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

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Presentation on theme: "Mole Calculations West Midlands Chemistry Teachers Centre Tuesday 23rd November 2010 Presenter: Dr Janice Perkins."— Presentation transcript:

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

2 Fe 2 O 3 (s) + 3CO(g) 2Fe(s) + 3CO 2 (g) Reacting ratio in equations 1 formula3 molecules 2 atoms 3 molecules 10 formulae30 molecules20 atoms30 molecules formulae molecules atoms molecules 1dozen formulae 3 dozen molecules 2 dozen atoms 3 dozen molecules

3 Funny numbers Dozen = 12 Gross = = Mole =Score =20 Thats

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

5 Fe 2 O 3 (s) + 3CO(g) 2Fe(s) + 3CO 2 (g) Reacting Ratio 1 formula3 molecules 2 atoms 3 molecules 10 formulae30 molecules20 atoms30 molecules formulae molecules atoms molecules 1 mole 3 moles 2 moles3 moles formulae molecules atoms molecules

6 3:2 Fe 2 O 3 : Fe = 1:2 Fe 2 O 3 (s) + 3CO(g) 2Fe(s) + 3CO 2 (g) 1:2 Mole Ratio (Reacting Ratio) CO : Fe = 3:2 CO : CO 2 = 1:1 3:3 Fe 2 O 3 : CO = 1:3 1:3

7 Sufficient data to calculate moles? Calculation based on Data needed Mass Solution Gases Mass Mass andM r MassMass andM r orFormulaMassMass andM r orFormula or Name SolutionVolumeSolutionVolume andConcentration GasesP P and TGasesP and T and VGasesP and T and V and R

8 Mole ratio from equation Moles of known substance Moles of unknown substance Mass Volume of solution Volume of gas Mass M r Mole Calculations Moles = Mass M r n = m M r n = v c n = pV RT Moles = volume conc Moles = P(in Pa) V(in m 3) R T(in Kelvin) Ideal Gas Equation pV = nRT

9 Mass = Moles M r Mass M r = Mass Moles Rearranging the formula Moles = Mass M r MolesMrMr Mass MolesMrMr Mass MolesMrMr

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

11 Vol = Moles Conc Moles Rearranging the formula Moles = Volume Concentration VolConc Conc = Moles Vol Moles VolConc Moles VolConc

12 Mole ratio from equation Moles of known substance Moles of unknown substance Mass Volume of solution Volume of gas Mass M r Vol Conc VPTVPT Mole Calculations n = m M r n = v c n = pV RT m = n M r M r = m n c = n v v = n c

13 Volume = nRT p Units are vital: V always in m 3 P always in Pa T always in Kelvin Rearranging the formula Moles = pV RT Pressure = nRT V Temperature = pV nR

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

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

16 First we need to calculate the number of moles of HCl. Moles = mass/M r Moles HCl = 19.6 Conc of HCl(aq)= moles = volume (in dm 3 ) = 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 cm 3. Conc of HCl(aq)= moles =0.537 volume (in dm 3 ) = 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 cm 3. Moles HCl = Moles HCl = 19.6 = mol 36.5 Conc of HCl(aq)= moles =0.537 volume (in dm 3 ) = Conc of HCl(aq)= moles =0.537 volume (in dm 3 ) = 2.15 mol dm -3

17 A metal carbonate MCO 3 reacts with HCl as in the following equation. MCO 3 + 2HCl MCl 2 + H 2 O + CO 2 A g sample of MCO 3 reacted completely with 30.7 cm 3 of mol dm -3 HCl. (a)Calculate the amount, in moles, of HCl which reacted with the g of MCO 3 Moles = vol x conc (in dm 3 ) = 30.7/1000 x Example 3: Calculating moles of solution and solid, then M r and A r (Jan 09 chem1) = mol

18 (b) Calculate the amount, in moles, of MCO 3 in g Look at equation again MCO 3 + 2HCl MCl 2 + H 2 O + CO 2 There is a 2:1 ratio of reactants. We have calculated that there are mol of HCl so there must be half that amount of MCO 3 = /2 = mol = 6.50 x mol

