# THE MOLE QUESTIONS AND ANSWERS-Empirical and Molecular Formula

1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen.

Chemical equation:   2H2 (g)  +       O2 (g)         ->      2H2O(l)

Volume ratios           2            :          1               :        0

Reacting volumes               50cm3   :        25cm3

50cm3 of Oxygen is used

2. Calculate the volume of air required to completely reacts with 50cm3 of Hydrogen.(assume Oxygen is 21% by volume of air)

Chemical equation:   2H2 (g)  +       O2 (g)         ->      2H2O(l)

Volume ratios           2            :          1               :        0

Reacting volumes               50cm3   :        25cm3

50cm3 of Oxygen is used

21%     =     25cm3

100%   =   100 x 25    =

21

3.If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of Carbon(IV)oxide and 10cm3 of water vapour/steam, obtain the equation for the reaction and hence find the value of x and y in CxHy.

Chemical equation:        CxHy (g)    +       O2 (g)         ->      H2O(g)  + CO2(g)

Volumes                          5cm3        :        15cm3                  :         10cm3  :  10cm3

Volume ratios                  5cm3        :        15cm3                  :         10cm3  :  10cm3   (divide by lowest volume)  5                     5                           5                5

Reacting volume ratios          1volume             3 volume            2 volume  2 volume

Balanced chemical equation:            CxHy (g) +          3O2 (g)  ->   2H2O(g) + 2CO2(g)

If “4H” are in 2H2O(g) the y=4

If “2C” are in 2CO2 (g) the x=2

Thus(i) chemical formula of hydrocarbon = C2H4

(ii) chemical name of hydrocarbon = Ethene

4.100cm3 of nitrogen (II)oxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen. All volumes measured at the same temperature and pressure. Obtain the equation for the reaction and name the gaseous product.

Chemical equation:        NO (g)       +       O2 (g)         ->      NOx

Volumes                        100cm3      :        50cm3                  :          100

Volume ratios                  100cm3    :         50cm3        :         100cm3         (divide by lowest volume)   50                          50                         50

Reacting volume ratios          2volume             1 volume              2 volume

Balanced chemical equation:           2 NO (g) +  O2 (g)  ->    2NO x(g)

Thus(i) chemical formula of the nitrogen compound = 2 NO2

(ii) chemical name of compound = Nitrogen(IV)oxide

5.When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen ,the resulting gaseous mixture occupied70cm3 at  room temperature and pressure. When the gaseous mixture was passed through, potassium hydroxide its volume decreased to 25cm3.

(a)What volume of Oxygen was used during the reaction.(1mk)

Volume of Oxygen used =100-25 =75cm3√

(P was completely burnt)

(b)Determine the molecular formula of the hydrocarbon(2mk)

CxHy  +  O2  ->  xCO2   +   yH2O

15cm3 : 75cm3

15         15

1        :   3√

=> 1 atom of C react with 6 (3×2)atoms of Oxygen

Thus x = 1 and y = 2   => P has molecula formula CH4

(g) Ionic equations

An ionic equation is a chemical statement showing the movement of ions (cations and anions ) from reactants to products.

Solids, gases and liquids do not ionize/dissociate into free ions. Only ionic compounds in aqueous/solution or molten state ionize/dissociate into free cations and anions (ions)

An ionic equation is usually derived from a stoichiometric equation by using the following guidelines

Guidelines for writing ionic equations

1.Write the balanced stoichiometric equation

2.Indicate the state symbols of the reactants and products

3.Split into cations and anions all the reactants and products that exist in aqueous state.

4.Cancel out any cation and anion that appear on both the product and reactant side.

5. Rewrite the chemical equation. It is an ionic equation.

Practice

(a)Precipitation of an insoluble salt

All insoluble salts are prepared in the laboratory from double decomposition /precipitation. This involves mixing two soluble salts to form one soluble and one insoluble salt

1. When silver nitrate(V) solution is added to sodium chloride solution,sodium nitrate(V) solution and a white precipitate of silver chloride are formed.

Balanced stoichiometric equation

AgNO3(aq)          +       NaCl(aq)    ->     AgCl (s)     +      NaNO3 (aq)

Split reactants product existing in aqueous state as cation/anion

Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq)   ->     AgCl(s) +   Na+(aq)+ NO3(aq)

Cancel out ions appearing on reactant and product side

Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq)   ->     AgCl(s) +   Na+(aq)+ NO3(aq)

Rewrite the equation

Ag+(aq) + Cl(aq)         ->     AgCl(s) (ionic equation)

2. When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI)  are formed.

Balanced stoichiometric equation

Ba(NO3)2(aq)       +       CuSO4(aq)  -> BaSO4 (s)       +      Cu(NO3) 2 (aq)

Split reactants product existing in aqueous state as cation/anion

Ba2+(aq) + 2NO3(aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3(aq)+ Cu2+(aq)

Cancel out ions appearing on reactant and product side

Ba2+(aq) + 2NO3(aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3(aq) + Cu2+(aq)

Rewrite the equation

Ba2+(aq) + SO42-(aq) ->         BaSO4(s) (ionic equation)

3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide.

Balanced stoichiometric equation

Pb(NO3)2(aq)       +       2KI(aq)      -> PbI2 (s) +      2KNO3  (aq)

Split reactants product existing in aqueous state as cation/anion

Pb2+(aq) + 2NO3(aq) + 2K +(aq) + 2I(aq) -> PbI2 (s) + 2NO3(aq)+ 2K +(aq)

Cancel out ions appearing on reactant and product side

Pb2+(aq) + 2NO3(aq) + 2K +(aq) + 2I(aq) -> PbI2 (s) + 2NO3(aq)+ 2K +(aq)

Rewrite the equation

Pb2+(aq) + 2I(aq)   ->          PbI2 (s) (ionic equation)

(b)Neutralization

Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base.

