Purification, Qualitative, Quantitative Analysis, and Molecular Determination of Organic Compounds

Qualitative Analysis of Organic Compounds

Qualitative analysis involves identifying the elements present in an organic compound. The primary elements are carbon and hydrogen. In addition, organic compounds may contain oxygen, nitrogen, sulphur, halogens, and phosphorus.

Detection of Carbon and Hydrogen

Carbon and hydrogen are detected by heating the organic compound with copper(II) oxide ( $CuO$ ).

Carbon present in the compound is oxidised to carbon dioxide ( $CO_2$ ): $C + 2CuO \rightarrow 2Cu + CO_2$
Carbon dioxide is tested using lime water ( $Ca(OH)_2$ ), which turns turbid due to the formation of calcium carbonate precipitate: $CO_2 + Ca(OH)_2 \rightarrow CaCO_3(s) + H_2O$
Hydrogen present in the compound is oxidised to water ( $H_2O$ ): $2H + CuO \rightarrow Cu + H_2O$
Water is tested using anhydrous copper sulphate ( $CuSO_4$ ), which turns from white to blue upon hydration: $5H_2O + CuSO_4 \rightarrow CuSO_4 \times 5H_2O$

Detection of Nitrogen

Nitrogen is detected using the Sodium Fusion Extract (Lassaigne's test).

During sodium fusion, nitrogen and carbon from the organic compound react with sodium metal to form sodium cyanide ( $NaCN$ ): $Na + C + N \rightarrow NaCN$
The sodium fusion extract is treated with iron(II) sulphate ( $FeSO_4$ ), heated, and then acidified with concentrated sulphuric acid ( $H_2SO_4$ ) or hydrochloric acid ( $HCl$ ).
Sodium cyanide reacts with iron(II) ions to form hexacyanidoferrate(II): $6CN^- + Fe^{2+} \rightarrow [Fe(CN)_6]^{4-}$
On adding iron(III) ions ( $Fe^{3+}$ ), Prussian blue colouration or precipitate of iron(III) hexacyanidoferrate(II) (ferriferrocyanide) is formed: $Fe^{3+} + [Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3 \times xH_2O$

Detection of Sulphur

Sulphur in an organic compound can be detected through the following tests using the sodium fusion extract:

Lead Acetate Test: The sodium fusion extract is acidified with acetic acid ( $CH_3COOH$ ) and lead acetate solution ( $Pb(CH_3COO)_2$ ) is added. A black precipitate of lead sulphide ( $PbS$ ) confirms the presence of sulphur: $S^{2-} + Pb^{2+} \rightarrow PbS(s)$
Sodium Nitroprusside Test: Addition of sodium nitroprusside solution to the alkaline sodium fusion extract produces a purple/violet colouration.

Detection of Halogens

Halogens are detected using Lassaigne's extract.

Pre-treatment Note: The sodium fusion extract is first boiled with concentrated nitric acid ( $HNO_3$ ) to decompose any sodium cyanide ( $NaCN$ ) or sodium sulphide ( $Na_2S$ ) formed during Lassaigne's test. These cyanide and sulphide ions would otherwise interfere with the silver nitrate test by forming white ( $AgCN$ ) or black ( $Ag_2S$ ) precipitates.
After boiling with $HNO_3$ , silver nitrate ( $AgNO_3$ ) solution is added:
- A white precipitate soluble in ammonium hydroxide ( $NH_4OH$ ) indicates Chlorine ( $Cl$ ).
- A pale yellow precipitate sparingly soluble in $NH_4OH$ indicates Bromine ( $Br$ ).
- A yellow precipitate insoluble in $NH_4OH$ indicates Iodine ( $I$ ).

