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

Mole Calculations

West Midlands Chemistry Teachers Centre

Tuesday 23rd November 2010

Presenter: Dr Janice Perkins

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

Reacting ratio in equations

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

Funny numbers

Dozen = 12 Gross = 12 12 = 144

6.023 1023

602300000000000000000000

Mole = Score = 20

That’s

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(s) + 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(s) + 3CO2(g)

1:2

Mole Ratio (Reacting Ratio)

CO : Fe = 3:2

CO : CO2 = 1:1

3:3

Fe2O3 : CO = 1:3

1:3

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

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

Example 1: Calculating moles from masses

Use the equation Mass = Mr x moles

Mr = 100

Rearrange equation

Moles = mass/Mr

= 300 100 = 3 moles

Calculate the number of moles in 300g of CaCO3

First we need to calculate the number of moles of HCl.

Moles = mass/Mr

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: 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.

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

A metal carbonate MCO3 reacts with HCl as in the following equation.

MCO3 + 2HCl MCl2 + H2O + CO2

A 0.548 g sample of MCO3 reacted completely with 30.7 cm3 of 0.424 mol dm-3 HCl.

(a) Calculate the amount, in moles, of HCl which reacted with the 0.548 g of MCO3

Moles = vol x conc (in dm3)

= 30.7/1000 x 0.424

Example 3: Calculating moles of solution and solid, then Mr and Ar (Jan 09 chem1)

= 0.0130 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 0.0130 mol of HCl

so there must be half that amount of MCO3

= 0.0130/2

= 0.0065 mol = 6.50 x 10-3 mol

(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 12 + 48 = 60.

So this means M must have an Ar of 84.3 – 60 = 24.3

This is Magnesium.

(c) Calculate the Mr of MCO3

A mass of 0.548g of MCO3 contains 0.0065 mol.

Use the equation Mr = mass/moles

Mr = 0.548/0.0065 = 84.3

Example 4: Identity unknown – less structured calculationThe 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.

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==

The carbonate of metal M has the formula M2CO3. The





Calculate the Ar of metal M and deduce its identity.

Moles HCl(aq) = volume concentration= 21.7=

The carbonate of metal M has the formula M2CO3. The





Calculate the Ar of metal M and deduce its identity.

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

(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

=

Mr of M2CO3 = mass = 0.394 moles

=

Mr of M2CO3 = mass = 0.394 moles 2.85 10-3

=

Mr of M2CO3 = mass = 0.394 moles 2.85 10-3

= 138(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

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 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)

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

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 Omass 30.4 69.6 Ar 14 16= moles 2.17 4.35Ratio 2.17/2.17 = 1 4.35/2.17= 2

Empirical formula = NO2

Example 6: Empirical formula Jun 09

Mr(Na2CO3 • xH2O) = 250

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)


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

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 0.200 mol dm-3 hydrochloric acid for complete reaction

Na2CO3 + 2HCl 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 0.200 mol dm-3 hydrochloric acid for complete reaction


Calculate the Mr of Na2CO3 • xH2O(a) Calculate moles HCl since you can do this

Moles HCl =volume concentration = 24.3 10-3 0.200 = 4.86 10-3 mol

(b) Now calculate the moles of Na2CO3 in the 25 cm3 sample of solution.


Moles Na2CO3 =Mole Ratio Moles HCl

(c) 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


= 1/2 4.86 10-3


= ½ 4.86 10-3 = 2.43 10-3

(d) Calculate the Mr of hydrated Na2CO3 [Mass = 3.01 g]

Mr= Mass / MolesMr = 3.01/2.43 10-2 = 124

(a) Calculate the amount, in moles, of TiCl4 in 165 g

Moles = mass = 165

Mr 189.9 = 0.869

Example 9: Percentage yield Jan 09

(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 0.869 moles of TiCl4 gives 0.869 moles of TiO2

In an experiment 165 g TiCl4 were added to an excess of water.

The equation for the reaction is as follows

TiCl4 + 2H2O TiO2 + 4HCl

(c) Calculate the maximum mass of TiO2 formed in this experiment.

mass = Mr x moles

= 79.9 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%

% 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.

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

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 11: Moles HCl, moles & identity ‘Z’The chloride of an element Z reacts with water according to the following equation.






















































• 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


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

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

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

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