(i)Reaction of alkalis with acids

1.Reaction of nitric(V)acid with potassium hydroxide

Balanced stoichiometric equation

HNO3(aq)   +       KOH(aq)   -> H2O (l) +      KNO3  (aq)

Split reactants product existing in aqueous state as cation/anion

H+(aq) + NO3(aq) + K +(aq) + OH(aq) -> H2O (l) + NO3(aq)+  K +(aq)

Cancel out ions appearing on reactant and product side

H+(aq) + NO3(aq) + K +(aq) + OH(aq) -> H2O (l) + NO3(aq)+  K +(aq)

Rewrite the equation

H+ (aq) + OH(aq)   ->         H2O (l)  (ionic equation)

2.Reaction of sulphuric(VI)acid with ammonia solution

Balanced stoichiometric equation

H2SO4(aq) +       2NH4OH(aq)       -> H2O (l) +      (NH4) 2SO4  (aq)

Split reactants product existing in aqueous state as cation/anion

2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH(aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)

Cancel out ions appearing on reactant and product side

2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH(aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)

Rewrite the equation

2H+ (aq) + 2OH(aq)   ->     2H2O (l)

H+ (aq)  + OH(aq)     ->     H2O (l)  (ionic equation)

3.Reaction of hydrochloric acid with Zinc hydroxide

Balanced stoichiometric equation

2HCl(aq)    +       Zn(OH)2 (s) -> 2H2O (l)         +      ZnCl 2  (aq)

Split reactants product existing in aqueous state as cation/anion

2H+(aq) + 2Cl(aq) + Zn(OH)2 (s)  ->2H2O (l) + 2Cl (aq)+ Zn 2+ (aq)

Cancel out ions appearing on reactant and product side

2H+(aq) + 2Cl(aq) + Zn(OH)2 (s)  ->2H2O (l) + 2Cl (aq)+ Zn 2+ (aq)

Rewrite the equation

2H+(aq) + Zn(OH)2 (s)  ->2H2O (l) + Zn 2+ (aq) (ionic equation)

(h)Molar solutions

A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M.

Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution.

One cubic decimeter is equal to one litre and also equal to 1000cm3.

The higher the molarity the higher the concentration and the higher/more solute has been dissolved in the solvent  to make one cubic decimeter/ litre/1000cm3 solution.

Examples

2M sodium hydroxide means 2 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water.

0.02M sodium hydroxide means 0.02 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water.

“2M” is more concentrated than“0.02M”.

Preparation of molar solution

Procedure

Weigh accurately 4.0 g of sodium hydroxide pellets into a 250cm3 volumetric   flask.

Shake vigorously for three minutes.

Remove the stopper for a second then continue to shake for about another two minutes until all  the solid has dissolved.

Add more water slowly upto exactly the 250 cm3 mark.

Sample questions

1.Calculate the number of moles of sodium hydroxide pellets present in:

(i) 4.0 g.

Molar mass of NaOH = (23 + 16 + 1)  =  40g

Moles = Mass               =>       4.0         =     0.1   /  1.0 x 10 -1 moles

Molar mass                    40

(ii) 250 cm3 solution in the volumetric flask.

Moles in 250 cm3 =  0.1   /  1.0 x 10 -1 moles

(iii) one decimeter of solution

Method 1

Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3

Volume of solution

=> 1.0 x 10 -1 moles  x 1000cm3 =

250cm3

0.4 M  / 0.4 molesdm-3

Method 2

250cm3 solution contain 1.0 x 10 -1 moles

1000cm3 solution = Molarity  contain 1000  x 1.0 x 10 -1 moles

250 cm3

0.4 M  / 0.4 molesdm-3

Theoretical sample practice

1. Calculate the molarity of a solution containing:

(i) 4.0 g sodium hydroxide dissolved in 500cm3 solution

Molar mass of NaOH = (23 + 16 + 1)  =  40g

Moles = Mass               =>       4.0         =     0.1   /  1.0 x 10 -1 moles

Molar mass                    40

Method 1

Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3

Volume of solution

=> 1.0 x 10 -1 moles  x 1000cm3

500cm3

0.2 M  / 0.2 molesdm-3

Method 2

500 cm3 solution contain 1.0 x 10 -1 moles

1000cm3 solution = Molarity  contain 1000  x 1.0 x 10 -1 moles

500 cm3

=  0.2 M  / 0.2 molesdm-3

(ii) 5.3 g  anhydrous sodium carbonate dissolved in 50cm3 solution

Molar mass of Na2CO3 = (23 x 2 + 12 + 16 x 3)  =  106 g

Moles = Mass          =>       5.3         =     0.05 / 5. 0 x 10-2   moles

Molar mass              106

Method 1

Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3

Volume of solution

=> 1.0 moles  x 1000cm3 =

50cm3

=1.0 M

Method 2

50 cm3 solution contain 5.0 x 10 -2 moles

1000cm3 solution = Molarity  contain 1000  x 5.0 x 10 -2 moles

50 cm3

= 1.0M /  1.0 molesdm-3

(iii) 5.3 g  hydrated sodium carbonate decahydrate dissolved in 50cm3 solution

Molar mass of Na2CO3.10H2O  = (23 x 2 + 12 + 16 x 3 + 20 x 1 + 10 x 16)  =286g

Moles = Mass          =>       5.3         =     0.0185 / 1.85 x 10 -2   moles

Molar mass              286

Method 1

Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3

Volume of solution

=> 1.85 x 10 -2 moles  x 1000cm3 =

50cm3

= 0.37 M/0.37 molesdm-3

Method 2

50 cm3 solution contain 1.85 x 10 -2 moles

1000cm3 solution = Molarity  contain 1000  x  1.85 x 10 -2 moles

50 cm3

= 3.7 x 10-1 M / 3.7 x 10-1 molesdm-3

(iv) 7.1 g of anhydrous sodium sulphate(VI)was dissolved in 20.0 cm3 solution. Calculate the molarity of the solution.