Detection of Phosphorus

The organic compound is heated with an oxidising agent such as sodium peroxide ( $Na_2O_2$ ) to convert phosphorus to phosphate ( $PO_4^{3-}$ ).
The solution is boiled with concentrated nitric acid ( $HNO_3$ ) and then treated with ammonium molybdate (NH_4)_2MoO_4$.\n* A yellow colouration or yellow precipitate of ammonium phosphomolybdate (NH_4)_3PO_4 imes 12MoO_3 indicates the presence of phosphorus.\n\n# Partition Chromatography\n\nPartition chromatography is based on the continuous differential partitioning of components of a mixture between a stationary phase and a mobile phase.\n\n* **Paper Chromatography:** Paper chromatography is a specific type of partition chromatography.\n* Special quality paper known as chromatography paper is used, which contains water trapped within its fibres; this trapped water serves as the stationary phase.\n* A spot of the solution containing the mixture is applied near one end of the chromatography paper.\n* The paper strip is suspended in a solvent or a mixture of solvents, which acts as the mobile phase.\n* As the solvent rises up the paper by capillary action, it flows over the spot and selectively carries different components along based on their relative partition coefficients between the stationary water phase and the mobile solvent phase.\n* The developed strip is called a chromatogram. Spots corresponding to colourless compounds can be visualized under ultraviolet (UV) light or by spraying an appropriate reagent (such as ninhydrin or iodine vapours).\n\n# Quantitative Analysis: Estimation of Elements\n\nQuantitative analysis involves determining the percentage composition of various elements present in an organic compound.\n\n## Estimation of Carbon and Hydrogen (Liebig's Combustion Method)\n\n* **Principle:** A known mass of an organic compound is completely burnt in the presence of excess dry oxygen ( O_2 $, free from$ CO_2 $) and heated cupric oxide ($ CuO).\n* Carbon and hydrogen in the compound are quantitatively oxidised to carbon dioxide (CO_2 $) and water ($ H_2O), respectively.\n* **Procedure:** The produced H_2O $and$ CO_2 $are absorbed in pre-weighed U-tubes containing anhydrous calcium chloride ($ CaCl_2 $) or magnesium perchlorate ($ Mg(ClO_4)_2 $) for water absorption, and a concentrated solution of potassium hydroxide ($ KOH $) for$ CO_2 absorption.\n* **Formulas:**\n \%C =

