Chemistry & Matter
- Chemistry = Study of matter and energy
- Matter = Anything which occupies space and has mass (e.g. air, water, sand)
Classification of Matter
- Physical Classification: Solid | Liquid | Gas
- Chemical Classification: Pure → Element, Compound | Impure → Homogeneous mixture, Heterogeneous mixture
- Element = Purest form of matter; made of single type of atom; e.g. N, K, P
- Compound = Pure substance with fixed composition irrespective of source; e.g. Marble, NaCl, H₂O
- Homogeneous mixture = Uniform composition throughout; e.g. Sugar solution, Air, Alloy
- Heterogeneous mixture = Non-uniform composition; e.g. Blood, Soil, Granite
Atom
- Smallest unit of matter which cannot exist independently
- Derived from Greek word "atmos" (means: indivisible)
- Introduced by Dalton
- Subatomic particles = Particles which constitute the atom = Electron, Proton, Neutron
| S.No. | Particle | Charge | Mass | Discovered by |
|---|
| 01 | Electron | −1.6×10⁻¹⁹ C | 9.1×10⁻³¹ kg | Thomson |
| 02 | Proton | +1.6×10⁻¹⁹ C | 1.64×10⁻²⁷ kg | Goldstein |
| 03 | Neutron | — | 1.67×10⁻²⁷ kg | Chadwick |
Atomic Number & Mass Number
- Atomic No. (Z) = No. of protons (nₚ)
- Since atom is neutral: Z = nₚ = nₑ (for neutral atom)
- Mass No. (A) = nₚ + n_N (protons + neutrons)
- So: n_N = A − Z
Ions (Charged Species)
- Cations (+ve): made by loss of e⁻; nₚ > nₑ; e.g. Na⁺, Ca²⁺, Al³⁺
- Anions (−ve): made by gain of e⁻; nₑ > nₚ; e.g. Cl⁻, O²⁻, S²⁻
- Key Note: No. of protons and neutrons in an ion are equal to its parent atom (only electrons change)
Chemical Symbol: ᴬ_Z X where Z = Atomic no., A = Mass no., X = Symbol of Element
e.g. ¹²₆C, ¹⁴₇N, ¹⁶₈O
⚡ Ritrick #1 — For Ions
No. of electrons = No. of protons − Charge (with sign)
No. of e⁻ = Atomic No. − Charge
e.g. Al³⁺ [A=27]: Z=13, n_N=14, e⁻ = 13−3 = 10
e.g. S²⁻ [A=32]: Z=16, n_N=16, e⁻ = 16−(−2) = 18
Practice Questions
Q: Select correct statement? (a) Atoms made of two fundamental particles (b) mass of e⁻ is 9.1×10⁻³¹kg (c) Hydrogen has three isotopes (d) Protons & Neutrons are collectively called Nucleons (e) As per Dalton atom is divisible
Ans: (iv) B, C, D ✓
Q: Which of the following is compound? (a) Brass (b) Petrol (c) Oxygen (d) Marble
Brass = alloy (mixture of metals) | Petrol = mixture of hydrocarbons | Oxygen = element | Marble = compound (CaCO₃)
Ans: (d) Marble ✓
Q: Statement: Sugar Solution is a compound. Reason: composition of compound is same while mixture can have different composition.
Ans: (d) S×, R✓ — Sugar solution is a mixture (variable composition); Reason is correct but doesn't support statement
Q: No. of p, n and e⁻ in Al³⁺ are respectively? [A=27]
p = Z = 13; n = A−Z = 27−13 = 14; e⁻ = Z − charge = 13−3 = 10
Ans: (d) 13, 14, 10 ✓
Q: Statement: No. of protons are always equal to no. of electrons. Reason: Protons have +ve charge & electrons have −ve charge.
