 2 moles 3 moles
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Mole Ratio (Reacting Ratio) Fe2O3(s) + 3CO(g) 2Fe + 3CO2(g)
1:2 3:2 Fe2O3(s) + 3CO(g) 2Fe + 3CO2(g) 1:3 3:3 Fe2O3 : CO = 1:3 CO : CO2 = 1:1 Fe2O3 : Fe = 1:2 CO : Fe = 3:2
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Strategy for performing chemical calculations
Read the question CAREFULLY !!!!!! Identify the two substances required in the calculation Use the chemical equation to work out their Mole Ratio Decide which substance has sufficient data to allow you to calculate its moles (known substance) – calculate this Use the mole ratio to deduce the number of moles of the other (unknown) substance Moles unknown subst = Mole Ratio Moles known subst Is answer sensible - do you need more or less moles of the unknown substance - apply Mole Ratio accordingly
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Sufficient data to calculate moles??
Calculation based on Data needed Mass Mass and Mr Mass Mass and Mr or Formula or Name Mass Mass Mass and Mr or Formula Mass Solution Volume Solution Volume and Concentration Solution Gases P and T and V and R Gases P and T and V Gases P Gases Gases P and T
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THE GIVEN MASS MUST NOT BE USED TO CALCULATE MOLES
Beware In questions on percentage purity, the mass given is the mass of the impure substance, the mass of the pure substances is UNKNOWN. THE GIVEN MASS MUST NOT BE USED TO CALCULATE MOLES You calculate the mass of pure substance in the impure mixture. Then calculate the percentage of this mass compared to the original mass.
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Moles of known substance Moles of unknown substance
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 Ideal Gas Equation pV = nRT Moles = P(in Pa) V(in m3) R T(in Kelvin) Moles = volume conc Moles = Mass Mr
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Rearranging the formula
Moles = Mass Mr Mr = Mass Moles Mass = Moles Mr Mass Moles Mr Mass Moles Mr Mass Moles Mr
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Moles of known substance Moles of unknown substance
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
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Rearranging the formula
Moles = Volume Concentration Vol = Moles Conc Conc = Moles Vol Moles Vol Conc Moles Vol Conc Moles Vol Conc
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Moles of known substance Moles of unknown substance
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
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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
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Mole Calculations Mass Mr Mass Moles of known substance
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 Start on the left-hand side – calculate KNOWN moles Use the mole fraction to get the UNKNOWN moles Calculate the answer using the right-hand side options
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Calculation Menu Example 1: Reacting masses - Na2CO3
Example 2: Making hydrochloric acid Example 3: Gas volumes - ethanol volume Example 4: Mass M2CO3 & identity ‘M’ Example 5: Back titration and % purity Example 6: An extended gas law calculation Example 7: Water of crystallisation Example 8: Moles HCl, moles / identity ‘Z’ Example 9: % atom economy
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Example 1: Reacting masses - Na2CO3
Sodium carbonate is manufactured in a two-stage process, via sodium hydrogen carbonate, as shown below. NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl 2NaHCO3 Na2CO3 + H2O + CO2 Calculate the maximum mass of sodium carbonate which could be obtained from 800 g of sodium chloride. Sodium carbonate is manufactured in a two-stage process, via sodium hydrogen carbonate, as shown below. NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl 2NaHCO3 Na2CO3 + H2O + CO2 Calculate the maximum mass of sodium carbonate which could be obtained from 800 g of sodium chloride. Sodium carbonate is manufactured in a two-stage process, via sodium hydrogen carbonate, as shown below. NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl 2NaHCO3 Na2CO3 + H2O + CO2 Calculate the maximum mass of sodium carbonate which could be obtained from 800 g of sodium chloride.
