Chemistry Practical

Mass-Volume Relationship

The Mole

The mole is a unit that represents a number of particles of a substance — atoms, ions, molecules, or electrons. One mole contains 6.02 × 10²³ particles, known as Avogadro’s number.

It is defined as the amount of substance that contains as many elementary units as there are atoms in 12g of Carbon-12.

Relative Atomic Mass

The relative atomic mass of an element is the number of times the average mass of one atom of that element is heavier than one-twelfth the mass of a carbon-12 atom.

Examples: H = 1, O = 16, C = 12, Na = 23, Ca = 40

Relative Molecular Mass

This is the sum of the relative atomic masses of all atoms in a molecule. It is also called the formula mass.

Examples:

• Magnesium chloride (MgCl₂): 24 + (35.5 × 2) = 95 g/mol
• Sodium hydroxide (NaOH): 23 + 16 + 1 = 40 g/mol
• Calcium carbonate (CaCO₃): 40 + 12 + (16 × 3) = 100 g/mol

Molar Volume of Gases

At standard temperature and pressure (s.t.p.), 1 mole of any gas occupies 22.4 dm³.

Note: s.t.p. = 273K and 760 mmHg.

Formulas and Relationships

• Molar mass (g/mol) = Mass (g) / Amount (mol)
• Molar volume (dm³/mol) = Volume (dm³) / Amount (mol)
• Amount (mol) = Number of particles / Avogadro’s constant
• Reacting mass = (Number of particles × Molar mass) / Avogadro’s number

  Mass = Amount × Molar mass
       = 2.7 mol × 27 g/mol
       = 72.9 g
  

Stoichiometry of Reactions

Stoichiometry is the calculation of the amounts (in moles or grams) of reactants and products involved in a chemical reaction using a balanced chemical equation.

  2 × 84 g NaHCO₃ → 106 g Na₂CO₃
  16.8 g NaHCO₃ → X g Na₂CO₃

  X = (106 × 16.8) / (2 × 84)
    = 10.6 g

  Answer: Mass of solid product = 10.6 g
  

Volumetric Analysis

Volumetric analysis is a method used in analytical chemistry to determine the titre or concentration of an analyte in a solution. This is achieved by measuring the volume of a standard solution of a reagent whose concentration is already known.

Preparing a Standard Solution

A standard solution is one with a known concentration. The steps to prepare a standard solution are:

1. Determine the desired volume and concentration of the solution.
2. Calculate the mass of solute required for the specified concentration and volume.
3. Weigh the calculated amount of solute.
4. Dissolve the solute completely in distilled water and transfer it into a volumetric flask that is partially filled with distilled water.
5. Top up the volumetric flask with distilled water up to the calibration mark.
6. Invert and shake the flask to ensure thorough mixing of the solution.

Acid-Base Titration

During acid-base titration, several materials are used. Some of these and the precautions associated with their use include:

Materials Used

• Weighing balance
• Chemical balance
• Pipette
• Burette
• Retort stand
• Filter paper
• Funnel
• White tile
• Standard volumetric flask
• Conical flask

Pipette

• Rinse with the solution it will measure (e.g., a base).
• Avoid trapping air bubbles inside the pipette.
• Ensure your eye is level with the mark when taking readings.
• Do not blow out the last drop of solution.

Burette

• Rinse with the acid to be used or drain well after rinsing.
• Ensure the burette is filled properly and has no air bubbles.
• Make sure the burette does not leak.
• Remove the funnel before taking any readings.
• Take consistent and accurate readings.

Conical Flask

• Rinse only with distilled water, not with any titration solutions.
• Use distilled water to wash down any solution adhering to the inner walls.

Concentration of a Solution

Concentration refers to the amount of solute dissolved in a given volume of solution. Volume is typically measured in cubic decimeters (dm³), where 1 dm³ = 1000 cm³. Solute can be measured in grams or moles, with two main units of concentration:

• g dm^-3: Concentration in grams per cubic decimeter.
• mol dm^-3: Concentration in moles per cubic decimeter.

