### TOPIC 1 – CHEMICAL EQUATIONS

### TOPIC 2 – HARDNESS OF WATER

### TOPIC 3 – ACIDS, BASES AND SALTS

### TOPIC 4 – THE MOLE CONCEPT AND RELATED CALCULATIONS

### TOPIC 5 – VOLUMETRIC ANALYSIS

### TOPIC 6 – IONIC THEORY AND ELECTROLYSIS

### TOPIC 7 – CHEMICAL KINETICS, EQUILIBRIUM AND ENERGETICS

### TOPIC 8 – EXTRACTIONS OF METALS

### TOPIC 9 – COMPOUNDS OF METALS

THE MOLE CONCEPT AND RELATED CALCULATIONS

**TOPIC 4: THE MOLE CONCEPT AND RELATED CALCULATIONS**

**The Mole as a Unit of Measurement**

The Mole with Other Units of Measurements

Compare the mole with other units of measurements

When carrying out an experiment, a chemist cannot weigh out a single atom, ion, electron, proton or molecule of a substance. These particles are simply very small. A counting unit that is useful in practical chemistry must be used.

The standard unit is called one

**mole**of the substance. One mole of each of these different substances contains the same number of the particles (atoms, molecules, ions, electrons, protons, neutrons, etc). That number per mole has been worked by several different experimental methods and is found to be 6.0 × 10^{23}. The value 6.0 × 10^{23}is called**Avogadro’s constant**or Avogadro’s number and is abbreviated as L. It is named after the nineteenth-century Italian chemist, Amedeo Avogadro.The value 6.0 × 10

^{23}is obtained through the following relationship.The mass of one atom of carbon-12 is 1.993 × 10^{-23}g. Then, the number of atoms present in 12g of carbon-12 is derived as follows:1 atom = 1.993 × 10

^{-23}gX atoms = 12g

**\**X = 6.0 × 10

^{23}atoms.

Therefore, the number of atoms in 12g of carbon-12 and hence the number of particles in a mole are 6.02 × 10

^{23}atoms.Hence, Avogadro’s number is the number of atoms in exactly 12g of carbon-12 isotope.One mole of any substance contains as many as many elementary particles as the Avogadro’s number (constant).

So, from the above explanation, the

**mole**can be defined as*the amount of a substance that contains as many elementary particles as the number of atoms present in 12g of carbon-12 isotope.*Substance |
Formula |
Relative formula mass, M_{r} |
Mass of one mole (molar mass) |
This mass (1 mole) contains |

Carbon | C | 12 | 12g | 6.0 × 10^{23} carbon atoms |

Iron | Fe | 56 | 56g | 6.0 × 10^{23} iron atoms |

Hydrogen | H_{2} |
2 × 1 = 2 | 2g | 6.0 × 10^{23} molecules |

Oxygen | O_{2} |
2 × 16 = 32 | 32 | 6.0 × 10^{23} molecules |

Water | H_{2}O |
(2×1) + 16 = 18 | 18g | 6.0 × 10^{23} formula units |

Magnesium oxide | MgO | 24 + 16 = 40 | 40g | 6.0 × 10^{23} formula units |

Calcium carbonate | CaCO_{3} |
40+12+(3×16) = 100 | 100g | 6.0 × 10^{23} formula units |

Silicon oxide | SiO_{2} |
28 + (2 × 16) = 60 | 60g | 6.0 × 10^{23} formula units |

Fe^{3+} |
Fe^{3+} |
56 | 56g | 6.0 × 10^{23} iron(III) ions |

Cl^{–} |
Cl^{–} |
35.5 | 35.5g | 6.0 × 10^{23 }chloride ions |

e^{–} |
e^{–} |
– | – | 6.0 × 10^{23 }electrons |

The other substances, which also exist as molecules, include ozone molecule (gas), O

_{3}; phosphorus molecule (solid), P_{4}; sulphur molecule, S_{8}, etc.In real life, when dealing with large numbers of small objects, it is usual to count them in groups. The objects are grouped and counted in unit amounts. For example, we buy a carton of soap, a gallon of kerosene, a crate of soda, a dozen of pencils, a ream of papers, etc.