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Example 1: Reacting masses - Na2CO3
NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl 2NaHCO3 Na2CO3 + H2O + CO2 NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl 2NaHCO3 Na2CO3 + H2O + CO2 Tackle the 2 stages separately - less likely to make mistakes Stage 1 NaCl : NaHCO3 = 1:1 Stage 1 NaCl : NaHCO3 = Moles NaCl = mass = 800 Mr 58.5 Moles NaCl = mass = = mol Mr 58.5 Moles NaCl = mass Mr Moles NaHCO3 = 1 Moles NaCl = mol Moles NaHCO3 = 1 Moles NaCl = Moles NaHCO3 = Moles NaCl =
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Example 1: Reacting masses - Na2CO3
NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl 2NaHCO3 Na2CO3 + H2O + CO2 NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl 2NaHCO3 Na2CO3 + H2O + CO2 MORE LESS Now use the number of moles of NaHCO3 to calculate the moles and mass of Na2CO3 Stage 2 NaHCO3 : Na2CO3 = Stage 2 NaHCO3 : Na2CO3 = 2:1 Moles Na2CO3 = 1/2 or 2/1 Moles NaHCO3 = Moles Na2CO3 = / Moles NaHCO3 = ½ Moles Na2CO3 = / Moles NaHCO3 = ½ = mol Moles Na2CO3 = / Moles NaHCO3 = Moles Na2CO3 = Mole Ratio Moles NaHCO3 = Mass Na2CO3 = moles Mr = 6.84 106 = 725 g Mass Na2CO3 = moles Mr = = Mass Na2CO3 = moles Mr = 6.84 106 =
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Example 2: Making hydrochloric acid
We don’t always need to use the Mole Ratio Example 2: Making hydrochloric acid 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. This question only deals with HCl, so we first need to calculate the number of moles of HCl Moles HCl = 19.6 36.5 Moles HCl = = mol 36.5 Moles HCl = 19.6 Conc of HCl(aq) = moles = 0.537 volume (in dm3) 250 10-3 = mol dm-3 Conc of HCl(aq) = moles = 0.537 volume (in dm3) 250 10-3 = Conc of HCl(aq) = moles = 0.537 volume (in dm3) = Conc of HCl(aq) = moles = volume (in dm3) =
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Example 3: Gas volumes - again no mole ratio
A sample of ethanol 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 of ethanol 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 of ethanol 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 46 Moles ethanol = = n = mol 46 Moles ethanol = 1.36 Use pV = nRT V = nRT = 8.31 p = 10-4 m3 = 10-4 106 Use pV = nRT V = nRT = 8.31 p = 10-4 m3 = 10-4 = cm3 Use pV = nRT V = nRT = 8.31 p = 10-4 m3 = Use pV = nRT V = nRT = p = Use pV = nRT V = nRT = 8.31 p = 10-4 = Use pV = nRT V = nRT = 8.31 p = Use pV = nRT
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Example 4: Mass M2CO3 & identity ‘M’
The carbonate of metal M has the formula M2CO3. The equation for the reaction of this carbonate with hydrochloric acid is given below. M2CO HCl 2MCl CO H2O A sample of M2CO3, of mass g, required the addition of 21.7 cm3 of 0.263 mol dm-3 solution of hydrochloric acid for complete reaction. The carbonate of metal M has the formula M2CO3. The equation for the reaction of this carbonate with hydrochloric acid is given below. M2CO HCl 2MCl CO H2O A sample of M2CO3, of mass g, required the addition of 21.7 cm3 of 0.263 mol dm-3 solution of hydrochloric acid for complete reaction. The carbonate of metal M has the formula M2CO3. The equation for the reaction of this carbonate with hydrochloric acid is given below. M2CO HCl 2MCl CO H2O A sample of M2CO3, of mass g, required the addition of 21.7 cm3 of 0.263 mol dm-3 solution of hydrochloric acid for complete reaction. 1 - identify known substance, find data and find moles Moles HCl(aq) = volume concentration = 21.7 10-3 0.263 = 5.71 10-3 mol Moles HCl(aq) = volume concentration = 21.7 10-3 0.263 = Moles HCl(aq) = volume concentration = Moles HCl(aq) = volume concentration = 21.7 = Moles HCl(aq) = volume concentration = 21.7 10-3 =
