NSEC 2026 Masterclass 01 Complete Mole - Olympiad Chemistry

Mole Concept and Stoichiometry

Definition and Fundamental Idea of Mole

The mole is a fundamental unit in chemistry used to count entities such as atoms, molecules, ions, or any specified objects. One mole is defined as exactly $6.022 \times 10^{23}$ entities, known as Avogadro's number ($N_A$). The term "entity" can refer to atoms, molecules, ions, electrons, or even arbitrary objects, depending on context.

• $1$ mole $= 6.022 \times 10^{23}$ entities ($N_A$)
• $N_A$ is called Avogadro's number and is symbolized as $N_A$
• The entity type is determined by what is being counted (e.g., atoms in $1$…

Mole Concept and Stoichiometry

Definition and Fundamental Idea of Mole

The mole is a fundamental unit in chemistry used to count entities such as atoms, molecules, ions, or any specified objects. One mole is defined as exactly $6.022 \times 10^{23}$ entities, known as Avogadro's number ($N_A$). The term "entity" can refer to atoms, molecules, ions, electrons, or even arbitrary objects, depending on context.

• $1$ mole $= 6.022 \times 10^{23}$ entities ($N_A$)
• $N_A$ is called Avogadro's number and is symbolized as $N_A$
• The entity type is determined by what is being counted (e.g., atoms in $1$ mole of He, molecules in $1$ mole of $O_2$)

Methods of Mole Calculation

There are four primary methods to calculate the number of moles, each applicable under specific conditions. The correct formula depends on the data provided and the physical state of the substance.

• Given Mass Method: $\text{Moles} = \frac{\text{Given Mass (g)}}{\text{Molar Mass (g/mol)}}$

• Number of Entities Method: $\text{Moles} = \frac{\text{Number of Entities}}{N_A}$

• Volume of Gas at STP: $\text{Moles} = \frac{\text{Volume (L at STP)}}{22.7}$

  • Use $22.7$ L at STP (Standard Temperature and Pressure: $0^\circ$C, $1$ bar), not $22.4$ L (which is for NTP)

• Ideal Gas Equation: $\text{Moles} = \frac{PV}{RT}$

  • $P$ = pressure, $V$ = volume, $R$ = gas constant, $T$ = temperature in Kelvin

• The third and fourth formulas are only applicable for gases.

• The first and second formulas can be used for any physical state.

Atomic and Molar Mass Calculations

Atomic and molar masses are essential for mole calculations. Atomic mass is the mass of a single atom, while molar mass is the mass of one mole of a substance.

• Atomic numbers and mass numbers for elements up to $Z=20$ should be memorized.
• For compounds, molar mass is calculated by summing the atomic masses of constituent atoms.
  • Example: $H_2SO_4$ molar mass $= 2 \times 1 + 32 + 4 \times 16 = 98$ g/mol
  • Example: Glucose ($C_6H_{12}O_6$) molar mass $= 6 \times 12 + 12 \times 1 + 6 \times 16 = 180$ g/mol

Application of Mole Formulas: Sample Problems

To solve mole problems, identify the type of data given and select the appropriate formula.

• If given mass: Use $\text{Moles} = \frac{\text{Given Mass}}{\text{Molar Mass}}$
• If given number of molecules/atoms: Use $\text{Moles} = \frac{\text{Number}}{N_A}$
• If given gas volume at STP: Use $\text{Moles} = \frac{\text{Volume}}{22.7}$
• If given gas conditions: Use $\text{Moles} = \frac{PV}{RT}$

Example calculations:

• $980$ g $H_2SO_4$: $980/98 = 10$ moles
• $12.044 \times 10^{24}$ molecules $O_2$: $12.044 \times 10^{24} / 6.022 \times 10^{23} = 20$ moles
• $45.4$ L He at STP: $45.4/22.7 = 2$ moles

The Dabba (Box) Method for Counting Atoms, Molecules, Electrons, and Neutrons

The "dabba method" is a systematic approach to relate moles, molecules, atoms, electrons, and neutrons in a sample.

• Step 1: Calculate moles from given data.
• Step 2: Find total molecules: $\text{Moles} \times N_A$
• Step 3: Use the box method to relate molecules to atoms/electrons/neutrons:
  • For $n$ molecules, if each molecule contains $x$ atoms, total atoms $= n \times x$
  • For electrons: Use atomic numbers to determine electrons per atom/molecule

Example:

• $900$ g glucose ($C_6H_{12}O_6$): $900/180 = 5$ moles
• Total molecules: $5N_A$
• Each molecule has $24$ atoms ($6$ C, $12$ H, $6$ O): $5N_A \times 24 = 120N_A$ atoms

Stoichiometric Analysis and Limiting Reagent

Stoichiometry involves quantitative relationships in balanced chemical equations. The limiting reagent is the reactant that is completely consumed first, determining the maximum amount of product formed.