\frac{12}{44} \times \frac{\text{weight of } CO_2 \text{ formed}}{\text{weight of organic compound}} \times 100\n \%H = \frac{2}{18} \times \frac{\text{weight of } H_2O \text{ formed}}{\text{weight of organic compound}} \times 100\n\n* **Worked Example (E-45):**\n * On complete combustion, 0.246\,g $of an organic compound gave$ 0.198\,g $of carbon dioxide and$ 0.1014\,g of water. Determine the percentage composition of carbon and hydrogen.\n * \%C = \frac{12 \times 0.198 \times 100}{44 \times 0.246} = 21.95\%\n * \%H = \frac{2 \times 0.1014 \times 100}{18 \times 0.246} = 4.58\%\n\n## Estimation of Nitrogen\n\nNitrogen present in an organic compound is estimated quantitatively using either the Dumas method or Kjeldahl's method.\n\n### Dumas Method\n\n* **Principle:** The nitrogen-containing organic compound is converted into molecular nitrogen (N_2).\n* **Procedure:** A weighed amount of the organic compound is heated with cupric oxide (CuO $) in an atmosphere of carbon dioxide ($ CO_2).\n* Carbon and hydrogen are oxidised to CO_2 $and$ H_2O $, while nitrogen is liberated as free$ N_2 gas.\n* Traces of nitrogen oxides formed during combustion are reduced back to molecular nitrogen by passing the gas mixture over heated copper gauze:\n \text{Oxides of nitrogen} + Cu \rightarrow N_2 + CuO\n* The mixture of gases is collected over a caustic potash (KOH $) solution in a nitrometer. The$ KOH $solution absorbs$ CO_2, and pure dinitrogen gas is collected in the upper graduated portion of the nitrometer.\n* **Formula:**\n \%N = \frac{28}{22400} \times \frac{\text{Volume of } N_2 \text{ in } ml \text{ at STP}}{\text{Weight of organic compound}} \times 100\n\n* **Worked Example (E-06):**\n * 0.25\,g $of an organic compound gave$ 30\,cm^3 $of moist dinitrogen at$ 288\,K $and$ 745\,mm $pressure. Calculate the percentage of nitrogen (Aqueous tension at$ 288\,K = 12.7\,mm).\n * Mass of substance = 0.25\,g\n * Volume of moist dinitrogen = 30\,cm^3\n * Dry gas pressure P_1 = 745 - 12.7 = 732.3\,mm\n * Volume of dinitrogen at STP (V_{STP}):\n V_{STP} = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{732.3 \times 30 \times 273}{288 \times 760} = 27.4\,cm^3\n * \%N = \frac{28}{22400} \times \frac{27.4}{0.25} \times 100 = 13.6\%\n\n* **Worked Example (W.E-07):**\n * 0.3\,g $of an organic compound gave$ 50\,ml $of nitrogen at$ 27^\circ C $($ 300\,K $) and$ 715\,mm $pressure. If the aqueous tension at$ 27^\circ C $is$ 15\,mm, calculate the percentage composition of nitrogen.\n * Mass of substance = 0.3\,g\n * Dry pressure P_1 = 715 - 15 = 700\,mm\n * Volume at STP:\n V_{STP} = \frac{700 \times 50 \times 273}{760 \times 300} = 41.9\,ml\n * \%N = \frac{28}{22400} \times \frac{41.9}{0.3} \times 100 = 17.46\%\n\n### Kjeldahl's Method\n\n* **Principle:** Nitrogen present in the organic compound is quantitatively converted into ammonia (NH_3).\n* **Procedure:** A known mass of the organic compound is digested with concentrated sulphuric acid (H_2SO_4 $) in the presence of potassium sulphate ($ K_2SO_4 $) and copper sulphate ($ CuSO_4) in a Kjeldahl flask.\n * K_2SO_4 $elevates the boiling point of$ H_2SO_4.\n * CuSO_4 acts as a catalyst.\n * Reaction during digestion: Organic compound + H_2SO_4 \rightarrow (NH_4)_2SO_4\n* The digested mixture is distilled with an excess solution of sodium hydroxide (NaOH):\n (NH_4)_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2NH_3 + 2H_2O\n* The evolved ammonia (NH_3 $) is absorbed in a known, excess volume of standard$ HCl $or$ H_2SO_4 solution.\n* The residual unreacted acid is back-titrated against a standard sodium hydroxide solution.\n* **Formulas:**\n \%N = \frac{1.4 \times N_1 \times (V - V_1)}{\text{weight of organic compound}}\n where N_1 $is the normality of the acid,$ V $is the total initial volume of acid taken, and$ V_1 is the volume of base required to neutralize excess unreacted acid.\n Alternatively:\n \%N = \frac{1.4 \times \text{m.eq. of } H_2SO_4 \text{ used}}{\text{weight of organic compound}}\n* **Applications:** kjeldahl's method is simple, highly convenient, and widely used to estimate nitrogen content in foodstuffs, soils, fertilizers, and agricultural products.\n* **Limitations:** Kjeldahl's method is **not applicable** to organic compounds containing:\n * Nitro groups (-NO_2)\n * Nitroso groups (-NO)\n * Azo groups (-N=N-)\n * Azoxy groups (-N=N(O)-)\n * Nitrogen present inside heterocyclic rings (e.g., pyridine, quinoline)\n * *Reason:* Nitrogen present in these structural functional groups is not quantitatively converted into ammonium sulphate under standard digestion conditions.\n\n* **Worked Example (W.E-08):**\n * In Kjeldahl's estimation, ammonia evolved from 0.5\,g $of an organic compound neutralized$ 10\,ml $of$ 1\,MH_2SO_4. Calculate the percentage of nitrogen.\n * 10\,ml $of$ 1\,MH_2SO_4 \equiv 20\,ml $of$ 1\,MNH_3\n * 1000\,ml $of$ 1\,MNH_3 $contains$ 14\,g nitrogen.\n * Mass of nitrogen in 20\,ml $of$ 1\,MNH_3 = \frac{14 \times 20}{1000} = 0.28\,g\n * \%N = \frac{0.28}{0.5} \times 100 = 56.0\%\n\n* **Worked Example (W.E-09):**\n * Ammonia obtained from 0.5\,g $of an organic substance was passed into$ 100\,cm^3 of $\frac{M}{10}$ H_2SO_4 $. The excess acid required$ 154\,cm^3 of $\frac{M}{10}$ NaOH for neutralization. Calculate the percentage of nitrogen.