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Example 4: Mass M2CO3 & identity ‘M’
LESS MORE M2CO HCl 2MCl CO H2O M2CO HCl 2MCl CO H2O 2 - find mole ratio and find moles unknown substance Moles M2CO3 = Mole Ratio Moles HCl = / 10-3 = Moles M2CO3 = Mole Ratio Moles HCl = 10-3 = Moles M2CO3 = Mole Ratio Moles HCl = Moles M2CO3 = Mole Ratio Moles HCl = 1/2 or 2/1 10-3 = Moles M2CO3 = Mole Ratio Moles HCl = ½ 10-3 = 2.85 10-3 3 - find Mr of M2CO3 Mr of M2CO3 = mass = moles 10-3 = Mr of M2CO3 = mass = moles 10-3 = 138 Mr of M2CO3 = mass = moles = Mr of M2CO3 = mass = moles =
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Example 4: Mass M2CO3 & identity ‘M’
4 - 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 2 Ar(M) = Mr(M2CO3) - Mr(‘CO3’) = 138 – 60 = 78 Ar(M) = 78/2 2 Ar(M) = Mr(M2CO3) - Mr(‘CO3’) = 138 – 60 = 78 2 Ar(M) = Mr(M2CO3) - Mr(‘CO3’) = 138 – 60 = 78 Ar(M) = 78/2 = 39 M = Potassium (from Periodic Table) 2 Ar(M) = Mr(M2CO3) - Mr(‘CO3’) = 138 – 60 2 Ar(M) = 2 Ar(M) = Mr(M2CO3) - Mr(‘CO3’) A similar approach can be used in calculating the number of molecules of water of crystallisation in CuSO4•xH2O. Calculate the Mr of the hydrated salt [Mr(hydr) = 249.6] Deduct from this the Mr of the salt. [= – = 90] Divide what is left by the Mr of water (18) to get ‘x’. x = 90 18 = 5 So, formula = CuSO4.5H2O Calculate the Mr of the hydrated salt [Mr(hydr) = 249.6] Deduct from this the Mr of the salt. Calculate the Mr of the hydrated salt Calculate the Mr of the hydrated salt [Mr(hydr) = 249.6] Calculate the Mr of the hydrated salt [Mr(hydr) = 249.6] Deduct from this the Mr of the salt. [= – = 90] Calculate the Mr of the hydrated salt [Mr(hydr) = 249.6] Deduct from this the Mr of the salt. [= – = 90] Divide what is left by the Mr of water (18) to get ‘x’. Calculate the Mr of the hydrated salt [Mr(hydr) = 249.6] Deduct from this the Mr of the salt. [= – = 90] Divide what is left by the Mr of water (18) to get ‘x’. x = 90 18 = 5
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Example 5 – back titration and % purity
A 1.00 g sample of limestone is reacted with 100 cm3 of mol dm-3 hydrochloric acid as shown below. CaCO3(s) + 2HCl(aq) CaCl2 + H2O + CO2 The excess acid required 24.8 cm3 of mol dm-3 sodium hydroxide solution for complete neutralisation. Calculate the percentage of calcium carbonate in the limestone. A 1.00 g sample of limestone is reacted with 100 cm3 of mol dm-3 hydrochloric acid as shown below. CaCO3(s) + 2HCl(aq) CaCl2 + H2O + CO2 The excess acid required 24.8 cm3 of mol dm-3 sodium hydroxide solution for complete neutralisation. Calculate the percentage of calcium carbonate in the limestone. Original moles HCl(aq) = volume concentration = 100 10-3 0.200 = 2.00 10-2 mol Original moles HCl(aq) = volume concentration = 100 10-3 0.200 = Original moles HCl(aq) = volume concentration = 100 = Original moles HCl(aq) = volume concentration = Original moles HCl(aq) = volume concentration = 100 10-3 =
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Example 5 – back titration and % purity
The excess acid required 24.8 cm3 of mol dm-3 sodium hydroxide solution for complete neutralisation. The excess acid required 24.8 cm3 of mol dm-3 sodium hydroxide solution for complete neutralisation. Moles NaOH(aq) = volume concentration = 24.8 10-3 0.100 = 2.48 10-3 mol Moles NaOH(aq) = volume concentration = 24.8 10-3 0.100 Moles NaOH(aq) = volume concentration = 24.8 10-3 Moles NaOH(aq) = volume concentration Excess moles of HCl(aq) i.e. neutralised by NaOH(aq) = moles NaOH(aq) = 10-3 mol Excess moles of HCl(aq) i.e. neutralised by NaOH(aq) = moles NaOH(aq) Excess moles of HCl(aq) i.e. neutralised by NaOH(aq) = Moles of HCl(aq) reacted with CaCO3(s) = initial moles HCl - excess moles HCl = 10-3 Moles of HCl(aq) reacted with CaCO3(s) = initial moles HCl - excess moles HCl = 10-3 = 10-2 mol Moles of HCl(aq) reacted with CaCO3(s) = initial moles HCl - excess moles HCl = 10-2 Moles of HCl(aq) reacted with CaCO3(s) = initial moles HCl Moles of HCl(aq) reacted with CaCO3(s) Moles of HCl(aq) reacted with CaCO3(s) = initial moles HCl - excess moles HCl
