In-depth Notes on Principles of Chemistry for IGCSE

Principle of Chemistry

• Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions.

Learning Outcomes

• Core:

  • 1. State the formulae of elements and compounds.

  • 2. Define molecular formula (number/type of atoms in a molecule).

  • 3. Deduce formula from models/diagrams.

  • 4. Construct word/symbol equations and include state symbols.

• Supplement:

  • 5. Define empirical formula (simplest whole number ratio of atoms).

  • 6. Deduce formula of ionic compounds from models or charges of ions.

  • 7. Construct symbol equations, including ionic equations.

  • 8. Deduce balanced symbol equations from information.

  • 9. Describe relative atomic mass (A) based on isotopes compared to $\frac{1}{12}^{th}$ of $^{12}C$.

  • 10. Define relative molecular mass (M) as the sum of relative atomic masses; relative formula mass (M) is for ionic compounds.

  • 11. Calculate reacting masses without mole concepts.

  • 12. State mole (mol) as unit of amount and its constant $6.02 \times 10^{23}$ particles.

  • 13. Use relationships for calculating:

  • (a) Amount of substance

  • (b) Mass

  • (c) Molar mass

  • (d) Relative atomic/molecular/formula mass

  • (e) Number of particles using Avogadro's constant.

  • 14. Concentration measured in $g/dm^3$ or $mol/dm^3$.

  • 15. Molar gas volume at r.t.p. is 24 dm³.

  • 16. Calculate stoichiometric masses and concentrations including conversion between $cm^3$ and $dm^3$.

  • 17. Use titration data to find moles of solute/concentration/volume.

  • 18. Calculate empirical and molecular formulae from data.

  • 19. Determine percentage yield, composition, and purity.

Chemical Formulae

• Chemical formulae represent chemical substances using symbols/ratios to denote elements and their quantities in a compound.

• Understand valency (combining power) to write formulae:

  • Metals: usually determined by group

  • Non-metals: based on the number of electrons to achieve an octet.

• Examples of monoatomic ions and their charges from groups (e.g., Group 1: $M^+$, Group 7: $X^-$, etc.).

• Ionic Compounds: Write cations and anions based on their charges and balance to form neutral compounds.

  • Example: Barium oxide ($BaO$):

  • Barium ($Ba^{2+}$), Oxygen ($O^{2-}$)

  • Formula: $BaO$.

  • Cross over valences to find subscripts.

Chemical Equations

• A representation of chemical reactions using words/formulae.

• Word Equations vs Chemical Equations:

  • Word equations describe reactions using names only.

  • Chemical equations require correct symbols and balancing.

• Balancing Equations: Use coefficients to reflect the number of moles of reactants/products.

  • Example: $aA + bB \rightarrow cC + dD$.

• State symbols indicate the physical state of substances (s, l, g, aq).

Reacting Quantities

• Calculate molar masses using atomic masses from the Periodic Table.

• Law of Conservation of Mass: Total mass of reactants = total mass of products.

• Percentage Composition: Determine % by mass of an element in a compound using:

  • % by mass = ($\frac{mass\ of\ element}{Mr\ of\ compound}$) $\times 100$.

The Mole Concept

• A mole is a quantity that contains $6.02 \times 10^{23}$ particles regardless of type.

• Molar volume of gases at r.t.p. = 24 dm³.

• Calculations involving moles: Use moles to determine mass or volume of substances and vice versa.

• Examples of calculations for gas volumes and solutions based on known concentrations and molar mass.

• Empirical and molecular formula derivation through experimental data.

Percentage Yield & Purity

• % yield indicates the efficiency of a reaction: % yield = ($\frac{actual\ yield}{theoretical\ yield}$) $\times 100$.

• % purity evaluates the quality of substances, especially in pharmaceuticals: % purity = ($\frac{mass\ of\ pure\ substance}{mass\ of\ impure\ sample}$) $\times 100$.