\n * \text{m.eq. of } H_2SO_4 \text{ taken} = \text{molarity} \times \text{basicity} \times \text{volume } (ml) = \frac{1}{10} \times 2 \times 100 = 20\n * \text{m.eq. of } NaOH \text{ used} = \frac{1}{10} \times 1 \times 154 = 15.4\n * \text{m.eq. of } H_2SO_4 \text{ unused} = 20 - 15.4 = 4.6\n * \%N = \frac{1.4 \times 4.6}{0.5} = 12.88\%\n\n## Estimation of Halogens (Carius Method)\n\n* **Principle:** A weighed amount of organic compound is heated with fuming nitric acid (HNO_3 $) in the presence of silver nitrate ($ AgNO_3) inside a heavy-walled glass tube known as a Carius tube.\n* Carbon and hydrogen are oxidised to CO_2 $and$ H_2O.\n* Halogen (X $) present in the compound is quantitatively converted into insoluble silver halide ($ AgX):\n X \rightarrow AgX(s)\n* The resulting precipitate of silver halide is filtered, thoroughly washed, dried, and weighed.\n* **Formulas:**\n \%X = \frac{\text{Atomic weight of halogen}}{\text{Molar weight of silver halide}} \times \frac{\text{weight of } AgX \text{ formed}}{\text{weight of organic compound}} \times 100\n * For Chlorine:\n \%Cl = \frac{35.5}{143.5} \times \frac{\text{weight of } AgCl}{\text{weight of organic compound}} \times 100\n * For Bromine:\n \%Br = \frac{80}{188} \times \frac{\text{weight of } AgBr}{\text{weight of organic compound}} \times 100\n * For Iodine:\n \%I = \frac{127}{235} \times \frac{\text{weight of } AgI}{\text{weight of organic compound}} \times 100\n\n* **Worked Example (E-10):**\n * In Carius method, 0.1890\,g $of an organic compound gave$ 0.2870\,g of silver chloride. Calculate the percentage of chlorine.\n * \%Cl = \frac{0.2870 \times 35.5 \times 100}{0.1890 \times 143.5} = 37.8\%\n\n* **Worked Example (W.E-11):**\n * 1.0\,g $of a bromoalkane on heating with excess silver nitrate in a Carius tube gave$ 0.94\,g $of yellow precipitate ($ AgBr). Calculate the percentage of bromine.\n * \%Br = \frac{0.94 \times 80 \times 100}{1.0 \times 188} = 40.0\%\n\n* **Worked Example (E-12):**\n * In Carius estimation, 0.15\,g $of an organic compound gave$ 0.12\,g $of$ AgBr. Find out the percentage of bromine.\n * \%Br = \frac{80 \times 0.12 \times 100}{188 \times 0.15} = 34.04\%\n\n## Estimation of Sulphur (Carius Method)\n\n* **Principle:** A weighed amount of organic compound is heated in a Carius tube with fuming nitric acid (HNO_3 $) or sodium peroxide ($ Na_2O_2).\n* Sulphur present in the compound is oxidised to sulphuric acid (H_2SO_4).\n* Addition of excess barium chloride (BaCl_2 $) solution precipitates the sulphuric acid as barium sulphate ($ BaSO_4):\n H_2SO_4 + BaCl_2 \rightarrow BaSO_4(s) + 2HCl\n* The precipitate is filtered, washed, dried, and weighed.\n* **Formula:**\n \%S = \frac{32}{233} \times \frac{\text{weight of } BaSO_4 \text{ formed}}{\text{weight of organic compound}} \times 100\n\n* **Worked Example (W.E-13):**\n * In sulphur estimation, 0.157\,g $of an organic compound gave$ 0.4813\,g of barium sulphate. What is the percentage of sulphur?\n * 233\,gBaSO_4 $contains$ 32\,g sulphur.\n * \%S = \frac{32 \times 0.4813 \times 100}{233 \times 0.157} = 42.10\%\n\n* **Worked Example (W.E-14):**\n * On heating 0.2\,g $of an organic compound with a mixture of barium chloride and nitric acid,$ 0.466\,g of barium sulphate was obtained. Calculate the percentage of sulphur.\n * \%S = \frac{0.466 \times 32 \times 100}{0.2 \times 233} = 32.0\%\n\n## Estimation of Phosphorus (Carius Method)\n\n* **Principle:** A weighed amount of organic compound is heated with fuming nitric acid (HNO_3 $) to oxidise phosphorus into phosphoric acid ($ H_3PO_4).\n* Phosphorus can be estimated via one of two precipitate forms:\n * **Magnesium Pyrophosphate Method:** Phosphoric acid is precipitated as magnesium ammonium phosphate (MgNH_4PO_4 $) by adding magnesia mixture ($ MgCl_2 + NH_4OH + NH_4Cl $). The precipitate is washed, dried, and strongly ignited to yield magnesium pyrophosphate ($ Mg_2P_2O_7):\n 2MgNH_4PO_4 \xrightarrow{\Delta} Mg_2P_2O_7 + 2NH_3 + H_2O\n \%P = \frac{62}{222} \times \frac{\text{weight of } Mg_2P_2O_7 \text{ formed}}{\text{weight of organic compound}} \times 100\n * **Ammonium Phosphomolybdate Method:** Phosphoric acid is precipitated directly as ammonium phosphomolybdate (NH_4)_3PO_4 \times 12MoO_3 by adding ammonia and ammonium molybdate solution.\n * Molecular mass of (NH_4)_3PO_4 \times 12MoO_3 = 1877\n \%P = \frac{31}{1877} \times \frac{\text{weight of } (NH_4)_3PO_4 \times 12MoO_3 \text{ formed}}{\text{weight of organic compound}} \times 100\n\n## Estimation of Oxygen\n\nOxygen percentage is determined either indirectly by difference or directly using Aluise's method.\n\n* **Method of Difference:**\n \%O = 100 - (\text{sum of percentages of all other constituent elements})\n* **Aluise's Method (Direct Estimation):**\n * A known amount of organic compound is subjected to pyrolysis in a stream of nitrogen gas.\n * The oxygenated gaseous decomposition products are passed over red-hot coke (C $) to quantitatively convert all oxygen into carbon monoxide ($ CO):\n 2C + O_2 \rightarrow 2CO\n * The generated carbon monoxide is then quantitatively converted into carbon dioxide (CO_2 $) by passing over warm iodine pentoxide ($ I_2O_5):\n 5CO + I_2O_5 \rightarrow I_2 + 5CO_2\n * The resulting gas mixture (CO_2 $and liberated$ I_2 $) is passed through potassium iodide ($ KI $) solution to absorb$ I_2 $, and then through$ KOH $solution to absorb$ CO_2$. * $\%O = \frac{16}{44} \times \frac{\text{weight of } CO_2 \text{ formed}}{\text{weight of organic compound}} \times 100$

