Mole Calculations West Midlands Chemistry Teachers Centre Tuesday 11th November 2008 Presenter: Dr...

Mole Calculations

West Midlands Chemistry Teachers Centre

Tuesday 11th November 2008

Presenter: Dr Janice Perkins

Extracting Iron – The Blast Furnace

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


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


Moles = Volume Concentration

Vol Conc

Conc = Moles Vol

Moles

Vol Conc

Moles

Vol Conc

Mole ratio

from equation

Moles of known

substance


Mass

Volume of solution

Volume of gas

Mass

Mr

Vol

Conc

V

P

T


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


Moles = pVRT

Pressure = nRTV

Temperature = pVnR

Mole ratio

from equation

Moles of known

substance


Mass

Volume of solution

Volume of gas

Mass

Mr

Vol

Conc

V

P

T


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


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.












Tackle the 2 stages separately - less likely to make mistakes

Stage 1 NaCl : NaHCO3 =



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





Mass Na2CO3 = moles Mr = 6.84 106

=

Mass Na2CO3 = moles Mr = 6.84 106

= 725 g


MORE LESS

Stage 2 NaHCO3 : Na2CO3 = 2:1

Moles Na2CO3 = 1/2 or 2/1 Moles NaHCO3

=

Moles Na2CO3 = 1/2 Moles NaHCO3

=


= ½ 13.68


= 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





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.




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.




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

==


= 5.71 10-3 =

3 - find Mr of M2CO3

Mr of M2CO3 = mass = moles

=

Mr of M2CO3 = mass = 0.394 moles

=

MORELESS


= 1/2 or 2/1 5.71 10-3 =


= 1/2 5.71 10-3 =


= 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=



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


Remember, 1.752 10-2 mol of HCl reacted with the calcium carbonate in the impure limestone


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


= 1/2 1.752 10-2


= ½ 1.752 10-2 = 8.76 10-3 mol

% purity CaCO3 =


Mass of CaCO3 = moles Mr

==



= 8.76 10-3 =



= 8.76 10-3 100=


= 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 =


= 2/3 1.92


= 2/3 1.92 = 1.28 mol


= 2/3 1.92 = 1.28 mol Mass NH4NO3 = moles Mr


= 2/3 1.92 = 1.28 mol Mass NH4NO3 = moles Mr = 1.28 80


= 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


= 4.86 10-3


= 1/2 or 2/1 4.86 10-3


= 1/2 4.86 10-3


= ½ 4.86 10-3 = 2.43 10-3


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


Mr = 3.01

4 – Calculate the Mr of hydrated Na2CO3 [Mass = 3.01 g]


Mr = 3.01/2.43 10-2


Mr = 3.01/2.43 10-2 = 124

Mr(Na2CO3 • xH2O) = 250


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)


x Mr(H2O) = Mr(Na2CO3 • xH2O) - Mr(Na2CO3)



= 250 – 106



= 250 – 106 = 144



= 250 – 106 = 144

x 18 = 144



= 250 – 106 = 144

x 18 = 144

x = 144/18



= 250 – 106 = 144

x 18 = 144

x = 144/18 = 8



= 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.




















































Use this information to calculate the number of moles of HCl produced and hence the number of moles of ZCl4 present in the sample.


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(250) = 2.43 10-3 250/25


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’


Find Mr of ZCl4 – then Ar of Z and hence identify Z

MORELESS

Moles ZCl4 = Mole Ratio Moles HCl

Mr of ZCl4 = mass = moles



= 0.0243


= 1/4 or 4/1 0.0243


= 1/4 0.0243


= 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


Ar of Z = Mr (ZCl4)


Ar of Z = Mr (ZCl4) – 4 Ar (Cl)


Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = 214.6 - 142


Ar of Z = Mr (ZCl4) - 4 Ar (Cl) = 214.6 - 142 = 72.6


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


• 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


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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Mole Calculations West Midlands Chemistry Teachers Centre Tuesday 11th November 2008 Presenter: Dr...

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