🔹 Key Formulas

1. Number of Particles
   Number of particles=Moles×NANumber of particles=Moles×NAMoles=Number of particlesNAMoles=NANumber of particles​

2. Mole-Mass Relationship
   Moles=Mass of substance (g)Molar mass (g/mol)Moles=Molar mass (g/mol)Mass of substance (g)​
   Mass=Moles×Molar massMass=Moles×Molar mass

3. Mole-Volume Relationship (for gases at STP)
   Moles=Volume of gas (L)22.4Moles=22.4Volume of gas (L)​
   Volume=Moles×22.4Volume=Moles×22.4

4. Concentration (Molarity)
   Molarity (M)=Moles of soluteVolume of solution (L)Molarity (M)=Volume of solution (L)Moles of solute​
   Moles=Molarity×Volume (L)Moles=Molarity×Volume (L)

5. Empirical and Molecular Formula
   Molecular formula=(Empirical formula)×nMolecular formula=(Empirical formula)×n
   n=Molar mass of compoundMolar mass of empirical formulan=Molar mass of empirical formulaMolar mass of compound​

6. Percentage Composition
   \text{% of element} = \left( \frac{\text{Total mass of element in compound}}{\text{Molar mass of compound}} \right) \times 100

7. Gas Laws (related to moles)
   Ideal Gas Law:
   PV=nRTPV=nRT
   Where:

   • P = Pressure (atm)

   • V = Volume (L)

   • n = Moles

   • R = 0.0821 L·atm/mol·K

   • T = Temperature (K)

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🧪 MOLES: Full Notes and Formulas

🔹 Basic Definitions

1. Mole (mol):

   • A mole is a unit that measures the amount of substance.

   • 1 mole = 6.022 × 10²³ particles (Avogadro's Number).

   • These particles can be atoms, molecules, ions, or electrons depending on the substance.

2. Avogadro’s Number (NA):

   • NA=6.022×1023N_A = 6.022 \times 10^{23}NA​=6.022×1023 particles/mol

🔹 Key Formulas

1. Number of Particles

Number of particles=Moles×NA\text{Number of particles} = \text{Moles} \times N_ANumber of particles=Moles×NA​ Moles=Number of particlesNA\text{Moles} = \frac{\text{Number of particles}}{N_A}Moles=NA​Number of particles​

2. Mole-Mass Relationship

Moles=Mass of substance (g)Molar mass (g/mol)\text{Moles} = \frac{\text{Mass of substance (g)}}{\text{Molar mass (g/mol)}}Moles=Molar mass (g/mol)Mass of substance (g)​ Mass=Moles×Molar mass\text{Mass} = \text{Moles} \times \text{Molar mass}Mass=Moles×Molar mass

3. Mole-Volume Relationship (for gases at STP)

• STP: Standard Temperature and Pressure (0°C, 1 atm)

• 1 mole of any gas at STP = 22.4 L

Moles=Volume of gas (L)22.4\text{Moles} = \frac{\text{Volume of gas (L)}}{22.4}Moles=22.4Volume of gas (L)​ Volume=Moles×22.4\text{Volume} = \text{Moles} \times 22.4Volume=Moles×22.4

4. Concentration (Molarity)

Molarity (M)=Moles of soluteVolume of solution (L)\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}Molarity (M)=Volume of solution (L)Moles of solute​ Moles=Molarity×Volume (L)\text{Moles} = \text{Molarity} \times \text{Volume (L)}Moles=Molarity×Volume (L)

5. Empirical and Molecular Formula

• Empirical formula: Simplest ratio of elements

• Molecular formula:

Molecular formula=(Empirical formula)×n\text{Molecular formula} = (\text{Empirical formula}) \times nMolecular formula=(Empirical formula)×n n=Molar mass of compoundMolar mass of empirical formulan = \frac{\text{Molar mass of compound}}{\text{Molar mass of empirical formula}}n=Molar mass of empirical formulaMolar mass of compound​

6. Percentage Composition

% of element=(Total mass of element in compoundMolar mass of compound)×100\%\text{ of element} = \left( \frac{\text{Total mass of element in compound}}{\text{Molar mass of compound}} \right) \times 100% of element=(Molar mass of compoundTotal mass of element in compound​)×100

7. Gas Laws (related to moles)

• Ideal Gas Law:

PV=nRTPV = nRTPV=nRT

Where:

• PPP = Pressure (atm)

• VVV = Volume (L)

• nnn = Moles

• RRR = 0.0821 L·atm/mol·K

• TTT = Temperature (K)

🔹 Stoichiometry (Mole Ratios in Reactions)

From a balanced chemical equation:

• Use mole ratios to convert between reactants and products.

