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Mole Calculations
West Midlands Chemistry Teachers Centre
Tuesday 11th November 2008
Presenter: Dr Janice Perkins
Chemistry of the Blast Furnace Process
At 2000 C
At 1500 C
At 1000 C
1 atom
1 molecule 1 molecule
1 atom
1 molecule 2 molecules
1 ‘formula’ 3 molecules 2 atoms 3 molecules
C(s) + O2(g) CO2(g)
C(s) + CO2(g) 2CO(g)
Fe2O3(s) + 3CO(g) 2Fe + 3CO2(g)
Fe2O3(s) + 3CO(g) 2Fe + 3CO2(g)
Reacting Ratio
1 ‘formula’ 3 molecules 2 atoms 3 molecules
10 ‘formulae’ 30 molecules 20 atoms 30 molecules
1106 ‘formulae’
3106 molecules
2106 atoms
3106 molecules
1dozen ‘formulae’
3 dozen molecules
2 dozen atoms
3 dozen molecules
That’s
Funny numbers
Dozen = 12 Gross = 12 12 = 144
Ton = 100
Score = 20
Monkey = £500
Mole = 6.023 1023
602300000000000000000000
602300000000000000000000
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 + 3CO2(g)
Reacting Ratio
1 formula 3 molecules 2 atoms 3 molecules
10 ‘formulae’ 30 molecules 20 atoms 30 molecules
1106 ‘formulae’
3106 molecules
2106 atoms
3106 molecules
1 mole 3 moles 2 moles 3 moles
6.021023 ‘formulae’
18.06 1023
molecules12.04 1023
atoms18.06 1023 molecules
3:2
Fe2O3 : Fe = 1:2
Fe2O3(s) + 3CO(g) 2Fe + 3CO2(g)
1:2
Mole Ratio (Reacting Ratio)
CO : Fe = 3:2
CO : CO2 = 1:1
3:3
Fe2O3 : CO = 1:3
1:3
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
Sufficient data to calculate moles??
Calculation based on
Data needed
Mass
Solution
Gases
Mass Mass Mass Mass and Mr Mass Mass and Mr or Formula Mass Mass and Mr or Formula or Name
Solution Volume Solution Volume and Concentration
Gases PGases P and T Gases P and T and VGases P and T and V and R
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.
Mole ratio
from equation
Moles of known
substance
Moles of unknown substance
Mass
Volume of solution
Volume of gas
Mass
Mr
Mole Calculations
Moles = MassMr
n = mMr
n = v c
n = pVRT
Moles = volume concMoles = P(in Pa) V(in m3)
R T(in Kelvin)Ideal Gas Equation pV = nRT
Mass = Moles Mr
Mass
Mr = MassMoles
Rearranging the formula
Moles = MassMr
Moles Mr
Mass
Moles Mr
Mass
Moles Mr
Mole ratio
from equation
Moles of known
substance
Moles of unknown substance
Mass
Volume of solution
Volume of gas
Mass
Mr
Vol
Conc
Mole Calculationsn = m
Mr
n = v c
n = pVRT
m = n Mr
Mr = m n
Vol = MolesConc
Moles
Rearranging the formula
Moles = Volume Concentration
Vol Conc
Conc = Moles Vol
Moles
Vol Conc
Moles
Vol Conc
Mole ratio
from equation
Moles of known
substance
Moles of unknown substance
Mass
Volume of solution
Volume of gas
Mass
Mr
Vol
Conc
V
P
T
Mole Calculationsn = m
Mr
n = v c
n = pVRT
m = n Mr
Mr = m n
c = n
v
v = n c
Volume = nRTp
Units are vital:
‘V’ always in m3
‘P’ always in Pa
‘T’ always in Kelvin
Rearranging the formula
Moles = pVRT
Pressure = nRTV
Temperature = pVnR
Mole ratio
from equation
Moles of known
substance
Moles of unknown substance
Mass
Volume of solution
Volume of gas
Mass
Mr
Vol
Conc
V
P
T
Mole Calculationsn = m
Mr
n = v c
n = pVRT
m = n Mr
Mr = m n
c = n
v
v = n c
V = nRTp
p = nRTV
T = pVnR
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
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 calculationExample 7: Water of crystallisation
Example 8: Moles HCl, moles / identity ‘Z’
Example 9: % atom economy
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.
NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl
2NaHCO3 Na2CO3 + H2O + CO2
Example 1: Reacting masses - Na2CO3
Tackle the 2 stages separately - less likely to make mistakes
Stage 1 NaCl : NaHCO3 =
NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl
2NaHCO3 Na2CO3 + H2O + CO2
Moles NaCl = mass Mr
Moles NaHCO3 = Moles NaCl =
Stage 1 NaCl : NaHCO3 = 1:1
Moles NaCl = mass = 800 Mr 58.5
Moles NaCl = mass = 800 = 13.68 mol Mr 58.5
Moles NaHCO3 = 1 Moles NaCl =Moles NaHCO3 = 1 Moles NaCl = 13.68 mol
Moles Na2CO3 = Mole Ratio Moles NaHCO3
=
Stage 2 NaHCO3 : Na2CO3 =
Mass Na2CO3 = moles Mr =
=
Now use the number of moles of NaHCO3 to calculate the moles and mass of Na2CO3
NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl
2NaHCO3 Na2CO3 + H2O + CO2
NaCl + NH3 + CO2 + H2O NaHCO3 + NH4Cl
2NaHCO3 Na2CO3 + H2O + CO2
Mass Na2CO3 = moles Mr = 6.84 106
=
Mass Na2CO3 = moles Mr = 6.84 106
= 725 g
Example 1: Reacting masses - Na2CO3
MORE LESS
Stage 2 NaHCO3 : Na2CO3 = 2:1
Moles Na2CO3 = 1/2 or 2/1 Moles NaHCO3
=
Moles Na2CO3 = 1/2 Moles NaHCO3
=
Moles Na2CO3 = 1/2 Moles NaHCO3
= ½ 13.68
Moles Na2CO3 = 1/2 Moles NaHCO3
= ½ 13.68 = 6.84 mol
This question only deals with HCl, so we first need to calculate the number of moles of HCl
Moles HCl = 19.6
Conc of HCl(aq) = moles =volume (in dm3)
=
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.
Conc of HCl(aq) = moles = 0.537volume (in dm3)
=
Example 2: Making hydrochloric acidExample 2: Making hydrochloric acidWe don’t always need to use the Mole Ratio
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.
Moles HCl = 19.6 36.5
Moles HCl = 19.6 = 0.537 mol36.5
Conc of HCl(aq) = moles = 0.537volume (in dm3) 250 10-3
=
Conc of HCl(aq) = moles = 0.537volume (in dm3) 250 10-3
= 2.15 mol dm-3
Use pV = nRTUse 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
=
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)
Moles ethanol = 1.36
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 = 1.3646
Moles ethanol = 1.36 = n = 0.0296 mol46
Use pV = nRT
V = nRT = 0.0296 8.31 366 p 100000
= 8.992 10-4 m3
=
Use pV = nRT
V = nRT = 0.0296 8.31 366 p 100000
= 8.992 10-4 m3
= 8.992 10-4 106
Use pV = nRT
V = nRT = 0.0296 8.31 366 p 100000
= 8.992 10-4 m3
= 8.992 10-4 106 = 899 cm3
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.
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. 1 - identify known substance, find data and find moles
Moles HCl(aq) = volume concentration==
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.
Moles HCl(aq) = volume concentration= 21.7=
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.
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
2 - find mole ratio and find moles unknown substance
Example 4: Mass M2CO3 & identity ‘M’ M2CO3 + 2HCl 2MCl + CO2 + H2O M2CO3 + 2HCl 2MCl + CO2 + H2O
Moles M2CO3 = Mole Ratio Moles HCl
==
Moles M2CO3 = Mole Ratio Moles HCl
= 5.71 10-3 =
3 - find Mr of M2CO3
Mr of M2CO3 = mass = moles
=
Mr of M2CO3 = mass = 0.394 moles
=
MORELESS
Moles M2CO3 = Mole Ratio Moles HCl
= 1/2 or 2/1 5.71 10-3 =
Moles M2CO3 = Mole Ratio Moles HCl
= 1/2 5.71 10-3 =
Moles M2CO3 = Mole Ratio Moles HCl
= ½ 5.71 10-3 = 2.85 10-3
Mr of M2CO3 = mass = 0.394 moles 2.85 10-3
=
Mr of M2CO3 = mass = 0.394 moles 2.85 10-3
= 138
A similar approach can be used in calculating the number of molecules of water of crystallisation in CuSO4•xH2O.
4 - find Ar of metal M and hence deduce its identity
Example 4: Mass M2CO3 & identity ‘M’
2 Ar(M) =2 Ar(M) = Mr(M2CO3) - Mr(‘CO3’)2 Ar(M) = Mr(M2CO3) - Mr(‘CO3’)
= 138 – 60
• 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. • Calculate the Mr of the hydrated salt [Mr(hydr) = 249.6]• Deduct from this the Mr of the salt. [= 249.6 – 159.6 = 90]• Calculate the Mr of the hydrated salt [Mr(hydr) = 249.6]• Deduct from this the Mr of the salt. [= 249.6 – 159.6 = 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. [= 249.6 – 159.6 = 90]• Divide what is left by the Mr of water (18) to get ‘x’.
x = 90 18 = 5
• Calculate the Mr of the hydrated salt [Mr(hydr) = 249.6]• Deduct from this the Mr of the salt. [= 249.6 – 159.6 = 90]• Divide what is left by the Mr of water (18) to get ‘x’.
x = 90 18 = 5 So, formula = CuSO4.5H2O
2 Ar(M) = Mr(M2CO3) - Mr(‘CO3’)
= 138 – 60 = 78
2 Ar(M) = Mr(M2CO3) - Mr(‘CO3’)
= 138 – 60 = 78
Ar(M) = 78/2
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 = 39
M = Potassium (from Periodic Table)
Example 5 – back titration and % purity
A 1.00 g sample of limestone is reacted with 100 cm3 of 0.200 mol dm-3 hydrochloric acid as shown below.