Method 1

20.0cm3 solution ->7.1 g

1000cm3 solution ->    1000    x   71   =   3550 g dm-3

20

Molar mass Na2SO4  = 142 g

Moles dm-3   =    Molarity   =   Mass          3550  = 2.5 M/ molesdm-3

Molar mass    142

Method 2

Molar mass Na2SO4  = 142 g

Moles = Mass          =>       7.1         =     0.05 / 5.0 x 10 -2   moles

Molar mass              142

Method 2(a)

Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3

Volume of solution

=> 5.0 x 10 -2   moles  x 1000cm3

20cm3

= 2.5 M/2.5 molesdm-3

Method 2(b)

20 cm3 solution contain 5.0 x 10 -2 moles

1000cm3 solution = Molarity  contain 1000  x  5.0 x 10 -2 moles

20 cm3

= 2.5 M/2.5 molesdm-3

(iv) The density of sulphuric(VI) is 1.84gcm-3 Calculate the molarity of the acid.

Method 1

1.0cm3 solution ->1.84 g

1000cm3 solution ->    1000    x   1.84   =   1840 g dm-3

1

Molar mass H2SO4  = 98 g

Moles dm-3  = Molarity = Mass                               = 1840

Molar mass                  98

= 18.7755 M/ molesdm-3

Method 2

Molar mass H2SO4  = 98 g

Moles = Mass          =>       1.84         =   0.0188 / 1.88 x 10 -2   moles

Molar mass               98

Method 2(a)

Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3

Volume of solution

=> 1.88 x 10 -2   moles  x 1000cm3

1.0cm3

= 18.8M/18.8 molesdm-3

Method 2(b)

20 cm3 solution contain 1.88 x 10 -2   moles

1000cm3 solution = Molarity  contain 1000  x  1.88 x 10 -2   moles

1.0 cm3

= 18.8M/18.8 molesdm-3

2. Calculate the mass of :

(i) 25 cm3 of 0.2M sodium hydroxide solution(Na =23.0.O =16.0, H=1.0)

Molar mass NaOH = 40g

Moles  in 25 cm3  =  Molarity x volume   => 0.2  x   25 = 0.005/5.0 x 10-3moles

1000                     1000

Mass of NaOH =Moles x molar mass = 5.0 x 10-3   x   40 = 0.2 g

(ii) 20 cm3 of 0.625 M sulphuric(VI)acid (S =32.0.O =16.0, H=1.0)

Molar mass H2SO4 = 98g

Moles  in 20 cm3 = Molarity x volume=> 0.625 x 20 = 0.0125/1.25.0 x 10-3moles

1000             1000

Mass of H2SO4 =Moles x molar mass => 5.0 x 10-3   x   40 = 0.2 g

(iii) 1.0 cm3 of 2.5 M Nitric(V)acid (N =14.0.O =16.0, H=1.0)

Molar mass HNO3 = 63 g

Moles  in 1 cm3 = Molarity x volume => 2.5 x 1 = 0.0025 / 2.5. x 10-3moles

1000                 1000

Mass of HNO3 =Moles x molar mass => 2.5 x 10-3   x   40 = 0.1 g

3. Calculate the volume required to dissolve :

(a)(i) 0.25moles of  sodium hydroxide solution to form a 0.8M solution

Volume (in cm3)   =   moles x 1000 =>  0.25  x 1000           = 312.5cm3

Molarity                0.8

(ii) 100cm3 was added to the sodium hydroxide solution above. Calculate the concentration of the solution.

C1 x V1 = C2 x V2  where:

C1   =   molarity/concentration before diluting/adding water

=> 0.8M  x 312.5cm3   =  C2    x  (312.5 + 100)

C= 0.8M  x 312.5cm3 = 0.6061M

412.5

(b)(ii) 0.01M solution  containing 0.01moles of  sodium hydroxide solution .

Volume (in cm3)   =   moles x 1000 =>  0.01  x 1000           = 1000 cm3

Molarity                0.01

(ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution.

C1 x V1 = C2 x V2  where:

C1   =   molarity/concentration before diluting/adding water

=> 0.01M  x 1000 cm3   =  0.008    x V2

V= 0.01M  x 1000cm3 = 1250cm3

0.008

Volume added  = 1250  – 1000  =   250cm3

(c)Volumetric analysis/Titration

Volumetric analysis/Titration is the process of determining unknown concentration of one reactant from a known concentration and volume of another.

Reactions take place in simple mole ratio of reactants and products.