Example: If the reaction is:
2H2+O2→2H2O2H_2 + O_2 \rightarrow 2H_2O2H2​+O2​→2H2​O
Then:

• 2 moles of H₂ react with 1 mole of O₂ to produce 2 moles of H₂O

🔹 Limiting Reactant Concept

• Find moles of all reactants.

• Compare using mole ratios to determine which is limiting (the one that runs out first).

• The amount of product formed depends on the limiting reactant.

🧠 Quick Tips

• Always check if conditions are at STP when dealing with gases.

• Molar mass = atomic mass from periodic table (in g/mol).

• For solutions, always convert volume to liters before using in molarity formula.

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🧪 MOLES CHEAT SHEET

🔹 1. Basic Concepts

• Mole: Unit for amount of substance.
  1 mole = 6.022 × 10²³ particles (Avogadro’s number)

• Particles = atoms, molecules, ions, electrons, etc.

🔹 2. Key Formulas

🔸 A. Number of Particles

Particles=Moles×6.022×1023\text{Particles} = \text{Moles} \times 6.022 \times 10^{23}Particles=Moles×6.022×1023Moles=Particles6.022×1023\text{Moles} = \frac{\text{Particles}}{6.022 \times 10^{23}}Moles=6.022×1023Particles​

🔸 B. Mass ↔ Moles

Moles=Mass (g)Molar Mass (g/mol)\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}Moles=Molar Mass (g/mol)Mass (g)​Mass (g)=Moles×Molar Mass\text{Mass (g)} = \text{Moles} \times \text{Molar Mass}Mass (g)=Moles×Molar Mass

🔸 C. Volume of Gases at STP

Moles=Volume (L)22.4\text{Moles} = \frac{\text{Volume (L)}}{22.4}Moles=22.4Volume (L)​Volume (L)=Moles×22.4\text{Volume (L)} = \text{Moles} \times 22.4Volume (L)=Moles×22.4

🔸 D. Molarity (Concentration)

Molarity (M)=Moles of soluteVolume of solution (L)\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}Molarity (M)=Volume of solution (L)Moles of solute​Moles=Molarity×Volume (L)\text{Moles} = \text{Molarity} \times \text{Volume (L)}Moles=Molarity×Volume (L)

🔹 3. Empirical & Molecular Formulas

• Empirical formula: Simplest ratio of atoms

• Molecular formula:

Molecular Formula=(Empirical Formula)×n\text{Molecular Formula} = (\text{Empirical Formula}) \times nMolecular Formula=(Empirical Formula)×nn=Molar Mass of compoundEmpirical formula massn = \frac{\text{Molar Mass of compound}}{\text{Empirical formula mass}}n=Empirical formula massMolar Mass of compound​

🔹 4. Percentage Composition

% of element=(Mass of element in 1 molMolar Mass of compound)×100\%\text{ of element} = \left( \frac{\text{Mass of element in 1 mol}}{\text{Molar Mass of compound}} \right) \times 100% of element=(Molar Mass of compoundMass of element in 1 mol​)×100

🔹 5. Gas Law (Ideal Gas Equation)

PV=nRTPV = nRTPV=nRT

Where:

• PPP = Pressure (atm)

• VVV = Volume (L)

• nnn = Moles

• RRR = 0.0821 L·atm/mol·K

• TTT = Temperature (K)

🔹 6. Stoichiometry

• Use balanced equations to convert between substances via mole ratios.

🔹 7. Limiting Reactant

• Calculate moles of all reactants.

• Use mole ratio to find which reactant limits the reaction.

• The limiting reactant determines the maximum amount of product.