Mole Concept

Chapter 1: Mole Concept

What is Chemistry?

Science is humankind’s’ continuing effort to understand and describe the nature.

• Chemistry: A branch of Science that studies preparation, properties, structure, and reactions of material substances. Essentially dealing with the compositions, properties and interaction of matter.

Few important Chemicals

• Cis-platin and taxol: Effective for cancer treatment.
• AZT (Azidothymidine) drug: Used to help AIDS patients.

Matter

• Matter is something that ha s mass and occupies space. (Ex- Pen board, electron)
  • ! Electron is matter but Electricity is not cause flow of electron is a phenomena so it has no mass or occupying space

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Classification of Chemistry

mindmap
  root("Classification of Chemistry")
    ("Physical")
      ("Solid")
      ("Liquid")
      ("Gas")
      ("Plasma")
      ("BEC aka Bose-Einstein Condensate")
    ("Chemical")
      ("Pure substance")
        ("Element")
        ("Compound")
      ("Mixture")
        ("Homogeneous")
        ("Heterogeneous")

Physical Classifications

FYI Plasma and BEC is’t in our syllabus

Physical Attributes	Solid	Liquid	Gas
Shape	Fixed	Not Fixed	Not Fixed
Volume	Fixed	Fixed	Not Fixed
Inter-particle Distance	Least (<)	Moderate	Max (>)
Inter-Particle Force of Attraction	Max (>)	Moderate	Least (<)
Compressibility	Least (<)	Negligible	Max (>)
Density	Max (>)	Moderate	Least (<)
Motion of Particles	Least (<)	Moderate	Max (>)
Diffusion aka ব্যাপন	Least (<)	Moderate	Max (>)
Energy (Thermal)	Least (<)	Moderate	Max (>)

Inter-conversion of States of Matter

graph LR
    S[Solid]
    L[Liquid]
    G[Gas]

    classDef stateNode fill:#24292e,stroke:#444d56,stroke-width:2px,color:#ffffff,font-weight:bold;
    class S,L,G stateNode;

    S -->|Melting / Fusion <br/> + Heat| L
    L -->|Vaporisation <br/> + Heat| G

    G -->|Condensation <br/> - Heat| L
    L -->|Freezing <br/> - Heat| S

    S -.->|Sublimation <br/> + Heat| G
    G -.->|Deposition <br/> - Heat| S

    linkStyle 0,1 stroke:#ff7675,stroke-width:2.5px;
    linkStyle 2,3 stroke:#74b9ff,stroke-width:2.5px;
    linkStyle 4 stroke:#e84393,stroke-width:2px;
    linkStyle 5 stroke:#0984e3,stroke-width:2px;

Sublimation vs. Deposition

Sublimation and Deposition are direct, opposite phase changes that completely skip the liquid state.

• Sublimation (Solid $\to$ Gas): An endothermic process where a solid absorbs heat to vaporize directly.
  • Examples: Dry ice (${\text{CO}}_2$), Camphor, Naphthalene (mothballs), and Iodine crystals.
• Deposition (Gas $\to$ Solid): An exothermic process where a gas rapidly cools and solidifies directly.
  • Examples: Sub-zero frost on windows, snowflake formation in clouds, and soot accumulation in chimneys.

Chemical Classifications

Pure Substance

A pure substance is a form of matter that has a constant, uniform composition and distinct chemical properties.

• It’s made up of only one substance. So, all samples have same properties and composition.
• Cannot be separated into other materials by physical methods (Ex- Filtration, Magnetic Separation, Hand Picking, Sublimation, Distillation, Chromatography, etc).
  • Ex: $Na,Mg,H_{2}O,NaCl,etc$

1. Element

Pure substance only made up of one type of element.

• These type of pure substance cant also be decomposed using chemical method along with physical method.
  • Ex: $Na,Mg,Al,H_2,N_2,etc$

2. Compound

Pure substance made up of two or more different elements combined in a fixed proportion by mass.

• The mass ratio is fixed*.
• Their constituents can be separated using chemical methods only.
• The properties of a compound is completely different from their constituents.
  • Ex: $H_{2}O\to 1:8,N{H}_3\to 14:3$

What is Atomicity?

The number of atoms present in a molecule.

• Ex: $O_2\to 2,H_{2}O\to 3,HCl\to 2,H_{2}S{O}_4\to 7$

Mixture

Mixture is a form of matter that is made up of two or more elements or compounds or both in any proportion by mass.

• Can be separated using Physical Methods.
  • Ex: Sugar Solution, RedBull, Tea, Fog, Bread, Air, Sand, Water, Blood, etc

1. Homogeneous Mixture

The mixture which has uniform composition throughout the mixture

• Ex: Air, Alloy, Alcohol + Water, Salt Solution, Sugar Solution

Solution

Solution is a type of homogeneous mixture with two or more components.

• Binary Solution: Solution with two components (1 Solute+ 1 Solvent).
• @ In a solution the Solute is comparatively lesser in amount than Solvent.

2. Heterogeneous Mixture

The mixture which doesn’t have uniform composition and sometimes different components are visible.

• Ex: Oil + Water, Chalk Powder +Water, Sand + Water, etc
  • ! Special Example: Milk, Blood, etc. These are colloids which often appear as Homogeneous to the naked eye but microscopically it’s a Heterogeneous Mixture.
    • Blood comprises cells (RBCs, WBCs, platelets) dispersed in plasma. These cellular components can be separated and are not uniformly distributed throughout the fluid.

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Classification of Properties of Matter

mindmap
  root("Properties of Matter")
	  ("Physical")
	  ("Chemical")

Physical Properties

Any property of a substance that can be observed or measured or altered without changing the identity or chemical composition of the substance.

• Ex: Length, Mass, Volume, Density, Colour, Taste, Lustre, Taste, Hardness, Melting Point, Boiling Point etc.

Chemical Properties

Any property of a substance that can be observed or measured or altered when the substance undergoes a chemical reaction that changes its chemical identity.

• Ex: Flammability, Rusting (corrosion), Toxicity, Reactivity, Acidity, Basicity, Combustibility, etc.

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Measurement of Physical Quantities

Any measured Physical Quantity is expressed in two parts: a numerical coefficient and a unit. A unit is the standard reference chosen to measure that quantity.

Overlapping section with Physics to avoid redundancy

• Understanding Physical Quantities (Base vs. Derived)
• The 7 SI Base Quantities (Length, Mass, Time, etc.)
• SI Prefixes (from yocto to yotta)
• Accuracy vs Precision
• Significant Figures

Key Measurements in Chemistry

While physics covers the broad spectrum of units, in chemistry, we frequently deal with a specific subset of measurements for lab calculations.

1. Mass vs. Weight

Though often used interchangeably in daily life, they are fundamentally different:

• Mass: The actual amount of matter present in a substance. Mass is constant anywhere in the universe.
• Weight: The force exerted by gravity on a substance. Weight can vary depending on the local gravity.
• Relation: $W=mg$

2. Volume

The amount of space occupied by a substance.

• SI Unit: ${\text{m}}^3$
• Common Chemistry Conversions: In the lab, we rarely use cubic meters. We use litres and millilitres.
  • $1\text{ ml}=1{\text{ cm}}^3=1\text{ cc}$
  • $1\text{ L}=1000\text{ ml}=1000{\text{ cm}}^3=1{\text{ dm}}^3=10^{-3}{\text{ m}}^3$

3. Temperature

There are three common scales used to measure temperature: ${}^{\circ }\text{C}$, ${}^{\circ }\text{F}$, and $\text{K}$.

• Conversions:
  • Celsius to Kelvin: $K{=}^{\circ }\text{C}+273.15$ (Note: In rough calculations, we often just use $+273$).
  • Celsius to Fahrenheit: $({}^{\circ }\text{F}-32)\times \frac{5}{9}{=}^{\circ }\text{C}$
• Absolute Zero: The minimum possible temperature is $0\text{ K}$ (or $-273.15^{\circ }\text{C}$).
  • ! Negative temperatures are possible on the Celsius scale, but not on the Kelvin scale.

4. Density

The amount of mass per unit volume.

• Formula: $\text{Density}=\frac{\text{Mass}}{\text{Volume}}$
• SI Unit: ${\text{kg/m}}^3$

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Scientific Notation

Because chemistry deals with the microscopic (atoms) and the macroscopic (moles), we use exponential notation to handle extremely large or small numbers.

Every number is represented in the form:
$N\times 10^n$

• $N$ (Digit Term): A number between $\mathrm{1.000...}$ and $\mathrm{9.999...}$.
• $n$ (Exponent): Can be a positive or negative integer.

Moving the Decimal

• Shifting the decimal to the Left $\to$ Exponent becomes Positive ($+ve$).
  • Ex: $100.2345\to 1.002345\times 10^2$
• Shifting the decimal to the Right $\to$ Exponent becomes Negative ($-ve$).
  • Ex: $0.001356\to 1.356\times 10^{-3}$

Calculations in Scientific Notation

• Multiplication: Add the exponents.→ $10^x\times 10^y=10^{x+y}$
  • Ex: $(1.2\times 10^3)\times (1.2\times 10^6)=1.44\times 10^9$
• Division: Subtract the exponents→ $\frac{10^x}{10^y}=10^{x-y}$
  • Ex: $\frac{1.69\times 10^8}{1.3\times 10^4}=1.3\times 10^4$

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Atoms

An atom is the smallest particle of an element which may or may not have independent existence, but always retains all the chemical properties of that element takes part in a chemical reaction.

