Matter and Its Classification
Matter is anything that occupies space and has mass. It can be classified as:
Mixtures: Two or more substances physically mixed (variable composition).
- Homogeneous: Uniform composition (e.g., salt solution).
- Heterogeneous: Non-uniform composition (e.g., sand and water).
Pure Substances: Fixed composition.
- Elements: Cannot be broken down chemically (e.g., H, O, Fe).
- Compounds: Two or more elements chemically combined (e.g., H2O, CO2).
Physical and Chemical Properties
Physical properties: Can be measured without changing composition (melting point, density).
Chemical properties: Describe chemical behaviour (reactivity, flammability).
Atomic and Molecular Masses
Atomic Mass
The mass of an atom relative to 1/12th the mass of carbon-12 atom.
1 amu = 1.66 x 10^(-24) g
Average Atomic Mass
Most elements occur as isotopes. The average atomic mass accounts for natural abundance.
Example: Chlorine has isotopes Cl-35 (75.77%) and Cl-37 (24.23%).
Average atomic mass = 35*75.77/100 + 37*24.23/100 = 35.48 u.
Molecular Mass
Sum of atomic masses of all atoms in a molecule.
Example: H_2O = 2*1 + 16 = 18 u.
Formula Mass
Used for ionic compounds (no discrete molecules).
Example: NaCl = 23 + 35.5 = 58.5 u.
Mole Concept
A mole is the amount of substance containing as many elementary entities as atoms in exactly 12 g of carbon-12.
1 mole = 6.022 x 10^23 entities (Avogadro number, N_A).
Molar Mass
Mass of one mole of a substance.
H_2: 2 g/molH_2O: 18 g/molNaCl: 58.5 g/mol
Mole Calculations
Number of moles (n) = text(Mass in grams)/text(Molar mass) = text(Number of particles)/N_A = text(Volume of gas at STP)/22.4 L
Percentage Composition
Mass percentage of an element in a compound:
% of element = (text(Mass of element in compound))/text(Molar mass of compound) x 100
Example: Find % composition of water (H2O).
Molar mass = 18 g/mol.
%H = 2/18 x 100 = 11.11%
%O = 16/18 x 100 = 88.89%
Empirical and Molecular Formulas
Empirical formula: Simplest whole-number ratio of atoms in a compound.
Molecular formula: Actual number of atoms in a molecule.
MF = (EF)_n where n = text(Molecular mass)/text(Empirical formula mass)
Steps to Find Empirical Formula
- Convert percentages to grams (assuming 100 g sample).
- Convert mass to moles using atomic masses.
- Divide by the smallest number of moles.
- Multiply by appropriate factor to get whole numbers.
Example: A compound has 40% C, 6.67% H, 53.33% O. Molar mass = 60 g/mol.
C: 40/12 = 3.33, H: 6.67/1 = 6.67, O: 53.33/16 = 3.33.
Divide by 3.33: C: 1, H: 2, O: 1. EF = CH_2O.
EF mass = 12 + 2 + 16 = 30. n = 60/30 = 2. MF = C_2H_4O_2.
Stoichiometry
The quantitative relationship between reactants and products in a balanced chemical equation.
Example: N_2 + 3H_2 -> 2NH_3
1 mol N2 reacts with 3 mol H2 to give 2 mol NH3.
28 g N2 + 6 g H2 -> 34 g NH3 (mass conservation verified).
Mass-Mass Relationships
Example: How many grams of NH3 are produced from 14 g N2?
14 g N_2 = 14/28 = 0.5 mol N_2.
From equation, 1 mol N2 gives 2 mol NH3.
So 0.5 mol N2 gives 1 mol NH3 = 17 g NH3.
Limiting Reagent
The reactant that is completely consumed in a chemical reaction and determines the amount of product formed.
Example: 2H_2 + O_2 -> 2H_2O. If 4 g H2 and 32 g O2 react, which is limiting?
H2: 4/2 = 2 moles. O2: 32/32 = 1 mole.
2 mol H2 needs 1 mol O2 (from equation ratio 2:1).
Both are present in exact stoichiometric ratio. No limiting reagent here.
Example with limiting: 6 g H2 and 32 g O2.
H2: 6/2 = 3 moles. O2: 32/32 = 1 mole.
3 mol H2 needs 1.5 mol O2, but only 1 mol O2 available. So O2 is limiting.
Product: 1 mol O2 gives 2 mol H2O = 36 g H2O.
Worked Examples
Example 1: Calculate the number of moles in 11 g of CO2.
Solution: Molar mass of CO2 = 44 g/mol. n = 11/44 = 0.25 mol.
Example 2: How many molecules are in 9 g of water?
Solution: n = 9/18 = 0.5 mol. Molecules = 0.5 * 6.022 x 10^23 = 3.011 x 10^23.
Common Mistakes
- amu vs g/mol: Numerically same but different concepts. 1 atom of C-12 = 12 amu. 1 mole of C-12 = 12 g.
- STP conditions: STP = 0 C (273 K) and 1 atm. Molar volume = 22.4 L.
- Empirical vs molecular: Two different compounds can have the same empirical formula.
- Limiting reagent identification: Always compare actual mole ratios with equation ratios.
ISC Exam Focus
- Theory (70%): Mole concept, stoichiometry, percentage composition, empirical/molecular formula.
- Application (30%): Numerical problems on mole calculations, limiting reagent, gas volumes.
- ISC frequently asks: "Find empirical formula given percentage composition" and "limiting reagent" problems.
- Mole concept numericals carry significant marks.
Self-Test Questions
Q1: Calculate the molar mass of H2SO4.
Answer: 2*1 + 32 + 4*16 = 98 g/mol.
Q2: How many moles are in 50 g of CaCO3? (Ca=40, C=12, O=16).
Answer: Molar mass = 100 g/mol. n = 50/100 = 0.5 mol.
Q3: Calculate the number of atoms in 0.5 mol of carbon.
Answer: 0.5 * 6.022 x 10^23 = 3.011 x 10^23 atoms.
Q4: A compound contains 75% C and 25% H. Find EF. If molar mass is 16 g/mol, find MF.
Answer: C: 75/12 = 6.25, H: 25/1 = 25. Divide by 6.25: C: 1, H: 4. EF = CH_4. MF = CH_4 (already matches molar mass 16).
Q5: 5.6 g of N2 reacts with H2 to produce NH3. Find mass of NH3 produced.
Answer: N_2 + 3H_2 -> 2NH_3. 28 g N2 gives 34 g NH3. So 5.6 g gives 5.6/28 * 34 = 6.8 g NH_3.
Q6: Define limiting reagent with an example. Answer: The reactant consumed first limiting the product amount. Example: 2 mol H2 and 1 mol O2 produce 2 mol H2O. If 3 mol H2 and 1 mol O2, O2 is limiting.
