The Mole Concept Pure-1

The Mole Concept
A mole is a counting unit, (abbreviation; mol) used by chemist to keep track of large
numbers of atoms, ions and/or molecules.
1 mole = 6.023 × 1023 particles (atoms, ions and molecules).
6.02 × 1023 is called the Avogadro’s constant and it has the symbol NA.
Example
1 mol of Na = 6.02 × 1023 atoms of Na = 23g

1 mol of Cl- = 6.02 × 1023 ions of chloride ions = 35.5g
1 mol of H2O = 6.02 × 1023 molecules of water = 18g
A mole of atoms / ions has a mass that is equal to the Relative Atomic Mass
(Ar) of the element in grams.
A mole of molecules has a mass which is equal to the Relative Molecular
Mass (Mr) in grams.
Exercise
a) What is the mass of one Na atom
1 mol of Na: 23g
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b) 1 mol of Mg = 24g = 6.023 × 1023 atoms of Mg

What is the mass of one Mg atom?
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c) 1 mol of Ca =40g = 6.023 × 1023 atoms of Ca
What is the mass of one Ca atom?
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d) 1 mol of C =12g= 6.023 × 1023 atoms of C

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Atomic Mass Unit = 12 of mass of a C-12 atom OR mass of C-12 atom
12
1
= 12 × 1.99×10-23 g
1.99×10exp−23
= 𝑔
12
Atomic Mass Unit =1.66 × 10-23 g
Exercise
Determine the Ar of the following
I. Ar of Na
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II. Ar of Mg
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III. Ar of Ca
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RELATIVE ATOMIC MASS (Ar) OF AN ELEMENT

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# It is the average mass of an atom of an element when compared to 12 𝑡ℎ of the mass of a
Carbon-12 atom. It is a dimensionless physical quantity i.e. it is unitless.
Atom/Ion Ar
Li 7
Li+2 7
Cl- 35.5
Cl 35.5
Fe+3 56
N 14
S-2 32
Mg 24
Mg2+ 24
NB; IONS HAVE THE SAME Ar AS THE ATOM BECAUSE ELECTRONS HAVE A VERY
SMALL MASS HENCCE LOSING OR GAINING DOES NOT MAKE MUCH OF A
DIFFERENCE on the mass of an atom.
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RELATIVE MOLECULA MASS (Mr)
# It is the average mass of elements of in a compound or molecule when compared

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to 12 𝑡ℎ of the mass of a carbon-12 atom.
It is a dimensionless physical quantity hence it has no units. It is calculated by adding

up all the Ar values for the atom in the formula of a molecule or a compound.
Exercise
1 Calculate the Relative molecular Mass of the following;
E.g.
Copper (ii) sulphate, CuSO4
Mr of CuSO4 = Cu+S+O+O+O+O
=Cu+S+(4×O)
=16 +32 +(16×4)
=160
a) Aluminum oxide, Al2O3

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b) Sodium Hydroxide, NaOH

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c) Iron(iii) oxide decahydrate, Fe2O3.10H2O

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d) Sodium Carbonate pentahydrate, Na2CO3.5H2O
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e) Ammonia, NH3
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f) Ethanol, C2H5OH
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g) Aluminum Nitrate, Al (NO3)3

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h) Magnesium Phosphate, Mg3(PO4)2

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MOLAR MASS - IS THE MASS OF ONE MOLE OF A SUBSTANCE, ITS UNITS ARE g/mol.
Molar mass of an element = Ar in grams per mol
Molar mass of a molecule / compound= Mr in grams per mol
Calculate the molar mass of the following substances;

a) Zinc Nitrate, Zn (NO3)2
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b) Ammonium Phosphate, (NH4)3PO4

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c) Copper (ii) Sulphate, CuSO4

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d) Sulphuric acid, H2SO4

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e) Magnesium, Mg
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f) Hydrochloric acid, HCl
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g) Sodium hydroxide, NaOH