Ans: (d) S×, R✓ — Statement false for ions; Reason is a correct fact but doesn't explain S
Q: Blood is a — (a) Homogeneous mixture (b) Heterogeneous mixture (c) Pure Liquid (d) Pure solid
Ans: (b) Heterogeneous mixture ✓
Q: Find the incorrect relation: (a)[nₚ]Fe = [nₚ]Fe²⁺ (b)[n_N]Al = [n_N]Al³⁺ (c)[A]cu = [A]cu⁻ (d)[nₑ]Fe = [nₑ]Fe²⁺
In ions only electrons change, not protons or neutrons. So [nₑ]Fe ≠ [nₑ]Fe²⁺
Ans: (d) ✓ — [nₑ]Fe ≠ [nₑ]Fe²⁺ is incorrect relation
2
Measurement of Atomic Mass — AAM, RAM, GAM
Atomic Mass
- Atomic mass = mass of 1 atom of any element
- Atomic mass = Total mass of p + Total mass of n = nₚ × mass of 1p + n_N × mass of 1N
- Since mass of proton ≈ mass of neutron: Atomic mass = (nₚ + n_N) × mass of 1 proton
AAM (Actual Atomic Mass) = Mass No. × 1.6 × 10⁻²⁴ g
e.g. AAM of N = 14 × 1.6×10⁻²⁴g | AAM of Na = 16 × 1.6×10⁻²⁴g
1 amu / 1u (Unified mass)
- amu = Atomic Mass Unit | u = unified mass
- 1 amu = 1.6 × 10⁻²⁴ g
- 1 amu = 1/12 of mass of 1 C-12 atom
- 1 amu = 1/Nₐ (where Nₐ = Avogadro Number)
- C-12 is taken as standard because: on C-12 scale, atomic masses of most elements are integers; C-12 is very abundant in nature
RAM (Relative Atomic Mass) = Mass No. (in amu or u)
e.g. RAM of N = 14u | RAM of Na = 23u | RAM of Ca = 40u (not shown as 40g)
GAM (Gram Atomic Mass) = Mass of 1 mol atoms = Mass No. (in grams)
1 mol = 6.022 × 10²³ = Avogadro No. (Nₐ) | GAM of N = 14g | GAM of Na = 23g
Summary — Mass No. leads to three different quantities
- Mass No. × 1.6×10⁻²⁴ g → AAM (mass of 1 atom)
- Mass No. in amu → RAM
- Mass No. in grams → GAM (mass of 1 mol atoms)
| Element | RAM | GAM | AAM |
|---|
| H | 1u | 1g | 1×1.6×10⁻²⁴g |
| N | 14u | 14g | 14×1.6×10⁻²⁴g |
| O | 16u | 16g | 16×1.6×10⁻²⁴g |
| Na | 23u | 23g | 23×1.6×10⁻²⁴g |
Practice Questions
Q: No. of neutrons in F [A=19]?
n_N = A−Z = 19−9 = 10
Ans: (b) 10 ✓
Q: An element has 7e⁻ and 8n, mass no. of element?
For neutral atom: Z = nₑ = 7; A = Z + n_N = 7+8 = 15
Ans: (b) 15 ✓
Q: If no. of protons in O atom are doubled and no. of neutrons decreased to 3/4th, new mass no.?
Original: nₚ=8, n_N=8; New: nₚ'=16, n_N'=8×3/4=6; A'=16+6=22; %change=(22−16)/16×100=37.5%
Ans: (a) 22 ✓ | % change = ↑ by 37.5% ✓
Q: In Mg [A=24] no. of protons halved and neutrons increased by 2/3rd. New mass no.?
nₚ=12, n_N=12; nₚ'=6; n_N'=12+12×2/3=12+8=20; A'=6+20=26
Ans: 26 ✓
Q: Which is correct [Na=23, Mg=24, Al=27]? (a)[n_N]Mg=[nₚ]Na (b)[n_N]Mg=[n_N]Na (c)[n_N]Al=[nₚ]P (d)[n_N]Al=[nₚ]Si
[n_N]Mg=12; [nₚ]Na=11; [n_N]Na=12; [n_N]Al=14; [nₚ]Si=14 → options (b) and (d) correct
Ans: (b) & (d) ✓
Q: Atomic mass of Ca is not shown by? (a)40 (b)40u (c)40×1.6×10⁻²⁴g (d)40g
40g = GAM (mass of 1 mol atoms), NOT atomic mass of 1 atom
Ans: (d) 40g ✓ — 40g represents GAM not atomic mass
Q: 1 amu is equal to? (a)1.6×10⁻²⁴g (b)1/12th of C-12 mass (c)1/Nₐ (d)All of above
Ans: (d) All of the above ✓
Q: Atomic mass of Mg given by? (a)24g (b)24u (c)24/Nₐ g (d)both (b)&(c)
RAM = 24u ✓; AAM = 24×1.6×10⁻²⁴g = 24/Nₐ g ✓; 24g = GAM (not atomic mass)
Ans: (d) both (b) & (c) ✓
Q: Mass of 1 atom of element is 8×10⁻²⁴g. Its atomic mass?
AAM = RAM × 1.6×10⁻²⁴; 8×10⁻²⁴ = RAM × 1.6×10⁻²⁴; RAM = 8/1.6 = 5u
Ans: (b) 4.8u — wait: 8/1.6 = 5u ✓
Q: No. of atoms in 108u Al [Al=27]?
108u / 27u per atom = 4 atoms (in amu, not grams)
Ans: (a) 4 atoms ✓
3
Average Atomic Mass & Isotopes
Isotopes
- Atoms of same element which have same atomic no. but different mass no.
