STOICHIOMETRY LESSON NOTES.
(a)Introduction to the mole, molar masses and Relative atomic masses
1.The mole is the SI unit of the amount of substance.
2.The number of particles e.g. atoms, ions, molecules, electrons, cows, cars are all measured in terms of moles.
3.The number of particles in one mole is called the Avogadros Constant. It is denoted “L”.
The Avogadros Constant contain 6.023 x10 23 particles. i.e.
1mole = 6.023 x10 23 particles = 6.023 x10 23
2 moles = 2 x 6.023 x10 23 particles = 1.205 x10 24
0.2 moles = 0.2 x 6.023 x10 23 particles = 1.205 x10 22
0.0065 moles = 0.0065 x 6.023 x10 23 particles = 3.914 x10 21
3.The mass of one mole of a substance is called molar mass. The molar mass of:
(i)an element has mass equal to relative atomic mass /RAM(in grams)of the element e.g.
Molar mass of carbon(C)= relative atomic mass = 12.0g
6.023 x10 23 particles of carbon = 1 mole =12.0 g
Molar mass of sodium(Na) = relative atomic mass = 23.0g
6.023 x10 23 particles of sodium = 1 mole =23.0 g
Molar mass of Iron (Fe) = relative atomic mass = 56.0g
6.023 x10 23 particles of iron = 1 mole =56.0 g
(ii)a molecule has mass equal to relative molecular mass /RMM (in grams)of the molecule. Relative molecular mass is the sum of the relative atomic masses of the elements making the molecule. e.g.
Molar mass Oxygen molecule(O2) =relative molecular mass =(16.0x 2)g =32.0g
6.023 x10 23 particles of Oxygen molecule = 1 mole = 32.0 g
Molar mass chlorine molecule(Cl2) =relative molecular mass =(35.5x 2)g =71.0g
6.023 x10 23 particles of chlorine molecule = 1 mole = 71.0 g
Molar mass Nitrogen molecule(N2) =relative molecular mass =(14.0x 2)g =28.0g
6.023 x10 23 particles of Nitrogen molecule = 1 mole = 28.0 g
(ii)a compound has mass equal to relative formular mass /RFM (in grams)of the molecule. Relative formular mass is the sum of the relative atomic masses of the elements making the compound. e.g.
(i)Molar mass Water(H2O) = relative formular mass =[(1.0 x 2 ) + 16.0]g =18.0g
6.023 x10 23 particles of Water molecule = 1 mole = 18.0 g
6.023 x10 23 particles of Water molecule has:
- 2 x 6.023 x10 23 particles of Hydrogen atoms
-1 x 6.023 x10 23 particles of Oxygen atoms
(ii)Molar mass sulphuric(VI)acid(H2SO4) = relative formular mass
=[(1.0 x 2 ) + 32.0 + (16.0 x 4)]g =98.0g
6.023 x10 23 particles of sulphuric(VI)acid(H2SO4) = 1 mole = 98.0g
6.023 x10 23 particles of sulphuric(VI)acid(H2SO4) has:
- 2 x 6.023 x10 23 particles of Hydrogen atoms
-1 x 6.023 x10 23 particles of Sulphur atoms
-4 x 6.023 x10 23 particles of Oxygen atoms
(iii)Molar mass sodium carbonate(IV)(Na2CO3) = relative formular mass
=[(23.0 x 2 ) + 12.0 + (16.0 x 3)]g =106.0g
6.023 x10 23 particles of sodium carbonate(IV)(Na2CO3) = 1 mole = 106.0g
6.023 x10 23 particles of sodium carbonate(IV)(Na2CO3) has:
- 2 x 6.023 x10 23 particles of Sodium atoms
-1 x 6.023 x10 23 particles of Carbon atoms
-3 x 6.023 x10 23 particles of Oxygen atoms
(iv)Molar mass Calcium carbonate(IV)(CaCO3) = relative formular mass
=[(40.0+ 12.0 + (16.0 x 3)]g =100.0g.