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𝑴𝑨𝑺𝑺
NUMBER OF MOLES=𝑀𝑂𝐿𝐴𝑅 𝑀𝐴𝑆𝑆
NB; ONLY TRUE FOR PURE SUBSTANCES
1kg=1000g 1tonne = 1000kg =1 000 000g
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Questions
1.Calculate the number of moles of the following;

a) 5g of Lithium Chloride, LiCl
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b) 2Kg of Aluminum Oxide, Al2O3

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c) 200Kg of Silver Chloride, AgCl

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d) 0.1 tones Copper (ii) oxide, CuO
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e) 30 g of Iron(ii) Sulphate, FeSO4

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f) 0.78g of Calcium Carbonate, CaCO3

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g) 10g of Sodium Carbonate, Na2CO3

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h) 200kg of Sodium Chloride Pentahydrate, NaCl. 5H2O

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i) 10g of Sodium Phosphate, Na3PO4
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j) 10g of Ammonium Sulphate, (NH4)2S

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k) 10g of Sodium Carbonate, Na2CO3

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2. Calculate the mass of the following substances;
a) 0.5mol of Hydrogen Chloride, HCl

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b) 2.05 mols of Hydrogen Sulphide, H2S
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c.) 0.317 mols of Iron (iii) Sulphate, FeSO4

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d.) 0.323 mols of Potassium Dichromate(vii), K2Cr2O7

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e.) 0.0558 mol of calcium, CaCl2

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f) 0.0125 moles of Copper (ii) Sulphate pentahydrate, CuSO4.5H2O

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g) 0.1275 mols of Iron (iii) oxide decahydrate, Fe2O3.10H2O

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h) 0.5 mols OF Aluminum Metal, Al
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F) 0.313 mols of Aluminum Phosphate, AlPO4

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J) 0.125 mols of Ammonia, NH4

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MOLAR VOLUME OF A GAS

Molar Volume of a gas is the volume of 1 mole of a gas at room temperature and pressure.
It is 24dm3 per mole/ 24 Cubic decimeter per molecule.
 1 mole of any gas at room temperature and pressure has a volume of 24dm 3or 24
000cm3
 1 mole of a gas (6.02 × 1023 particles) occupies a space of 24 dm3 at room
temperature and pressure.
1dm3=1000cm3=1000ml=1L
24dm3=24 000cm3=24 000ml=24L
1dm3=1L
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 1 mole of oxygen gas at r.t.p has a volume of 24dm3 or 24 000cm3
 1 mole of nitrogen gas at r.t.p has a volume of 24dm3 or 24 000cm3
 1 mole of carbon dioxide at r.t.p has a volume of 24dm3 or 24 000cm3
 1 mole of Ammonia at r.t.p has a volume of 24dm3 or 24 000cm3
𝑽𝑶𝑳𝑼𝑴𝑬(𝑮𝑨𝑺)
NUMBER OF MOLES(GAS)=𝑀𝑂𝐿𝐴𝑅 𝑉𝑂𝐿𝑈𝑀𝐸
 THIS PRINCIPLE IS ONLY TRUE FOR GASES AT ROOM

TEMPERATURE AND PRESSURE.
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EXERCISE
Q1. Calculate the volume of the following gases at r.t.p;
a) 0.5 moles of hydrogen gas, H2
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b) 0.033 moles of methane, CH4

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c) 6.02 × 1023 molecules of oxygen gas, O2

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d) 5.73g of ammonia gas, NH3

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e) 2.24g of carbon dioxide, CO2

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Q2. Calculate the number of moles for the following substances at room temperature and
pressure;
a.) 180cm3 of Nitrogen gas, N2

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b.) 240 cm3of carbon dioxide, CO2
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c.) 102 dm3 of Helium gas, He

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d.) 480dm3 of Sulphur dioxide gas, SO2

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e.) 4.7 Kg of Ammonia gas, NH3

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f.) 10L of 0xygen gas, O2

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g.) 24.08 × 1023 molecules of hydrogen gas, H2

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h.) 250cm3 of hydrogen chloride gas, HCl
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i.) 3.93g of methane gas, CH4