• ! Noble gases(He, Ne, Ar, Kr, Xe, Rn) can exist in nature Independently

Representation of Atom

An atom of element X having atomic number A and mass number Z is represented by
${}_{\text{Atomic No }=Z}^{\text{Mass Num }=A}X$

Key Formulas:

• Mass Number ($A$) = Protons + Neutrons (also called Nucleons).
• Atomic Number ($Z$) = Number of protons.
  • For Neutral Atoms: Z = Num of Protons = Num of Electrons
  • For Ions:
    • Cations: (+ve) aka Loss of Electron
    • Anions: (-ve) aka Gain of Electron

Atomic Number	Element	Element Symbol	Atomic Mass (Rounded)
1	Hydrogen	H	1
2	Helium	He	4
3	Lithium	Li	7
4	Beryllium	Be	9
5	Boron	B	11
6	Carbon	C	12
7	Nitrogen	N	14
8	Oxygen	O	16
9	Fluorine	F	19
10	Neon	Ne	20
11	Sodium	Na	23
12	Magnesium	Mg	24
13	Aluminium	Al	27
14	Silicon	Si	28
15	Phosphorus	P	31
16	Sulphur	S	32
17	Chlorine	Cl	35.5
18	Argon	Ar	40
19	Potassium	K	39
20	Calcium	Ca	40

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Molecules

A molecule is the smallest particle of an element or a compound that is capable of independent existence and shows all the chemical properties of that substance.

Homonuclear & Heteronuclear

mindmap
root ("Molecules")
  ("Heteronuclear")
  ("Homonuclear")

• Homo-nuclear: Only one type of atoms. Ex- $H_2,O_2,N_2,P_4,S_8,etc$
• Hetero-nuclear: Two or more types of atoms. Ex- $Hcl,N{H}_3,C{O}_2,H_{2}S{O}_4,etc$

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Ions and its Types

An ion is an atom or a molecule that has a net electric charge due to the loss or gain of one or more electrons.

• Mono-atomic Ions: Single atom. Ex- $N{a}^{+},M{g}^{2+},O^{2-},N^{3-},\text{etc.}$
• Poly-atomic Ions: Two or More atoms. Ex- $C{O}_3^{2-},P{O}_4^{3-},Mn{O}_4^{-},\text{etc.}$

graph TD
    Ions("Ions")
    
    %% First Branch
    Ions --> Charge("Classification by Charge")
    Charge --> Cations("Cations (+ve)")
    Charge --> Anions("Anions (-ve)")
    
    %% Second Branch
    Ions --> Atoms("Classification by Number of Atoms")
    Atoms --> Mono("Monoatomic Ions")
    Atoms --> Poly("Polyatomic Ions")
    
    %% Styling
    style Ions fill:#f9f,stroke:#333,stroke-width:2px
    style Charge fill:#bbf,stroke:#333,stroke-width:1px
    style Atoms fill:#bbf,stroke:#333,stroke-width:1px

Important Ions to Remember

Cations (+ve)	Symbol	Anions (-ve)	Symbol
Hydrogen ion	$H^{+}$	Hydride ion	$H^{-}$
Sodium ion	$N{a}^{+}$	Fluoride ion	$F^{-}$
Potassium ion	$K^{+}$	Chloride ion	$C{l}^{-}$
Calcium ion	$C{a}^{2+}$	Bromide ion	$B{r}^{-}$
Magnesium ion	$M{g}^{2+}$	Iodide ion	$I^{-}$
Aluminium ion	$A{l}^{3+}$	Oxide ion	$O^{2-}$
Barium ion	$B{a}^{2+}$	Sulphide ion	$S^{2-}$
Cuprous / Copper (I) ion	$C{u}^{+}$	Nitride ion	$N^{3-}$
Cupric / Copper (II) ion	$C{u}^{2+}$	Hydroxide ion	$O{H}^{-}$
Mercurous / Mercury (I) ion	$H{g}^{+}$	Cyanide ion	$C{N}^{-}$
Mercuric / Mercury (II) ion	$H{g}^{2+}$	Nitrite ion	$N{O}_2^{-}$
Ferrous / Iron (II) ion	$F{e}^{2+}$	Nitrate ion	$N{O}_3^{-}$
Ferric / Iron (III) ion	$F{e}^{3+}$	Sulphite ion	$S{O}_3^{2-}$
Ammonium ion	$N{H}_4^{+}$	Sulphate ion	$S{O}_4^{2-}$
Zinc ion	$Z{n}^{2+}$	Bisulphate / Hydrogen sulphate ion	$HS{O}_4^{-}$
Plumbus (Plumbous) ion	$P{b}^{2+}$	Carbonate ion	$C{O}_3^{2-}$
Plumbic ion	$P{b}^{4+}$	Bicarbonate / Hydrogen carbonate ion	$HC{O}_3^{-}$
		Phosphate ion	$P{O}_4^{3-}$
		Acetate ion	$C{H}_{3}CO{O}^{-}$
		Oxalate ion	$C_2{O}_4^{2-}$
		Dichromate ion	$C{r}_2{O}_7^{2-}$
		Permanganate ion	$Mn{O}_4^{-}$

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Formula of a Compound

The chemical formula of a compound is derived by crossing over the valencies (charges) of the constituent ions.

Examples of Compound Formation

Compound Name	Cation	Anion	Chemical Formula
Sodium chloride	$N{a}^{+}$	$C{l}^{-}$	$NaCl$
Magnesium sulphate	$M{g}^{2+}$	$S{O}_4^{2-}$	$MgS{O}_4$
Zinc chloride	$Z{n}^{2+}$	$C{l}^{-}$	$ZnC{l}_2$
Potassium sulphate	$K^{+}$	$S{O}_4^{2-}$	$K_{2}S{O}_4$
Ammonium sulphate	$N{H}_4^{+}$	$S{O}_4^{2-}$	$(N{H}_4{)}_{2}S{O}_4$
Calcium carbonate	$C{a}^{2+}$	$C{O}_3^{2-}$	$CaC{O}_3$
Silver nitrite	$A{g}^{+}$	$N{O}_2^{-}$	$AgN{O}_2$
Calcium phosphate	$C{a}^{2+}$	$P{O}_4^{3-}$	$C{a}_3(P{O}_4{)}_2$
Sodium nitrate	$N{a}^{+}$	$N{O}_3^{-}$	$NaN{O}_3$
Potassium dichromate	$K^{+}$	$C{r}_2{O}_7^{2-}$	$K_{2}C{r}_2{O}_7$
Ferric sulphate	$F{e}^{3+}$	$S{O}_4^{2-}$	$F{e}_2(S{O}_4{)}_3$
Ferrous chloride	$F{e}^{2+}$	$C{l}^{-}$	$FeC{l}_2$
Ferrous oxalate	$F{e}^{2+}$	$C_2{O}_4^{2-}$	$Fe{C}_2{O}_4$
Potassium permanganate	$K^{+}$	$Mn{O}_4^{-}$	$KMn{O}_4$
Ammonium chloride	$N{H}_4^{+}$	$C{l}^{-}$	$N{H}_{4}Cl$
Cupric chloride	$C{u}^{2+}$	$C{l}^{-}$	$CuC{l}_2$

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🚀Dalton’s Atomic Theory

John Dalton proposed his atomic theory to explain the nature of matter and the laws of chemical combination.

Postulates

• Indivisibility: Matter consists of indivisible particles called atoms.
• Identical Properties: All atoms of a given element have identical properties, including identical mass.
• Different Masses: Atoms of different elements differ in mass.
• Fixed Ratio Combinations: Compounds are formed when atoms of different elements combine in a fixed ratio.
  • Ex: In water ($H_{2}O$), the ratio of Hydrogen to Oxygen by mass is $1:8$.
• Reorganisation in Reactions: Chemical reactions involve the reorganisation of atoms. Atoms are neither created nor destroyed in a chemical reaction.

Drawbacks of Dalton's Theory

Modern atomic structure and discoveries revealed several limitations in Dalton’s original postulates:

1. Subatomic Particles: Atoms are not indivisible; they can be subdivided into protons, neutrons, and electrons.
2. Isotopes: This theory cannot explain the existence of isotopes (atoms of the same element with different masses).
3. Isobars: This theory cannot explain the existence of isobars (atoms of different elements with the same mass).
4. Chemical Bonding: It could not provide the underlying reason for why or how atoms combine to form molecules.

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Atomic Species: Isotopes, Isobars, and Isotones

• Isotopes: Atoms of the same element having the same atomic number ($Z$) but different mass numbers ($A$).
  • Ex: ${}_6^{12}C$ and ${}_6^{14}C$
• Isobars: Atoms of different elements having the same mass number ($A$) but different atomic numbers ($Z$).
  • Ex: ${}_6^{14}C$ and ${}_7^{14}N$
• Isotones: Atoms of different elements having the same number of neutrons ($A-Z$).
  • Ex: ${}_6^{14}C$ and ${}_8^{16}O$

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Laws of Chemical Combination

1. Law of Mass Conservation

In any physical or chemical change, mass can neither be created or nor be destroyed.