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

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

21 (b) Calculate the moles of the other substance- in this case M 2 CO 3 The mole ratio is 2:1 We have calculated the moles HCl = mol So the moles M 2 CO 3 = /2 = 2.85 x mol M 2 CO 3 + 2HCl 2MCl + CO 2 + H 2 O (c) Now find M r of M 2 CO 3 M r of M 2 CO 3 = mass = moles = M r of M 2 CO 3 = mass = moles = M r of M 2 CO 3 = mass = moles = M r of M 2 CO 3 = mass = moles =138 (d) find A r of metal M and hence deduce its identity 2 A r (M) = M r (M 2 CO 3 ) - M r (CO 3 ) = 138 – 60 = 78 A r (M) = 78/2=39 M = Potassium

22 Use pV = nRT V = nRT = p = Use pV = nRT V = nRT = p = Use pV = nRT V = nRT = p = = Example 5: Gases – calculating the volume A sample of ethanol vapour, C 2 H 5 OH (M r = 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 cm 3, 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 A sample of ethanol vapour, C 2 H 5 OH (M r = 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 cm 3, 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 = Moles ethanol = 1.36 = n = mol 46 Use pV = nRT V = nRT = p = m 3 = Use pV = nRT V = nRT = p = m 3 = Use pV = nRT V = nRT = p = m 3 = = 899 cm 3

23 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 mass A r = moles Ratio 2.17/2.17 = /2.17= 2 Empirical formula = NO 2 Example 6: Empirical formula Jun 09

24 M r ( Na 2 CO 3 xH 2 O)= 250 Example 7: Water of crystallisation The M r 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, Na 2 CO 3 xH 2 O. M r ( Na 2 CO 3 xH 2 O)= 250 M r (Na 2 CO 3 ) = 106 M r ( Na 2 CO 3 xH 2 O)= 250 M r (Na 2 CO 3 ) = 106 x M r (H 2 O) = M r ( Na 2 CO 3 xH 2 O) M r ( Na 2 CO 3 xH 2 O)= 250 M r (Na 2 CO 3 ) = 106 x M r (H 2 O) = M r ( Na 2 CO 3 xH 2 O) - M r (Na 2 CO 3 ) M r ( Na 2 CO 3 xH 2 O)= 250 M r (Na 2 CO 3 ) = 106 x M r (H 2 O) = M r ( Na 2 CO 3 xH 2 O) - M r (Na 2 CO 3 ) = 250 – 106 M r ( Na 2 CO 3 xH 2 O)= 250 M r (Na 2 CO 3 ) = 106 x M r (H 2 O) = M r ( Na 2 CO 3 xH 2 O) - M r (Na 2 CO 3 ) = 250 – 106 = 144 M r ( Na 2 CO 3 xH 2 O)= 250 M r (Na 2 CO 3 ) = 106 x M r (H 2 O) = M r ( Na 2 CO 3 xH 2 O) - M r (Na 2 CO 3 ) = 250 – 106 = 144 x 18 = 144 M r ( Na 2 CO 3 xH 2 O)= 250 M r (Na 2 CO 3 ) = 106 x M r (H 2 O) = M r ( Na 2 CO 3 xH 2 O) - M r (Na 2 CO 3 ) = 250 – 106 = 144 x 18 = 144 x= 144/18 M r ( Na 2 CO 3 xH 2 O)= 250 M r (Na 2 CO 3 ) = 106 x M r (H 2 O) = M r ( Na 2 CO 3 xH 2 O) - M r (Na 2 CO 3 ) = 250 – 106 = 144 x 18 = 144 x= 144/18= 8 M r ( Na 2 CO 3 xH 2 O)= 250 M r (Na 2 CO 3 ) = 106 x M r (H 2 O) = M r ( Na 2 CO 3 xH 2 O) - M r (Na 2 CO 3 ) = 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 Na 2 CO 3 xH 2 O. A 3.01 g sample of one of these hydrates was dissolved in water and the solution made up to 250 cm 3. In titration, a 25 cm 3 portion of this solution required 24.3 cm 3 of mol dm -3 hydrochloric acid for complete reaction Na 2 CO 3 + 2HCl 2NaCl + H 2 O + CO 2 Calculate the Mr of Na 2 CO 3 xH 2 O Sodium carbonate forms a number of hydrates of general formula Na 2 CO 3 xH 2 O. A 3.01 g sample of one of these hydrates was dissolved in water and the solution made up to 250 cm 3. In titration, a 25 cm 3 portion of this solution required 24.3 cm 3 of mol dm -3 hydrochloric acid for complete reaction Na 2 CO 3 + 2HCl 2NaCl + H 2 O + CO 2 Calculate the Mr of Na 2 CO 3 xH 2 O (a)Calculate moles HCl since you can do this Moles HCl =volume concentration = = mol