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CONCENTRATION
Concentration is the amount of solute dissolved in a certain volume
of solution.
Solute + Solvent Solution
𝑨𝑴𝑶𝑼𝑵𝑻 𝑶𝑭 𝑺𝑶𝑳𝑼𝑻𝑬
CONCENTRATION=𝑉𝑂𝐿𝑈𝑀𝐸 𝑂𝐹 𝑆𝑂𝐿𝑈𝑇𝐼𝑂𝑁
 If we have dissolved a lot of solute in a small volume of solvent,

then we say that the solution is concentrated.
 If we have dissolved a small amount of solute in a large volume of solvent
then we say that the solution is dilute.
STANDARD SOLUTION- it is a solution of known concentration. It is prepared

using a volumetric flask.
MOLARITY- is the concentration of a solution when given in mol/dm3.
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mol/dm3=M, where M=Molar
𝑵𝑼𝑴𝑩𝑬𝑹 𝑶𝑭 𝑴𝑶𝑳𝑬𝑺 (𝑺𝑶𝑳𝑼𝑻𝑬)

MOLARITY= 𝑉𝑂𝐿𝑈𝑀𝐸 𝑂𝐹 𝑆𝑂𝐿𝑈𝑇𝐼𝑂𝑁(𝑑𝑚3)
N.B; THIS FORMULA CAN ONLY BE USED IF YOU ARE DEALING WITH
SOLUTIONS. E.g; SALT SOLUTIONS, ACIDS AND ALKALINE
SOLUTIONS.
Exercise
Q1. Calculate the Morality of the following;
a.) Copper (ii) Sulphate solution prepared from 0.4135g of Copper (ii) Sulphate in 500cm3,
CuSO4.
1. Measure 0.4135g of Copper (ii) Sulphate and add it into a beaker and
then add enough water to dissolve the solute.
2. Transfer the solution into a 500cm3 volumetric flask and then rinse the
beaker with water and add the solution into the volumetric flask.
3. Add water until near the line marked on the volumetric flask.
4. A pipette or dropper is used to fill the volumetric flask to the line on the
flask.
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5. Seal the volumetric flask and invert it to thoroughly mix the solution.
b) 0.5 moles of sodium chloride (NaCl) dissolved in a 1 dm3 solution.
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c.) 0.25 moles of Sulphuric acid, (H2SO4) dissolved in a 500ml of solutions.

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d) 0.2g of sodium hydroxide (NaOH) dissolved in 2dm3 of the solution.

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e) 15.2 g of copper (ii) chloride dissolved in 20 liters of the solution

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f). A solution has a volume of 250ml and contains 42.5 g of NaCl.
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Q2. Calculate the concentration of the following solution;
a). 0.125 mols of hydrochloric acid (HCl) dissolved in 25cm3 solution

i) in mol/cm3
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ii) mol/dm3
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b). 0.5 moles of sodium hydroxide (NaOH) dissolved in 250cm3

i) mole/cm3
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ii) mol/dm3
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c). 5.95g of Potassium Bromide (KBr) dissolved in 1 liter solution
i) g/L
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ii) mol/dm3
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c) 2.00g of potassium permanganate (KMnO4) dissolved in 500ml solution.

i) in g/ml
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ii) mol/dm3
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c). Calculate the concentration of sodium hydroxide solution (NaOH) prepared

by dissolved 0.133g of sodium hydroxide 100ml solution.
i). in g/ml
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ii) mol/dm3
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d). Calculate the concentration of Barium Nitrate solution (Ba (NO3)2) prepared
by dissolving 0.811g of Barium Nitrate in water to make a 250ml solution.
i) in g/ml
ii) mol/dm3
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Q2. Calculate the Molarity of the following solutions;

a) A solution chloride (NaCl) contains 11.7g of NaCl in 500cm3 solution
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b) A solution of Copper (ii) Sulphate (CuSO4) contains 8.00g of CuSO4 in

250ml solution.
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Q3. Calculate the number of moles of the following;
a) Hydrochloric acid (HCl) in 25.0ml of 0.25 mol/dm3 of hydrochloric

acid solution.
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b) Calculate the number of moles of silver nitrate (AgNO3) in 1800ml of

