Organic Chemistry – Some Basic Principles and TechniquesClass 11 Chemistry Notes

Here are your study notes for the chapter Organic Chemistry – Some Basic Principles and Techniques.

GENERAL INTRODUCTION

Organic chemistry is the branch of chemistry that studies compounds of carbon. Carbon has a unique property called catenation, which is the ability to form strong covalent bonds with other carbon atoms, creating long chains and rings. Carbon also forms strong bonds with other elements like hydrogen, oxygen, nitrogen, and halogens. These compounds are essential for life and are found in everything from our DNA and proteins to fuels, medicines, and plastics.

Initially, it was believed that organic compounds could only be produced by living organisms due to a 'vital force'. This idea, known as the vital force theory, was disproved in 1828 when Friedrich Wöhler synthesized urea (an organic compound) from an inorganic compound, ammonium cyanate.

$\underset{\text{Ammonium cyanate}}{\mathrm{NH}_{4} \mathrm{CNO}} \xrightarrow{\text{Heat}} \underset{\text{Urea}}{\mathrm{NH}_{2} \mathrm{CONH}_{2}}$

This breakthrough, along with later syntheses of acetic acid and methane, established that organic compounds could be created in a laboratory, paving the way for modern organic chemistry.

TETRAVALENCE OF CARBON: SHAPES OF ORGANIC COMPOUNDS

The Shapes of Carbon Compounds

Carbon has four valence electrons, allowing it to form four covalent bonds, a property known as tetravalence. The shapes of organic molecules are determined by the type of hybridisation of the carbon atoms' orbitals.

• $sp^3$ Hybridisation: Found in molecules like methane ($CH_4$), where carbon forms four single bonds. The shape is tetrahedral with bond angles of $109.5^\circ$.
• $sp^2$ Hybridisation: Found in molecules like ethene ($C_2H_4$), where carbon forms a double bond. The shape is trigonal planar with bond angles of $120^\circ$.
• $sp$ Hybridisation: Found in molecules like ethyne ($C_2H_2$), where carbon forms a triple bond. The shape is linear with a bond angle of $180^\circ$.

Hybridisation also affects bond properties:

• Bond Length and Strength: The more 's' character in the hybrid orbital, the closer it is to the nucleus, resulting in shorter and stronger bonds.
  • Strength/Length: $sp$ > $sp^2$ > $sp^3$
• Electronegativity: The greater the 's' character, the higher the electronegativity of the carbon atom.
  • Electronegativity: $sp$ (50% s) > $sp^2$ (33.3% s) > $sp^3$ (25% s)

Some Characteristic Features of $\pi$ Bonds

A pi ($\pi$) bond is formed by the sideways overlap of two parallel p-orbitals.

• In a double bond (like in ethene, $H_2C=CH_2$), all atoms lie in the same plane.
• The electron cloud of the $\pi$ bond is located above and below the plane of the atoms.
• Rotation around a carbon-carbon double bond is restricted because it would break the sideways overlap of the p-orbitals.
• The electrons in a $\pi$ bond are more exposed and easily available, making $\pi$ bonds the most reactive centers in molecules with multiple bonds.

Solution

To solve this, remember:

• Every single bond is one $\sigma$ bond.
• Every double bond is one $\sigma$ and one $\pi$ bond.
• Every triple bond is one $\sigma$ and two $\pi$ bonds.
• Don't forget the C-H bonds!

(a) For $HC \equiv C-CH=CH-CH_3$

• C-C single bonds: 2
• C=C double bond: 1
• C≡C triple bond: 1
• Total C-C sigma bonds ($\sigma_{C-C}$): $2 + 1 + 1 = 4$
• Total C-H sigma bonds ($\sigma_{C-H}$): 6
• Total $\pi$ bonds from C=C: 1
• Total $\pi$ bonds from C≡C: 2

Final Answer:

• $\sigma$ bonds: $4 (\text{C-C}) + 6 (\text{C-H}) = 10$
• $\pi$ bonds: $1 + 2 = 3$

(b) For $CH_2=C=CH-CH_3$

• C-C single bonds: 1
• C=C double bonds: 2
• Total C-C sigma bonds ($\sigma_{C-C}$): $1 + 1 + 1 = 3$
• Total C-H sigma bonds ($\sigma_{C-H}$): 6
• Total $\pi$ bonds from two C=C: 2

Final Answer:

• $\sigma$ bonds: $3 (\text{C-C}) + 6 (\text{C-H}) = 9$
• $\pi$ bonds: 2

Solution

Count the number of sigma bonds and lone pairs around each carbon atom.

