Mole Concept: Notes and Important Questions by PW

Mole Concept numericals can become confusing when a question involves several conversions, such as moving from mass to moles, moles to particles or gas volume to the number of molecules. Problems involving limiting reagents, percentage purity and concentration terms also require you to identify the correct formula before calculating the answer.

PW’s Mission 11th Comeback: Mole Concept video brings together important concepts, calculation tricks and practice questions for students preparing for NEET. Use the chapter notes alongside the video to revise essential formulas, understand how to approach numerical problems and practise applying the concepts to different question types.

Mole And Avogadro’s Number

A mole is a counting unit used for atoms, molecules, ions, electrons, and other particles.

1 mole = 6.022 × 10^23 entities

This value is called Avogadro’s Number, represented by N_A.

• 1 mole of atoms = 6.022 × 10^23 atoms

• 2 moles of molecules = 2N_A molecules

• 3 moles of ions = 3N_A ions

(*Memory Tip: Whatever entity follows “mole” is multiplied by N_A.)*

The notation must be read carefully:

• O represents one oxygen atom.

• O2 represents one oxygen molecule.

• 2 mol O represents 2N_A oxygen atoms.

• 2 mol O2 represents 2N_A oxygen molecules.

Atomicity And Molecule-To-Atom Conversion

Atomicity is the number of atoms of a particular element present in one molecule.

For example, one molecule of H2O contains:

• 2 hydrogen atoms

• 1 oxygen atom

• Total atomicity = 3

The basic relations are:

• Number of atoms = Number of molecules × Atomicity

• Number of molecules = Number of atoms / Atomicity

For CO2, each molecule contains one carbon atom and two oxygen atoms. Therefore, 3N_A CO2 molecules contain:

• 3N_A carbon atoms

• 6N_A oxygen atoms

These relations can also be used to calculate total electrons, protons, and neutrons in a given number of molecules.

Molar Mass And Mole Calculations

Molar mass is the mass of one mole of a substance. It is expressed in grams per mole.

The number of moles can be calculated using:

• Moles = Given mass / Molar mass

• Moles = Number of particles / N_A

• Moles = Gas volume / Molar volume

The reverse relations are:

• Mass = Moles × Molar mass

• Number of particles = Moles × N_A

For water, H2O:

Molar mass = 2(1) + 16 = 18 g/mol

Therefore, 90 g of water contains:

90 / 18 = 5 moles

Number of water molecules:

5N_A molecules

Since each water molecule contains two hydrogen atoms and one oxygen atom:

• Hydrogen atoms = 10N_A

• Oxygen atoms = 5N_A

(*Memory Tip: One atom or molecule is measured in amu, while one mole is measured in grams. The numerical value remains the same, but the unit changes.)*

Gas Volume And Molar Volume

For gases, moles can be calculated from volume:

Moles = Volume / Molar Volume

At 273 K and 1 atm:

Molar volume = 22.4 L/mol

At 273 K and 1 bar:

Molar volume = 22.7 L/mol

For example, 2.24 L of a gas at the older STP condition contains:

2.24 / 22.4 = 0.1 mole

The number of molecules is:

0.1N_A

For gases, the volume occupied is equal to the volume of the container.

Important volume conversions include:

• 1 L = 1000 mL

• 1 L = 1000 cm3

• 1 L = 1 dm3

Water must not be treated like a gas. It is a liquid, and its density is approximately 1 g/mL.

Ideal Gas Equation

The ideal gas equation is:

PV = nRT

Here:

• P = Pressure

• V = Volume

• n = Number of moles

• R = Universal gas constant

• T = Temperature in Kelvin

Temperature must always be converted into Kelvin:

Temperature in Kelvin = Temperature in Celsius + 273

For a fixed amount of gas:

P1V1/T1 = P2V2/T2

The value of R depends on the units of pressure and volume. For example:

R = 0.082 L atm K−1 mol−1

or

R = 8.314 J K−1 mol−1

Density Of A Gas

Using n = mass/molar mass in the ideal gas equation:

Density = PM / RT

Here, M is the molar mass of the gas.

Vapour Density

Vapour density is:

Vapour Density = Molar Mass of Gas / 2

Therefore:

Molar Mass = 2 × Vapour Density

Laws Of Chemical Combination

Law Of Conservation Of Mass

Matter can neither be created nor destroyed in a chemical reaction.

Total mass of reactants = Total mass of products

Chemical equations must be balanced to follow this law.

Law Of Constant Proportions

A pure compound always contains the same elements in a fixed proportion by mass, regardless of its source or method of preparation.

For CO2:

• Carbon = 12 parts

• Oxygen = 32 parts

Mass ratio of carbon to oxygen = 12:32

Law Of Multiple Proportions

When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in small whole-number ratios.

Gay-Lussac’s Law Of Combining Volumes

The volumes of reacting gases, measured at the same temperature and pressure, combine in small whole-number ratios.

Avogadro’s Law

Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules and equal numbers of moles.

Summary Table

Empirical And Molecular Formulae

The molecular formula shows the actual number of atoms in one molecule.

Examples:

• Glucose: C6H12O6

• Ethane: C2H6

• Benzene: C6H6

The empirical formula shows the simplest whole-number ratio of atoms.