Knowing the concentration/ volume of one reactant, the other can be determined from the relationship:

M1V1  = M2V2   where:

n1            n2

M1  = Molarity of 1st reactant

M2  = Molarity of 2nd  reactant

V1   = Volume of 1st reactant

V1     = Volume of 2nd  reactant

n1     = number of moles of 1st reactant from stoichiometric equation

n2     = number of moles of 2nd reactant from stoichiometric equation

Examples

1.Calculate the molarity of MCO3 if 5.0cm3 of MCO3 react with 25.0cm3 of 0.5M hydrochloric acid.(C=12.0 ,O =16.0)

Stoichiometric equation:MCO3(s) + 2HCl(aq)  -> MCl2(aq) + CO2(g) + H2O(l)

Method 1

M1V1  = M2V2       ->       M1  x  5.0cm3  = 0.5M  x  25.0cm3

n1            n2                                          1                          2

=> M=         0.5  x  25.0 x1 =   1.25M / 1.25 moledm-3

5.0 x 2

Method 2

Moles of HCl used = molarity x volume

1000

=> 0.5   x   25.0  =  0.0125 /1.25 x 10-2moles

1000

Mole ratio MCO3 : HCl  =  1:2

Moles MCO3 =  0.0125 /1.25 x 10-2moles = 0.00625 / 6.25 x 10-3 moles

2

Molarity MCO3 = moles x 1000   =>   0.00625 / 6.25 x 10-3   x  1000

Volume                                5

= 1.25M / 1.25 moledm-3

2. 2.0cm3 of 0.5M hydrochloric acid react with 0.1M of M2CO3. Calculate the volume of 0.1M M2CO3 used.

Stoichiometric equation:M2CO3 (aq) + 2HCl(aq)  -> 2MCl (aq) + CO2(g) + H2O(l)

Method 1

M1V1  = M2V2       ->       0.5  x  2.0cm3  = 0.1M  x  V2 cm3

n1            n2                                          2                          1

=> V=         0.5  x  2.0 x1 =   1.25M / 1.25 moledm-3

0.1 x 2

Method 2

Moles of HCl used = molarity x volume

1000

=>  0.5     x     2.0  =      0.0125 /1.25 x 10-2moles

1000

Mole ratio M2CO3 : HCl  =  1:2

Moles M2CO3 =    0.0125 /1.25   x   10-2moles =    0.00625 / 6.25 x 10-3 moles

2

Molarity M2CO3 = moles x 1000   =>   0.00625 / 6.25 x 10-3   x  1000

Volume                                5

= 1.25M / 1.25 moledm-3

3. 5.0cm3 of 0.1M sodium iodide react with 0.1M of Lead(II)nitrate(V). Calculate(i) the volume of Lead(II)nitrate(V) used.

(Pb=207.0, I =127.0)

Stoichiometric equation:    2NaI(aq)  + Pb(NO3)2(aq) -> 2NaNO3(aq)  + PbI2(s)

Method 1

M1V1  = M2V2       ->        5 x  0.1cm3  =   0.1M  x  V2 cm3

n1            n2                                          2                          1

=> V=    0.1  x  5.0 x 1 =   1.25M / 1.25 moledm-3

0.1  x  2

Method 2

Moles of HCl used = molarity x volume

1000

=>  0.1     x     5.0  =      0.0125 /1.25 x 10-2moles

1000

Mole ratio M2CO3 : HCl  =  1:2

Moles M2CO3 =    0.0125 /1.25   x   10-2moles =    0.00625 / 6.25 x 10-3 moles

2

Molarity M2CO3 = moles x 1000   =>   0.00625 / 6.25 x 10-3   x  1000

Volume                                5

= 1.25M / 1.25 moledm-3

4. 0.388g of a monobasic organic acid B required 46.5 cm3 of 0.095M sodium hydroxide for complete neutralization. Name and draw the structural formula of B

Moles of NaOH used = molarity x volume

1000

=>  0.095     x   46.5 =  0.0044175 /4.4175 x 10-3moles

1000

Mole ratio B: NaOH  =  1:1

Moles B=    0.0044175 /4.4175 x 10-3moles

Molar mass B                 = mass         =>                0.388

moles                  0.0044175 /4.4175 x 10-3moles

= 87.8324 gmole-1

X-COOH = 87.8324 where X is an alkyl group

X =87.8324- 42 = 42.8324=43

By elimination:  CH3  = 15     CH3CH=  29     CH3CH2 CH2  = 43

Molecula formula : CH3CH2 CH2COOH

Molecule name :  Butan-1-oic acid

Molecular structure

H      H      H      O

H      C       C       C       C       O      H

H       H       H       H

5.  10.5 g of an impure sample containing ammonium sulphate (VI) fertilizer was warmed with 250cm3 of o.8M sodium hydroxide solution.The excess of the alkali was neutralized by 85cm3 of 0.5M hydrochloric acid. Calculate the % of impurities in the ammonium sulphate (VI)fertilizer. (N=14.0,S=32.0,O=16.0, H=1.0)

Equation for neutralization

NaOH(aq)  +       HCl(aq)      ->      NaOH(aq)  +       H2O(l)

Mole ratio NaOH(aq):HCl(aq)= 1:1

Moles of HCl = Molarity x volume   =>   0.5  x  85  =     0.0425 moles

1000                              1000

Excess moles of NaOH(aq)= 0.0425 moles

Equation for reaction with ammonium salt

2NaOH(aq) +       (NH4) 2SO4(aq)    -> Na 2SO4(aq) + 2NH3 (g)+  2H2O(l)

Mole ratio NaOH(aq): (NH4) 2SO4(aq)= 2:1

Total moles of  NaOH = Molarity x volume   =>   0.8  x  250  =     0.2 moles

1000                      1000

Moles of NaOH that reacted with(NH4) 2SO4  =  0.2  –  0.0425  =  0.1575moles

Moles (NH4) 2SO4   =   ½   x  0.1575moles   =  0. 07875moles

Molar mass (NH4) 2SO4= 132 gmole-1

Mass of in impure sample = moles  x  molar mass  =>0. 07875 x 132 = 10.395 g

Mass of impurities =     10.5 -10.395     =   0.105 g

% impurities           =         0.105   x  100  =    1.0 %

10.5

Practically volumetric analysis involves titration.