Worked Example (W.E-15):
- $0.2\,g$ of an organic compound gave $0.147\,g$ of $CO_2$ , $0.12\,g$ of $H_2O$ , and $74.6\,c.c.$ of $N_2$ at STP. Calculate the percentage composition of all constituents.
- Mass of compound = $0.2\,g$
- $\%C = \frac{0.147 \times 12 \times 100}{0.2 \times 44} = 20.04\%$
- $\%H = \frac{0.12 \times 2 \times 100}{0.2 \times 18} = 6.66\%$
- $\%N = \frac{74.6 \times 28 \times 100}{0.2 \times 22400} = 46.63\%$
- $\%O = 100 - (20.04 + 6.66 + 46.63) = 100 - 73.33 = 26.67\%$

Modern Automated Elemental Analysis

In modern laboratories, estimation of elements ( $C$ , $H$ , and $N$ ) is carried out automatically using an automated Elemental Analyser.
The instrument requires very small quantities of substance ( $1\text{--}3\,mg$ ) and rapidly displays analytical results on a digital screen.

Determination of Molecular Mass and Formulas

Chemical Methods for Molecular Mass Determination

Silver Salt Method for Acids

Organic carboxylic acids react with silver salts to form insoluble silver salts ( $RCOOAg$ ).
Upon thermal decomposition/ignition, the silver salt leaves a residue of pure metallic silver: $2RCOOAg \xrightarrow{\Delta} 2Ag(s) + \text{other products}$
Mathematical Derivation: $\frac{\text{Equivalent weight of silver salt}}{\text{Equivalent weight of silver}} = \frac{\text{Mass of silver salt}}{\text{Mass of silver}}$ $\frac{E + 108 - 1}{108} = \frac{\text{Mass of silver salt}}{\text{Mass of silver}}$ $E = \left( \frac{\text{Mass of silver salt}}{\text{Mass of silver}} \times 108 \right) - 107$ where $E$ is the equivalent weight of the acid, and the atomic weight of silver is taken as $108\,g/mol$ .
Molecular Weight: $\text{Molecular weight of acid} = \text{Equivalent weight of acid } (E) \times \text{basicity}$