- e.g. (i) ¹H, ²H, ³H (ii) ¹⁰B, ¹¹B (iii) ¹²C, ¹⁴C (iv) ³⁵Cl, ³⁷Cl
- Average atomic mass = fractional atomic mass = weighted average of all isotopes
\[A_{avg} = \frac{A_1 x_1 + A_2 x_2}{x_1 + x_2} = \frac{A_1 x_1 + A_2 x_2}{100}\]
\[A_{avg} = A_1 f_1 + A_2 f_2\]
x₁, x₂ = % abundance | f₁, f₂ = fractional abundance | x₁ + x₂ = 100 | f₁ + f₂ = 1
⚡ Ritrick #2 — Equal Abundance
If isotopes are equally abundant (50% each):
\(A_{avg} = \dfrac{A_1 + A_2}{2}\)
If not equal → use line method: closer to heavier isotope means higher % of heavier isotope
Practice Questions
Q: Cl has ³⁵Cl and ³⁷Cl. If ³⁵Cl is found 75% on earth, find avg atomic mass of Cl.
A_avg = (35×75 + 37×25)/100 = (2625+925)/100 = 3550/100 = 35.5u
Ans: 35.5u ✓
Q: Two isotopes of element M have atomic masses 16u & 18u, ratio 1:3. Find avg atomic mass.
f₁=1/4, f₂=3/4; A_avg = 16×1/4 + 18×3/4 = 4+13.5 = 17.5u
Ans: 17.5u ✓
Q: Two isotopes 36u & 38u are equally abundant. Avg atomic mass?
A_avg = (36+38)/2 = 37u
Ans: 37u ✓
Q: Boron has isotopes 10u & 11u, avg atomic mass 10.8. % of both isotopes?
10x₁ + 11x₂ = 10.8×100 = 1080; x₁+x₂=100 → x₁=20%, x₂=80%
Ans: (d) ¹⁰B=20%, ¹¹B=80% ✓
Q: Element has isotopes 50u & 52u. Avg atomic mass = 51.7. % of isotopes?
Avg of equal = 51; 51.7 > 51 → heavier(52) has more %. By line method: ⁵²=85%, ⁵⁰=15%
Ans: (d) 50(15%), 52(85%) ✓
4
Gram Atomic Mass (GAM) & Mole Concept
Gram Atomic Mass
- GAM = Mass of 1 mol atoms of any element in gram
- 1 mol = 6.022 × 10²³ = Avogadro No. (Nₐ)
- Mass of 1 mol atoms = mass of 1 atom × Nₐ = mass no. × 1.6×10⁻²⁴ × Nₐ = mass no. × 1/Nₐ × Nₐ = mass no. (g)
- So: GAM = mass no. (g)
Calculation of Mole — All Four Methods
- No. of mol = Given particles / Nₐ
- No. of mol = Given mass / Molar mass
- Mol of ideal gas = Volume at STP (L) / 22.4
- Mol of ideal gas = Volume at STP (mL) / 22400
Important Notes
- 1 mol atoms = Nₐ atoms = also called 1g atoms
- Avogadro's Law: 1 mol of any ideal gas occupies 22.4L at STP/NTP [T=273K=0°C, P=1atm]
- Gases always exist in molecular form: O₂ [M=32], N₂ [M=28], H₂ [M=2]
- Note: 1 mol atoms is also called 1g atoms
⚡ Ritrick — Mol ↔ Atom/Molecule Conversion
Mol. Molecule × Atomicity → Mol. Atom
Mol. Atom ÷ Atomicity → Mol. Molecule
Practice Questions
Q: No. of atoms in 128g O [multiple correct]?
128g O / 16g = 8 mol O atoms; 1 mol = Nₐ atoms; 8 mol = 8Nₐ atoms
Ans: (b) 8 mol atoms & (c) 8Nₐ atoms ✓
Q: No. of N atoms in 56g N [mass no.=14]?
mol = 56/14 = 4 mol atoms = 4Nₐ atoms
Ans: (d) All of the above ✓
Q: Charge on 64 amu S²⁻ ion [atomic mass S=32]?
mass of 1 S²⁻ = 32 amu; 64 amu = 2 S²⁻ ions; charge = 2×2×1.6×10⁻¹⁹ = 6.4×10⁻¹⁹ C
Ans: (d) 6.4×10⁻¹⁹ C ✓
Q: Mass of 3.01×10²² atoms of element is 0.7g. Element is?
GAM = 0.7×Nₐ/3.01×10²² = 0.7×6.022×10²³/3.01×10²² = 0.7×20 = 14g → N
Ans: (b) N ✓
Q: Mass of 12.04×10²³ atoms of Na [A=23]?
mol = 12.04×10²³/6.02×10²³ = 2 mol; mass = 2×23 = 46g
Ans: (d) 46g ✓
Q: No. of mol of C in 3.01×10²² atoms of C?
mol = 3.01×10²²/6.02×10²³ = 0.05 mol
Ans: (d) 0.05 mol ✓
Q: Question mark has 12.04×10²¹ atoms of C. Find wt of C used?
mol of C = 12.04×10²¹/6.02×10²³ = 0.02; W = 0.02×12 = 0.24g
Ans: W = 0.24g ✓
Q: From 4.4g CO₂, 3.01×10²² molecules removed. Remaining mass?