• For a reaction $aA + bB \rightarrow cC + dD$:
  • If only one reactant's amount is given, use the box method to relate moles of reactant to product.
  • If more than one reactant is given, calculate $\frac{\text{moles}}{\text{stoichiometric coefficient}}$ for each; the smallest value identifies the limiting reagent.
  • All calculations for product and remaining reactants are based on the limiting reagent.

Example:

• $2A + 3B \rightarrow 4C$, given $3$ mol $A$, $10$ mol $B$:
  • $A$: $3/2 = 1.5$, $B$: $10/3 \approx 3.33$; $A$ is limiting.
  • Product $C$ formed: $3$ mol $A$ corresponds to $3 \times 4/2 = 6$ mol $C$

Percentage Composition and Minimum Molar Mass

Percentage composition by mass of an element in a compound is calculated as:

• $\%$ of element $= \frac{\text{mass of element in formula}}{\text{molar mass of compound}} \times 100$

For minimum molar mass calculation when percentage by mass is given:

• Assume total mass $= 100$ g, set up a ratio with the atomic mass and number of atoms, solve for minimum molar mass.

Example:

• Urea ($NH_2CONH_2$), molar mass $= 60$ g/mol, N content: $2 \times 14/60 \times 100 = 46.67\%$

Empirical and Molecular Formula Determination

Empirical formula gives the simplest whole-number ratio of atoms; molecular formula gives the actual number of atoms.

Steps to determine empirical formula:

1. List elements and their given masses or atom counts.
2. Convert to moles.
3. Divide all by the smallest mole value.
4. If necessary, multiply to get whole numbers.

Molecular formula:

• $x = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$
• Multiply empirical subscripts by $x$

Example:

• Given $72$ g C, $12N_A$ H atoms, $96$ g O:
  • Moles: $6$ C, $12$ H, $6$ O
  • Simplest ratio: $1:2:1$ (empirical formula $CH_2O$)
  • If molar mass $= 180$, $x = 180/30 = 6$, molecular formula $C_6H_{12}O_6$

Concentration Terms and Their Interrelations

Several concentration terms are used to express solution composition:

• Molarity ($M$): Moles of solute per liter of solution.
• Molality ($m$): Moles of solute per kg of solvent.
• Mole Fraction ($\chi$): Moles of component divided by total moles.
• Mass Fraction: Mass of component divided by total mass.
• Percentage by Mass/Volume:
  • $w/w$: $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$
  • $w/v$: $\frac{\text{mass of solute}}{\text{volume of solution (ml)}} \times 100$
  • $v/v$: $\frac{\text{volume of solute}}{\text{volume of solution}} \times 100$
• PPM/PPB: For very dilute solutions, $PPM = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$, $PPB = \times 10^9$

Key relationships (as stated):

• $M = m \times \frac{1000 \times d}{1000 - M \times \text{molar mass}}$
• $M = \frac{\% w/w \times d \times 10}{\text{molar mass}}$
• $M = \frac{\% w/v \times 10}{\text{molar mass}}$
• For dilution: $M_1V_1 = M_2V_2$ (moles of solute constant)

Density and Related Concepts

Density is defined as mass per unit volume. There are several types:

• Absolute Density: $\frac{\text{mass}}{\text{volume}}$ (g/ml or kg/m$^3$)
• Relative Density:
  • Vapour Density (for gases): $\frac{\text{density of gas}}{\text{density of H}_2}$, or $\frac{\text{molar mass}}{2}$
  • Specific Gravity (for liquids/solids): $\frac{\text{density of substance}}{\text{density of water at }4^\circ C}$
• Density of Gas: $\rho = \frac{PM}{RT}$

Laws of Chemical Combination

Several fundamental laws govern chemical reactions:

• Law of Conservation of Mass: Mass is neither created nor destroyed in a chemical reaction; total mass of reactants equals total mass of products. Not applicable to nuclear reactions.
• Law of Definite Proportion: A compound always contains the same elements in the same proportion by mass, regardless of source. Not applicable to isotopes.
• Law of Multiple Proportion: When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in simple whole-number ratios.
• Law of Reciprocal Proportion: If element A combines with B and C separately, and B and C also combine, the ratio of masses of B and C that combine with a fixed mass of A is the same as the ratio in which B and C combine with each other.

Special Cases, Pitfalls, and Exam Traps

• For gases, always check if conditions are STP (use $22.7$ L) or NTP (use $22.4$ L).
• For limiting reagent, always base all calculations on the limiting reactant.
• In dilution problems, distinguish between "by" (additive) and "to" (final volume) language.
• For empirical/molecular formula, always reduce to the simplest ratio and check for integer subscripts.
• For concentration conversions, use the correct formula and units as per the data provided.
• In advanced problems, avoid relying solely on memorized formulas; use systematic approaches as formulas may not always apply directly.