# West Midlands Chemistry Teachers Centre Tuesday 23rd November 2010

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

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 1106 ‘formulae’ 3106 molecules 2106 atoms 1dozen ‘formulae’ 3 dozen molecules 2 dozen atoms

Funny numbers Dozen = 12 Gross = 12  12 = 144 Score = 20 Mole =  1023 That’s

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!

Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
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’ 3106 molecules 2106 atoms 3106 molecules 6.021023 ‘formulae’ 18.06 1023 molecules 12.04 1023 atoms 18.06 1023 molecules 1 mole 3 moles 2 moles 3 moles

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

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

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

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

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

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

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

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

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 V = nRT p Volume of gas p = nRT V n = pV RT T = pV nR

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

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 = = mol 36.5 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 = volume (in dm3) = Conc of HCl(aq) = moles = 0.537 volume (in dm3) =

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

(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 mol of HCl so there must be half that amount of MCO3 = /2 = mol = 6.50 x 10-3 mol

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

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.  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. 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.  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. 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.  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. 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 = 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 = 21.7  10-3 = Moles HCl(aq) = volume  concentration = Moles HCl(aq) = volume  concentration = 21.7 =

M2CO HCl  2MCl CO 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  10-3 = Mr of M2CO3 = mass = moles  10-3 = 138 Mr of M2CO3 = mass = moles = Mr of M2CO3 = mass = moles = (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

Example 5: Gases – calculating the volume
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

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 mass Ar = moles Ratio /2.17 = /2.17= 2 Empirical formula = NO2

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 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 = 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

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 mol dm-3 hydrochloric acid for complete reaction  Na2CO HCl  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 cm3 of mol dm-3 hydrochloric acid for complete reaction  Na2CO HCl  2NaCl + H2O + CO2 Calculate the Mr of Na2CO3 • xH2O Calculate moles HCl since you can do this Moles HCl = volume  concentration =  10-3  0.200 =  10-3 mol

(b) Now calculate the moles of Na2CO3 in the 25 cm3 sample of solution.
Na2CO HCl  2NaCl + H2O + CO2 Moles Na2CO3 = Mole Ratio  Moles HCl = ½   10-3 = 2.43  10-3 Moles Na2CO3 = Mole Ratio  Moles HCl Moles Na2CO3 = Mole Ratio  Moles HCl = /   10-3 (c) deduce the moles of Na2CO3 in 250 cm3 of solution Moles Na2CO3 (250) = Moles Na2CO3 (25)  250 25 Moles Na2CO3 (250) = Moles Na2CO3 (25)  250 25 = 2.43  10-2 Moles Na2CO3 (250) = Moles Na2CO3 (25)  (d) Calculate the Mr of hydrated Na2CO3 [Mass = 3.01 g] Mr = Mass / Moles Mr = /2.43  = 124

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 TiCl H2O  TiO HCl (a) Calculate the amount, in moles, of TiCl4 in 165 g Moles = mass = 165 Mr = (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 of TiO2 So moles of TiCl4 gives moles of TiO2

Calculate the maximum mass of TiO2 formed in this experiment.
mass = Mr x moles = 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%

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

Desired product = CH2Cl2 Mr =(12+ 2 +71) = 85
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

Example 11: Moles HCl, moles & identity ‘Z’

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  0.112 = 2.43  10-3 mol 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 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

ZCl4(l) + 2H2O(l)  ZO2(s) + 4HCl(aq)
Find HCl : ZCl4 mole ratio and hence find moles ZCl4 Moles ZCl4 = Mole Ratio  Moles HCl =  Moles ZCl4 = Mole Ratio  Moles HCl = 1/4 or 4/1  Moles ZCl4 = Mole Ratio  Moles HCl = /  Moles ZCl4 = Mole Ratio  Moles HCl = /  =  10-3 Moles ZCl4 = Mole Ratio  Moles HCl Find Mr of ZCl4 – then Ar of Z and hence identify Z 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 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 Ar of Z = Mr (ZCl4) – 4  Ar (Cl) 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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