• Proposed by the French Chemist Antoine Laurent Lavoisier.
• Formulae: $\text{Total Mass of Reactants}=\text{Total Mass of Products}$
• ! This is not applicable for Nuclear Reactions

2. Law of Definite/Constant Proportion/Composition ()

Any given chemical compound always contains exactly the *same* proportion of elements by mass.

• Proposed by French Chemist Joseph Proust
• @ Mass Percentage of elements in a compound is fixed
  • Mass% of Hydrogen = 11.11%
  • Mass% of Oxygen = 88.88%
• ! Not applicable for Isotopes and Non-stoichiometric compound

3. Law of Multiple Proportion

When two elements combine with each other to form two or more compounds, the masses of one of the element that combines with the fixed mass of the other, always bear simple whole number ratio.

• Proposed by English Chemist John Dalton

Examples & Exceptions

Basic Examples

• H + O:
  • $H_{2}O$ (2:16)
  • $H_2{O}_2$ (2:32)
  • Ratio of Oxygen (for 2g H) = 16 : 32 $\to$ 1 : 2
• C + O:
  • $CO$ (12:16)
  • $C{O}_2$ (12:32)
  • Ratio of Oxygen (for 12g C) = 16 : 32 $\to$ 1 : 2

Tricky Examples (Nitrogen Oxides)

• Group A (Fixed N = 28g):
  • $N_{2}O$ (28:16), $N_2{O}_3$ (28:48), $N_2{O}_4$ (28:64), $N_2{O}_5$ (28:80)
  • Ratio of Oxygen = 16:48:64:80 $\to$ 1 : 3 : 4 : 5
• Group B (Mixed Formulas):
  To find the ratio, you must first mathematically “fix” the mass of Nitrogen.
  • $NO$ (14:16)
  • $N{O}_2$ (14:32)
  • $N_{2}O$ (28:16 $\to$ convert to 14:8)
  • Ratio of Oxygen (for 14g N) = 16 : 32 : 8 $\to$ 2 : 4 : 1

Exceptions (The “Fail” Cases)

• Isotopes: Law fails if different isotopes are used (e.g., $C^{12}O$ vs $C^{14}O$).
• Non-Stoichiometric Compounds: Fails for minerals like $F{e}_{0.95}O$.
• Large Organic Molecules: Ratios like 20:21:22 are whole numbers but aren’t considered “simple.”

4. Gay Lussac’s Law of Gaseous Volume

When gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume provided that all gases are at same temperature and pressure.

• Proposed by French Chemist Joseph Louis Gay-Lussac.
• @ The volume ratio of gases matches their stoichiometric coefficients in a balanced equation.
• ! This is only applicable for gases.

Examples & Practice Cases

Basic Example ($H_2+C{l}_2\to HCl$)

• Balanced Equation: $\text{1 }{H}_{2(g)}+\text{1 }C{l}_{2(g)}\to \text{2 }HC{l}_{(g)}$
• Ratio of Volumes = 1 : 1 : 2
• Application: 20 L of $H_2$ reacts with 20 L of $C{l}_2$ to form 40 L of $HCl$.

Haber’s Process ($N_2+H_2\to N{H}_3$)

• Balanced Equation: $\text{1 }{N}_{2(g)}+\text{3 }{H}_{2(g)}\to \text{2 }N{H}_{3(g)}$
• Ratio of Volumes = 1 : 3 : 2
• Application: 50 L of $N_2$ reacts with 150 L of $H_2$ to form 100 L of $N{H}_3$.

5. Avogadro’s Hypothesis

Under similar conditions of temperature and pressure, equal volumes of all gases contain an equal number of molecules.

• Proposed by Italian Scientist Amedeo Avogadro.
• Formulae: $V\propto n$ (where $V$ is volume and $n$ is the number of moles/molecules and T, P is constant)

Examples & Molecular Relationships

Case 1: Hydrogen Chloride Synthesis

• Reaction: $\text{1 }{H}_{2(g)}+\text{1 }C{l}_{2(g)}\to \text{2 }HC{l}_{(g)}$
• By Volume: $50\text{ mL}$ of $H_2$ + $50\text{ mL}$ of $C{l}_2$ $\to$ $100\text{ mL}$ of $HCl$
• By Molecules: $N\text{ molecules}$ of $H_2$ + $N\text{ molecules}$ of $C{l}_2$ $\to$ $2N\text{ molecules}$ of $HCl$

Case 2: Haber’s Process

• Reaction: $\text{1 }{N}_{2(g)}+\text{3 }{H}_{2(g)}\to \text{2 }N{H}_{3(g)}$
• By Volume: $50\text{ mL}$ of $N_2$ + $150\text{ mL}$ of $H_2$ $\to$ $100\text{ mL}$ of $N{H}_3$
• By Molecules: $x\text{ molecules}$ of $N_2$ + $3x\text{ molecules}$ of $H_2$ $\to$ $2x\text{ molecules}$ of $N{H}_3$

Concept Application: Atom Ratios via Avogadro's Law

Q: Four $1\text{ L}$ flasks are separately filled with the gases $H_2$, $He$, $O_2$, and $O_3$ at the same temperature and pressure. The ratio of the total number of atoms of these gases present in the different flasks would be:

1. $1:1:1:1$
2. $1:2:2:3$
3. $2:1:2:3$ $✓$
4. $3:2:2:1$

Solution / Breakdown

1. Apply Avogadro’s Law: Since all flasks have equal volume ($1\text{ L}$) at identical $T$ and $P$, they contain the exact same number of molecules ($N$).
2. Determine Atomicity: Calculate the number of atoms in $N$ molecules for each gas:
   • $H_2$ (Diatomic): $N\text{ molecules}\to 2N\text{ atoms}$
   • $He$ (Monoatomic): $N\text{ molecules}\to 1N\text{ atom}$
   • $O_2$ (Diatomic): $N\text{ molecules}\to 2N\text{ atoms}$
   • $O_3$ (Triatomic): $N\text{ molecules}\to 3N\text{ atoms}$
3. Find the Ratio:
   $\text{Ratio of Atoms}=2N:1N:2N:3N\to 2:1:2:3$

6. Law of Reciprocal Proportion

If two different elements ($B$ and $C$) combine separately with the same weight of a third element ($A$), the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine directly ($B$ and $C$).

• Proposed by German Chemist Jeremias Richter in 1792.
• @ Also known as the Law of Equivalent Proportions and three elements are required

Examples & Visualization

The $\text{Na}-\text{Cl}-\text{H}$ Triangle (Case: Same Ratio)

• Element A (Fixed): Hydrogen ($H$)
• Element B: Sodium ($Na$)
• Element C: Chlorine ($Cl$)

Step 1: Combine separately with Fixed Element A ($H$)

• $Na+H\to NaH$
  • Mass ratio = $23:1$ (23g of $Na$ combines with 1g of $H$)
• $Cl+H\to HCl$
  • Mass ratio = $1:35.5$ (1g of $H$ combines with $35.5\text{g}$ of $Cl$)
• Ratio of $Na:Cl$ combining with fixed $H$ = $23:35.5$

Step 2: Combine directly with each other ($B+C$)

• $Na+Cl\to NaCl$
  • Mass ratio of $Na:Cl$ = $23:35.5$

Conclusion

Both ratios are exactly the same ($23:35.5$), perfectly matching the law.

The $\text{S}-\text{O}-\text{H}$ Triangle (Case: Simple Multiple Ratio)

• Element A (Fixed): Hydrogen ($H$)
• Element B: Sulfur ($S$)
• Element C: Oxygen ($O$)

Step 1: Combine separately with Fixed Element A ($H$)

• $S+H\to H_{2}S$
  • Mass ratio = $2:32$ (2g of $H$ combines with $32\text{g}$ of $S$)
• $O+H\to H_{2}O$
  • Mass ratio = $2:16$ (2g of $H$ combines with $16\text{g}$ of $O$)
• Ratio of $S:O$ combining with fixed $H$ = $32:16\to 2:1\to \left(\frac{2}{1}\right)\to 2\times \left(\frac{1}{1}\right)$

Step 2: Combine directly with each other ($B+C$)

• $S+O\to S{O}_2$
  • Mass ratio of $S:O$ = $32:32\to 1:1\to \left(\frac{1}{1}\right)$

Step 3: Compare Ratio 1 and Ratio 2

• $\frac{\text{Ratio 1}}{\text{Ratio 2}}=\frac{2/1}{1/1}=\frac{2}{1}\to 2$

Conclusion

The two ratios are not identical, but they form a simple whole number multiple ($2$), validating the law.

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Measurement of Mass of Atoms & Molecules

1. Relative Atomic Mass

The RAM number of an element indicates how many times the mass of one atom of an element is heavier in comparison to $\frac{1}{12}th$ mass of $C^{12}$ atom

• Formula: $\boxed{RAM=\frac{\text{Mass of an Element}}{\frac{1}{12}\times [\text{Mass of }{C}^{12}\text{ atom}]}}$
• @ $\frac{1}{12}th\text{ mass of }{C}^{12}\text{ atom}=1amu=1u=1.66\times 10^{-24}g$

It indicates how many times one molecule of a substance is heavier in comparison to $\frac{1}{12}$ of $C^{12}$ atom

2. Average Atomic Mass

Some elements are found in different isotopes in nature with various abundance. For this reason we need an average atomic mass of that element.