removed mol = 3.01×10²²/6.02×10²³ = 0.05 mol; removed wt = 0.05×44 = 2.2g; remaining = 4.4−2.2 = 2.2g
Ans: (b) 2.2g ✓
5
Molecules, Molecular Mass & Molar Mass
Molecules
- Molecules are made up of combination of atoms
- Smallest independent particle of matter
- Homoatomic (Element): H−H, O=O, N≡N (same type of atoms)
- Heteroatomic (Compound): H₂O, O=C=O, HCl (different types of atoms)
Molecular Mass = Σ Mass Numbers (in amu or u) = mass of 1 molecule
e.g. N₂→28u | O₂→32u | HCl→36.5u | SO₂→64u | H₂SO₄→98u | C₆H₁₂O₆→180u | NH₂CONH₂→60u | C₁₂H₂₂O₁₁→342u
Molar Mass = Σ GAM of atoms = Σ mass no. (in grams) = mass of 1 mol molecules
1 mol molecule = 1g molecule | Molar mass = Σ mass numbers (g)
Atomicity
- No. of atoms of any element in a molecule
- He, Ne, Ar, Fe, Cu, Na, Mg → Atomicity = 1 (monoatomic)
- O₂ → Atomicity of O = 2 | NH₃ → N atomicity=1, H atomicity=3
- 1 mol NH₃ = 1 mol N atom + 3 mol H atom
- Nₐ molecules NH₃ = Nₐ atoms N + 3Nₐ atoms H
⚡ Ritrick — Mol Molecule ↔ Mol Atom
Mol. Molecule × Atomicity → Mol. Atom
Mol. Atom ÷ Atomicity → Mol. Molecule
Practice Questions
Q: Find Molecular mass of K₄[Fe(CN)₆] [K=40, Fe=56, C=12, N=14]?
= 4×40 + 56 + 6×(12+14) = 160+56+156 = 372u
Ans: 372u ✓
Q: Mass of 1g molecule O₂?
Ans: (c) 32g ✓ [1g molecule = 1 mol; molar mass O₂ = 32g]
Q: Which is heaviest? (a)1 molecule H₂O (b)1g H₂O (c)1g molecule H₂O (d)All equal
1g molecule = 1 mol H₂O = 18g; 1g H₂O = 1g; 1 molecule = 18u only
Ans: (c) 1g molecule H₂O = 18g ✓
Q: Molar mass of H₂C₂O₄·2H₂O?
= 2+24+64+2×18 = 90+36 = 126g/mol
Ans: (d) 126g ✓
Q: Atomicity of H in (NH₄)₃PO₄?
(NH₄)₃PO₄: H = 3×4 = 12 H atoms per molecule
Ans: (d) 12 ✓
Q: How many moles of Mg₃(PO₄)₂ contain 0.8 mol O atoms?
1 mol Mg₃(PO₄)₂ = 8 mol O; 0.8/8 = 0.1 mol compound
Ans: (a) 0.1 mol ✓
Q: No. of mol of O atoms in 0.2 mol Na₂SO₄·10H₂O?
1 mol = 4+10 = 14 mol O; 0.2×14 = 2.8 mol
Ans: (d) 2.8 mol ✓
Q: (NH₄)₃PO₄ has 1.08 mol H atoms. No. of O atoms in same quantity?
1 mol compound = 12 mol H; compound = 1.08/12 = 0.09 mol; 1 mol = 4 mol O; O = 0.09×4 = 0.36 mol = 0.36Nₐ
Ans: (d) 0.36Nₐ ✓
Q: Mol of O atoms in 24g O₃?
mol O₃ = 24/48 = 0.5; mol O = 0.5×3 = 1.5 mol
Ans: (c) 1.5 mol ✓
Q: Total no. of atoms in 4.25g NH₃?
mol NH₃ = 4.25/17 = 0.25; 1 mol NH₃ = 4 mol atoms; atoms = 0.25×4 = 1 mol = Nₐ
Ans: (c) Nₐ ✓
Q: Which will have max no. of O atoms? (a)1g O (b)1g O₂ (c)1g O₃ (d)All equal
mol O atoms: 1g O=1/16; 1g O₂=1/32×2=1/16; 1g O₃=1/48×3=1/16 → all equal
Ans: (d) All equal ✓ [Ritrick #5: equal mass → equal atoms]
\[\text{Mass\%} = \frac{\text{mass of element}}{\text{molar mass}} \times 100 = \frac{\text{Atomic mass} \times \text{Atomicity}}{\text{Molar mass of molecule}} \times 100\]
Example — Glucose C₆H₁₂O₆
- 1 mol C₆H₁₂O₆ = 6 mol C + 12 mol H + 6 mol O
- 180g C₆H₁₂O₆ = 72g C + 12g H + 96g O
- Mass% of C = 72/180 × 100 = 40%
- Mass% of O = 96/180 × 100 = 53.33%
⚡ Ritrick #5
Equal mass of any element in all forms will have equal no. of atoms.
e.g. 2g Cl and 2g Cl₂ → both have equal no. of atoms ✓
Q: Mass% of O in metal carbonate M₂CO₃ is 60%. Find atomic mass of M.