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Example 5 – back titration and % purity
Remember, 10-2 mol of HCl reacted with the calcium carbonate in the impure limestone LESS MORE CaCO3(s) + 2HCl(aq) CaCl2 + H2O + CO2 Moles CaCO3 = Mole Ratio Moles HCl = / 10-2 Moles CaCO3 = Mole Ratio Moles HCl = ½ 10-2 = 8.76 10-3 mol Moles CaCO3 = Mole Ratio Moles HCl = 10-2 Moles CaCO3 = Mole Ratio Moles HCl Moles CaCO3 = Mole Ratio Moles HCl = 1/2 or 2/1 10-2
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Example 5 – back titration and % purity
A 1.00 g sample of limestone is reacted with 100 cm3 of mol dm-3 hydrochloric acid as shown below. A 1.00 g sample of limestone is reacted with 100 cm3 of mol dm-3 hydrochloric acid as shown below. Mass of CaCO3 = moles Mr = 10-3 100 = g Mass of CaCO3 = moles Mr = 10-3 100 = Mass of CaCO3 = moles Mr = 10-3 = Mass of CaCO3 = moles Mr = % purity CaCO3 = mass pure CaCO3 100 mass of limestone = 100 1.00 = % % purity CaCO3 = mass pure CaCO3 100 mass of limestone = 100 1.00 % purity CaCO3 = mass pure CaCO3 100 mass of limestone = 100 % purity CaCO3 = mass pure CaCO3 mass of limestone % purity CaCO3 = % purity CaCO3 = mass pure CaCO3 % purity CaCO3 = mass pure CaCO3 100 mass of limestone
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Example 6 – An extended gas law calculation
A sample of ammonium nitrate decomposed on heating as shown in the equation below. NH4NO3 2H2O N ½O2 On cooling the resulting gases to 298 K, the volume of nitrogen and oxygen together was found to be m3 at a pressure of 95.0 kPa. A sample of ammonium nitrate decomposed on heating as shown in the equation below. NH4NO3 2H2O N ½O2 On cooling the resulting gases to 298 K, the volume of nitrogen and oxygen together was found to be m3 at a pressure of 95.0 kPa. First, we need to use the data in the question to calculate the number of moles of gas formed. Use pV = nRT n = pV = 95.0 1000 RT 298 Use pV = nRT n = pV = 95.0 1000 RT 298 = mol Use pV = nRT n = pV = 95.0 1000 RT Use pV = nRT n = pV = 95.0 RT Use pV = nRT n = pV RT Use pV = nRT Use pV = nRT n = pV = 95.0 1000 RT
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Example 6 – An extended gas law calculation
Remember, the amount of gas formed = 1.92 mol. This is the total moles of N2 and O2 gases To calculate the number of moles of NH4NO3 we need the mole ratio of NH4NO3 : Total gas (N2 + O2) NH4NO3 2H2O N ½O2 LESS MORE 1 mole ½ mole 1:1½ or 2:3 Moles NH4NO3 = Mole Ratio Moles gas = /3 = mol Mass NH4NO3 = moles Mr Moles NH4NO3 = Mole Ratio Moles gas = /3 = mol Mass NH4NO3 = moles Mr = 1.28 80 Moles NH4NO3 = Mole Ratio Moles gas = /3 = mol Mass NH4NO3 = moles Mr = 1.28 80 = 102 g Moles NH4NO3 = Mole Ratio Moles gas = /3 = mol Moles NH4NO3 = Mole Ratio Moles gas Moles NH4NO3 = Mole Ratio Moles gas = 1.92 Moles NH4NO3 = Mole Ratio Moles gas = 2/3 or 3/2 = Moles NH4NO3 = Mole Ratio Moles gas = /3 1.92
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Example 7: Water of crystallisation
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 mol dm-3 hydrochloric acid for complete reaction Na2CO HCl 2NaCl + H2O + CO2 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 mol dm-3 hydrochloric acid for complete reaction Na2CO HCl 2NaCl + H2O + CO2 1 – calculate the number of moles of HCl used. Moles HCl(aq) = volume concentration = 24.3 10-3 0.200 Moles HCl(aq) = volume concentration = 24.3 10-3 0.200 = 4.86 10-3 mol Moles HCl(aq) = volume concentration = 24.3 Moles HCl(aq) = volume concentration Moles HCl(aq) = volume concentration = 24.3 10-3