• Formula: $\boxed{AAM=\sum (\text{Relative Isotopic Mass}\times \text{Fractional Abundance})}$

3. Relative Molecular Mass

It indicates how many times one molecule of a substance is heavier in comparison to $\frac{1}{12}$ of $C^{12}$ .

• Formula: $\boxed{RMM=\frac{\text{Mass of one molecule of Compound}}{\frac{1}{2}\times [\text{Mass of one atom of}{C}^{12}]}=\text{The sum of Atomic Masses of Elements present in a Compound}}$
• ! The Relative Mass doesn’t have unit. Ex - H has relative mass of 1 and mass of 1 amu

4. Gram Atomic Mass

It indicates the mass of $N_A$ number of atoms of an element.

• It’s expressed in Grams
• Formula: $\boxed{GAM=\text{Atomic mass}\times N_{A}g}$

5. Gram Molecular Mass

It indicates the mass of $N_A$ number of molecules of a compound.

• It’s expressed in Grams
• Formula: $\boxed{GAM=\text{Molecular mass}\times N_{A}g}$

6. Formula Mass (Formula is just the Ionic counterpart of Molecule)

For ionic compounds, discrete molecules do not exist. Instead, their constituent units are ions arranged in a continuous three-dimensional lattice. Therefore, we calculate Formula Mass instead of Molecular Mass.

• It indicates the sum of atomic masses of all constituent atoms present in a formula unit of an ionic compound.
• Formula Unit: The simplest whole-number ratio of ions represented in the compound (e.g., ${\text{Na}}^{+}:{\text{Cl}}^{-}=1:1$).
• ! Measured in unified mass units ($\text{u}$) or atomic mass units ($\text{amu}$).
• @ Note on Nomenclature: Terms like Relative Formula Mass and Gram Formula Mass are simply the ionic counterparts to Relative Molecular Mass and Gram Molecular Mass, used because ionic compounds lack distinct individual molecules.

Calculations & Formula Units

Case 1: Sodium Chloride ($\text{NaCl}$)

• Ionic Ratio: ${\text{Na}}^{+}:{\text{Cl}}^{-}=1:1$ (Formula Unit = $\text{NaCl}$)
• Formula Mass: $23+35.5=58.5\text{ u}$
• Gram Formula Mass: $58.5\text{ g}$ (Mass of $N_A$ or $6.022\times 10^{23}$ formula units)

Case 2: Calcium Chloride (${\text{CaCl}}_2$)

• Ionic Ratio: ${\text{Ca}}^{2+}:{\text{Cl}}^{-}=1:2$ (Formula Unit = ${\text{CaCl}}_2$)
• Formula Mass: $40+2(35.5)=40+71=111\text{ u}$
• Gram Formula Mass: $111\text{ g}$ (Mass of $6.022\times 10^{23}$ formula units)

Case 3: Sodium Oxide (${\text{Na}}_2\text{O}$)

• Ionic Ratio: ${\text{Na}}^{+}:{\text{O}}^{2-}=2:1$ (Formula Unit = ${\text{Na}}_2\text{O}$)
• Formula Mass: $2(23)+16=46+16=62\text{ u}$
• Gram Formula Mass: $62\text{ g}$ (Mass of $6.022\times 10^{23}$ formula units)

Case 4: Zinc Sulphide ($\text{ZnS}$)

• Ionic Ratio: ${\text{Zn}}^{2+}:{\text{S}}^{2-}=1:1$ (Formula Unit = $\text{ZnS}$)

Difference between Measuring Atomic Mass

RAM of Oxygen = 16
Mass of One atom of Oxygen = 16u or 16amu
GAM of Oxygen = 16g

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Mole Concept

A mole is the amount of substance that contains as many species (Atoms, molecules, ions or other particles) as there are atoms in exactly 12g of C-12.

• The number atoms in exactly 12g of C-12 is: $N_A=\frac{1}{amu}=6.022\times 10^{23}$
• 1 mole is the collection of $6.022\times 10^{23}$ particles (atoms, molecules, ions, formula unit, etc)

Molar Mass

Mass of one mole of a substance.

• Formula: $\boxed{\text{Molar Mass}=\frac{\text{Mass of the Substance}}{\text{Mole Number}}}$
• Ex: Molar mass of C is 12g/mol
• Equivalent Terms:
  • For atoms: Gram Atomic Mass
  • For molecules: Gram Molecular Mass

Tricky Terms

• 1 Gram-Atom = 1 Mole of an Atom
• 1 Gram-Molecule = 1 Mole of a Molecule
• 1 Gram-Ions = 1 Mole of an Ion

Average Molar Mass or Molecular Mass ($M_{\text{av}}$)

The average molar mass of a mixture represents the total mass of the mixture divided by the total number of moles present.

• General Formula: $\text{Molar mass}=\frac{\text{Total Given mass}}{\text{Total No. of moles}}$
• For a mixture containing component A and component B*
  • $n_A,n_B=$ number of moles of A and B respectively
  • $M_A,M_B=$ molar mass of A and B respectively
• Mixture Formula: $\boxed{M_{\text{av}}=\frac{n_A{M}_A+n_B{M}_B}{n_A+n_B}}$
  • ! For Gases, if only % is given, consider it mol % if no other information is given.

Methods to Calculate Moles

• In terms of Mass: $\text{Mole Number}=\frac{\text{Given Mass of Substance}}{\text{Molar Mass of Substance}}$

• In terms of Volume: $\text{Mole Number}=\frac{\text{Given Volume of Substance}}{\text{Molar Volume of Substance}}$

• In terms of Particles: $\boxed{\text{Mole Number}=\frac{\text{Given number of Particles}}{N_A}}$

Calculating the volume of 1mole Gas using $PV=nRT$: STP Confusion

STP(New Updated aka 1bar): When temperature is $0^{\circ }C$ or $273.15K$ and pressure is $1bar=0.987atm$

• The volume of N mole Gas is $N\times 22.7L$

1atm (Still Used Deprecated Sys): When When temperature is $0^{\circ }C$ or $273.15K$ and pressure is $1atm$.

• The volume of N mole Gas in old system is $N\times 22.4L$

Calculating charge of an Ion using $q=n{e}^{-}$

The total charge ($q$) of an ion is calculated by multiplying the number of excess or missing electrons ($n$) by the elementary charge ($e=1.602\times 10^{-19}\text{ C}$).

• ${\text{Na}}^{+}$: Has 1 unit of positive charge $\to q=+1(1.6\times 10^{-19}\text{ C})=+1.6\times 10^{-19}\text{ C}$

• ${\text{Mg}}^{2+}$: Has 2 units of positive charge $\to q=+2(1.6\times 10^{-19}\text{ C})=+3.2\times 10^{-19}\text{ C}$

• ${\text{Al}}^{3+}$: Has 3 units of positive charge $\to q=+3(1.6\times 10^{-19}\text{ C})=+4.8\times 10^{-19}\text{ C}$

• ${\text{F}}^{-}$: Has 1 unit of negative charge $\to q=-1(1.6\times 10^{-19}\text{ C})=-1.6\times 10^{-19}\text{ C}$

• ${\text{O}}^{2-}$: Has 2 units of negative charge $\to q=-2(1.6\times 10^{-19}\text{ C})=-3.2\times 10^{-19}\text{ C}$

• ! Avoid the common mistake of writing the value as $1.66\times 10^{-19}\text{ C}$, which mistakenly confuses the elementary charge with the atomic mass unit constant.

Charge on 1 Mole of Electrons ( $1\text{ Faraday}$)

The total charge carried by exactly one mole of electrons is known as 1 Faraday ($1\text{ F}$).

• Calculation: $\text{Charge}=(1.6\times 10^{-19}\text{ C})\times N_A\approx 96500\text{ C}$
• Formula: $\boxed{\text{Charge on }1\text{ mol }{e}^{-}=N_A\times 1.6\times 10^{-19}\text{ C}=96500\text{ C}=1\text{ F}}$

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Percentage Composition

The percentage composition of a compound is the relative mass of each constituent present in 100 parts of it.

• Formula: $\boxed{\text{Mass \% of an element}=\frac{\text{Mass of element in 1 mol of substance}}{\text{Molar mass of substance}}\times 100\%}$

Mass % of Hydrogen ( $\text{H}$) in Water (${\text{H}}_2\text{O}$)

1. Total Molar Mass of Substance

• Molar mass of $\text{H}$: $1.008\text{ g/mol}$
• Molar mass of $\text{O}$: $16.00\text{ g/mol}$
  $\text{Total Molar Mass of }{\text{H}}_2\text{O}=(2\times 1.008)+16.00=18.016\text{ g/mol}$

2. Mass of Hydrogen in 1 Mol of Substance

• Water contains 2 moles of $\text{H}$ atoms (${\text{H}}_2$).
  $\text{Mass of H}=2\times 1.008\text{ g}=2.016\text{ g}$

3. Calculate Mass Percentage

$\text{Mass \% of H}=\frac{2.016\text{ g}}{18.016\text{ g}}\times 100\%=11.19\%$

Minimum Molecular Mass

Question: Insulin contains $3.2\%$ of $\text{S}$ by mass. Find out the minimum molecular mass of insulin. $[\text{S}=32]$

Concept:

Key Rule

For minimum molecular mass, assume at least 1 atom of the element is present in one molecule.