• 4 sigma bonds = $sp^3$
• 3 sigma bonds = $sp^2$
• 2 sigma bonds = $sp$

(a) $CH_3Cl$: The carbon is bonded to 3 H's and 1 Cl (4 single bonds). Answer: $sp^3$

(b) $(CH_3)_2CO$: This is propanone ($CH_3-CO-CH_3$). The two methyl ($CH_3$) carbons each have 4 single bonds. The central carbonyl carbon has a double bond with oxygen and two single bonds to the other carbons (3 sigma bonds). Answer: The two $CH_3$ carbons are $sp^3$; the central C is $sp^2$.

(c) $CH_3CN$: The methyl ($CH_3$) carbon has 4 single bonds. The nitrile carbon has a triple bond with N and a single bond with C (2 sigma bonds). Answer: The $CH_3$ carbon is $sp^3$; the CN carbon is $sp$.

(d) $HCONH_2$: The carbon is double-bonded to O and single-bonded to H and N (3 sigma bonds). Answer: $sp^2$

(e) $CH_3CH=CHCN$:

• $C_1$ ($CH_3$): 4 single bonds $\rightarrow sp^3$
• $C_2$ (in $CH=$): double bond and two single bonds $\rightarrow sp^2$
• $C_3$ (=CH): double bond and two single bonds $\rightarrow sp^2$
• $C_4$ (in CN): triple bond and one single bond $\rightarrow sp$ Answer: From left to right: $sp^3, sp^2, sp^2, sp$

STRUCTURAL REPRESENTATIONS OF ORGANIC COMPOUNDS

Complete, Condensed and Bond-line Structural Formulas

Organic molecules can be drawn in several ways:

1. Complete Structural Formula (Dash Structure): Shows all atoms and represents every covalent bond as a dash.

   • Ethane ($C_2H_6$): Shows all 6 C-H bonds and the C-C bond.
   • Ethene ($C_2H_4$): Shows the C=C double bond as two dashes.

2. Condensed Structural Formula: Omits some or all of the dashes representing bonds and groups identical atoms together.

   • Ethane: $CH_3CH_3$
   • Octane: $CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_3$ can be condensed to $CH_3(CH_2)_6CH_3$.

3. Bond-line Structural Formula: A very simple representation where lines are used to represent carbon-carbon bonds.

   • Carbon and hydrogen atoms are not shown.
   • Lines are drawn in a zig-zag fashion.
   • Terminals (ends of lines) represent methyl ($CH_3$) groups.
   • Junctions (where lines meet) represent carbon atoms with the appropriate number of hydrogens to satisfy carbon's valency of four.
   • Atoms other than C and H (like O, N, Cl) are explicitly written.

Solution

• Condensed Formula: $CH_3CH_2CH(CH_3)CH_2CH_2CH_2CH_2CH_3$
• Bond-line Formula: An eight-carbon chain is the main structure. A one-carbon branch (methyl group) is on the third carbon. <img src="https://i.imgur.com/k9m6XyX.png" alt="Bond-line structure of 3-Methyloctane" width="200"/>

Three-Dimensional Representation of Organic Molecules

To show the 3-D structure of molecules on a 2-D surface like paper, we use the wedge-and-dash formula.

• A normal line (-) represents a bond in the plane of the paper.
• A solid wedge (▬) represents a bond projecting out towards the observer.
• A dashed wedge (···) represents a bond projecting away from the observer.

CLASSIFICATION OF ORGANIC COMPOUNDS

Organic compounds are classified based on their carbon skeleton.

1. Acyclic or Open Chain Compounds

• Also called aliphatic compounds.
• Consist of straight or branched chains of carbon atoms.
• Examples: Ethane ($CH_3CH_3$), Acetic acid ($CH_3COOH$).

2. Cyclic or Closed Chain (Ring) Compounds

• Alicyclic Compounds: Contain carbon atoms joined in a ring (homocyclic). They behave like aliphatic compounds.
  • Examples: Cyclopropane, Cyclohexane.
  • If an atom other than carbon is in the ring, it's called a heterocyclic compound (e.g., Tetrahydrofuran).
• Aromatic Compounds: Special types of ring compounds, typically containing a benzene ring.
  • Benzenoid: Compounds that contain one or more benzene rings (e.g., Benzene, Aniline).
  • Non-Benzenoid: Aromatic compounds that do not contain a benzene ring (e.g., Tropone).
  • Heterocyclic Aromatic: Aromatic rings containing at least one atom other than carbon (e.g., Pyridine, Furan).