26 (b) Now calculate the moles of Na 2 CO 3 in the 25 cm 3 sample of solution. Na 2 CO 3 + 2HCl 2NaCl + H 2 O + CO 2 Moles Na 2 CO 3 =Mole Ratio Moles HCl (c) deduce the moles of Na 2 CO 3 in 250 cm 3 of solution Moles Na 2 CO 3 (250)=Moles Na 2 CO 3 (25) Moles Na 2 CO 3 (250)=Moles Na 2 CO 3 (25) Moles Na 2 CO 3 (250)=Moles Na 2 CO 3 (25) = Moles Na 2 CO 3 =Mole Ratio Moles HCl = 1/ Moles Na 2 CO 3 =Mole Ratio Moles HCl = ½ = (d) Calculate the M r of hydrated Na 2 CO 3 [Mass = 3.01 g] M r = Mass / Moles M r = 3.01/ = 124

27 (a) Calculate the amount, in moles, of TiCl 4 in 165 g Moles = mass = 165 M r = Example 9: Percentage yield Jan 09 (b) Calculate the maximum amount, in moles, of TiO 2 which can be formed in this experiment. Look at the equation. 1 mole of TiCl 4 gives 1 mole of TiO 2 So moles of TiCl 4 gives moles of TiO 2 In an experiment 165 g TiCl 4 were added to an excess of water. The equation for the reaction is as follows TiCl 4 + 2H 2 O TiO 2 + 4HCl

28 (c)Calculate the maximum mass of TiO 2 formed in this experiment. mass = M r x moles = 79.9 x = 69.4 g (d) In this experiment only 63.0 g of TiO 2 were produced. Calculate the percentage yield. Percentage yield = actual (experimental) mass of product theoretical ( calculated) mass of product = = 90.8%

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

30 Calculate the % atom economy for the formation of CH 2 Cl 2 in this reaction. CH 4 + 2Cl 2 CH 2 Cl 2 +2HCl There are no numbers given You only need the M r values Desired product = CH 2 Cl 2 Mr =( ) = 85 Reactants =CH 4 +2Cl 2 Mr = (12+4)+(2 x71)= 158 %atom economy = 85 x 100 = 53.8% 158