0.1mol/dm3 of silver nitrate solution.
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c) Calculate the number of Sulphuric acid (H2SO4) in 100cm3 of 0.0100 mol/dm3

solution of Sulphuric acid.
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Stoichiometry
This is the calculation of amounts of reactants and products in chemical reactions in
chemistry by making use of stoichiometric ratios/mole ratios from a balanced
equations. Stoichiometry is founded on the law of conservation of mass where the total
mass of the reactants equals the total mass of the products.
1. Aluminium reacts with chlorine gas to form aluminium chloride. In this reaction 35.0g of
aluminium was reacted with excess chlorine gas.
2Al(s) + 3Cl2(g)  2AlCl3(s)
A. Calculate the number of moles Aluminium in the 35.0g sample.
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B. Use your answer to (a) to calculate then number of moles of Aluminium chloride produced.
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C. Calculate the mass of aluminium chloride produced.
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D. Use your answer to (a) to calculate the number of moles chlorine gas that reacted with
aluminium.
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E. Calculate the mass of chlorine gas
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2. In an experiment, 100cm3 of 0.100 mol/dm3 of hydrochloric acid was reacted with excess
zinc powder.
2HCl(aq) + Zn(s)  ZnCl2(aq) + H2(g)
A. Calculate the number of moles in 100cm3 of 0.100mol/dm3 of hydrochloric acid
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B. Use your answer to (a) to calculate the number of moles of zinc that reacted with the acid
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C. Calculate the mass of zinc that reacted with the acid
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D. Calculate the volume of hydrogen gas measured at room temperature and pressure.
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3. Propane reacts with oxygen gas during complete combustion to form carbon dioxide gas and
steam. During this combustion reaction 117.6g of propane reacted with excess oxygen gas.
a. Calculate the number of moles of propane in 117.6g of propane.
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b. Use your answer to (a) to calculate the number of moles of steam.
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c. Calculate the mass of steam in the reaction.
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d. Use your answer in (a) to calculate the number of moles of carbon dioxide gas produced.
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e. Calculate the mass of carbon dioxide gas produced
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f. Use your answer to (a) to calculate the number of moles of oxygen gas that completely reacted
with propane.
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g. Calculate the mass of oxygen gas that reacted with propane.
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4. Excess magnesium was added to 25.0cm3 of dilute hydrochloric acid.