Functional Group

A functional group is an atom or group of atoms within a molecule that is responsible for its characteristic chemical properties. Compounds with the same functional group react in similar ways.

• Examples: Hydroxyl group (–OH), Aldehyde group (–CHO), Carboxylic acid group (–COOH).

Homologous Series

A homologous series is a series of compounds with the same functional group and similar chemical properties, in which successive members differ by a $–CH_2$ group.

• Examples: Alkanes ($CH_4, C_2H_6, C_3H_8, ...$), Alcohols ($CH_3OH, C_2H_5OH, C_3H_7OH, ...$).

NOMENCLATURE OF ORGANIC COMPOUNDS

The systematic method of naming organic compounds is called the IUPAC (International Union of Pure and Applied Chemistry) system. Before this, compounds were given common or trivial names, often based on their source (e.g., formic acid from ants, Latin: formica).

The IUPAC System of Nomenclature

An IUPAC name generally consists of three parts:

• Word Root: Indicates the number of carbon atoms in the main chain.
• Suffix: Indicates the principal functional group or type of C-C bonds.
• Prefix: Indicates any substituent groups or side chains.

IUPAC Nomenclature of Alkanes

Alkanes are saturated hydrocarbons (contain only C-C single bonds).

1. Straight-Chain Alkanes:

• The name consists of a prefix for the number of carbons (meth-, eth-, prop-, but-, pent-, etc.) and the suffix -ane.
• Example: $C_6H_{14}$ is Hexane.

2. Branched-Chain Alkanes:

• Branches are called alkyl groups (e.g., methyl $–CH_3$, ethyl $–CH_2CH_3$), named by replacing '-ane' with '-yl'.
• Rules for Naming:
  1. Identify the longest continuous carbon chain (parent chain).
  2. Number the parent chain from the end that gives the substituent(s) the lowest possible number(s).
  3. Name the substituents and indicate their position with the number of the carbon they are attached to.
  4. List substituents alphabetically. Prefixes like di-, tri-, tetra- (for multiple identical groups) are used but ignored for alphabetizing.
  5. If two different substituents are at equivalent positions, the one that comes first alphabetically gets the lower number.

Solution

1. Longest Chain: The longest continuous chain has 9 carbons (nonane).
2. Numbering: Numbering from left to right gives substituents at positions 2 and 6. Numbering from right to left gives positions 4 and 8. The lowest set is 2,6.
3. Substituents: There is a methyl ($–CH_3$) group at position 2 and an ethyl ($–CH_2CH_3$) group at position 6.
4. Alphabetical Order: Ethyl comes before methyl.

Final Answer: 6-Ethyl-2-methylnonane

Nomenclature of Organic Compounds having Functional Group(s)

1. Identify the principal functional group to determine the suffix (e.g., -ol for alcohol, -al for aldehyde, -oic acid for carboxylic acid).
2. Find the longest carbon chain that includes the principal functional group.
3. Number the chain to give the principal functional group the lowest possible number.
4. Name other functional groups as prefixes (e.g., hydroxy- for –OH, oxo- for =O).

Solution

1. Principal Functional Group: There are two functional groups: hydroxyl (–OH) and ketone (>C=O). According to the priority order, ketone is the principal group. The suffix will be -one. The –OH group will be named as a prefix, hydroxy-.
2. Longest Chain: The longest chain containing the ketone has 7 carbons (heptane).
3. Numbering: Number the chain from the right to give the ketone group the lowest number (position 2). This places the hydroxy group at position 7.
4. Assemble the Name: The name is based on heptanone.

Final Answer: 7-Hydroxyheptan-2-one

Nomenclature of Substituted Benzene Compounds

• For monosubstituted benzene, the name is the substituent's name as a prefix to "benzene" (e.g., Bromobenzene, Nitrobenzene). Many have common names (e.g., Toluene for methylbenzene, Aniline for aminobenzene).
• For disubstituted benzene, positions are indicated by numbers or prefixes:
  • ortho (o-) for 1,2 positions.
  • meta (m-) for 1,3 positions.
  • para (p-) for 1,4 positions.
• Numbering should give the substituents the lowest possible numbers.