Solution:
$\text{Mass \% of S}=\frac{\text{Mass of S in 1 molecule}}{\text{Molecular mass of Insulin}}\times 100\%$

Let the minimum molecular mass of Insulin be $M$:
$3.2=\frac{32}{M}\times 100$

$M=\frac{32\times 100}{3.2}$

$\mathbf{M}=1000\text{ u}$

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Concepts Related to Density

1. Absolute Density

For Solids and Liquids:

• Formula: $\boxed{\text{Density}=\frac{\text{Mass}}{\text{Volume}}}$

For Gases

Derived from the Ideal Gas Law: $P=\frac{dRT}{M_w}$

• Gas Density Formula: $\boxed{d_{\text{gas}}=\frac{M_{w}P}{RT}}$
• Variables:
  • $P$ = Pressure
  • $d$ = Density
  • $M_w$ = Molar mass
  • $R$ = Gas constant
  • $T$ = Temperature
• Key Proportionality:
  • At a particular (constant) temperature and pressure: $d_{\text{gas}}\propto M_w$
  • This means the density of a gas is directly proportional to its molar mass.

2. Relative Density

It is the density of a substance with respect to any other substance.

• Formula: $\boxed{\text{Relative Density}=\frac{\text{Density of A}}{\text{Density of B}}}$
• For Gases (Special Case):
  • Since the density of a gas is directly proportional to its molar mass ($d_{\text{gas}}\propto M_w$), we can substitute molar mass directly into the relative density formula.
  • Formula: $\boxed{{\text{Relative Density}}_{\text{gas}}=\frac{M_w(\text{Gas A})}{M_w(\text{Gas B})}}$

Specific Gravity:

A special case of relative density where the reference substance is water at $4^{\circ }\text{C}$.

• Formula: $\boxed{\text{Specific gravity}=\frac{\text{Density of substance}}{\text{Density of }{\text{H}}_2\text{O}\text{ at }{4}^{\circ }\text{C}}}$
• Reference value denominator: $1\text{ g/mL}$
• It is completely unitless because it is a ratio of two identical physical quantities.

Vapour Density (V.D.):

It is defined as the density of a gas wrt hydrogen gas ($\displaystyle \text{H}_2$) at the same temperature and pressure.

• Formula: $\boxed{\text{V.D.}=\frac{\text{Density of gas}}{\text{Density of }{\text{H}}_2}}$
• Molar Mass Relationship: Using the gas shortcut, the densities are replaced by their respective molar masses ($M_w$ of ${\text{H}}_2=2\text{ g/mol}$):
  • $\text{V.D.}=\frac{(M_w{)}_{\text{gas}}}{(M_w{)}_{{\text{H}}_2}}$
  • Final Working Formula: $\boxed{\text{V.D.}=\frac{(M_w{)}_{\text{gas}}}{2}}$

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Empirical and Molecular Formula

Basic Definitions

• Molecular Formula: It is the formula which gives the exact number of atoms of different elements present in a compound.
• Empirical Formula: It is the formula which gives the simplest whole-number ratio of atoms of different elements present in a compound.

Mathematical Relationship

• Formula: $\boxed{\text{Molecular Formula}=n\times [\text{Empirical Formula}]}$
• Calculating the value of $n$:
  • $\boxed{n=\frac{\text{Molecular Formula Mass}}{\text{Empirical Formula Mass}}}$

Benzene as an Example

• Formulas:
  • Molecular Formula of Benzene = ${\text{C}}_6{\text{H}}_6$
  • Simplest ratio of $\text{C}:\text{H}⟹6:6⟹1:1$
  • Empirical Formula of Benzene = $\text{CH}$
• Mass Calculations:
  • Molecular formula mass of ${\text{C}}_6{\text{H}}_6=12\times 6+6\times 1=78$
  • Empirical formula mass of $\text{CH}=12+1=13$
• Finding $n$ and Verifying:
  • $n=\frac{\text{Molecular For. Mass}}{\text{Empirical For. Mass}}=\frac{78}{13}=6$
  • $\text{Molecular Formula}=6\times [\text{CH}]={\text{C}}_6{\text{H}}_6$

Calculating Empirical Formula from % Composition

To calculate the empirical formula from percentage mass, we use a systematic tabular method to find the simplest whole-number ratio of moles.

Steps for Tabular Method

1. Mass: Assume the total mass is $100\text{ g}$ (so the % given becomes the mass in grams).
2. Moles: Divide the mass by the Atomic Mass to get the number of moles ($n=\frac{w}{M_w}$).
3. Simplest Ratio: Divide all the calculated mole values by the smallest mole value obtained.
4. Whole Number: If the resulting ratio isn’t a whole number, multiply all values by a suitable integer to make them whole (e.g., if you get $3.5$, multiply all by $2$).

Example: Unknown Potassium Compound

Given: A compound contains $26.6\%\text{ K}$, $35.4\%\text{ Cr}$, and $38\%\text{ O}$.

Element	Mass (g)	At. Mass	Moles ($n=\frac{w}{M_w}$)	Simplest Ratio	Whole No. Ratio
K	$26.6$	$39$	$\frac{26.6}{39}=0.68$	$\frac{0.68}{0.68}=1$	$1\times 2=2$
Cr	$35.4$	$52$	$\frac{35.4}{52}=0.68$	$\frac{0.68}{0.68}=1$	$1\times 2=2$
O	$38$	$16$	$\frac{38}{16}=2.38$	$\frac{2.38}{0.68}=3.5$	$3.5\times 2=7$

Empirical Formula: $K_{2}C{r}_2{O}_7$ (Potassium dichromate)

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Balancing Chemical Reactions

A balanced chemical equation obeys the Law of Mass Conservation, ensuring the number of atoms of each element is exactly equal on both the reactant and product sides.

• Ex: $N_{2(g)}+3H_{2(g)}\to 2N{H}_{3(g)}$
• Ex: $2NaOH+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2H_{2}O$
• Ex: $F{e}_2{O}_3+2Al\to A{l}_2{O}_3+2Fe$
• @ FYI you can write fractional Stoichiometric Coefficients.

Combustion of Hydrocarbons (General Formula)

Hydrocarbons (compounds of carbon and hydrogen) burn in the presence of oxygen to produce $C{O}_2$ and $H_{2}O$. You can use a direct algebraic formula to balance these reactions instantly:

$\boxed{C_x{H}_y+\left(x+\frac{y}{4}\right)O_2\to xC{O}_2+\frac{y}{2}{H}_{2}O}$

• Example ($C_3{H}_6$): Here $x=3$, $y=6$.
  • $O_2$ coefficient $=3+\frac{6}{4}=3+1.5=4.5=\frac{9}{2}$
  • $C{O}_2$ coefficient $=3$
  • $H_{2}O$ coefficient $=\frac{6}{2}=3$
  • Balanced Equation: $C_3{H}_6+\frac{9}{2}{O}_2\to 3C{O}_2+3H_{2}O$ (Note: You can multiply the entire equation by 2 to remove the fraction: $2C_3{H}_6+9O_2\to 6C{O}_2+6H_{2}O$)

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Stoichiometry & Stoichiometric Calculations

The calculations based on the quantitative relationship between reactants and products in a balanced chemical equation are known as stoichiometric calculations.

• @ The coefficients in a balanced equation are called Stoichiometric Coefficients. They represent the exact mole ratio of the reacting species.

Multi-Dimensional Interpretation of Stoichiometry

Let’s analyze Haber’s Process: $1{\mathbf{N}}_2(\mathbf{g})+3{\mathbf{H}}_2(\mathbf{g})\to 2N{\mathbf{H}}_3(\mathbf{g})$

Interpretation	Nitrogen ($1N_2$)	Hydrogen ($3H_2$)	Ammonia ($2N{H}_3$)
In terms of Moles	$1\text{ mol}$	$3\text{ mol}$	$2\text{ mol}$
In terms of Mass	$28\text{ g}$	$6\text{ g}$	$34\text{ g}$
In terms of Volume (STP)	$22.7\text{ L}$	$3\times 22.7\text{ L}$	$2\times 22.7\text{ L}$
In terms of Particles	$N_A\text{ molecules}$	$3N_A\text{ molecules}$	$2N_A\text{ molecules}$

Stoichiometric Calculations (Unitary Method)

Problem 1: Mass-Mass Relationship

Q: $16\text{ g}$ of $O_2$ reacts with $H_2$ to produce water. Find the amount of water formed.