2HCl(aq) + Mg(s)  MgCl2(aq) + H2(g)
The hydrogen produced was completely reacted with 2.0g of copper (ii) oxide as shown by the
equation.
CuO(s) + H2(g)  Cu(s) + H2O(g)
a. (i) Calculate the Mr of copper (ii) oxide
(ii) Write the ionic equation of, 2HCl(aq) + Mg(s)  MgCl2(aq) + H2(g)
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(iii) Calculate the number of moles in 2.0g of copper (ii) oxide
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(iv) Calculate the number of moles of hydrogen that reacted with the copper (ii) oxide
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(v) Calculate the number of hydrochloric acid used to make the amount of hydrogen in your answer
to (iii)
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(vi) What is the concentration of the acid in g/cm3?
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5. Reaction 5.00g of calcium carbonate with excess hydrochloric acid gives calcium chloride, carbon
dioxide and water.
A. write a balanced equation for the reaction
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(i) Calculate the number of moles of calcium carbonate used in the reaction.
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(ii) Use your answer to (i) to calculate the number of moles of carbon dioxide produced
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(iii) Calculate the volume of carbon dioxide gas produced at room temperature and pressure
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(iv) Calculate the mass of carbon dioxide
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Dilution
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1. Tetlo diluted 0.05dm3 of a 0.1 mol/dm3 copper (ii) sulphate (CuSO4) solution to 0.250dm3. Calculate
the concentration of the solution after dilution.
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2. Calculate the concentration of the solution formed when 2.00litres of 0.500 solution of nitric acid
is diluted to 3.50litres of solution.
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3. Calculate the volume of the solution formed when 25.0cm3 of 0.150mol/dm3 of hydrochloric acid
is diluted to make a solution of 0.500mol/dm3
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(i) What volume of water was added when diluting the solution?
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Titration
1. 25.0cm3 of 0.100mol/dm3 sodium hydroxide solution (NaOH) was titrated with sulphuric acid
using methyl orange indicator. The results of the titration are shown below.
Titration 1 2 3 4 5
final 25.8 30.0 30.2 30.6 25.9
volume(cm3)
Initial 0.6 4.6 5.0 2.0 0.0
volume(cm3)
Volume(cm3) 25.2 25.4 25.2 28.6 25.9
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(i) Calculate the average volume of the acid used.
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(ii) Calculate the number of moles of sodium hydroxide (NaOH) that reacted with the acid.
NaOH (aq) + H2SO4 (aq)  Na2SO4 (aq) + 2H2O (l)
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(iii) Use the equation to calculate the number of moles of the acid that reacted with sodium
hydroxide.
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(iv) Use your answer to questions (i) and (ii) to calculate the concentration of sulphuric acid.
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2. A student determines the concentration of dilute sulphuric acid (H2SO4) by titrating the acid
against 25.0cm3 portions of 0.100mol/dm3 sodium hydroxide (NaOH) solution. Methyl orange
indicator was used.
(a) What is the colour of methyl orange in sodium hydroxide solution?
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(b) The diagrams show burettes at the start and at the end of three titrations.
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Please print and
paste these
diagrams or
draw on a maths
paper and paste
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25
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c. complete the table below to show the volume of sulphuric acid.
Titration number 1 2 3
Final reading (cm3)
Initial reading (cm3)
Volume of acid used (cm3)
Tick the best results
d. using the ticked values, calculate the average volume of the acid used.
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e. using the equation given, calculate the concentration of sulphuric acid.
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(f) State any two sources of inaccuracy when doing the experiment.
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(g) The experiment was repeated without adding the indicator, mixing 25.0cm3 of 0.100mol/dm3
NaOH with the average volume of H2SO4 acid found in (ii) b. one of the products is sodium sulphate,
Na2SO4.
(h)Suggest how crystals of Na2SO4 are obtained from the mixture.
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% mass composition
1. Calculate the % composition of copper in copper (ii) sulphate (CuSO4)
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2. Calculate the % composition of Iron in Iron (iii) oxide decahydrate, Fe2O3. 10H2O
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3. Calculate the % composition of copper in hydrated copper (ii) sulphate, CuSO4
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4. Calculate the % composition of copper in hydrated copper (ii) sulphate, CuSO4.5H2O
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5. Calculate the % composition of oxygen in Aluminium sulphate, Al2(SO4)3
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6. Calculate % of water in hydrated magnesium sulphate, MgSO4.7H2O
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% Purity
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1. Limestone is an ore of calcium carbonate. 6.00grams of limestone was found to contain 5.00grams
of calcium carbonate. Find the percentage of the ore.
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2. An impure sample of calcium carbonate CaCO3 contains calcium sulphate as an impurity. When
excess hydrochloric acid was added to 6g of the sample, 1200ml of carbon dioxide gas was produced
at r.t.p
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
a. find the number of moles of carbon dioxide, CO2 produced at r.t.p
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b. use your answer to (a) to find the number of moles of pure CaCO3
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c. find the mass of pure CaCO3
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d. find the % purity
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3. Solution X contains 5.00g of impure sulphuric acid dissolved in 1dm3 of solution. 25.0ml of
solution X required 23.5ml of 0.100mol/dm3 NaOH for the reaction in a titration.
Calculate the % purity of the acid.
H2SO4(aq) + 2NaOH(aq) → NaSO4(aq) + 2H2O(l)