Titration generally involves filling a burette with known/unknown concentration of a solution then adding the solution to unknown/known concentration of another solution in a conical flask until there is complete reaction. If the solutions used are both colourless, an indicator is added to the conical flask. When the reaction is over, a slight/little excess of burette contents change the colour of the indicator. This is called the end point.

The titration process involve involves determination of titre. The titre is the volume of burette contents/reading before and after the end point. Burette contents/reading before titration is usually called the Initial burette reading. Burette contents/reading after titration is usually called the Final burette reading. The titre value is thus a sum of the Final less Initial burette readings.

To reduce errors, titration process should be repeated at least once more.

The results of titration are recorded in a titration table as below

Sample titration table

As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration.

For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council.

As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided.

Calculate the average volume of solution used

24.0 + 24.0 + 24.0   =  24.0 cm3

3

As evidence of understanding the degree of accuracy of burettes , all readings must be recorded to a decimal point.

As evidence of accuracy in carrying the out the titration , candidates value should be within 0.2 of the school value .

The school value is the teachers readings presented to the examining body/council based on the concentrations of the solutions s/he presented to her/his candidates.

Bonus mark is awarded for averaged reading within 0.1 school value as Final answer.

Calculations involved after the titration require candidates thorough practical and theoretical practice mastery on the:

(i)relationship among the mole, molar mass, mole ratios, concentration, molarity.

(ii) mathematical application of 1st principles.

Very useful information which candidates forget appears usually in the beginning of the question paper as:

“You are provided with…

All calculation must be to the 4th decimal point unless they divide fully to a lesser decimal point.

Candidates are expected to use a non programmable scientific calculator.

(a)Sample Titration Practice 1 (Simple Titration)

You are provided with:

0.1M sodium hydroxide solution A

Hydrochloric acid solution B

You are required to determine the concentration of solution B in moles per litre.

Procedure

Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask. Titrate solution A with solution B using phenolphthalein indicator to complete the titration table 1

Sample results Titration table 1

Sample worked questions

1. Calculate the average volume of solution B used

Average titre = Titre 1 + Titre 2 +Titre 3  => ( 20.0 +20.0 +20.0 ) = 20.0cm3

3                                    3

2. How many moles of:

(i)solution A were present in 25cm3 solution.

Moles of solution A =   Molarity x volume  =  0.1 x 25 =  2.5 x 10-3 moles

1000                   1000

(ii)solution B were present in the average volume.

Chemical equation:  NaOH(aq) +  HCl(aq)  ->  NaCl(aq) +  H2O(l)

Mole ratio 1:1   =>  Moles of A = Moles of B =  2.5 x 10-3 moles

(iii) solution B in moles per litre.

Moles of B per litre =    moles x 1000    = 2.5 x 10-3 x 1000  =  0.1M

Volume                    20

(b)Sample Titration Practice 2 (Redox Titration)

You are provided with:

Acidified Potassium manganate(VII) solution A

0.1M of an iron (II)salt  solution B

8.5g of ammonium iron(II)sulphate(VI) crystals(NH4)2 SO4FeSO4.xH2O   solid C

You are required to

(i)standardize acidified potassium manganate(VII)

(ii)determine the value of x in the formula (NH4)2 SO4FeSO4.xH2O.

Procedure 1

Fill the burette with solution A. Pipette 25.0cm3 of solution B into a conical flask. Titrate solution A with solution B until a pink colour just appears.

Record your results to complete  table 1.

Table 1:Sample results

Sample worked questions

1. Calculate the average volume of solution A used

Average titre = Titre 1 + Titre 2 +Titre 3  => ( 20.0 +20.0 +20.0 ) = 20.0cm3

3                                        3

2. How many moles of:

(i)solution B were present in 25cm3 solution.

Moles of solution A =   Molarity x volume  =  0.1 x 25 =  2.5 x 10-3 moles

1000                   1000

(ii)solution A were present in the average volume. Assume one mole of B react with five moles of B

Mole ratio A : B = 1:5

=>  Moles of A = Moles of B =      2.5 x 10-3 moles     =    5.0 x 10 -4 moles

5                         5

(iii) solution B in moles per litre.

Moles of B per litre =    moles x 1000    = 2.5 x 10-3 x 1000

Volume                    20

0.025 M /moles per litre /moles l-1

Procedure 2

Place all the solid C into the 250cm3 volumetric flask carefully. Add about 200cm3 of distilled water. Shake to dissolve. Make up to the 250cm3 of solution by adding more distilled water. Label this solution C. Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2.

Table 2:Sample results

Sample worked questions

1. Calculate the average volume of solution A used

Average titre = Titre 1 + Titre 2 +Titre 3  => ( 20.0 +20.0 +20.0 ) = 20.0cm3

3                                        3

2. How many moles of:

(i)solution A were present inin the average titre.

Moles of solution A =   Molarity x volume  =  0.025 x 20 =  5.0 x 10-4 moles

1000                    1000

(ii)solution C in 25cm3 solution given the equation for the reaction:

MnO4(aq) +  8H+(aq) +  5Fe2+ (aq)  ->   Mn2+(aq)  +  5Fe3+(aq) + 4H2O(l)

Mole ratio MnO4(aq): 5Fe2+ (aq) =  1:5     =>                                              Moles of 5Fe2+ (aq) = Moles of MnO4(aq) = 5.0 x 10-4 moles = 1.0 x 10 -4 moles

5                           5

(iii) solution B in 250cm3.