1. Balanced Equation: $2H_2+1O_2\to 2H_{2}O$
2. Standard Stoichiometry (Mass): $1\text{ mol }{O}_2\to 2\text{ mol }{H}_{2}O$
   • $32\text{ g }{O}_2\to 36\text{ g }{H}_{2}O$
3. Unitary Method:
   • $16\text{ g }{O}_2\to \frac{36}{32}\times 16\text{ g}=18\text{ g }{\mathbf{H}}_2\mathbf{O}$

Problem 2: Calcium Combustion

Q: What amount of $CaO$ is produced by $10\text{ g}$ calcium? $[Ca=40,O=16]$

1. Balanced Equation: $2Ca+1O_2\to 2CaO$
2. Standard Stoichiometry:
   • Mass of $2Ca=2\times 40=80\text{ g}$
   • Mass of $2CaO=2\times (40+16)=2\times 56=112\text{ g}$
   • $80\text{ g }Ca\to 112\text{ g }CaO$
3. Unitary Method:
   • $10\text{ g }Ca\to \frac{112}{80}\times 10=14\text{ g }\mathbf{CaO}$

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Contraction in Volume

General Combustion Equation

• Balanced Chemical Equation:
  • ${\text{C}}_x{\text{H}}_y(g)+\left(x+\frac{y}{4}\right){\text{O}}_2(g)⟶x{\text{CO}}_2(g)+\frac{y}{2}{\text{H}}_2\text{O}(l)$
• Key Assumption:
  • At room temperature, water is in liquid state, and its volume is negligible compared to gases: $\boxed{V_{(l)}\ll V_{(g)}}$

Volume Relations (Eudiometry)

• By Gay-Lussac’s law of combining volumes, the mole ratios can be treated as volume ratios:
  • Volume of ${\text{C}}_x{\text{H}}_y=1\text{ mL}$
  • Volume of ${\text{O}}_2=\left(x+\frac{y}{4}\right)\text{ mL}$
  • Volume of ${\text{CO}}_2=x\text{ mL}$
  • Volume of ${\text{H}}_2\text{O}\approx 0\text{ mL}$ (considered negligible)
• Total Volume of Reactants:
  • $\boxed{{\text{Total Volume}}_{\text{Reactants}}=\left\{1+x+\frac{y}{4}\right\}\text{ mL}}$
• Total Volume of Gaseous Products:
  • $\boxed{{\text{Total Volume}}_{\text{Products}}=x\text{ mL}}$

Volume Contraction Formula

• Volume contraction is the change/reduction in gaseous volume after complete combustion:
  • $\boxed{\text{Contraction of Volume}=\text{Vol. of Reactants}-\text{Vol. of Products}}$
• Simplified Working Formula:
  • $\text{Contraction}=\left(1+x+\frac{y}{4}\right)-x$
  • $\boxed{\Delta V=1+\frac{y}{4}}$

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Limiting Reagent

Basic Definitions

• Limiting Reagent (L.R.):
  • The reactant which is completely consumed during the chemical reaction.
  • Key Rule: The amount of product formed depends on the quantity of limiting reagent.
• Excess Reagent:
  • The reactant that is left over after the reaction stops because the limiting reagent has been entirely consumed.

Method to Identify Limiting Reagent

• Rule: Divide the given moles of each reactant by their respective stoichiometric coefficient from the balanced chemical equation.
• Identification Formula:
  • $\boxed{\text{Ratio}=\frac{\text{Given Moles}}{\text{Stoichiometric Coefficient}}}$
• The reactant with the least ratio is the Limiting Reagent (L.R.).

When to look for L.R

If the question provides specific quantities (moles, mass, or volume) for two or more reactants, you must almost always identify the Limiting Reagent (L.R.) before calculating the amount of product formed.

Applications

Example 1: Identifying the L.R. in Haber’s Process

• Question: In the reaction of $3\text{ mol}$ of ${\text{N}}_2(g)$ with $8\text{ mol}$ of ${\text{H}}_2(g)$ to form ammonia, determine which reactant acts as the limiting reagent.
• Balanced Equation: $1{\text{N}}_2(g)+3{\text{H}}_2(g)⟶2{\text{NH}}_3(g)$
• L.R. Identification:
  • Ratio for ${\text{N}}_2$: $\frac{3}{1}=3$
  • Ratio for ${\text{H}}_2$: $\frac{8}{3}\approx 2.67$
• Conclusion:
  • Since $\frac{8}{3}<3$, $\boxed{{\text{H}}_2\text{ is the Limiting Reagent (L.R.)}}$
  • ${\text{N}}_2$ is the Excess Reagent.

Example 2: Finding Product Amount using L.R.

• Question: If $10\text{ mol}$ of ${\text{H}}_2(g)$ react with $8\text{ mol}$ of ${\text{Cl}}_2(g)$, find out the total moles of $\text{HCl}$ formed.
• Balanced Equation: ${\text{H}}_2(g)+1{\text{Cl}}_2(g)⟶2\text{HCl}(g)$
• L.R. Identification:
  • Ratio for ${\text{H}}_2$: $\frac{10}{1}=10$
  • Ratio for ${\text{Cl}}_2$: $\frac{8}{1}=8$
  • Since $8<10$, ${\text{Cl}}_2$ is the Limiting Reagent (L.R.).
• Product Calculation (Based on L.R.):
  • Since $1{\text{ mol Cl}}_2⟶2\text{ mol HCl}$
  • Therefore, $8{\text{ mol Cl}}_2⟶8\times 2=16\text{ mol HCl}$
  • Final Answer: $\boxed{\text{Moles of HCl formed}=16\text{ mol}}$

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Percentage Yield

In practical chemistry, reactions rarely go to 100% completion. This can happen due to side reactions, incomplete conversions, or loss of material during the experiment.

• Theoretical Yield: The maximum amount of product that should be formed according to stoichiometric calculations (always based on the Limiting Reagent).
• Actual Yield: The amount of product actually obtained in the experiment (this is usually given to you in the question).
• Formula:
  • $\boxed{\%\text{ Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%}$

Example: Calculating % Yield

Q: 10 g of ${\text{CaCO}}_3$ on heating gives 5 g of the residue (as $\text{CaO}$). The percent yield of the reaction is approximately?

1. Balanced Equation & Theoretical Yield:

• ${\text{CaCO}}_3(s)\to \text{CaO}(s)+{\text{CO}}_2(g)$
• Standard: 100 g of ${\text{CaCO}}_3\to$ 56 g of $\text{CaO}$
• Unitary Method: 10 g of ${\text{CaCO}}_3\to \frac{56}{100}\times 10=5.6\text{ g of CaO}$ (This is our Theoretical Yield)

2. Actual Yield:

• Given directly in the question = 5 g

3. % Yield Calculation:

• $\%\text{ Yield}=\frac{5}{5.6}\times 100\%\approx 89\%$

Pro-Tip for Multi-Reactant Problems

If a question gives you the quantities of multiple reactants AND asks for the % yield, you must first find the Limiting Reagent to calculate the Theoretical Yield.

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Percentage Purity

Samples of chemicals found in nature or synthesized in labs are rarely 100% pure; they often contain inert impurities (like dirt, unreacted remnants, etc.).

• Golden Rule of Purity: In any chemical reaction, *only the pure part* of the sample reacts. The impurities act as inert spectators and do not participate in the stoichiometry.
• Formula:
  • $\boxed{\%\text{ Purity}=\frac{\text{Mass of Pure Substance}}{\text{Mass of Impure Sample}}\times 100\%}$

Example 1: Finding Product from an Impure Sample

Q: Calculate the weight of lime ($\text{CaO}$) that can be prepared by heating 200 g of limestone (${\text{CaCO}}_3$) which is 95% pure.

1. Find the Pure Mass of Reactant:

• $\text{Pure Mass}=200\times \frac{95}{100}=190{\text{ g of pure CaCO}}_3$

2. Stoichiometry (using ONLY the pure mass):

• ${\text{CaCO}}_3\to \text{CaO}+{\text{CO}}_2$
• 100 g of ${\text{CaCO}}_3\to$ 56 g of $\text{CaO}$
• 190 g of ${\text{CaCO}}_3\to \frac{56}{100}\times 190=106.4\text{ g of CaO}$

Example 2: Finding % Purity from Product Formed (Reverse Calculation)

Q: 5 g of impure ${\text{KClO}}_3$ gave 0.03 mol of ${\text{O}}_2$. Find the % purity of the ${\text{KClO}}_3$ sample. ($M_w$ of ${\text{KClO}}_3$ = 122.5 g/mol)

1. Balanced Equation: $2{\text{KClO}}_3\to 2\text{KCl}+3{\text{O}}_2$

2. Reverse Stoichiometry (Find how much pure reactant was required):

• 3 mol of ${\text{O}}_2$ is produced by $\to 2\times 122.5\text{ g}$ (245 g) of pure ${\text{KClO}}_3$
• 0.03 mol of ${\text{O}}_2$ is produced by $\to \frac{245}{3}\times 0.03=2.45{\text{ g of pure KClO}}_3$

3. Calculate % Purity:

• $\%\text{ Purity}=\frac{2.45\text{ g (Pure)}}{5\text{ g (Total Impure Sample)}}\times 100\%=49\%$

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Concentration

Concentration of solution is the amount of solute dissolved in a known amount of solvent or solution.

Methods of Expressing Concentration of a Solution

Quantitatively

1. Percentage

It refers to the amount of the solute per 100 parts of the solution. It is also commonly called parts per hundred (PPh). There are four main ways to express percentage concentration:

1. Mass % or % (w/w)
The mass of solute (in grams) present in 100 g of the solution.

• Formula:

  • $\boxed{\%\text{ (w/w)}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100\%}$

• Interpretation: A 20% (w/w) $N{a}_{2}C{O}_3$ solution means 20 g of $N{a}_{2}C{O}_3$ is present in 100 g of the solution (which implies 80 g of solvent).

2. Volume % or % (v/v)
The volume of solute (in mL) present in 100 mL of the solution.

• Formula:

  • $\boxed{\%\text{ (v/v)}=\frac{\text{Volume of solute}}{\text{Volume of solution}}\times 100\%}$

• Interpretation: A 30% (v/v) $H_{2}S{O}_4$ solution means 30 mL of $H_{2}S{O}_4$ is present in 100 mL of the solution (which implies 70 mL of solvent).