a. calculate the number of moles of 23.5ml of 0.100mol/dm3 of NaOH solution
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b. use your answer in (a) to calculate number of moles of pure H2SO4 in 25.0 ml solution
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c. how many moles are in 1dm3 of the H2SO4. Use your answer to (b)
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d. calculate the mass of pure H2SO4
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4. A solution contains 4.00g of impure NaOH per dm3. In a titration reaction 25.0ml of the impure
alkali solution required 22.5ml of 0.100mol/dm3 HCl for reaction. Calculate the % purity of the alkali.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(L)
a. calculate the number of moles in 22.5ml of 0.100mol/dm3 HCl solution.
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b. use your answer in (a) to calculate the number of moles of pure NaOH
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c. calculate the number of moles in 1.00dm3
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d. calculate the mass of pure NaOH in the sample
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e. calculate % purity
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% yield
1. Calcium carbonate decomposes on heating according to the following
CaCO3(s) → CaO(S) + CO2(g)
i) calculate the relative molecular mass of pure calcium carbonate.
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ii) 20.0g of calcium carbonate was used. Calculate the number of moles of calcium carbonate used
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(ii) Calculate the theoretical mass of calcium oxide, CaO formed.
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iv) At the end of heating, the calcium oxide formed was weighed and only 10g was collected. This is
the actual yield/ actual mass.
Calculate the percentage yield of the product.
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(v) Name a substance that would react with calcium oxide to form calcium chloride
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(vi) Write a balanced equation for the reaction. (Include state symbols).
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2. When an excess of zinc was added to dilute sulphuric acid, aqueous zinc sulphate was formed and
hydrogen was evolved.
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
a. describe a test to confirm that the gas produced was hydrogen.
Test: ………………………………………………………………………………………………………………………………………….
Results: ……………………………………………………………………………………………………………………………………..
b. describe how you would obtain crystals of hydrated zinc sulphate from zinc sulphate solution.
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3. The Haber process is used in the manufacturing of ammonia (NH3). Ammonia is a highly useful
substance as it is used in the manufacturing of fertilizers, also used in the manufacturing of nitric
acids, explosives and domestic bleaches.
N2(g) + 3H2(g)  2NH3(g)
a. 12353g of nitrogen gas is used in this reaction. Calculate the number of moles of nitrogen
gas used.
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b. use your answer to (a) to calculate the number of moles of ammonia

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c. calculate the theoretical mass of ammonia
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d. the actual mass of ammonia obtained is 5000g. Calculate the % yield of the product (ammonia)
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e. give three potential reasons why it is not always possible to get a 100% percentage yield
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4. Calcium oxide is reacted with water to form calcium hydroxide. If the theoretical yield is 3.0g, but
only 1.40g of calcium hydroxide is produced.
Calculate the % yield of calcium hydroxide.
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5. Ethanol can be manufactured from ethene by catalytic addition of steam according to the
reaction.
C2H4(g) + H2O(g)  C2H5OH(l)
10g of ethene is used in this process.
a. calculate the number of mole of ethene
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b. use your answer to (a) to calculate the number of moles of ethanol
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c. calculate the theoretical mass of ethanol.
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d. the actual mass of ethanol obtained is 1.64kg. Calculate the % yield
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41
1. Calculating the value of X in hydrated salt Na2CO3∙XH2O. In a titration reaction, 25.0ml of solution
Q required 22.6ml of 0.100 M of Nitric Acid. Solution Q is an aqueous solution of sodium carbonate
containing 7.15g of hydrated sodium carbonate (Na2CO3∙XH2O) dissolved in 500ml of solution.
Na2CO3∙XH2O(aq) + 2HNO3(aq) → 2NaNO3(aq) + CO2(g) + XH2O(l)
i) Calculate the number of moles in Nitric acid
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ii) Use your answer to (i) to calculate the number of moles of Na2CO3∙XH2O in 25ml of sample
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iii)Calculate the number of moles of Na2CO3∙XH2O in 500 ml of the sample.
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iv)Calculate the molar mass of Na2CO3∙XH2O
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v) Work out X in Na2CO3∙XH2O
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EMPIRICAL FORMULA AND MOLECULAR FORMULA
42
Empirical formula-Shows the simplest ratios of atoms in the compound / molecule
Molecular formula-Shows the total number of atoms of each element in the compound / molecule
-Shows the total amount of each element in the compound.
The formula of a compound can be worked out by finding the masses of the elements present
experimentally and the masses can then be used to calculate the formula.
Molecular formula = n x empirical formula
Hence, Mr = n x Me
If n=1, then the molecular formula is equal to the empirical formula
Example
C6H12O6 CH2O
6:12:6 1:2:0
MOLECULAR FORMULA Empirical formula
Molecular formula= n x empirical formula
= 6 x (CH2O)
= C6H12O6
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Molecular formula=n x Empirical formula
Mass of the Crucible=56g
Mass of crucible + Iron =58.8g
Mass of crucible +oxide of iron=60g
Mass of iron=2.8g
Mass of oxygen=1.2g
Fe O
mass
# of moles
Divide all the # of moles by the