Solution

1. Base Compound: The compound can be named as a derivative of toluene (methylbenzene). The methyl group is at position 1.
2. Numbering: We can number clockwise or counter-clockwise. To give the other substituents the lowest numbers, we number so that chloro is at 2 and nitro is at 4. (The other way would give 1, 4, 6, which is higher).
3. Substituents: Chloro at 2, Nitro at 4.
4. Alphabetical Order: Chloro comes before methyl, which comes before nitro. However, since toluene is the base name, methyl is part of the parent name. We list the other substituents alphabetically.

Final Answer: 2-Chloro-4-nitrotoluene (or 1-Chloro-4-methyl-2-nitrobenzene if benzene is the parent). The problem in the source text has a slightly different example, leading to 2-Chloro-1-methyl-4-nitrobenzene.

ISOMERISM

Isomerism is the phenomenon where two or more compounds have the same molecular formula but different properties. These compounds are called isomers.

Structural Isomerism

Compounds with the same molecular formula but different structures (how atoms are connected) are structural isomers.

• Chain Isomerism: Isomers have different carbon skeletons (arrangements of the carbon chain).
  • Example ($C_5H_{12}$): Pentane (straight chain), Isopentane (2-methylbutane, branched), Neopentane (2,2-dimethylpropane, highly branched).
• Position Isomerism: Isomers differ in the position of a substituent or functional group on the same carbon skeleton.
  • Example ($C_3H_8O$): Propan-1-ol and Propan-2-ol.
• Functional Group Isomerism: Isomers have the same molecular formula but different functional groups.
  • Example ($C_3H_6O$): Propanal (an aldehyde) and Propanone (a ketone).
• Metamerism: Isomers have different alkyl chains on either side of the same functional group.
  • Example ($C_4H_{10}O$): Ethoxyethane ($C_2H_5–O–C_2H_5$) and Methoxypropane ($CH_3–O–C_3H_7$).

Stereoisomerism

Compounds with the same molecular formula and sequence of bonds but a different spatial arrangement of atoms are stereoisomers. The two main types are geometrical and optical isomerism.

FUNDAMENTAL CONCEPTS IN ORGANIC REACTION MECHANISM

An organic reaction involves a substrate (the main organic molecule) reacting with a reagent to form intermediates and finally products. A reaction mechanism is the step-by-step sequence of events that describes how the reaction occurs, including electron movement and bond breaking/formation.

Fission of a Covalent Bond

A covalent bond can break in two ways:

1. Heterolytic Cleavage (Heterolysis): The bond breaks unevenly, and one atom takes both electrons from the shared pair. This creates ions.

   • Breaking the C-Br bond in $CH_3Br$ gives a carbocation ($CH_3^+$) and a bromide ion ($Br^-$).
     • A carbocation is a species with a positively charged carbon atom. Their stability increases with the number of attached alkyl groups: tertiary > secondary > primary > methyl. This is due to inductive and hyperconjugation effects. Carbocations are $sp^2$ hybridized and have a trigonal planar shape.
   • This cleavage can also form a carbanion, a species with a negatively charged carbon atom (e.g., $H_3C:^-$). Carbanions are generally $sp^3$ hybridized with a distorted tetrahedral shape.

2. Homolytic Cleavage (Homolysis): The bond breaks evenly, and each atom gets one electron from the shared pair. This creates neutral species with an unpaired electron.

   • These species are called free radicals (e.g., $CH_3\cdot$).
   • The stability of free radicals also increases with alkyl substitution: tertiary > secondary > primary > methyl.

Nucleophiles and Electrophiles

• A nucleophile ("nucleus-loving") is a reagent that is electron-rich and donates an electron pair to form a new bond. It attacks an electron-deficient center (an electrophilic center).
  • Examples: $OH^−$, $CN^−$, $H_2O:$, $:NH_3$.
• An electrophile ("electron-loving") is a reagent that is electron-deficient and accepts an electron pair. It attacks an electron-rich center (a nucleophilic center).
  • Examples: $H^+$, $CH_3^+$, $NO_2^+$, neutral molecules like $BF_3$.

Electron Displacement Effects in Covalent Bonds

These are effects that cause a shift of electrons within a molecule, creating centers of high or low electron density.

Inductive Effect (I Effect)

• A permanent effect caused by the difference in electronegativity between two atoms in a $\sigma$-bond.
• The electron density shifts towards the more electronegative atom, creating partial positive ($\delta^+$) and partial negative ($\delta^−$) charges.
• This polarity is transmitted through the carbon chain but weakens rapidly with distance.
• Electron-withdrawing groups (-I effect): Pull electron density away from the carbon chain (e.g., $–NO_2, –CN, –COOH$, Halogens).
• Electron-donating groups (+I effect): Push electron density towards the carbon chain (e.g., alkyl groups like $–CH_3, –C_2H_5$).