CaCO3(s) + 2HCl(aq) CaCl2 + H2O + CO2
The excess acid required 24.8 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralisation. Calculate the percentage of calcium carbonate in the limestone.
Original moles HCl(aq) = volume concentration==
Original moles HCl(aq) = volume concentration= 100=
A 1.00 g sample of limestone is reacted with 100 cm3 of 0.200 mol dm-3 hydrochloric acid as shown below.
CaCO3(s) + 2HCl(aq) CaCl2 + H2O + CO2
The excess acid required 24.8 cm3 of 0.100 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
=
Original moles HCl(aq) = volume concentration= 100 10-3 0.200=
Original moles HCl(aq) = volume concentration= 100 10-3 0.200= 2.00 10-2 mol
Example 5 – back titration and % purityThe excess acid required 24.8 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralisation.
Excess moles of HCl(aq) i.e. neutralised by NaOH(aq)=
Moles of HCl(aq) reacted with CaCO3(s)
The excess acid required 24.8 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralisation.
Moles NaOH(aq) = volume concentrationMoles NaOH(aq) = volume concentration= 24.8 10-3
Moles of HCl(aq) reacted with CaCO3(s) = initial moles HCl
Moles NaOH(aq) = volume concentration= 24.8 10-3 0.100
Moles NaOH(aq) = volume concentration= 24.8 10-3 0.100= 2.48 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 NaOH(aq) = 2.48 10-3 mol
Moles of HCl(aq) reacted with CaCO3(s) = initial moles HCl - excess moles HCl
Moles of HCl(aq) reacted with CaCO3(s) = initial moles HCl - excess moles HCl= 2.00 10-2
Moles of HCl(aq) reacted with CaCO3(s) = initial moles HCl - excess moles HCl= 2.00 10-2 - 2.48 10-3
Moles of HCl(aq) reacted with CaCO3(s) = initial moles HCl - excess moles HCl= 2.00 10-2 - 2.48 10-3 = 1.752 10-2 mol
Example 5 – back titration and % purity
Remember, 1.752 10-2 mol of HCl reacted with the calcium carbonate in the impure limestone
CaCO3(s) + 2HCl(aq) CaCl2 + H2O + CO2
MORELESS
Moles CaCO3 = Mole Ratio Moles HClMoles CaCO3 = Mole Ratio Moles HCl
= 1.752 10-2
Moles CaCO3 = Mole Ratio Moles HCl
= 1/2 or 2/1 1.752 10-2
Moles CaCO3 = Mole Ratio Moles HCl
= 1/2 1.752 10-2
Moles CaCO3 = Mole Ratio Moles HCl
= ½ 1.752 10-2 = 8.76 10-3 mol
% purity CaCO3 =
Example 5 – back titration and % purity
Mass of CaCO3 = moles Mr
==
A 1.00 g sample of limestone is reacted with 100 cm3 of 0.200 mol dm-3 hydrochloric acid as shown below.
Mass of CaCO3 = moles Mr
= 8.76 10-3 =
A 1.00 g sample of limestone is reacted with 100 cm3 of 0.200 mol dm-3 hydrochloric acid as shown below.
Mass of CaCO3 = moles Mr
= 8.76 10-3 100=
Mass of CaCO3 = moles Mr
= 8.76 10-3 100= 0.876 g
% purity CaCO3 = mass pure CaCO3% purity CaCO3 = mass pure CaCO3
mass of limestone% purity CaCO3 = mass pure CaCO3 100
mass of limestone% purity CaCO3 = mass pure CaCO3 100
mass of limestone
= 0.876 100
% purity CaCO3 = mass pure CaCO3 100 mass of limestone
= 0.876 1001.00
% purity CaCO3 = mass pure CaCO3 100 mass of limestone
= 0.876 1001.00
= 87.6 %
First, we need to use the data in the question to calculate the number of moles of gas formed.
Use pV = nRT
A sample of ammonium nitrate decomposed on heating as shown in the equation below.
NH4NO3 2H2O + N2 + ½O2
On cooling the resulting gases to 298 K, the volume of nitrogen and oxygen together was found to be 0.0500 m3 at a pressure of 95.0 kPa.