smaller # of moles and find the
simplest ratio
Empirical formula; ……………………………………………………………………………………………………………………………………
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1. Deduce the empirical formula of the following compounds
a) 3.5g of Nitrogen combined with 8.0g Oxygen
N O
mass
# of moles

simplest ratio
b) 2.4g of Carbon, 6.4g Oxygen and 0.2g of Hydrogen
Fe O
mass
# of moles

simplest ratio
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c) 9.2g of Sodium, 12.8g of sulphur and 9.6g of oxygen
Na S O
mass
# of moles
Divide all the # of moles

by the smaller # of moles
and find the simplest
ratio
2. Determine the molecular formula of the following compounds;
a) 75% carbon and 25% hydrogen by mass
C H
# of moles

simplest ratio
46
Molecular formula; ………………………………………………………………………………………………………………………………
b) 46.7% Silicon and 53.3% Oxygen by Mass
Si O
# of moles

simplest ratio
c) 48.6% carbon, 43.4% Oxygen and 8% Hydrogen
C H O
# of moles

ratio
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d) 43.4% Sodium , Carbon 11.3% and Oxygen 45.3% by mass
Na C O
# of moles

ratio
e) 60.05% Carbon, 4.44% hydrogen and Oxygen 35.56%
C H O
# of moles

ratio
48
Molecular formula=n x Empirical formula
Mr = n x Me
Where Mr = relative molecular mass
Me = empirical formula mass
If n = 1, Mr = Me
Molecular formula gives us the relative molecular mass (Mr), while the
Empirical formula gives us the empirical formula mass (Me).
Calculate the molecular formula of the following compounds;
a) The empirical formula of the compound is CH3O and its relative molecular mass (Mr) is 62.
(iii) Molecular formula = n x empirical formula (ii) n = Mr

= 2 x (CH3O) Me
= C2H6O2 =62
31
=2
(i) Empirical formula mass (Me) of CH3O = C + (Hx3) + O
= 12 + (1x3) + 16
=31
b) The empirical formula of the compound is CH2O and its relative molecular mass (Mr) is 90.
Molecular formula = n x empirical formula n = Mr

= Me
= =
Empirical formula mass (Me) of CH2O =
49
=
c) The empirical formula of the compound is CH2O2 and its relative molecular mass (Mr) is 46.

= Me
= =
Empirical formula mass (Me) of CH2O2 =
d) A monosaccharide contains 6.0g of carbon, 1.0g of hydrogen and 8.0g of oxygen.
i) Calculate the empirical formula of this compound.
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ii) The relative molecular mass (Mr) of this compound is 180. Calculate the molecular formula
of this compound.

= Me
= =
Empirical formula mass (Me) of ……………………. =
50
1. Find the molecular formula of the following compounds;
a) The empirical formula of the compound is C3H3O and its relative molecular mass is 116.
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b) The empirical formula of the compound is CH and its relative molecular mass is 78
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c) The empirical formula of the compound is H2CO2 and its relative molecular mass is 46
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d). i) Calculate the empirical formula of the compound which contains 6g of carbon, 1 g of Hydrogen
and 8g of Oxygen
C H O
mass
# of moles

ratio
Empirical formula; ………………………………………………………………………………………………………………………………
(ii) The relative molecular mass of the above compound is 180. Find its molecular formula.
Skip 5 lines
2. The structural of caffeine is show below
i) Find the molecular formula
Skip 2 lines
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ii) Find the empirical formula
Skip 6 lines
f) Methyl Benzoate is an ester with the structure shown
i) What is the molecular formula of the compound?
Skip 2 lines
ii) What is the empirical formula of the compound?
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53
e) The structural formulae of palmitic acid and stearic acid is show below.
Stearic acid
a) Write the molecular formula of;
i) Palmitic acid ………………………………………………………………….
ii) Stearic acid ……………………………………………………………………
b) Write the empirical formula of;
i) Palmitic acid ………………………………………………………………….
ii) Stearic acid ……………………………………………………………….
54
c) Calculate the relative molecular molecule (Mr) of;
i) Palmitic acid ………………………………………………………………….
ii) Stearic acid ……………………………………………………………………
55