Let AM of M = x; molar mass M₂CO₃ = 2x+12+48 = 2x+60; %O = 48/(2x+60)×100 = 60; 4800 = 120x+3600; x=10
Ans: Atomic mass of M = 10 ✓
Q: Myoglobin is polymer of Fe with 4 Fe atoms per molecule. If Fe is 0.25% by mass [Fe=56], molecular mass of myoglobin?
%Fe = (4×56/M)×100 = 0.25; M = 4×56×100/0.25 = 89600
Ans: Molar mass = 89600 ✓
Q: Polymer of Fe contains 0.33% Fe. Find minimum molecular mass [Fe=56]?
For min M, put atomicity=1; %Fe = (56×1/M)×100 = 0.33; M_min = 56×100/0.33 = 56/33×10⁴ = 16000u ≈ 16970u
Ans: M_min = 16000u ✓ [Note: for min mol mass, put atomicity = 1]
7
Empirical & Molecular Formula
Molecular Formula
- Tells exact no. of atoms of different elements in a molecule
- Molecular formula is unique for any compound
- e.g. C₆H₁₂O₆ (Glucose), H₂O (Water)
Empirical Formula
- Formula which tells simplest ratio of atoms or moles of different elements
- More than one compound may have same empirical formula
- e.g. CH₂O (formaldehyde), C₂H₄O₂ (acetic acid), C₆H₁₂O₆ (glucose) — all have EF = CH₂O
\[n = \frac{\text{Molecular mass}}{\text{Empirical mass}}\]
Molecular Formula = Empirical Formula × n
Steps to Find Empirical Formula from % composition
- Assume 100g sample → % = grams
- Divide each by atomic mass → get mol ratio
- Divide by smallest mol ratio → get simple ratio
- If not whole number → multiply to get whole number
- Write EF from ratio
⚡ Ritrick #6 — By Mass%
\(\dfrac{p_1}{p_2} = \dfrac{A_1 \cdot n_1}{A_2 \cdot n_2}\) → Solve for n₁/n₂
mass% of C = 12×nC/M × 100
mass% of H = 1×nH/M × 100
\(\dfrac{n_C \times 12}{n_H \times 1} = \dfrac{p_C}{p_H}\)
Practice Questions
Q: EF = CH₂O. Molecular mass = 180. Compound is?
EF mass = 12+2+16 = 30; n = 180/30 = 6; MF = (CH₂O)×6 = C₆H₁₂O₆
Ans: (d) C₆H₁₂O₆ ✓
Q: EF = CH₂O. If 0.25 mol compound contains 1 mol H. MF?
1 mol compound = 1/0.25 = 4 mol H; EF has 2H → n=2; MF = C₂H₄O₂ = CH₃COOH
Ans: (b) CH₃COOH ✓
Q: Compound has 38.71%C, 9.67%H, rest O. Find EF.
%O=51.62%; mol%: C=38.71/12=3.3; H=9.67/1=9.67; O=51.62/16=3.3; ratio=1:3:1
Ans: CH₃O ✓
Q: Compound has 40%C, 53.28%O, rest H. EF?
%H=6.72%; pC/pO = (12nC)/(16nO); 40/53.28 = 12nC/16nO → nC:nO=1:1; nH=6.72/3.3≈2; C:H:O=1:2:1
Ans: (a) CH₂O ✓
Q: Organic compound contains 3C, 9H, 6N atoms per molecule. EF?
C:H:N = 3:9:6 = 1:3:2
Ans: (b) CH₃N₂ ✓
Q: Drug marijuana contains 70% C atoms as H atoms and 15× H atoms as O atoms, 0.00318 mol in 1g. MF?
nC=70%nH; nH=15nO → option (b) C₂₁H₃₀O₂ satisfies both
Ans: (b) C₂₁H₃₀O₂ ✓
Q: Element M makes two oxides. First MO₂ has M=50%, second has M=40%. Formula of 2nd oxide?
MO₂: 50/50=Aₘ×1/16×2 → Aₘ=32; 2nd oxide: 40/60=32×nₘ/16×nO → nₘ/nO=1/3 → MO₃
Ans: (d) MO₃ ✓
8
Ideal Gas & Vapour Density
Ideal Gas
- A hypothetical state of matter
- Avogadro Law: 1 mol of any ideal gas occupies 22.4L at STP/NTP [T=273K=0°C, P=1 atm]
- mol of ideal gas = Volume at STP (L) / 22.4
- mol of ideal gas = Volume at STP (mL) / 22400
Vapour Density
- Ratio of density of any gas to density of H₂ at same temperature and pressure
- VD is applicable only for gases
- VD is a unitless quantity
\[\text{Vapour Density (VD)} = \frac{\rho_{gas}}{\rho_{H_2}} = \frac{M_{gas}/2}{M_{H_2}/2}\]
Vapour Density = Molar mass of gas / 2
At STP: Density of gas = Molar mass / 22.4 g/L
Practice Questions
Q: Formula of gas is (CO)ₓ and VD = 70. Find x.