• ! For Gaseous Mixtures: The volume % is directly proportional to the mole % (based on Avogadro’s Law): $\%\text{ (v/v)}=\frac{n_{\text{solute}}}{n_{\text{solution}}}\times 100\%$

3. Mass/Volume % or % (w/v)
The mass of solute (in grams) present in 100 mL of the solution.

• Formula:
  • $\boxed{\%\text{ (w/v)}=\frac{\text{Mass of solute (in g)}}{\text{Volume of solution (in mL)}}\times 100\%}$

4. Volume/Mass % or % (v/w)
The volume of solute (in mL) present in 100 g of the solution.

• Formula:
  • $\boxed{\%\text{ (v/w)}=\frac{\text{Volume of solute (in mL)}}{\text{Mass of solution (in g)}}\times 100\%}$

Example: Mixing Two Solutions

Q: 300 g of 25% (w/w) solution of solute A is mixed with 400 g of 40% (w/w) solution of another solute B. What is the w/w percentage of the new mixture?

1. Calculate total mass of the final solution:

• $\text{Total Mass}=300\text{ g}+400\text{ g}=700\text{ g}$

2. Calculate the mass of each solute:

• Mass of Solute A = $300\times \frac{25}{100}=75\text{ g}$
• Mass of Solute B = $400\times \frac{40}{100}=160\text{ g}$
• $\text{Total Solute Mass}=75+160=235\text{ g}$

3. Calculate final % (w/w):

• $\%\text{ (w/w)}=\frac{235}{700}\times 100\%=33.57\%$

Example: Interconverting % (w/v) to % (w/w)

Q: Calculate the % (w/w) of an $NaOH$ solution containing 40% (w/v) $NaOH$. The density of the solution is 2 g/mL.

1. Interpret the given % (w/v):

• 40% (w/v) means 40 g of $NaOH$ is dissolved in 100 mL of solution.

2. Use density to find the mass of the solution:

• ${\text{Mass}}_{\text{sol}}=d_{\text{sol}}\times V_{\text{sol}}$
• ${\text{Mass}}_{\text{sol}}=2\text{ g/mL}\times 100\text{ mL}=200\text{ g}$

3. Calculate final % (w/w):

• $\%\text{ (w/w)}=\frac{40\text{ g}}{200\text{ g}}\times 100\%=20\%$

Effect of Temperature on Concentration Terms

• Mass is temperature independent.
• Volume is temperature dependent (it expands or contracts with heat).
• Therefore, any concentration term involving liquid volume (e.g., % (w/v), Molarity, Normality) is temperature dependent, while mass-based terms (e.g., % (w/w), Mole fraction, Molality) are temperature independent.
• ! Exception for Gaseous Mixtures: For gases, % (v/v) is directly proportional to mole % (based on Avogadro’s Law). Since moles do not change with temperature, % (v/v) for gaseous solutions is temperature independent.

2. Mole Fraction ($\chi$)

It is the ratio of the number of moles of a particular component to the total number of moles present in the solution.

For a Binary Solution containing component A (solvent) and component B (solute):

• Mole fraction of A:
  • ${\chi }_A=\frac{n_A}{n_A+n_B}$
• Mole fraction of B:
  • ${\chi }_B=\frac{n_B}{n_A+n_B}$

Golden Rule of Mole Fraction

The sum of the mole fractions of all components in a solution is always exactly equal to 1.
$\boxed{{\chi }_A+{\chi }_B+{\chi }_C\cdots =1}$

Example: Mole Fraction Calculation

Q: A solution is prepared by adding 360 g of glucose ($C_6{H}_{12}{O}_6$) to 864 g of water. Calculate the mole fraction of glucose. ($M_w$ of glucose = 180 g/mol, $M_w$ of water = 18 g/mol)

1. Calculate moles of each component:

• $n_{\text{glucose}}=\frac{360}{180}=2\text{ moles}$
• $n_{\text{water}}=\frac{864}{18}=48\text{ moles}$

2. Calculate mole fraction ($\chi$):

• ${\chi }_{\text{glucose}}=\frac{n_{\text{glucose}}}{n_{\text{glucose}}+n_{\text{water}}}$
• ${\chi }_{\text{glucose}}=\frac{2}{2+48}=\frac{2}{50}=\frac{1}{25}\text{ or }0.04$

(Optional check: ${\chi }_{\text{water}}$ would be $1-0.04=0.96$)

3. Molarity (M)

It is defined as the number of moles of solute present in 1 litre of the solution.

• ! Temperature Dependence: Because it depends on the volume of the solution, it is temperature dependent.

• Unit: $\text{mol/L}$ or ${\text{mol/dm}}^3$ (often just written as M).

• Formula:

  • $\boxed{M=\frac{n_{\text{solute}}}{V_{\text{solution}}\text{ (in L)}}}$

  • $\boxed{M=\frac{n_{\text{solute}}\times 1000}{V_{\text{solution}}\text{ (in mL)}}}$

• Interpretation: A 5 M $H_{2}S{O}_4$ solution means 5 moles of $H_{2}S{O}_4$ is present in 1 L of the solution.

Shortcut: Relationship between Molarity, Density, and Mass %

You can directly calculate Molarity if you are given the Mass % ($x$) and the density of the solution ($d$ in $\text{g/mL}$):

• $\boxed{M=\frac{10\times x\times d}{M_w}}$
  (where $M_w$ is the molar mass of the solute)

Note: If you are given % (w/v) instead, the density is already factored in, so the formula simplifies to:*

• $\boxed{M=\frac{10\times \%(\text{w/v})}{M_w}}$

Special Cases in Molarity

1. Case of Dilution
When we add more solvent (like water) to a solution, the total volume changes, but the total moles of the solute remain exactly the same.

• Formula:
  • $\boxed{M_1{V}_1=M_2{V}_2}$ (where $M_1,M_2$ and $V_1,V_2$ are respectively molarity and volume of solute before and after dilution)

2. Mixing of Two Solutions
When mixing two different solutions that contain the same solute, the total moles simply add up.

• Formula:
  • $\boxed{M_1{V}_1+M_2{V}_2+\cdots =M_R(V_1+V_2+\dots )}$ (where $M_R$ is the resultant molarity of the final mixture)

3. Molarity of Ions
When an ionic compound dissolves, it completely dissociates into its constituent ions. The molarity of these individual ions depends on the stoichiometry of the salt.

• General Dissociation: $\boxed{A_x{B}_y⟶x{A}^{+y}+y{B}^{-x}}$
• If the concentration of the parent salt is $M$, then:
  • Concentration of cation: $[A^{+y}]=xM$
  • Concentration of anion: $[B^{-x}]=yM$
  • @ In short the stoichiometric coefficients will be the molarity of those ions in the compound.

Example: Molarity of Dissociated Ions

Q: An aqueous solution of Barium Nitrate $Ba(N{O}_3{)}_2$ has a nitrate ion concentration of $0.80\text{ M}$. What is the molarity of the solution?

1. Write the dissociation equation:

• $Ba(N{O}_3{)}_2⟶B{a}^{2+}+2N{O}_3^{-}$

2. Relate stoichiometry to molarity:

• 1 mole of the salt produces 2 moles of $N{O}_3^{-}$.
• Therefore, $[N{O}_3^{-}]=2\times M_{\text{salt}}$

3. Calculate final molarity:

• $2M=0.80\text{ M}$
• $M=0.40\text{ M}$

4. Molality (m)

It is defined as the number of moles of solute dissolved in 1 kg (1000 g) of the solvent.

• Because it depends purely on the mass of the solvent (not volume), it is temperature independent.
• Unit: $\text{mol/kg}$ or molal (m).
• Formula:

  • $\boxed{m=\frac{n_{\text{solute}}}{\text{mass of solvent (in kg)}}}$

  • $\boxed{m=\frac{n_{\text{solute}}\times 1000}{\text{mass of solvent (in g)}}}$

Important Relationships for Interconversion

1. Molality and Mole Fraction:

• $\boxed{m=\frac{{\chi }_B}{{\chi }_A}\times \frac{1000}{M_A}}$
  (where ${\chi }_B$ is the mole fraction of the solute, ${\chi }_A$ is the mole fraction of the solvent, and $M_A$ is the molar mass of the solvent in g/mol).

2. Molarity, Molality, and Density:

• $\boxed{d=M\left[\frac{1}{m}+\frac{M_B}{1000}\right]}$
  (where $d$ is density in g/mL, $M$ is Molarity, $m$ is molality, and $M_B$ is the molar mass of the solute).