The Mole Concept Pure-1

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The Mole Concept

1 mol of Na = 6.02 × 1023 atoms of Na = 23g

b) 1 mol of Mg = 24g = 6.023 × 1023 atoms of Mg

d) 1 mol of C =12g= 6.023 × 1023 atoms of C

Atomic Mass Unit =1.66 × 10-23 g

RELATIVE ATOMIC MASS (Ar) OF AN ELEMENT

# It is the average mass of elements of in a compound or molecule when compared

It is a dimensionless physical quantity hence it has no units. It is calculated by adding

a) Aluminum oxide, Al2O3

b) Sodium Hydroxide, NaOH

c) Iron(iii) oxide decahydrate, Fe2O3.10H2O

g) Aluminum Nitrate, Al (NO3)3

h) Magnesium Phosphate, Mg3(PO4)2

Calculate the molar mass of the following substances;

b) Ammonium Phosphate, (NH4)3PO4

c) Copper (ii) Sulphate, CuSO4

d) Sulphuric acid, H2SO4

g) Sodium hydroxide, NaOH

NB; ONLY TRUE FOR PURE SUBSTANCES

1kg=1000g 1tonne = 1000kg =1 000 000g

1.Calculate the number of moles of the following;

b) 2Kg of Aluminum Oxide, Al2O3

c) 200Kg of Silver Chloride, AgCl

e) 30 g of Iron(ii) Sulphate, FeSO4

f) 0.78g of Calcium Carbonate, CaCO3

g) 10g of Sodium Carbonate, Na2CO3

h) 200kg of Sodium Chloride Pentahydrate, NaCl. 5H2O

j) 10g of Ammonium Sulphate, (NH4)2S

k) 10g of Sodium Carbonate, Na2CO3

2. Calculate the mass of the following substances;

a) 0.5mol of Hydrogen Chloride, HCl

c.) 0.317 mols of Iron (iii) Sulphate, FeSO4

d.) 0.323 mols of Potassium Dichromate(vii), K2Cr2O7

e.) 0.0558 mol of calcium, CaCl2

f) 0.0125 moles of Copper (ii) Sulphate pentahydrate, CuSO4.5H2O

g) 0.1275 mols of Iron (iii) oxide decahydrate, Fe2O3.10H2O

F) 0.313 mols of Aluminum Phosphate, AlPO4

J) 0.125 mols of Ammonia, NH4

MOLAR VOLUME OF A GAS

24dm3=24 000cm3=24 000ml=24L

 THIS PRINCIPLE IS ONLY TRUE FOR GASES AT ROOM

b) 0.033 moles of methane, CH4

c) 6.02 × 1023 molecules of oxygen gas, O2

d) 5.73g of ammonia gas, NH3

e) 2.24g of carbon dioxide, CO2

a.) 180cm3 of Nitrogen gas, N2

c.) 102 dm3 of Helium gas, He

d.) 480dm3 of Sulphur dioxide gas, SO2

e.) 4.7 Kg of Ammonia gas, NH3

f.) 10L of 0xygen gas, O2

g.) 24.08 × 1023 molecules of hydrogen gas, H2

i.) 3.93g of methane gas, CH4

Solute + Solvent Solution

 If we have dissolved a lot of solute in a small volume of solvent,

STANDARD SOLUTION- it is a solution of known concentration. It is prepared

MOLARITY- is the concentration of a solution when given in mol/dm3.

𝑵𝑼𝑴𝑩𝑬𝑹 𝑶𝑭 𝑴𝑶𝑳𝑬𝑺 (𝑺𝑶𝑳𝑼𝑻𝑬)

b) 0.5 moles of sodium chloride (NaCl) dissolved in a 1 dm3 solution.

c.) 0.25 moles of Sulphuric acid, (H2SO4) dissolved in a 500ml of solutions.