M = VD×2 = 140; M of (CO)ₓ = 28x = 140; x = 5
Ans: (c) 5 ✓
Q: VD of an alkane is 36. Formula of alkane?
M = 72; CₙH₂ₙ₊₂: 14n+2=72; n=5 → C₅H₁₂
Ans: (b) C₅H₁₂ ✓
Q: Find density of O₂ gas at STP?
Density = M/22.4 = 32/22.4 = 1.43 g/L
Ans: (d) 1.43 g/L ✓
Q: 5g of ideal gas occupies 1.12L at STP. VD of gas?
mol = 1.12/22.4 = 0.05; M = 5/0.05 = 100; VD = 100/2 = 50
Ans: (c) 50 ✓
Q: 5.6L of ideal gas weighs 7g at STP. Gas is?
mol = 5.6/22.4 = 0.25; M = 7/0.25 = 28 → N₂
Ans: (c) N₂ ✓
Q: Volume occupied by 0.5 mol N₂ at NTP?
V = 0.5 × 22.4 = 11.2L
Ans: (c) 11.2L ✓
Q: wg gas occupies Vml at STP. Molar mass?
mol = w/M = V/22400; M = w×22400/V
Ans: (d) 22400w/V ✓
Q: Which has max molecules? (a)56g N₂ (b)32g O₂ (c)64g SO₂ (d)8g H₂
mol: N₂=2; O₂=1; SO₂=1; H₂=4 mol
Ans: (d) 8g H₂ ✓
9
Stoichiometry of Chemical Reactions
Balanced Chemical Reaction
- A chemical reaction which follows: (i) Law of conservation of atoms (ii) Law of conservation of mass
- Quantitative analysis of balanced chemical reactions
- Stoichiometric coefficient = number written before formula in balanced equation
N₂(g) + 3H₂(g) → 2NH₃(g) — All Interpretations
- 1 molecule : 3 molecules : 2 molecules (molecular level)
- Nₐ molecule : 3Nₐ molecule : 2Nₐ molecule [×Nₐ]
- 1 mol : 3 mol : 2 mol → mol-mol relation
- 28g : 6g : 34g → mass-mass relation
- 1×22.4L : 3×22.4L : 2×22.4L (vol at STP)
- 1L : 3L : 2L → volume-volume relation (for gaseous rxns only)
⚡ Ritrick for Stoichiometry — aA + bB → cC + dD
\(\dfrac{\text{mol of A}}{a} = \dfrac{\text{mol of B}}{b} = \dfrac{\text{mol of C}}{c} = \dfrac{\text{mol of D}}{d}\)
\(\dfrac{\text{Vol of A}}{a} = \dfrac{\text{Vol of B}}{b} = \dfrac{\text{Vol of C}}{c}\) → for gaseous rxns only
Practice Questions
Q: Find mol of H₂ required to produce 10 mol NH₃. [N₂+3H₂→2NH₃]
mol H₂/3 = mol NH₃/2 = 10/2 = 5; mol H₂ = 15 mol
Ans: 15 mol ✓
Q: Find wt of residue after combustion of 2 mol Ag₂CO₃ [Ag=108].
Ag₂CO₃(s)→2Ag(s)+CO₂(g)+½O₂(g); only Ag solid remains; 2 mol Ag₂CO₃→4 mol Ag; W=4×108=432g
Ans: W_Ag = 432g ✓
Q: Volume of CO₂ at STP on combustion of 25g CaCO₃ [Ca=40,C=12,O=16]?
CaCO₃→CaO+CO₂; mol CaCO₃=25/100=0.25; mol CO₂=0.25; V=0.25×22.4=5.6L
Ans: (b) 5.6L ✓
Q: Wt of Al to produce 408g Al₂O₃ [Al=27,O=16]? 4Al+3O₂→2Al₂O₃
mol Al₂O₃=408/102=4; mol Al=4×4/2=8; W=8×27=216g
Ans: (c) 216g ✓
Q: Volume of O₂ to combust 5L C₃H₈? C₃H₈+5O₂→3CO₂+4H₂O
V_C₃H₈/1 = V_O₂/5; V_O₂=25L
Ans: (d) 25L ✓
Q: Volume of air for 10L CH₄? [air=20% O₂]
CH₄+2O₂→CO₂+2H₂O; V_O₂=20L; V_air=5×20=100L
Ans: 100L ✓
Q: 200mL hydrocarbon → 800mL CO₂ + 1000mL H₂O. Formula?
x=800/200=4; y=2×1000/200=10 → C₄H₁₀
Ans: (d) C₄H₁₀ ✓
Q: No. of mol of O₃ such that molecules equal those in 56g N₂?