Example: Calculating Molality from Mole Fraction

Question: The mole fraction of ${\text{I}}_2$ in benzene is $0.2$. Find the molality of the solution. (Molar mass of benzene, ${\text{C}}_6{\text{H}}_6=78\text{ g/mol}$)

1. Identify the given components:

• Solute ($B$): ${\text{I}}_2⟹{\chi }_B=0.2$
• Solvent ($A$): Benzene $⟹{\chi }_A=1-0.2=0.8$
• Molar mass of solvent ($M_A$): $78\text{ g/mol}$

2. Apply the relationship formula:

• $m=\frac{{\chi }_B}{{\chi }_A}\times \frac{1000}{M_A}$
• $m=\frac{0.2}{0.8}\times \frac{1000}{78}$
• $m=\frac{1}{4}\times \frac{1000}{78}=\frac{250}{78}=3.20\text{ m}$

Example: Calculating Molality from Molarity and Density

Question: Density of a $2\text{ M}$ ${\text{CH}}_3\text{COOH}$ solution is $1.2\text{ g/mL}$. Calculate the molality of the solution. (Molar mass of ${\text{CH}}_3\text{COOH}=60\text{ g/mol}$)

Method 1: Using the direct formula

• $d=M\left[\frac{1}{m}+\frac{M_B}{1000}\right]$
• $1.2=2\left[\frac{1}{m}+\frac{60}{1000}\right]$
• $0.6=\frac{1}{m}+0.06$
• $\frac{1}{m}=0.54⟹m=\frac{1}{0.54}=1.85\text{ m}$

Method 2: Basic deduction (Unitary Method)

• $2\text{ M}$ means $2\text{ moles}$ of solute in $1000\text{ mL}$ of solution.
• Mass of solution = $V\times d=1000\times 1.2=1200\text{ g}$
• Mass of solute = $n\times M_w=2\times 60=120\text{ g}$
• Mass of solvent ($W_A$) = $1200-120=1080\text{ g}$
• $m=\frac{2}{1080}\times 1000=1.85\text{ m}$

5. Formality (f)

It is specifically used for ionic compounds (like $\text{NaCl}$) because they do not exist as discrete molecules, but rather as 3D crystal lattices. Instead of “molecular mass”, we use Formula Mass.

• It is defined as the number of gram formula masses (or moles of formula units) dissolved per *1 litre* of the solution.
• Essentially, it is numerically identical to Molarity, just conceptually tailored for ionic substances.
• Formula:
  • $\boxed{f=\frac{\text{moles of formula units}}{V_{\text{solution}}(\text{in L})}}$

6. Parts Per Million (ppm)

Used when the solute is present in trace quantities (very small amounts).

• It is the mass of the solute present in $10^6$ (one million) parts by mass of the solution.
• Formula:
  • $\boxed{\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6}$

7. Parts Per Billion (ppb)

Similar to ppm, but used for even smaller trace quantities.

• It is the mass of the solute present in $10^9$ (one billion) parts by mass of the solution.
• Formula:
  • $\boxed{\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9}$

8. Strength of Solution

It is defined as the mass of the solute in grams dissolved per 1 litre of the solution.

• Unit: $\text{g/L}$
• Formula:
  • $\boxed{\text{Strength}=\frac{W_B\text{ (in g)}}{V_{\text{solution}}\text{ (in L)}}}$
• Relationship with Molarity:
  • $\boxed{\text{Strength (g/L)}=\text{Molarity (M)}\times \text{Molar mass of solute}}$

9. Normality (N)

(need to link)

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Advanced Stoichiometry

Sequential Reactions

These are chemical processes that occur in a series of steps, where the product of one reaction becomes the reactant for the next.

• To solve these problems, you can either balance all sequential equations and relate their stoichiometric coefficients, or use the POAC method to save time.

Principle of Atomic Conservation (POAC)

POAC is based entirely on the Law of Conservation of Mass. It states that the total number of atoms of any specific element on the reactant side must exactly equal the number of atoms of that element on the product side.

• ! The Golden Rule of POAC: You DO NOT need to Balance the chemical equation to apply this principle!
• Formula Principle:
  • $\text{Moles of atoms of an element in reactants}=\text{Moles of atoms of that element in products}$

Applying POAC (Without Balancing)

Reaction: ${\text{KClO}}_3⟶\text{KCl}+{\text{O}}_2$

1. POAC on Potassium (K) atoms:

• $1\times n_{{\text{KClO}}_3}=1\times n_{\text{KCl}}$

2. POAC on Chlorine (Cl) atoms:

• $1\times n_{{\text{KClO}}_3}=1\times n_{\text{KCl}}$

3. POAC on Oxygen (O) atoms:

• $3\times n_{{\text{KClO}}_3}=2\times n_{{\text{O}}_2}$

Applying POAC (Haber's Process)

Reaction: ${\text{N}}_2+{\text{H}}_2⟶{\text{NH}}_3$

1. POAC on Nitrogen (N) atoms:

• $2\times n_{{\text{N}}_2}=1\times n_{{\text{NH}}_3}$

2. POAC on Hydrogen (H) atoms:

• $2\times n_{{\text{H}}_2}=3\times n_{{\text{NH}}_3}$

Parallel Reactions

Unlike sequential reactions, parallel reactions occur when a single reactant undergoes two or more reactions *simultaneously* to form different products.

• Sequential: $A⟶B⟶C$
• Parallel: $A⟶B$ and $A⟶C$
• Key Logic: The total moles of reactant $A$ consumed is the sum of the moles consumed in each individual branch.

Example: Parallel Reaction Stoichiometry

Question: 10 moles of A are taken in a closed container. The reactions are $3A⟶2B$ and $4A⟶5C$. If 4 moles of B are formed, calculate the moles of C formed.

1. Analyze the branches:

• Branch 1: $3A⟶2B$
• Branch 2: $4A⟶5C$

2. Find moles of A consumed in Branch 1:

• $2\text{ moles}$ of B are formed from $3\text{ moles}$ of A.
• $4\text{ moles}$ of B are formed from $\frac{3}{2}\times 4=6\text{ moles of A}$.

3. Find remaining moles of A for Branch 2:

• Total A = $10\text{ moles}$.
• A left for Branch 2 = $10-6=4\text{ moles}$.

4. Calculate product C:

• $4\text{ moles}$ of A will produce exactly $5\text{ moles of C}$.

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Percentage Labelling of Oleum

Oleum is fuming sulphuric acid. It is essentially pure ${\text{H}}_2{\text{SO}}_4$ containing extra dissolved ${\text{SO}}_3$ gas. Its chemical formula is represented as ${\text{H}}_2{\text{S}}_2{\text{O}}_7$.

When water is added to oleum, it reacts with the free ${\text{SO}}_3$ to form even more sulphuric acid:
${\text{SO}}_3+{\text{H}}_2\text{O}⟶{\text{H}}_2{\text{SO}}_4$

• Meaning of the Label: A label of “y% Oleum” (e.g., 109% Oleum) means that if you take exactly $100\text{ g}$ of this oleum and add just enough water to convert all the free ${\text{SO}}_3$ into ${\text{H}}_2{\text{SO}}_4$, you will obtain a total of $y\text{ g}$ of pure ${\text{H}}_2{\text{SO}}_4$.
• Mass of water added = $(y-100)\text{ g}$

Formula: % of Free ${\text{SO}}_3$

You can directly calculate the percentage of free ${\text{SO}}_3$ in the oleum sample using the label value ($y$):

• $\boxed{\%{\text{ free SO}}_3=\frac{80(y-100)}{18}}$

Why?: Because 18g of water reacts with exactly 80g of ${\text{SO}}_3$ according to the balanced equation.

Example: Decoding an Oleum Label

Question: Calculate the % of free ${\text{SO}}_3$ in an oleum that is labelled 118%.

1. Identify the given values:

• Label ($y$) = 118
• This means $100\text{ g}$ of oleum requires $18\text{ g}$ of water.

2. Apply the formula:

• $\%{\text{ free SO}}_3=\frac{80(118-100)}{18}$
• $\%{\text{ free SO}}_3=\frac{80\times 18}{18}=80\%$
  (This means 100g of the oleum contains 80g of free ${\text{SO}}_3$ and 20g of original ${\text{H}}_2{\text{SO}}_4$).

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Mixture Analysis

In problems involving a mixture of two or more compounds reacting with a single reagent (or being heated), the trick is identifying that usually only one component of the mixture reacts, while the other acts as an inert spectator.

Example: Heating a Mixture

Question: 4 g of a mixture of ${\text{CaCO}}_3$ and ${\text{SiO}}_2$ is treated with an excess of $\text{HCl}$ and $0.88\text{ g}$ of ${\text{CO}}_2$ is produced. Calculate the percentage of ${\text{CaCO}}_3$ in the mixture.

1. Identify the active reactant:

• ${\text{SiO}}_2$ (sand) does not react with $\text{HCl}$. Only ${\text{CaCO}}_3$ reacts.
• ${\text{CaCO}}_3+2\text{HCl}⟶{\text{CaCl}}_2+{\text{H}}_2\text{O}+{\text{CO}}_2$

2. Stoichiometry from the product:

• $44\text{ g}$ of ${\text{CO}}_2$ is produced by $100\text{ g}$ of ${\text{CaCO}}_3$.
• $0.88\text{ g}$ of ${\text{CO}}_2$ is produced by $\frac{100}{44}\times 0.88=2\text{ g}$ of pure ${\text{CaCO}}_3$.

3. Calculate percentage composition:

• Total mixture mass = $4\text{ g}$. Pure ${\text{CaCO}}_3$ = $2\text{ g}$.
• $\%{\text{ CaCO}}_3=\frac{2}{4}\times 100\%=50\%$

(Note: Concepts like Volume Strength of ${\text{H}}_2{\text{O}}_2$, Degree of Dissociation, and Eudiometry will be covered in later chapters like Organic Chemistry and States of Matter).