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Example 7: Water of crystallisation
2 – from the moles of HCl used, deduce the moles of Na2CO3 in the 25 cm3 sample of solution LESS MORE Na2CO HCl 2NaCl + H2O + CO2 Moles Na2CO3 = Mole Ratio Moles HCl = 1/2 or 2/1 10-3 Moles Na2CO3 = Mole Ratio Moles HCl = / 10-3 Moles Na2CO3 = Mole Ratio Moles HCl = ½ 10-3 = 2.43 10-3 Moles Na2CO3 = Mole Ratio Moles HCl = 10-3 Moles Na2CO3 = Mole Ratio Moles HCl 3 – 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
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Example 7: Water of crystallisation
4 – Calculate the Mr of hydrated Na2CO3 [Mass = 3.01 g] 4 – Calculate the Mr of hydrated Na2CO3 Moles = Mass Mr = Mass Mr Moles Mr = /2.43 = 124 Moles = Mass Mr = Mass Mr Moles Mr = /2.43 10-2 Moles = Mass Mr = Mass Mr Moles Moles = Mass Mr = Mass Mr Moles Mr = Moles = Mass Mr The second part of this question provides the Mr value of a different hydrated sodium carbonate. You have to work out the number of molecules of water of crystallisation ‘x’ in this new hydrate. The question is split so that an error in calculating the Mr in the first part doesn’t stop you working out ‘x’.
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Example 7: Water of crystallisation
In an experiment. The Mr of a different 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. In an experiment. The Mr of a different 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 x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3) = – = 144 x 18 = 144 Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106 x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3) = – = 144 x 18 = 144 x = 144/18 Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106 x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3) = – = 144 x 18 = 144 x = 144/18 = Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106 x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3) = – = 144 x 18 = 144 x = 144/18 = Formula = Na2CO3 • 8H2O Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106 x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3) = – = 144 Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106 x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3) Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106 Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106 x Mr(H2O) = Mr(Na2CO3 • xH2O) Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3 • xH2O) = 250 Mr(Na2CO3) = 106 x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3) = – 106
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Example 8: Moles HCl, moles & identity ‘Z’
The chloride of an element Z reacts with water according to the following equation. ZCl4(l) H2O(l) ZO2(s) HCl(aq) A g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask. 25.0 cm3 of this solution was titrated against mol dm-3 NaOH(aq), and 21.7 cm3 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) H2O(l) ZO2(s) HCl(aq) A g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask. 25.0 cm3 of this solution was titrated against mol dm-3 NaOH(aq), and 21.7 cm3 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) H2O(l) ZO2(s) HCl(aq) A g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask. 25.0 cm3 of this solution was titrated against mol dm-3 NaOH(aq), and 21.7 cm3 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) H2O(l) ZO2(s) HCl(aq) A g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask. 25.0 cm3 of this solution was titrated against mol dm-3 NaOH(aq), and 21.7 cm3 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) H2O(l) ZO2(s) HCl(aq) A