Unit |
Number of objects per unit |

Pair | 1 pair = 2 objects, e.g. gloves, shoes, socks, scissors, etc are always sold in pairs. |

Dozen | 1 dozen = 12 objects e.g. a dozen of cups, plates, spoons, etc. |

Gross | 1 gross = 144 objects, e.g. a box of blackboard chalk contains 144 pieces of chalk. |

Ream | 1 ream = 500 objects, e.g. papers are sold in reams of 500 sheets. |

Mole | 1 mole = 6.02 ×10^{23} particles. In chemistry, extremely small particles are expressed in moles. For example:1 mole of atoms = 6.02 ×10^{23} atoms1 mole of electrons = 6.02 ×10^{23} electrons1 mole of protons = 6.02 ×10^{23} protons1 mole of ions = 6.02 ×10^{23} ions1 mole of molecules = 6.02 ×10^{23} molecules |

Molar Quantities of Different Substances

Measure molar quantities of different substances

_{2}H

_{5}OH, contains two carbon atoms, six hydrogen atoms and one oxygen atom. So, the molar mass of ethanol can be calculated thus: Molar mass of C

_{2}H

_{5}OH = (2 × 12) + (6×1) + 16 = 46g.

^{-1}).

*relative atomic mass*or

*relative molecular mass*has

*no unit*while molar masses are always expressed in grams or kilograms.

*molar mass*

*of a compound*is the same as the

*relative molecular mass*and the

**molar mass**

**of an**

**element**is the same as the

**relative atomic mass**(A

_{r}) of that element. The only difference lies in the units.

Example 1

- M(CO
_{2}) = 44g (or g mol^{-1}) = molar mass of carbon dioxide - M
_{r}(CO_{2}) = 44 = relative molecular mass of carbon dioxide - M(Fe) = 56g (or g mol
^{-1}) molar mass of iron - M
_{r}(Fe) = 56 = Relative atomic mass of iron

Similarly, the molar masses of each of the following substances can be calculated using values for the relative atomic masses of the elements.

Substance |
Formula |
Molar mass |

Ammonia | NH_{3} |
14 + 1×3 = 17g |

Ammonium chloride | NH_{4}Cl |
14 + (1×4) + 35.5 = 53.5g |

Lead (II) nitrate | Pb(NO_{3})_{2} |
207 + (14×2) + (16×6) = 331g |

Sulphuric acid | H_{2}SO_{4} |
(1×2) + 32 + (16×4) = 98g |

Calcium carbonate | CaCO_{3} |
40 + 12 + (16×3) = 100g |

Potassium dichromate | K_{2}Cr_{2}O_{7} |
(39×2) + (52 ×2) + (16×7) = 294g |

**Application of the Mole Concept**

Known Masses of Elements, Molecules or Ions to Moles

Convert known masses of elements, molecules or ions to moles

Example 2

Convert 49g of sulphuric acid, H

_{2}SO_{4}, into moles.*Given:*Mass = 49g; molar mass = 98g*Formula:*

*Solution:*49g of H

_{2}SO

_{4}= 49/98= 0.5 mol.

Known Volumes of Gases at S.T.P to Moles

Convert known volumes of gases at S.T.P to moles

^{3}. This volume is called the molar volume of a gas. The molar volume of a gas, therefore, has the value of 22.4 dm

^{3}at s.t.p. Remember that 1 dm

^{3}(1 litre) = 1000 cm

^{3}. One important thing about this value is that it applies to all gases. Therefore, at s.t.p. 32g of oxygen (O

_{2}) or 17g of ammonia (NH

_{3}) or 44g of carbon dioxide (CO

_{2}) or 40g of argon (Ar) will occupy a volume of 22.4 dm

^{3}. This makes it easy to convert the volume of any gas at s.t.p. into moles, or moles into volume. However, it is important to note that as the conditions of temperature and pressure change the molar volume will also change.

^{3}).

**,**4.4d m

^{3}of carbon dioxide gas at s.t.p. = 4.4/22.4= 0.196 mol.Similarly, 2.24 dm

^{3}of neon gas at s.t.p. = 2.24/22.4= 0.1 mol.

^{3}, then it should be divided by the molar volume of a gas expressed in cm

^{3}. For example, 560 cm

^{3}of nitrogen gas = 560cm

^{3}/22400cm

^{3}mol= 0.025 mol.

^{3}and then divides by the molar volume, expressed in dm

^{3}, that is, 0.46dm

^{3}/22.4dm

^{3}= 0.25mol

Masses of Solids or Volumes of Known Gases to Actual Number of Parties

Change masses of solids or volumes of known gases to actual number of parties

The number of particles in one mole of any substance is 6.02 × 10

^{23}. To find the number of particles in a substance, we use the expression:- N = n.L, where
- N = the number of particles in that substance;
- n = the amount of substance (moles); and
- L = the Avogadro’s constant (6.02 × 10
^{23}).