Concentration in g dm^-3

This tells us how many grams of solute are present in 1 dm³ of solution. For example, a concentration of 10 g dm^-3 means 10 grams of solute are dissolved in 1 dm³ of solution.

Formula:

Concentration = Mass of solute (g) / Volume of solution (dm³)

Concentration in mol dm^-3 (Molarity)

Molarity is the number of moles of solute per dm³ (litre) of solution.

A concentration of 2 mol dm^-3 means that 2 moles of solute are dissolved in 1 dm³ of solution.

Formula:
Molarity = Moles of solute (mol) / Volume of solution (dm³)

Standardization of a Solution

The concentration of an unknown acid or base can be determined by titrating it with a standard solution of a base or acid. A balanced chemical equation is needed to find the mole ratio.

Determination of Degree of Purity

The degree of purity of an acid or base can be determined through titration. For example, suppose a sample of anhydrous sodium carbonate (Na₂CO₃) is contaminated with sodium chloride (NaCl). A known mass of the impure mixture is dissolved in a known volume of solution and titrated against a standard solution of a pure acid.

During titration, only the pure Na₂CO₃ will react with the acid, while the impurity (NaCl) remains unreacted. The molar concentration determined from the titration corresponds to the pure sodium carbonate.

In general, the mass of pure substance is less than the total mass of the impure sample due to the presence of impurities.

Mathematical Expression

Mass of pure substance = Mass of impure substance − Mass of impurity

Note: You need more of an impure substance to complete a chemical reaction than if it were 100% pure.

(CA × VA) / nA = (CB × VB) / nB

(0.100 × 22.65) / 2 = CB × 25.00
CB = (0.100 × 22.65) / (2 × 25.00) = 0.0453 mol/dm³

Mass concentration = 0.0453 mol/dm³ × 106 g/mol = 4.80 g/dm³
Mass in 250 cm³ = (4.80 × 250) / 1000 = 1.20 g

% NaCl = (1.20 / 6.00) × 100 = 20.00%
% Purity = (4.80 / 6.00) × 100 = 80.00%

Determination of Water of Crystallisation

When a hydrated compound is used in a volumetric analysis, only the anhydrous part takes part in the chemical reaction. The water of crystallisation remains in solution like an impurity.

2HCl (aq) + Na₂CO₃·xH₂O (aq) → 2NaCl (aq) + xH₂O (l) + CO₂ (g)

Molar concentration = Mass / Molar mass
= 6.00 / 36.5
= 0.164 mol/dm³

= 0.164 mol/dm³ × (13.10 / 1000) dm³
= 0.00215 mol

2 mol HCl : 1 mol Na₂CO₃
⇒ 0.00215 mol HCl corresponds to 0.00108 mol Na₂CO₃

0.00108 mol in 25 cm³
= 0.00108 × (1000 / 25) = 0.0432 mol/dm³

Molar mass of Na₂CO₃·xH₂O = 279 g/mol
106 + 18x = 279
=> 18x = 173
=> x = 173 / 18 = 9.61 ≈ 10 (rounded to nearest whole number)
  

% Water of Crystallisation =
= (12.0 - 4.56) / 12.0 × 100
= 7.44 / 12.0 × 100 = 62.0%
  

Qualitative Analysis

Identification of Ions

There are 10 cations and 4 anions to be studied:

Cations

• Sodium (Na⁺)
• Calcium (Ca²⁺)
• Magnesium (Mg²⁺)
• Iron (II) (Fe²⁺)
• Iron (III) (Fe³⁺)
• Lead (II) (Pb²⁺)
• Aluminium (Al³⁺)
• Copper (II) (Cu²⁺)
• Zinc (Zn²⁺)
• Ammonium (NH₄⁺)

Anions

• Chloride (Cl⁻)
• Sulphate (SO₄²⁻)
• Nitrate (NO₃⁻)
• Carbonate (CO₃²⁻)

Colour of Ions

Heating Effect on Carbonate Salts

All carbonates except sodium and potassium decompose upon heating to release carbon dioxide.