Resonance Effect (R or M Effect)

• A permanent effect that occurs in conjugated systems (systems with alternating single and multiple bonds). It involves the delocalization of $\pi$-electrons or lone pairs.
• The actual structure of the molecule is a resonance hybrid of several contributing structures, called canonical forms or resonance structures.
• Positive Resonance Effect (+R effect): A group donates electrons to the conjugated system (e.g., $–OH, –NH_2, –OR$).
• Negative Resonance Effect (-R effect): A group withdraws electrons from the conjugated system (e.g., $–NO_2, –CHO, –COOH$).

Electromeric Effect (E Effect)

• A temporary effect that occurs only in the presence of an attacking reagent.
• It involves the complete transfer of a shared pair of $\pi$-electrons to one of the atoms in a multiple bond.
• Positive Electromeric Effect (+E effect): The $\pi$-electrons shift to the atom to which the attacking reagent gets attached.
• Negative Electromeric Effect (-E effect): The $\pi$-electrons shift to the atom to which the attacking reagent does not get attached.

Hyperconjugation

• Also known as no-bond resonance. It is a permanent effect.
• It involves the delocalization of $\sigma$-electrons from a C-H bond of an alkyl group into an adjacent empty p-orbital or a $\pi$-system.
• This effect is used to explain the stability of carbocations, free radicals, and alkenes. The more alkyl groups (and thus more C-H bonds) available for hyperconjugation, the greater the stability.

METHODS OF PURIFICATION OF ORGANIC COMPOUNDS

Purifying a compound is crucial after its synthesis or extraction. The method used depends on the nature of the compound and its impurities.

• Sublimation: Used to separate a sublimable compound (one that turns directly from solid to gas) from non-sublimable impurities.
• Crystallisation: The most common method for purifying solids. It relies on the difference in solubility of the compound and its impurities in a suitable solvent. The impure solid is dissolved in a hot solvent, and as the solution cools, the pure compound crystallizes out.
• Distillation: Used to separate volatile liquids from non-volatile impurities or to separate two liquids with significantly different boiling points.
  • Fractional Distillation: Used when the boiling points of two liquids are close. A fractionating column provides a large surface area for repeated vaporization and condensation cycles, enriching the vapor with the more volatile component.
  • Distillation under Reduced Pressure: Used for liquids with very high boiling points or those that decompose upon heating. Reducing the external pressure lowers the boiling point.
  • Steam Distillation: Used for separating substances that are steam-volatile and immiscible with water (e.g., aniline). The substance vaporizes at a lower temperature because its partial vapor pressure, along with that of water, equals the atmospheric pressure.
• Differential Extraction: Used to separate a compound from an aqueous solution by shaking it with an immiscible organic solvent in which the compound is more soluble.
• Chromatography: A powerful technique for separating, purifying, and testing the purity of compounds. It involves distributing the components of a mixture between a stationary phase and a mobile phase.
  • Adsorption Chromatography: Based on differential adsorption of substances on a solid stationary phase (like silica gel or alumina).
    • Column Chromatography: Stationary phase is packed in a column.
    • Thin Layer Chromatography (TLC): Stationary phase is a thin layer on a glass plate. The separation is quantified by the retardation factor ($R_f$). $R_f = \frac{\text{Distance moved by the substance}}{\text{Distance moved by the solvent}}$
  • Partition Chromatography: Based on differential partitioning (distribution) of components between a liquid stationary phase and a liquid or gas mobile phase.
    • Paper Chromatography: Uses special paper where trapped water acts as the stationary phase.

QUALITATIVE ANALYSIS OF ORGANIC COMPOUNDS

This involves identifying the elements present in a compound.