Moles of B per litre =  moles x 250   = 1.0 x 10 -4  x 250  = 1.0 x 10 -3 moles                                                  Volume                    25

3. Calculate the molar mass of solid C and hence the value of x in  the chemical formula (NH4)2SO4FeSO4.xH2O.

(N=14.0, S=32.0, Fe=56.0, H=1.0 O=16.0)

Molar mass =  mass perlitre  =         8.5                 =      8500 g

Moles per litre           1.0 x 10 -3 moles

NH4)2SO4FeSO4.xH2O = 8500

284 + 18x =8500

8500  – 284 =    8216     =       18x      =   454.4444

18               18

x =  454  (whole number)

(c)Sample Titration Practice 3 (Back  titration)

You are provided with:

(i)an impure calcium carbonate labeled M

(ii)Hydrochloric acid labeled solution N

(iii)solution L containing 20g per litre sodium hydroxide.

You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M.

Procedure 1

Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator. Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark)

Sample Table 1

Sample questions

(a) Calculate the average volume of solution N used

6.5  +  6.5  +  6.5    =     6.5 cm3

3

(b) How many moles of sodium hydroxide are contained in 25cm3of solution L

Molar mass NaOH =40g

Molarity of L   =  mass per litre           =>       20     = 0.5M

Molar mass NaOH              40

Moles NaOH in 25cm3 = molarity  x volume   =>  0.5M  x 25cm3 =  0.0125 moles

1000                            1000

(c)Calculate:

(i)the number of moles of hydrochloric acidthat react with sodium hydroxide in (b)above.

Mole ratio NaOH : HCl from stoichiometric equation= 1:1

Moles HCl =Moles NaOH => 0.0125 moles

(ii)the molarity of hydrochloric acid solution N.

Molarity =   moles x 1000   => 0.0125 moles x 1000 =1.9231M/moledm-3

6.5                            6.5

Procedure 2

Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette. Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2.

Sample Table 2

Sample calculations

(a)Calculate the average volume of solution L used(1mk)

24.5  +  24.5  +  24.5  = 24.5cm3

3

(b)How many moles of sodium hydroxide are present in the average volume of solution L used?

Moles = molarity x average burette volume => 0.5 x 24.5

1000                                        1000

= 0.01225 /1.225 x 10-2 moles

(c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K?

Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles

Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles

25cm3(volume pipetted)

=0.49 /4.9 x 10-1moles

(d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used

Original moles = Original molarity x pipetted volume  =>

1000cm3

1.9231M/moledm-3 x   25    =   0.04807/4.807 x 10-2 moles

1000

(e)How many moles of hydrochloric acid were used to react with calcium carbonate present?

Moles that reacted = original moles –moles in average titre  =>

0.04807/4.807 x 10-2moles  –  0.01225 /1.225 x 10-2moles

= 0.03582/3.582 x 10 -2moles

(f)Write the equation for the reaction between calcium carbonate and hydrochloric acid.

CaCO3(s)  + 2HCl(aq)   ->   CaCl2(aq)  +   CO2(g)  +   H2O(l)

(g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid.

From the equation CaCO3(s):2HCl(aq) = 1:2

=> Moles CaCO3(s) = 1/2moles HCl

= 1/2 x 0.03582/3.582 x 10 -2 moles

= 0.01791 /1.791 x 10-2moles

(h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0)

Molar mass CaCO3 = 100g

Mass CaCO3 = moles x molar mass   =>  0.01791 /1.791 x 10-2moles x 100g

= 1.791g

(i)Determine the % of calcium carbonate present in the mixture

% CaCO3 = mass of pure x 100%      =>     1.791g x 100%    =   44.775%

Mass of impure                             4.0

(d)Sample titration practice 4 (Multiple titration)

You are provided with:

(i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O.

(ii)solution M which is acidified potassium manganate(VII)

(iii)solution N a mixture of sodium ethanedioate and ethanedioic acid

(iv)0.1M sodium hydroxide solution P

(v)1.0M sulphuric(VI)

You are required to:

(i)standardize solution M using solution L

(ii)use standardized solution M and solution P to determine the % of sodium   ethanedioate in the mixture.

Procedure 1

Fill the burette with solution M. Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1.

Sample Table 1

Sample calculations

(a)Calculate the average volume of solution L used (1mk)

24.0  +  24.0  +  24.0  = 24.0cm3

3

(b)Given that the concentration of the dibasic acid is 0.05molesdm-3.determine the value of x in the formula H2X.2H2O (H=1.0,O=16.0)

Molar mass H2X.2H2O= mass per litre    =>       5.0g/litre           = 100g

Moles/litre                  0.05molesdm-3

H2X.2H2O =100

X = 100 – ((2 x1) + 2 x (2 x1)  + (2 x 16) => 100  –  34  =  66

(c) Calculate the number of moles of the dibasic acid H2X.2H2O.

Moles = molarity x pipette volume =>    0.5 x 25      =   0.0125/1.25 x10 -2 moles

1000                                1000

(d)Given the mole ratio manganate(VII)(MnO4): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4) in the average titre.

Moles H2X  =  2/5 moles of MnO4–

=>  2/5  x 0.0125/1.25 x10 -2 moles

= 0.005/5.0 x 10 -3moles

(e)Calculate the concentration of the manganate(VII)(MnO4) in moles per litre.

Moles per litre/molarity  =  moles   x   1000

average burette volume

=>0.005/5.0 x 10 -3moles x 1000  =  0.2083 molesl-1/M

24.0

Procedure 2

With solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2.