g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask. 25.0 cm3 of this solution was titrated against mol dm-3 NaOH(aq), and 21.7 cm3 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) H2O(l) ZO2(s) HCl(aq) A g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask. 25.0 cm3 of this solution was titrated against mol dm-3 NaOH(aq), and 21.7 cm3 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) H2O(l) ZO2(s) HCl(aq) A g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask. 25.0 cm3 of this solution was titrated against mol dm-3 NaOH(aq), and 21.7 cm3 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) H2O(l) ZO2(s) HCl(aq) A g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask. 25.0 cm3 of this solution was titrated against mol dm-3 NaOH(aq), and 21.7 cm3 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) H2O(l) ZO2(s) HCl(aq) A g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask. 25.0 cm3 of this solution was titrated against mol dm-3 NaOH(aq), and 21.7 cm3 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) H2O(l) ZO2(s) HCl(aq) A g sample of ZCl4 was added to water. The solid ZO2 was removed by filtration and the resulting solution was made up to 250 cm3 in a volumetric flask. 25.0 cm3 of this solution was titrated against mol dm-3 NaOH(aq), and 21.7 cm3 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.
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Example 8: Moles NaOH – moles HCl
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 = 21.7 10-3 Moles NaOH(aq) = volume concentration = 21.7 10-3 0.112 = 2.43 10-3 mol Moles NaOH(aq) = volume concentration = 21.7 10-3 0.112 Moles NaOH(aq) = volume concentration = 21.7 Moles NaOH(aq) = volume concentration Moles HCl(25) = mole ratio moles of NaOH = 1 2.43 10-3 = 2.43 10-3 Moles HCl(250) = 10-3 250/25 Moles HCl(25) = mole ratio moles of NaOH = 1 2.43 10-3 = 2.43 10-3 Moles HCl(250) = 10-3 250/25 = mol 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 = 10-3 Moles HCl(25) = mole ratio moles of NaOH Moles HCl(25) = mole ratio moles of NaOH = 1 2.43 10-3
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Example 8: moles / Mr ZCl4 - Ar & identity ‘Z’
LESS MORE ZCl4(l) H2O(l) ZO2(s) HCl(aq) ZCl4(l) H2O(l) ZO2(s) HCl(aq) Find HCl : ZCl4 mole ratio and hence find moles ZCl4 Moles ZCl4 = Mole Ratio Moles HCl = Moles ZCl4 = Mole Ratio Moles HCl = / Moles ZCl4 = Mole Ratio Moles HCl = / = 10-3 Moles ZCl4 = Mole Ratio Moles HCl = 1/4 or 4/1 Moles ZCl4 = Mole Ratio Moles HCl Find Mr of ZCl4 – then Ar of Z and hence identify Z Mr of ZCl4 = mass = = 214.6 moles 10-3 Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = Mr of ZCl4 = mass = = 214.6 moles 10-3 Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = = 72.6 Mr of ZCl4 = mass = (given in q) = 214.6 moles 10-3 Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = = 72.6 Element Z = Ge So, ZCl = GeCl4 Mr of ZCl4 = mass = = 214.6 moles 10-3 Ar of Z = Mr (ZCl4) – 4 Ar (Cl) Mr of ZCl4 = mass = = 214.6 moles 10-3 Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = = 72.6 Element Z = Ge Mr of ZCl4 = mass = = 214.6 moles 10-3 Mr of ZCl4 = mass = moles Mr of ZCl4 = mass = moles 10-3 Mr of ZCl4 = mass = moles Mr of ZCl4 = mass = = 214.6 moles 10-3 Ar of Z = Mr (ZCl4)
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Example 9: % atom economy
New addition to chem1 % atom economy = mass of desired product x total mass of reactants
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Example 9: % atom economy
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 = CH2Cl2 Mr =( ) = 85 Reactants =CH4+2Cl2 Mr = (12+4)+(2 x71)= 158 %atom economy = 85 x 100 = 53.8% 158
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