Platinic Chloride Method for Organic Bases

Organic bases ( $B$ ) combine with chloroplatinic acid ( $H_2PtCl_6$ ) to form insoluble chloroplatinate salts.
For a mono-acidic base $B$ , the formula of the salt is $B_2 H_2 PtCl_6$
Ignition of a known mass of chloroplatinate salt leaves a residue of pure metallic platinum ( $Pt$ ).
Mathematical Derivation: $\frac{\text{Mass of platinum salt}}{\text{Mass of platinum}} = \frac{2E + 410}{195}$ $E = \frac{1}{2} \left( \frac{\text{Mass of platinum salt}}{\text{Mass of platinum}} \times 195 - 410 \right)$ where $E$ is the equivalent weight of the base, atomic mass of $Pt = 195$ , and formula weight of $H_2PtCl_6$ minus $Pt$ is 410$.\n* **Molecular Mass:**\n \text{Molecular mass of base} = \text{Equivalent mass of base } (E) \times \text{acidity}\n\n### Victor Meyer's Method for Volatile Substances\n\n* A known mass of a volatile liquid or solid is rapidly vaporized in a Victor Meyer tube.\n* The generated vapours displace an equal volume of air into a graduated tube collected over water or mercury.\n* The displaced volume of air is converted to STP conditions using the ideal gas relationship:\n \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\n* **Formula:**\n \text{Molecular mass} = \frac{\text{Mass of substance taken}}{\text{Volume of air displaced in } ml \text{ at STP}} \times 22400\n\n## Empirical Formula and Molecular Formula\n\n* **Empirical Formula:** The simplest whole-number ratio of atoms of various elements present in one molecule of a compound.\n* **Steps to Calculate Empirical Formula:**\n 1. Divide the mass percentage of each constituent element by its respective atomic mass to obtain the relative number of atoms (moles).\n 2. Divide each calculated relative number by the lowest numerical value obtained to determine the simple ratio.\n 3. If the simple ratios are non-integers, multiply all ratios by a suitable integer to obtain the simplest whole-number ratio.\n 4. Write the chemical symbols of elements side-by-side with their corresponding whole numbers as subscripts.\n* **Molecular Formula:** Represents the exact actual number of atoms of each element present in one molecule of a compound.\n \text{Molecular formula} = (\text{Empirical formula})_n\n where\n n = \frac{\text{Molecular weight of the compound}}{\text{Empirical formula weight}}\n* If vapour density (VD) is given:\n \text{Molecular weight} = 2 \times \text{Vapour density}\n\n## Eudiometry (Analysis of Gaseous Hydrocarbons)\n\nEudiometry is a direct volumetric method used to determine the molecular formula of gaseous hydrocarbons without prior quantitative element percentage estimation or separate mass measurement.\n\n* **Procedure:**\n * A known volume of gaseous hydrocarbon (C_x H_y $) is mixed with an excess known volume of pure dry oxygen ($ O_2) in a eudiometer tube inverted over a mercury trough.\n * Combustion/explosion is triggered by passing an electric spark between platinum electrodes.\n * Carbon and hydrogen oxidise according to the balanced stoichiometric equation:\n C_x H_y + \left(x + \frac{y}{4}\right) O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O\n * The eudiometer tube is cooled back to room temperature. Water vapour condenses into liquid water, whose physical volume contribution is negligible.\n * The remaining gas volume inside the tube consists of produced CO_2 $and unreacted excess$ O_2$.
- Caustic potash ( $KOH$ or $NaOH$ ) solution is introduced into the tube to absorb $CO_2$ completely: $2KOH + CO_2 \rightarrow K_2CO_3 + H_2O$
- The decrease in gas volume after adding $KOH$ represents the volume of $CO_2$ produced.
- Pyrogallol solution may subsequently be added to absorb any unreacted excess $O_2$ and measure its residual volume.
Stoichiometric Volumetric Relationships:
- For $1\,volume$ of hydrocarbon $C_x H_y$ :
- Volume of $O_2$ consumed = (\left(x + \frac{y}{4}\right)) volumes
- Volume of $CO_2$ produced = $x\,volumes$
- Volume contraction on explosion and cooling = (1 + \left(x + \frac{y}{4}\right) - x = 1 + \frac{y}{4}) volumes (neglecting liquid water volume).