^{3}of ammonia gas to the actual number of ammonia (NH

_{3}) molecules, change 5.6 dm

^{3}of ammonia to moles =0.46dm

^{3}/22.4dm

^{3}=0.25 mol. Then multiply by the Avogadro’s constant to get the total number of molecules0.25 × 6.02 × 10

^{23}= 1.5 × 10

^{23}molecules

^{3}of hydrogen gas = 1.12/22.4= 0.05 mol. This is equal to 0.05 × 6.02 × 10

^{23}= 3.0 ×10

^{22}molecules

^{3}) of a gas at s.t.p. = 6.02 × 10

^{23}molecules. So, 5.6 dm

^{3}= 5.6×6.02 × 10

^{23}/22.4= 1.5 × 10

^{23}molecules

Molar Solutions of Various Soluble Substances

Prepare molar solutions of various soluble substances

^{3}or 1000 cm

^{3}) of the solution.Let us consider the case of sodium hydroxide, NaOH. The molecular weight of this compound is 40g. Therefore, a molar solution of sodium hydroxide will contain 40g in 1000 cm

^{3 }(1 dm

^{3}) of the solution.

_{2}CO

_{3}. 1 mole of this carbonate weights 106g. Hence, its molar solution will contain 106g of the anhydrous salt in 1000 cm

^{3}of solution.If, however, 0.1 moles (10.6g) of the solute is dissolved in 1.0 dm

^{3}, the solution is 0.1 molar. But if 0.1 moles is dissolved in 0.1 dm

^{3}of the solution, the solution is still 1.0 molar (since 1 dm

^{3}of solution would contain 1.0 mole of the solute).

The molecular weights of some common substances are shown below:

Compound |
Molecular weight (1 mole) |

Potassium hydroxide, NaOH | 56g |

Hydrochloric acid, HCl | 36.5g |

Sulphuric acid, H_{2}SO_{4} |
98g |

Sodium chloride, NaCl | 58.5g |

Sodium bicarbonate, NaHCO_{3} |
84g |

Calcium hydroxide, Ca(OH)_{2} |
74g |

^{3}(1 dm

^{3}) of distilled water. We see, therefore, that 40g of sodium hydroxide in 1000 cm

^{3}of solution will give a 1.0M solution. Hence, 20g of the hydroxide should give a 0.5M solution. In a similar way, we can make derivative solution concentrations ranging as follows: 0.1M, 0.2M, 0.3M, 0.4M….1M, 2M, etc.

^{3}. The concentration ranges like these are known as

*molarities*of solutions. Hence, 0.5M sodium carbonate can also be read as “a sodium carbonate solution with a molarity of 0.5M.”

**The Concentration of Solutions**

**mass**(in grams) or its

**amount**(in moles). The final volume of the solution is usually measured in dm

^{3}.

When we measure the

**mass**of the solute in**grams**, we obtainthe**mass concentration**in g/dm^{3}Example 3

^{3}) of sodium chloride solution (NaCl) that contains 20g of sodium chloride in a final solution of 100 cm

^{3}

*Solution*

First, convert the given volume to dm

^{3}Volume (dm

^{3}) = 100/1000= 0.1 dm^{3}^{3}).

=20g/0.1dm

^{3}= 200g/dm

^{3}^{3}, e.g.:If 20g of the solution are contained in 100 cm

^{3}of the solution, then the amount of solute in 1000 cm

^{3}(1 dm

^{3}) of the solution would be

1000×20/100 = 200g/dm

^{3}Calculations Based on the Mole Concept

Perform calculations based on the mole concept

Consider an equation for the reaction between hydrogen and nitrogen to produce ammonia:

3H

_{2(g)}+ N_{2(g)}→ 2NH_{3}This can be read as follows:

*three**moles of hydrogen reacts with*

**one**mole of nitrogen to yield**two**moles of ammonia.Example 4

*What volume of carbon dioxide (CO*2NaHCO

_{2}) measured at s.t.p. will be produced when 21.0g of sodium hydrogencarbonate (NaHCO_{3}) is completely decomposed according to the equation._{3(s)}→ Na

_{2}CO

_{3(s) }+ CO

_{2(g)}+ H

_{2}O

_{(l)}

*Solution*First, find the weight of carbon dioxide that will be produced by the hydrogencarbonate.

- Mass of 2NaHCO
_{3}= 2 × 84 = 168g - Mass of CO
_{2}= 44g

The weight of carbon dioxide produced can be obtained from the following relation:

168g ≡ 44g

21g ≡ X

X = 21×44/168 = 5.5g

The weight of carbon dioxide produced = 5.5g

_{2}to volume at s.t.p.We know that one mole (44g) of carbon dioxide at s.t.p. occupies 22.4 dm

^{3}

That is, 44g ≡ 22.4dm

^{3}5.5g ≡ X dm

^{3}X = 5.5×22.4/44 = 2.8dm

^{3}