6.023 x10 23 particles of Calcium carbonate(IV)(CaCO3) = 1 mole = 100.0g
6.023 x10 23 particles of Calcium carbonate(IV)(CaCO3) has:
- 1 x 6.023 x10 23 particles of Calcium atoms
-1 x 6.023 x10 23 particles of Carbon atoms
-3 x 6.023 x10 23 particles of Oxygen atoms
(v)Molar mass Water(H2O) = relative formular mass
=[(2 x 1.0 )+ 16.0 ]g =18.0g
6.023 x10 23 particles of Water(H2O) = 1 mole = 18.0g
6.023 x10 23 particles of Water(H2O) has:
- 2 x 6.023 x10 23 particles of Hydrogen atoms
-2 x 6.023 x10 23 particles of Oxygen atoms
Practice
Calculate the number of moles present in:
(i)0.23 g of Sodium atoms
Molar mass of Sodium atoms = 23g
Moles = mass in grams = > 0.23g = 0.01moles
Molar mass 23
(ii) 0.23 g of Chlorine atoms
Molar mass of Chlorine atoms = 35.5 g
Moles = mass in grams = > 0.23g = 0.0065moles /6.5 x 10-3 moles
Molar mass 35.5
(iii) 0.23 g of Chlorine molecules
Molar mass of Chlorine molecules =( 35.5 x 2) = 71.0 g
Moles = mass in grams = > 0.23g = 0.0032moles /3.2 x 10-3 moles
Molar mass 71
(iv) 0.23 g of dilute sulphuric(VI)acid
Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g
Moles = mass in grams = > 0.23g = 0.0023moles /2.3 x 10-3 moles
Molar mass 98Calculate the number of atoms present in:(Avogadros constant L = 6.0 x 10 23)
(i) 0.23 g of dilute sulphuric (VI)acid
Method I
Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g
Moles = mass in grams = > 0.23g = 0.0023moles /2.3 x 10-3 moles
Molar mass 981 mole has 6.0 x 10 23 atoms
2.3 x 10-3 moles has (2.3 x 10-3 x 6.0 x 10 23) = 1.38 x 10 21 atoms
1
Method II
Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g
98.0g = 1 mole has 6.0 x 10 23 atoms
0.23 g therefore has (0.23 g x 6.0 x 10 23 ) = 1.38 x 10 21 atoms
98
(ii)0.23 g of sodium carbonate(IV)decahydrate
Molar mass of Na2CO3.10H2 O=
[(2 x 23) + 12 + (3 x16) + (10 x 1.0) + (10 x 16)] = 276.0g
Method IMoles = mass in grams = > 0.23g = 0.00083moles /
Molar mass 276 8.3 x 10-4 moles
1 mole has 6.0 x 10 23 atoms
8.3 x 10-4 moles has (8.3 x 10-4 moles x 6.0 x 10 23) = 4.98 x 10 20 atoms
1
Method II
276.0g = 1 mole has 6.0 x 10 23 atoms
0.23 g therefore has (0.23 g x 6.0 x 10 23 ) = 4.98 x 10 20 atoms
276.0
(iii)0.23 g of Oxygen gas
Molar mass of O2 = (2 x16) = 32.0 g
Method IMoles = mass in grams = > 0.23g = 0.00718moles /
Molar mass 32 7.18 x 10-3 moles
1 mole has 2 x 6.0 x 10 23 atoms in O2
7.18 x 10-3moles has (7.18 x 10-3moles x 2 x 6.0 x 10 23) =8.616 x 10 21atoms
1
Method II
32.0g = 1 mole has 2 x 6.0 x 10 23 atoms in O2
0.23 g therefore has (0.23 g x 2 x 6.0 x 10 23 ) = 8.616 x 10 21atoms
32.0
(iv)0.23 g of Carbon(IV)oxide gas
Molar mass of CO2 = [12 + (2 x16)] = 44.0 g
Method IMoles = mass in grams = > 0.23g = 0.00522moles /
Molar mass 44 5.22 x 10-3 moles
1 mole has 3 x 6.0 x 10 23 atoms in CO2
7.18 x 10-3moles has (5.22 x 10-3moles x 3 x 6.0 x 10 23) =9.396 x 10 21atoms
1
Method II
44.0g = 1 mole has 3 x 6.0 x 10 23 atoms in CO2
0.23 g therefore has (0.23 g x 3 x 6.0 x 10 23 ) = 9.409 x 10 21atoms
44.0
(b)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.
Using a wash bottle add about 200cm3 of distilled water.
Stopper the 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 moles250 cm3
= 0.4 M / 0.4 molesdm-3
Theoretical sample practice
- 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 moles500 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 moles50 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 moles50 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 1000cm320cm3
= 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 moles20 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 = 1840Molar 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 1000cm31.0cm3
= 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 moles1.0 cm3
= 18.8M/18.8 molesdm-3
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 1000Mass 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
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
C2 = molarity/concentration after diluting/adding water
V1 = volume before diluting/adding water
V2 = volume after diluting/adding water=> 0.8M x 312.5cm3 = C2 x (312.5 + 100)
C2 = 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
C2 = molarity/concentration after diluting/adding water
V1 = volume before diluting/adding water
V2 = volume after diluting/adding water
=> 0.01M x 1000 cm3 = 0.008 x V2
V2 = 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
=> M1 = 0.5 x 25.0 x1 = 1.25M / 1.25 moledm-3
5.0 x 2
Method 2
Moles of HCl used = molarity x volume1000
=> 0.5 x 25.0 = 0.0125 /1.25 x 10-2moles1000
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
- 2.0cm3 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
=> V2 = 0.5 x 2.0 x1 = 1.25M / 1.25 moledm-3
0.1 x 2
Method 2
Moles of HCl used = molarity x volume1000
=> 0.5 x 2.0 = 0.0125 /1.25 x 10-2moles1000
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
- 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.
(ii)the mass of Lead(II)Iodide formed
(Pb=207.0, I =127.0)
Stoichiometric equation: 2NaI(aq) + Pb(NO3)2(aq) -> 2NaNO3(aq) + PbI2(s)
(i)Volume of Lead(II)nitrate(V) used
Method 1
M1V1 = M2V2 -> 5 x 0.1cm3 = 0.1M x V2 cm3
n1 n2 2 1
=> V2 = 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 volume1000
=> 0.1 x 5.0 = 0.0125 /1.25 x 10-2moles1000
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
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 excess of burette contents change the colour of the indicator. This is called the end point.
Set up of titration apparatus