Sample Table 2

Sample calculations

(a)Calculate the average volume of solution L used (1mk)

12.5  +  12.5  +  12.5  =12.5cm3

3

(b)Calculations:

(i)How many moles of manganate(VII)ions are contained in the average volume of solution M used?

Moles = molarity of solution M x average burette volume

1000

=>    0.2083 molesl-1/ M  x  12.5    =   0.0026 / 2.5 x 10-3 moles

1000

(ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation:

2MnO4(aq) +  5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l)

Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M.

From the stoichiometric equation,mole ratio MnO4(aq): C2O42- (aq) = 2:5

=> moles C2O42-  =  5/2 moles MnO4 => 5/2 x 0.0026 / 2.5 x 10-3 moles

= 0.0065 /6.5 x10-3 moles

(iii)Calculate the number of moles of ethanedioate ions contained in 250cm3 solution N.

25cm3 pipette volume    ->   0.0065 /6.5 x10-3 moles

250cm3                         ->

0.0065 /6.5 x10-3 moles  x  250     =    0.065 / 6.5 x10-2 moles

25

Procedure 3

Remove solution M from the burette and rinse it with distilled water. Fill the burette with sodium hydroxide solution P. Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3.

Sample Table 2

Sample calculations

(a)Calculate the average volume of solution L used (1mk)

24.9  +  24.9  +  24.9  =  24.9 cm3

3

(b)Calculations:

(i)How many moles of sodium hydroxide solution P were contained in the average volume?

Moles = molarity of solution P  x average burette volume

1000

=>    0.1 molesl-1   x  24.9    =   0.00249 / 2.49 x 10-3 moles

1000

(ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is:

2NaOH(aq)  +   H2C2O4 (aq)    ->   Na2C2O4(g)   +   2H2O(l)

Calculate the number of moles of ethanedioic acid that were used in the reaction

From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1

=> moles H2C2O4  =  1/2 moles NaOH => 1/2 x  0.00249 / 2.49 x 10-3 moles

= 0.001245/1.245 x10-3 moles.

(iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N?

25cm3 pipette volume    ->  0.001245/1.245 x10-3 moles

250cm3                         ->

0.001245/1.245 x10-3 moles x  250     =   0.01245/1.245 x10-2 moles

25

(iii)Determine the % by mass of sodium ethanedioate in the micture             (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution)

Molar mass H2C2O4 = 90.0g

Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4

=>0.01245/1.245 x10-2 moles x 90.0

= 1.1205g

% by mass of sodium ethanedioate

=(Mass of mixture   –  mass of H2C2O4) x 100%

Mass of mixture

=>   2.0  – 1.1205 g   =  43.975%

2.0

Note

(i) L is 0.05M Oxalic acid

(ii) M is 0.01M KMnO4

(iii) N is 0.03M oxalic acid(without sodium oxalate)

Practice example 5.(Determining equation for a reaction)

You are provided with

-0.1M hydrochloric acid solution A

-0.5M sodium hydroxide solution B

You are to determine the equation for thereaction between solution A and B

Procedure

Fill the burette  with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1

Table 1(Sample results)

Sample questions

Calculate the average volume of solution A used.

12.5+12.5+12.5  =       12.5cm3

3

Theoretical Practice examples

1. 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution. 25.0 cm3 of this solution reacted with 30.0cm3 of 0.06M sodium hydroxide solution. Calculate the value of x in HOOC(CH2)xCOOH. (C=12.0,H=1.0,O=16.)

Chemical equation

2NaOH(aq) + H2X(aq) -> Na2X (aq) + 2H2O(aq)

Mole ratio NaOH(aq) :H2X(aq) = 2:1

Method 1

Ma Va = na          =>          Ma  x  25.0     = 1  =>  Ma =0.06  x  30.0 x1

Mb Vb = nb              0.06  x  30.0      2               25.0 x 2

Molarity of acid = 0.036M/Mole l-1

Mass of acid per lite =   1.0 x1000       =       4.0 g/l

250

0.036M/  Mole l-1     ->  4.0 g /l

1 mole= molar mass of HOOC(CH2)xCOOH = 4.0 x 1        = 111.1111 g

0.036

Molar mass (CH2)x =  111.1111 – (HOOCCOOH  = 90.0) = 21.1111

(CH2)x = 14x  =  21.1111  =  1.5  = 1 (whole number)

14

Method 2

Moles of sodium hydroxide =    Molarity x volume  =  0.06 x 30 = 1.8 x 10 -3moles

1000

Moles of Hydrochloric acid = 1/x  1.8 x 10 -3moles = 9.0 x10 -4moles

Molarity of Hydrochloric acid = moles x 1000 = 9.0 x10 -4moles x1000

Volume                     25

Molarity of acid = 0.036M/Mole l-1

Mass of acid per lite =   1.0 x1000       =       4.0 g/l

250

0.036M/  Mole l-1     ->  4.0 g /l

1 mole= molar mass of HOOC(CH2)xCOOH = 4.0 x 1        = 111.1111 g

0.036

Molar mass (CH2)x =  111.1111 – (HOOCCOOH  = 90.0) = 21.1111

(CH2)x = 14x  =  21.1111  =  1.5  = 1 (whole number)

14

2. 20.0cm3 of 0.05 M acidified potassium manganate(VII)solution oxidized  25.0cm3 of Fe2+(aq) ions in 40.0g/l of impure Iron (II)sulphate(VI) to Fe3+(aq) ions. Calculate the  percentage  impurities in the  Iron (II)sulphate(VI).

MnO4 (aq) + 8H+(aq)+ 5Fe2+(aq)-> 5Fe3+(aq)  + Mn2+(aq) + 4H2O(aq)

Fe=56.0,S= 32.0, O=16.0).