For glucose:

C6H12O6 becomes CH2O

Therefore:

• Molecular formula = C6H12O6

• Empirical formula = CH2O

The relation is:

Molecular Formula = n × Empirical Formula

where:

n = Molecular mass / Empirical formula mass

Steps To Find An Empirical Formula

1. Write the percentage or mass of each element.

2. Divide each value by its atomic mass.

3. Divide all results by the smallest value.

4. Convert the ratio into the simplest whole numbers.

5. Write the formula using these numbers as subscripts.

Stoichiometry

Stoichiometry involves quantitative calculations in a balanced chemical reaction.

A balanced equation can be read in terms of:

• Moles

• Mass

• Gas volume

For example:

CH4 + 2O2 → CO2 + 2H2O

The equation represents:

• 1 mole CH4 reacts with 2 moles O2.

• 1 mole CH4 produces 1 mole CO2.

• 1 mole CH4 produces 2 moles H2O.

The coefficients provide the mole ratio. The unitary method is then used to calculate the required quantity.

(*Memory Tip: First balance the reaction, then read the coefficients as mole ratios.)*

Purity, Yield, And Limiting Reagent

Percentage Purity

Only the pure portion of a sample takes part in a reaction.

Percentage Purity = Mass of Pure Reactant / Total Sample Mass × 100

The stoichiometric calculation must use the mass of the pure reactant, not the total impure sample.

Percentage Yield

Theoretical yield is the amount predicted by calculation. Actual yield is the amount obtained experimentally.

Percentage Yield = Actual Yield / Theoretical Yield × 100

(*Memory Tip: Theoretical means calculated; actual means obtained in the experiment.)*

Limiting Reagent

The limiting reagent is consumed completely first and determines the amount of product formed.

To identify it:

1. Convert each reactant quantity into moles.

2. Divide the moles by the stoichiometric coefficient.

3. The reactant with the lowest value is the limiting reagent.

The other reactant is the excess reagent.

Concentration Terms

Mole Fraction

Mole fraction is the ratio of moles of one component to total moles.

Mole Fraction of Solute = n2 / (n1 + n2)

Mole Fraction of Solvent = n1 / (n1 + n2)

The sum of mole fractions is:

Mole Fraction of Solute + Mole Fraction of Solvent = 1

Molality

Molality is the number of moles of solute present in one kilogram of solvent.

Molality = Moles of Solute / Mass of Solvent in kg

Molality uses the mass of solvent.

Molarity

Molarity is the number of moles of solute present in one litre of solution.

Molarity = Moles of Solute / Volume of Solution in L

Molarity uses the volume of solution.

Molarity And Dilution

During dilution or concentration, the number of moles of solute remains constant:

M1V1 = M2V2

Comparison Table

Other Concentration Terms

• Mass percentage = Mass fraction × 100

• Parts per million = Fraction × 10^6

• Parts per billion = Fraction × 10^9

• Parts per trillion = Fraction × 10^12

Watch Mission 11th Comeback: Mole Concept || Arjuna NEET 2027 by PW to revisit mole conversions, stoichiometry, gas laws, concentration terms and other important concepts. The video can help you revise calculation methods, learn useful shortcuts and practise applying formulas to Chemistry questions.

Important Questions on Mole Concept

These five questions cover different numerical and conceptual areas from the chapter notes. Try solving each question before checking the solution.

Question 1: Calculate the number of hydrogen atoms present in 18 g of water. (Molar mass of H₂O = 18 g/mol; Avogadro’s number = 6.022 × 10²³ mol⁻¹.)

Solution: Moles of H₂O = 18/18 = 1 mol. Number of water molecules = 6.022 × 10²³. Each H₂O molecule contains two hydrogen atoms, so hydrogen atoms = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ atoms.

Question 2: How many moles of a gas are present in 5.6 L at 273 K and 1 atm? Use a molar volume of 22.4 L/mol.

Solution: Moles = Gas volume/Molar volume = 5.6/22.4 = 0.25 mol.

Question 3: A compound contains 40% carbon, 6.67% hydrogen and 53.33% oxygen by mass. Find its empirical formula. (Atomic masses: C = 12, H = 1, O = 16.)

Solution: Assume 100 g of the compound. Moles: C = 40/12 = 3.33, H = 6.67/1 = 6.67, O = 53.33/16 = 3.33. Dividing by the smallest value gives the ratio C:H = 1:2:1. Empirical formula = CH₂O.

Question 4: For the reaction N₂ + 3H₂ → 2NH₃, 2 mol of N₂ reacts with 3 mol of H₂. Identify the limiting reagent and calculate the maximum moles of NH₃ formed.

Solution: For N₂, moles/coefficient = 2/1 = 2. For H₂, it is 3/3 = 1. H₂ has the lower value, so it is the limiting reagent. From the equation, 3 mol H₂ produces 2 mol NH₃. Therefore, the maximum NH₃ formed is 2 mol.

Question 5: What volume of a 2 M solution is required to prepare 500 mL of a 0.5 M solution?

Solution: Using M₁V₁ = M₂V₂, V₁ = (0.5 × 500)/2 = 125 mL.

Mole Concept becomes easier to apply when you understand the relationships between mass, moles, particles, gas volume and concentration. Practise the formulas regularly, balance equations before solving stoichiometry questions, and check your units at every step. PW’s Mission 11th Comeback: Mole Concept video can support your revision with important concepts, calculation tricks and practice questions.

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