mol N₂=2; mol O₃=2; mass=2×48=96g
Ans: (b) 96g ✓
10
Percentage Purity & Limiting Reagent
% Purity = (wt of pure reactant / wt of impure sample) × 100
⚠️ Important Note
- Any wt of reactant (either given or asked) is always wt of IMPURE SAMPLE
- Wt of pure reactant is always calculated in question and is NEVER GIVEN directly
Limiting & Excess Reagent
- Limiting Reagent (LR) = reactant which gets completely consumed during reaction
- Excess Reagent (ER) = reactant which remains after reaction
- Amount of product in a reaction is always calculated by Limiting Reagent
⚡ Ritrick for LR Problems
(i) Find LR: ratio = Given mol / Stoichiometric coefficient → Reactant with smaller ratio = LR
(ii) Find mol of product:
\(\dfrac{\text{mol of LR}}{\text{St. coeff of LR}} = \dfrac{\text{mol of product}}{\text{St. coeff of product}}\)
(iii) Find mol of ER remaining:
n(remaining)_ER = n_initial − n_reacted
\(\dfrac{\text{mol of LR}}{\text{St. coeff}} = \dfrac{\text{n reacted of ER}}{\text{St. coeff}}\)
Practice Questions
Q: Wt of CO₂ from heating 200g of 90% pure CaCO₃?
CaCO₃→CaO+CO₂; W_pure=200×90/100=180g; mol CaCO₃=180/100=1.8; mol CO₂=1.8; W=1.8×44=79.2g
Ans: ≈79.2g ✓
Q: 20g sample of MgCO₃ produces 8g MgO. % purity of MgCO₃ [Mg=24]?
MgCO₃→MgO+CO₂; mol MgO=8/40=0.2; mol MgCO₃=0.2; W_pure=0.2×84=16.8g; %=16.8/20×100=84%
Ans: (d) 84% ✓
Q: Find wt of 95% pure CaCO₃ reacting with 0.05 mol HCl. CaCO₃+2HCl→CaCl₂+H₂O+CO₂
mol CaCO₃ pure=0.05/2=0.025; W_pure=2.5g; W_sample=2.5×100/95=2.63g
Ans: 2.63g ✓
Q: AlCl₃ formed: 2Al+3Cl₂→2AlCl₃ (5 mol Al, 6 mol Cl₂). Find LR, ER, mol AlCl₃, ER remaining.
Ratio: Al=5/2=2.5; Cl₂=6/3=2 → Cl₂ is LR; mol AlCl₃: 6/3=AlCl₃/2→4 mol; Al reacted: 6/3=Al/2→4 mol; Al remaining=5−4=1 mol
Ans: LR=Cl₂, ER=Al, AlCl₃=4 mol, Al remaining=1 mol ✓
Q: 3 mol Zn + 5 mol S + 4 mol Fe → ZnFeS₂. Mol of ZnFeS₂ formed?
Zn+Fe+2S→ZnFeS₂; ratio: Zn=3; Fe=4; S=5/2=2.5 → S is LR; mol product=5/2=2.5 mol
Ans: (d) 2.5 mol ✓
Q: 16g H₂ and 80g O₂ react: 2H₂+O₂→2H₂O. Find wt of water.
mol H₂=8; mol O₂=2.5; ratio: H₂=8/2=4; O₂=2.5/1=2.5 → O₂ is LR; mol H₂O=2×2.5=5; W=5×18=90g
Ans: W_H₂O = 90g ✓
Q: Strongly heating 50g CaCO₃ sample produces 4.48L CO₂. % impurity?
mol CO₂=4.48/22.4=0.2; W_pure CaCO₃=0.2×100=20g; %purity=40%; %impurity=60%
Ans: (c) 60% impurity ✓
11
Laws of Chemical Combination
① Law of Conservation of Mass (Lavoisier)
- Total mass of reactants in a chemical reaction = total mass of products
- e.g. N₂(g) + 3H₂(g) → 2NH₃(g): 28g + 6g = 34g ✓
- This law is the basis of balancing chemical reactions
- Limitation: Not applicable on nuclear reactions. On those reactions, law of mass and energy conservation is applicable
② Law of Definite Proportion (Joseph Proust)
- The ratio of masses of different elements in a compound is always same irrespective of its source of origin
- e.g. 1 mol H₂O = 2g H + 16g O → W_H/W_O = 2/16 = 1/8 (always fixed)
- Limitation: This law is not followed when isotopes of an element make a compound
③ Law of Multiple Proportions (Dalton)
- When two elements make more than one compound, the ratio of mass of one element combining with fixed mass of other element is a whole number ratio
- e.g. H & O → H₂O (16g O) and H₂O₂ (32g O per 2g H) → [W_O]_H₂O / [W_O]_H₂O₂ = 16/32 = 1/2
④ Gay Lussac's Law of Constant Volume
- In gaseous reactions, ratio of volumes of reactant molecules are always in whole number ratio
- e.g. N₂(g) + 3H₂(g) → 2NH₃(g): V_N₂ : V_H₂ : V_NH₃ = 1:3:2
- Applicable ONLY for gaseous reactions! All reactants and products must be gases
Practice Questions
Q: Copper forms two oxides CuO & Cu₂O. This demonstrates?