Moles of MnO4 (aq)   = Molarity  x  volume =   0.05  x  20.0     = 0.001 Moles

1000                                1000

Mole ratio MnO4 (aq): 5Fe2+(aq)= 1:5

Moles 5Fe2+(aq)   =   5 x0.001   =   0.005 Moles

Moles of 5Fe2+(aq) per litre/molarity = Moles x 1000   =     0005 x  1000

Volume                      25.0

= 0.2 M/ Moles/litre

Molar mass =FeSO4=152 g

Mass of  in the mixture  =  Moles x molar mass  => 0.2 x 152  = 30.4 g

Mass of impurity = 40.0 – 30.4  =9.6 g

% impurity     =      9.6 g x100 = 24.0 % impurity

40.0

3.9.7 g of a mixture of Potassium hydroxide and Potassium chloride was dissolved to make one litre solution.20.0cm3 of this solution required 25.0cm3 of 0.12M hydrochloric acid for completed neutralization. Calculate the percentage by mass of Potassium chloride.(K=39.0,Cl= 35.5)

Chemical equation

KOH(aq) +  HCl(aq) ->  KCl(aq) +  H2O(l)

Moles of HCl = Molarity  x  volume => 0.12 x 25.0   = 0.003/3.0 x 10 -3 moles

1000                         1000

Mole ratio KOH(aq) : HCl(aq) -= 1:1

Moles KOH =0.003/3.0 x 10 -3 moles

Method 1

Molar mass KOH  =56.0g

Mass KOH in 25cm3 =0.003/3.0 x 10 -3 moles x56.0 = 0.168g

Mass KOH in 1000cm3/1 litre = 0.168 x1000= 8.4 g/l

20

Mass of KCl =  9.7g  – 8.4g   =  1.3 g

% of KCl =      1.3 x 100       =   13.4021%

9.7

Method 2

Moles KOH in 1000cm3 /1 litre = Moles in 20cm3 x 1000 =>0.003 x 1000

20                                 20

=0.15M/Moles /litre

Molar mass KOH  =56.0g

Mass KOH in 1000/1 litre = 0.15M/Moles /litre  x 56.0  = 8.4g/l

Mass of KCl =  9.7g  – 8.4g   =  1.3 g

% of KCl =      1.3 x 100       =   13.4021%

9.7

4.A certain carbonate, GCO3, reacts with dilute hydrochloric acid according to the equation given below:

GCO3(s) + 2HCl(aq)              ->      GCl2 (aq)  +   CO2 (g)  + H2O(l)

If 1 g of the carbonate reacts completely with 20 cm3 of 1 M hydrochloric acid ,calculate the relative atomic mass of  G (C = 12.0 = 16.0)

Moles of HCl = Molarity x volume=> 1 x20  = 0.02 moles

1000                 1000

Mole ratio HCl; GCO3  = 2:1

Moles of GCO3= 0.02 moles = 0.01moles

2

Molar mass of GCO3    =   mass   =>      1         =   100 g

moles       0.01moles

G= GCO3 – CO3 =>100g – (12+ 16 x3 = 60) = 40(no units)

5. 46.0g of a metal carbonate MCO3 was dissolved 160cm3 of 0.1M excess hydrochloric acid and the resultant solution diluted to one litre.25.0cm3 of this solution required 20.0cm3 of 0.1M sodium hydroxide solution for complete neutralization. Calculate the atomic mass of ‘M’

Equation

Chemical equation

NaOH(aq) +  HCl(aq) ->  KCl(aq) +  H2O(l)

Moles of NaOH = Molarity x volume=> 0.1 x20  = 0.002 moles

1000                 1000

Mole ratio HCl; NaOH = 1:1

Excess moles of HCl     =  0.002 moles

25cm3       ->      0.002 moles

1000cm3    ->         1000 x 0.002     =    0.08moles

25cm3

Original moles of HCl   = Molarity x volume  => 1M  x 1litre = 1.0 moles

Moles of HCl reacted with MCO3   =  1.0 – 0.08 moles  =  0.92moles

Chemical equation

MCO3(s) + 2HCl(aq)              ->      MCl2 (aq)  +   CO2 (g)  + H2O(l)

Mole ratio MCO3(s) : HCl(aq) =1:2

Moles of MCO3    =   0.92moles   => 0.46moles

2

Molar mass of MCO3=       mass     =>         46g          =  100 g

moles                0.46moles

M= MCO3 – CO3 =>100g – (12+ 16 x3 = 60) = 40

6. 25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction.

A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction. Calculate the concentration of Fe2+ and Fe3+ ion in the solution on moles per litre.

Mole ratio Fe2+ :Mn04 = 5:1

Moles Mn04used          =       0.02 x 15                      =       3.0 x 10-4 moles

1000

Moles Fe2+                    =       3.0 x 10-4 moles           =       6.0 x 10-5 moles

5

Molarity of Fe2+  =       6.0 x 10-4 moles x 1000   =     2.4 x 10-3 moles l-1

25

Since Zinc reduces  Fe3+  to  Fe2+  in the mixture:

Moles Mn04 that reacted with all  Fe2+=    0.02 x 19       =   3.8 x 10-4 moles

1000

Moles of  all  Fe2+                        =   3.8 x 10-4 moles              =  7.6 x 10-5 moles

5

Moles of   Fe3+                                          =   3.8 x 10-4 – 6.0 x 10-5       =     1.6 x 10-5 moles

Molarity of Fe3+                 =  1.6 x 10-5 moles x 1000   =     4.0 x 10-4 moles l-1