• Detection of Carbon and Hydrogen: The compound is heated with copper(II) oxide ($CuO$). Carbon is oxidized to $CO_2$ (turns lime-water milky), and hydrogen is oxidized to $H_2O$ (turns anhydrous copper sulfate blue). $C + 2CuO \xrightarrow{\Delta} 2Cu + CO_2$ $2H + CuO \xrightarrow{\Delta} Cu + H_2O$
• Detection of Other Elements (N, S, Halogens, P) - Lassaigne's Test: The organic compound is fused with sodium metal to convert the elements from their covalent form into ionic sodium salts (NaCN, $Na_2S$, NaX). This is called the sodium fusion extract.
  • Test for Nitrogen: The extract is treated with $FeSO_4$ and then acidified with conc. $H_2SO_4$. The formation of a Prussian blue color confirms nitrogen.
  • Test for Sulphur:
    1. Acidify the extract with acetic acid and add lead acetate. A black precipitate of $PbS$ confirms sulfur.
    2. Add sodium nitroprusside to the extract. A violet color confirms sulfur.
  • Test for Halogens: Acidify the extract with $HNO_3$ and add $AgNO_3$.
    • White ppt, soluble in $NH_4OH \rightarrow$ Chlorine
    • Yellowish ppt, sparingly soluble in $NH_4OH \rightarrow$ Bromine
    • Yellow ppt, insoluble in $NH_4OH \rightarrow$ Iodine
  • Test for Phosphorus: The compound is oxidized to phosphate, which is then treated with ammonium molybdate. A yellow precipitate confirms phosphorus.

QUANTITATIVE ANALYSIS

This involves determining the percentage composition of elements in a compound.

Carbon and Hydrogen

A known mass of the compound is completely burned. The mass of $H_2O$ and $CO_2$ produced is measured by absorbing them in anhydrous $CaCl_2$ and $KOH$ solution, respectively.

• Percentage of Carbon = $\frac{12 \times \text{mass of } CO_2 \times 100}{44 \times \text{mass of compound}}$
• Percentage of Hydrogen = $\frac{2 \times \text{mass of } H_2O \times 100}{18 \times \text{mass of compound}}$

Given

• Mass of organic compound, $m = 0.246$ g
• Mass of carbon dioxide, $m_2 = 0.198$ g
• Mass of water, $m_1 = 0.1014$ g

To Find

• Percentage of Carbon
• Percentage of Hydrogen

Formula

• Percentage of Carbon $= \frac{12 \times m_2 \times 100}{44 \times m}$
• Percentage of Hydrogen $= \frac{2 \times m_1 \times 100}{18 \times m}$

Solution

For Carbon: $\text{Percentage of Carbon} = \frac{12 \times 0.198 \times 100}{44 \times 0.246} = 21.95\%$

For Hydrogen: $\text{Percentage of Hydrogen} = \frac{2 \times 0.1014 \times 100}{18 \times 0.246} = 4.58\%$

Final Answer: The compound contains 21.95% Carbon and 4.58% Hydrogen.

Nitrogen

• Dumas Method: The compound is heated with $CuO$ to produce $N_2$ gas, which is collected and its volume is measured.
  • Percentage of Nitrogen = $\frac{28 \times V \times 100}{22400 \times m}$ (where V is the volume of $N_2$ at STP in mL, and m is the mass of the compound in g).
• Kjeldahl's Method: The compound is heated with conc. $H_2SO_4$ to convert nitrogen into ammonium sulfate. This is then treated with excess NaOH to liberate ammonia ($NH_3$), which is absorbed in a standard acid solution. The amount of unreacted acid is determined by titration.
  • Percentage of Nitrogen = $\frac{1.4 \times M \times 2(V - V_1/2)}{m}$ (where m = mass of compound, M = Molarity of acid, V = volume of acid taken, $V_1$ = volume of NaOH used for back titration).

Halogens (Carius Method)

A known mass of the compound is heated with fuming $HNO_3$ and $AgNO_3$. The halogen is converted into silver halide (AgX), which is precipitated, dried, and weighed.

• Percentage of Halogen = $\frac{\text{atomic mass of X} \times m_1 \times 100}{\text{molecular mass of AgX} \times m}$ (where $m_1$ is the mass of AgX formed and m is the mass of the compound).

Sulphur (Carius Method)

The compound is oxidized to sulfuric acid ($H_2SO_4$), which is then precipitated as barium sulfate ($BaSO_4$) by adding $BaCl_2$.

• Percentage of Sulphur = $\frac{32 \times m_1 \times 100}{233 \times m}$ (where $m_1$ is the mass of $BaSO_4$ formed and m is the mass of the compound).

Phosphorus

The compound is oxidized to phosphoric acid ($H_3PO_4$), which is precipitated as ammonium phosphomolybdate or as $Mg_2P_2O_7$.

• Percentage of Phosphorus = $\frac{62 \times m_1 \times 100}{222 \times m}$ (when estimated as $Mg_2P_2O_7$, where $m_1$ is the mass of $Mg_2P_2O_7$).

Oxygen

The percentage of oxygen is usually determined by difference:

• % Oxygen = 100 – (Sum of percentages of all other elements).