Examples:

• CaCO₃ → CaO + CO₂
• Al₂(CO₃)₃ → Al₂O₃ + 3CO₂
• CuCO₃ → CuO + CO₂
• 2Ag₂CO₃ → 4Ag + 2CO₂ + O₂
• (NH₄)₂CO₃ → 2NH₃ + 2CO₂ + H₂O

Heating Effect on Nitrate Salts

All nitrates decompose upon heating, with varying products depending on the metal cation. (Table of nitrate decomposition to follow.)

Heating Effect on Nitrate Salts

• NH₄NO₃ → N₂O + 2H₂O
• 2KNO₃ → 2KNO₂ + O₂
• 2NaNO₃ → 2NaNO₂ + O₂
• 2Mg(NO₃)₂ → 2MgO + 4NO₂ + O₂
• 4Fe(NO₃)₃ → 2Fe₂O₃ + 12NO₂ + 3O₂
• 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂
• 2AgNO₃ → 2Ag + 2NO₂ + O₂

Heating Effect on Sulphate Salts

• 2FeSO₄•7H₂O → Fe₂O₃ + SO₂ + SO₃ + 14H₂O
• ZnSO₄ → ZnO + SO₃
• CuSO₄ → CuO + SO₃
• Fe₂(SO₄)₃ → Fe₂O₃ + 3SO₃
• (NH₄)₂SO₄ → NH₃ + H₂SO₄

Identification of Gases

Identifying Anions

Carbonate (CO₃²⁻)

Add dilute HCl: effervescence observed. Gas turns limewater milky (CO₂).

Sulphate (SO₄²⁻)

Add HCl and BaCl₂: white precipitate forms if sulphate is present.

Chloride (Cl⁻)

Add nitric acid and silver nitrate: white precipitate of AgCl forms.

Nitrate (NO₃⁻)

Test 1: Add NaOH and aluminium powder. If nitrate is present, ammonia is evolved.

Test 2: Add H₂SO₄, FeSO₄, then drop concentrated H₂SO₄. Brown ring forms at the interface.

Test for Cations Using NaOH and NH₃ Solutions

Identifying Cations Using Anions

Test with Chloride Ions

Among the 10 common cations, only Lead(II) ions (Pb²⁺) form a precipitate with chloride ions because lead(II) chloride (PbCl₂) is insoluble in water. However, PbCl₂ dissolves in hot water.

Reaction: Pb²⁺ + 2Cl⁻ → PbCl₂

Test with Sulphate Ions

Only Calcium (Ca²⁺) and Lead(II) (Pb²⁺) ions form precipitates with sulphate ions because CaSO₄ and PbSO₄ are insoluble in water.

Reactions:

• Pb²⁺ + SO₄²⁻ → PbSO₄
• Ca²⁺ + SO₄²⁻ → CaSO₄

Test with Carbonate Ions

All cations except sodium (Na⁺) and ammonium (NH₄⁺) form precipitates with carbonate ions, since Na₂CO₃ and (NH₄)₂CO₃ are soluble in water.

Test with Iodide Ions

Iodide ions form precipitates with Lead(II) (Pb²⁺) and Copper(II) (Cu²⁺) ions. Lead(II) iodide is a yellow precipitate that dissolves in hot water.

Reaction: Pb²⁺ + 2I⁻ → PbI₂

Distinguishing Between Iron(II) and Iron(III) Ions

You can differentiate Fe²⁺ and Fe³⁺ using the following reagents:

• Potassium hexacyanoferrate(II)
• Potassium hexacyanoferrate(III)
• Potassium thiocyanate