31 The chloride of an element Z reacts with water according to the following equation. ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) A g sample of ZCl 4 was added to water. The solid ZO 2 was removed by filtration and the resulting solution was made up to 250 cm 3 in a volumetric flask cm 3 of this solution was titrated against mol dm -3 NaOH(aq), and 21.7 cm 3 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 ZCl 4 present in the sample. Calculate the M r of ZCl 4. From your answer deduce the A r of element Z and hence its identity. Example 11: Moles HCl, moles & identity Z The chloride of an element Z reacts with water according to the following equation. ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) A g sample of ZCl 4 was added to water. The solid ZO 2 was removed by filtration and the resulting solution was made up to 250 cm 3 in a volumetric flask cm 3 of this solution was titrated against mol dm -3 NaOH(aq), and 21.7 cm 3 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 ZCl 4 present in the sample. Calculate the M r of ZCl 4. From your answer deduce the A r of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) A g sample of ZCl 4 was added to water. The solid ZO 2 was removed by filtration and the resulting solution was made up to 250 cm 3 in a volumetric flask cm 3 of this solution was titrated against mol dm -3 NaOH(aq), and 21.7 cm 3 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 ZCl 4 present in the sample. Calculate the M r of ZCl 4. From your answer deduce the A r of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) A g sample of ZCl 4 was added to water. The solid ZO 2 was removed by filtration and the resulting solution was made up to 250 cm 3 in a volumetric flask cm 3 of this solution was titrated against mol dm -3 NaOH(aq), and 21.7 cm 3 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 ZCl 4 present in the sample. Calculate the M r of ZCl 4. From your answer deduce the A r of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) A g sample of ZCl 4 was added to water. The solid ZO 2 was removed by filtration and the resulting solution was made up to 250 cm 3 in a volumetric flask cm 3 of this solution was titrated against mol dm -3 NaOH(aq), and 21.7 cm 3 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 ZCl 4 present in the sample. Calculate the M r of ZCl 4. From your answer deduce the A r of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) A g sample of ZCl 4 was added to water. The solid ZO 2 was removed by filtration and the resulting solution was made up to 250 cm 3 in a volumetric flask cm 3 of this solution was titrated against mol dm -3 NaOH(aq), and 21.7 cm 3 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 ZCl 4 present in the sample. Calculate the M r of ZCl 4. From your answer deduce the A r of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) A g sample of ZCl 4 was added to water. The solid ZO 2 was removed by filtration and the resulting solution was made up to 250 cm 3 in a volumetric flask cm 3 of this solution was titrated against mol dm -3 NaOH(aq), and 21.7 cm 3 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 ZCl 4 present in the sample. Calculate the M r of ZCl 4. From your answer deduce the A r of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) A g sample of ZCl 4 was added to water. The solid ZO 2 was removed by filtration and the resulting solution was made up to 250 cm 3 in a volumetric flask cm 3 of this solution was titrated against mol dm -3 NaOH(aq), and 21.7 cm 3 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 ZCl 4 present in the sample. Calculate the M r of ZCl 4. From your answer deduce the A r of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) A g sample of ZCl 4 was added to water. The solid ZO 2 was removed by filtration and the resulting solution was made up to 250 cm 3 in a volumetric flask cm 3 of this solution was titrated against mol dm -3 NaOH(aq), and 21.7 cm 3 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 ZCl 4 present in the sample. Calculate the M r of ZCl 4. From your answer deduce the A r of element Z and hence its identity. The chloride of an element Z reacts with water according to the following equation. ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) A g sample of ZCl 4 was added to water. The solid ZO 2 was removed by filtration and the resulting solution was made up to 250 cm 3 in a volumetric flask cm 3 of this solution was titrated against mol dm -3 NaOH(aq), and 21.7 cm 3 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 ZCl 4 present in the sample. Calculate the M r of ZCl 4. From your answer deduce the A r 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 cm 3 We scale this to give the number of moles of HCl, in 250 cm 3, formed in the reaction Moles HCl(25)=mole ratio moles of NaOH Moles NaOH(aq) =volume concentration =21.7 Moles NaOH(aq) =volume concentration = Moles NaOH(aq) =volume concentration = Moles NaOH(aq) =volume concentration = = mol Moles HCl(25)=mole ratio moles of NaOH = Moles HCl(25)=mole ratio moles of NaOH = Moles HCl(25)=mole ratio moles of NaOH = = Moles HCl(25)=mole ratio moles of NaOH = = Moles HCl(250)= /25 Moles HCl(25)=mole ratio moles of NaOH = = Moles HCl(250)= /25 = mol

33 Find HCl : ZCl 4 mole ratio and hence find moles ZCl 4 ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) Find M r of ZCl 4 – then A r of Z and hence identify Z Moles ZCl 4 =Mole Ratio Moles HCl M r of ZCl 4 = mass = moles ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) Moles ZCl 4 =Mole Ratio Moles HCl = Moles ZCl 4 =Mole Ratio Moles HCl = 1/4 or 4/ Moles ZCl 4 =Mole Ratio Moles HCl = 1/ Moles ZCl 4 =Mole Ratio Moles HCl = 1/ = M r of ZCl 4 = mass = moles M r of ZCl 4 = mass = moles M r of ZCl 4 = mass = =214.6 moles M r of ZCl 4 = mass = =214.6 moles A r of Z = M r (ZCl 4 ) M r of ZCl 4 = mass = =214.6 moles A r of Z = M r (ZCl 4 ) – 4 A r (Cl) M r of ZCl 4 = mass = =214.6 moles A r of Z = M r (ZCl 4 ) - 4 A r (Cl) = M r of ZCl 4 = mass = =214.6 moles A r of Z = M r (ZCl 4 ) - 4 A r (Cl) = = 72.6 M r of ZCl 4 = mass = =214.6 moles A r of Z = M r (ZCl 4 ) - 4 A r (Cl) = = 72.6 Element Z =Ge M r of ZCl 4 = mass = (given in q) =214.6 moles A r of Z = M r (ZCl 4 ) - 4 A r (Cl) = = 72.6 Element Z =GeSo, ZCl 4 =GeCl 4

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