Ans: (c) Law of Multiple Proportions ✓
Q: On which rxn Gay Lussac's Law applicable? (a)CaCO₃(s)→CaO(s)+CO₂(g) (b)N₂(g)+O₂(g)→2NO(g) (c)H₂(g)+I₂(g)→2HI(g) (d)H₂(g)+½O₂(g)→H₂O(l)
Ans: (b) & (c) ✓ — all species must be gaseous
Q: 100kg ethylene polymerizes: nCH₂=CH₂→[CH₂−CH₂]ₙ. Mass of polythene formed?
Ans: (d) 100kg ✓ — by Law of Conservation of Mass
Q: Sample CaCO₃ from Bengal has 40% Ca. % Ca in sample from Rajasthan?
Ans: (d) 40% ✓ — by Law of Definite Proportion, composition is always fixed
12
Scale Shift Problems — AAM, RAM, GAM Variation
Key Rules — Which Changes and Which Doesn't
- AAM — NEVER changes (it's actual physical mass)
- GAM = AAM × Nₐ → GAM changes ONLY when Nₐ is changed; GAM NEVER changes when amu changes
- RAM changes when: (i) fraction of amu is changed OR (ii) reference element is changed
⚡ Ritrick — Type I (Fraction of amu changed)
Standard: 1 amu = 1/12 of mass of 1 C-12 atom
If new: 1 amu' = 1/x of mass of 1 C-12 atom
RAM' = RAM × x / 12
⚡ Ritrick — Type II (Reference element changed)
\(\dfrac{\text{Atomic wt of old ref.}}{\text{Atomic wt of new ref.}} = \dfrac{\text{old RAM}}{\text{new RAM}}\)
Alternate: Old amu / New amu = Old RAM / New RAM
Practice Questions
Q: If 1/24th of mass of C-12 is new amu, atomic mass of Ag[108]?
x=24; RAM' = 108×24/12 = 216
Ans: (c) 216 ✓
Q: If 1/5th of mass of C-12 is new amu, atomic mass of Mg[24]?
x=5; RAM' = 24×5/12 = 10
Ans: (a) 10 ✓
Q: If O-16 taken as new standard in place of C-12, new atomic mass of Ag[108]?
Old amu/New amu = 12/16; 108/[Ag]new = 12/16; [Ag]new = 108×16/12 = 144
Ans: (a) 144 ✓
Q: If Mg-24 taken as new standard, new atomic mass of Fe[56]?
A_C-12/A_Mg-24 = A_Fe_old/A_Fe_new; 12/24 = 56/A_new; A_new = 112
Ans: (b) 112 ✓
Q: If Nₐ changed from 6.022×10²³ to 6.022×10²⁰, which changes?
GAM = AAM × Nₐ → changes; AAM = fixed; RAM = fixed; So mass of 1 mol C atoms changes
Ans: (d) mass of 1 mol C atoms ✓
13
Water Drop Problems & Combustion
Density of water = 1 g/mL → mass of water (g) = volume of water (mL)
Combustion of Hydrocarbons
\(C_xH_y(g) + \left[x + \dfrac{y}{4}\right]O_2(g) \rightarrow xCO_2(g) + \dfrac{y}{2}H_2O(l)\)
- e.g. CH₄+2O₂→CO₂+2H₂O | C₃H₈+5O₂→3CO₂+4H₂O
- Air contains 20% O₂ by volume → V_air = 5 × V_O₂
Practice Questions
Q: No. of electrons in 36mL water?
36mL water=36g; mol H₂O=36/18=2; 1 H₂O molecule has 10e⁻; mol e⁻=2×10=20 mol
Ans: (d) 20 mol e⁻ ✓
Q: Volume of 1 molecule of water?
mass of 1 molecule=18 amu=18×1.6×10⁻²⁴g=28.8×10⁻²⁴g; density=1g/mL; V=28.8×10⁻²⁴ mL
Ans: 28.8×10⁻²⁴ mL ✓
Q: On strongly heating 50g sample CaCO₃ → 4.48L CO₂. % impurity?
mol CO₂=4.48/22.4=0.2; W pure CaCO₃=20g; %purity=40%; %impurity=60%
Ans: (c) 60% ✓
Q: Find volume of air required to combust 10L CH₄?
CH₄+2O₂→CO₂+2H₂O; V_O₂=20L; V_air=5×20=100L
Ans: 100L ✓
Q: 16g H₂ and 80g O₂ react → H₂O. Find wt of water [multiple reagent problem].
mol H₂=8; mol O₂=2.5; 2H₂+O₂→2H₂O; ratio H₂=8/2=4; O₂=2.5/1=2.5 → O₂ LR; mol H₂O=2×2.5=5; W=90g
Ans: 90g ✓