Example 6 – An extended gas law calculation
Use pV = nRT
n = pVRT
A sample of ammonium nitrate decomposed on heating as shown in the equation below.
NH4NO3 2H2O + N2 + ½O2
On cooling the resulting gases to 298 K, the volume of nitrogen and oxygen together was found to be 0.0500 m3 at a pressure of 95.0 kPa.
Use pV = nRT
n = pV = 95.0RT
Use pV = nRT
n = pV = 95.0 1000RT
Use pV = nRT
n = pV = 95.0 1000 0.0500 RT
Use pV = nRT
n = pV = 95.0 1000 0.0500 RT 8.31 298
Use pV = nRT
n = pV = 95.0 1000 0.0500 RT 8.31 298
= 1.92 mol
• 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 + N2 + ½O2
Example 6 – An extended gas law calculation
LESS MORE
1:1½ or 2:31 mole 1 mole ½ mole
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 1.92 =
Moles NH4NO3 = Mole Ratio Moles gas
= 2/3 1.92
Moles NH4NO3 = Mole Ratio Moles gas
= 2/3 1.92 = 1.28 mol
Moles NH4NO3 = Mole Ratio Moles gas
= 2/3 1.92 = 1.28 mol Mass NH4NO3 = moles Mr
Moles NH4NO3 = Mole Ratio Moles gas
= 2/3 1.92 = 1.28 mol Mass NH4NO3 = moles Mr = 1.28 80
Moles NH4NO3 = Mole Ratio Moles gas
= 2/3 1.92 = 1.28 mol Mass NH4NO3 = moles Mr = 1.28 80 =102 g
Moles HCl(aq) = volume concentration
Example 7: Water of crystallisationSodium 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
1 – calculate the number of moles of HCl used.
Moles HCl(aq) = volume concentration= 24.3
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
Moles HCl(aq) = volume concentration= 24.3 10-3
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
Example 7: Water of crystallisation
2 – from the moles of HCl used, deduce the moles of Na2CO3 in the 25 cm3 sample of solution
MORELESSNa2CO3 + 2HCl 2NaCl + H2O + CO2
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
Moles Na2CO3 = Mole Ratio Moles HCl
= 4.86 10-3
Moles Na2CO3 = Mole Ratio Moles HCl
= 1/2 or 2/1 4.86 10-3
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
Example 7: Water of crystallisation
4 – Calculate the Mr of hydrated Na2CO3
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’.
Moles = Mass Mr = Mass Mr Moles
Moles = Mass Mr = Mass Mr Moles
Mr = 3.01
4 – Calculate the Mr of hydrated Na2CO3 [Mass = 3.01 g]
Moles = Mass Mr = Mass Mr Moles
Mr = 3.01/2.43 10-2
Moles = Mass Mr = Mass Mr Moles
Mr = 3.01/2.43 10-2 = 124
Mr(Na2CO3 • xH2O) = 250
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
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
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
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) = 106
x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3)
= 250 – 106 = 144
x 18 = 144
x = 144/18 = 8
Formula = Na2CO3 • 8H2O
The chloride of an element Z reacts with water according to the following equation.
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 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 0.112 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.
Example 8: 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 1.304 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 0.112 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) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 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 0.112 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) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 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 0.112 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) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 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 0.112 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) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 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 0.112 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) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 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 0.112 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) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 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 0.112 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) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 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 0.112 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) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 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 0.112 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.
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 HCl(25) = mole ratio moles of NaOH
Moles NaOH(aq) = volume concentrationMoles 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= 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
Find HCl : ZCl4 mole ratio and hence find moles ZCl4
Example 8: moles / Mr ZCl4 - Ar & identity ‘Z’
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
Find Mr of ZCl4 – then Ar of Z and hence identify Z
MORELESS
Moles ZCl4 = Mole Ratio Moles HCl
Mr of ZCl4 = mass = moles
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
Moles ZCl4 = Mole Ratio Moles HCl
= 0.0243
Moles ZCl4 = Mole Ratio Moles HCl
= 1/4 or 4/1 0.0243
Moles ZCl4 = Mole Ratio Moles HCl
= 1/4 0.0243
Moles ZCl4 = Mole Ratio Moles HCl
= 1/4 0.0243 = 6.076 10-3
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 = 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 (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 = Ge So, ZCl4 = GeCl4
Example 9: % atom economy
• New addition to chem1
% atom economy = mass of desired product x 100 total mass of reactants
Calculate the % atom economy for the formation of CH2Cl2 in this reaction.
CH4 + 2Cl2 CH2Cl2 +2HCl
Example 9: % atom economy
There are no numbers given
You only need the Mr values
Desired product = CH2Cl2 Mr =(12+ 2 +71) = 85
Reactants =CH4+2Cl2 Mr = (12+4)+(2 x71)= 158
%atom economy = 85 x 100 = 53.8% 158
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