This page puts the organic chemistry class 11 formulas from NCERT Chapter 8 (Organic Chemistry – Some Basic Principles and Techniques) in one place: the rule for counting \( \sigma \) and \( \pi \) bonds, the steam-distillation condition and \( R_f \) value used in purification, and the quantitative-analysis formulas that give the percentage of carbon, hydrogen, nitrogen, halogens, sulphur, phosphorus and oxygen in a compound.
Each formula is grouped by topic, with the meaning of every symbol and its unit, and one line on when to use it. Three worked examples show the substitution step by step with original numbers. For the derivations and detailed explanations, start from the Class 11 chemistry formulas index; the official NCERT Class 11 Chemistry textbook is available on the NCERT website.
Organic Chemistry Class 11 Formulas at a Glance
This table is the index of the whole page: one row per formula. Symbols are defined in the next section, and the conditions for using each formula follow after that.
| Purpose (what you are finding) | Formula |
|---|---|
| Counting \( \sigma \) and \( \pi \) bonds in a structure | \( \mathrm{C-C}: 1\sigma;\ \mathrm{C=C}: 1\sigma+1\pi;\ \mathrm{C\equiv C}: 1\sigma+2\pi \) |
| General formula of alkanes (derived from the alkane series in Table 8.2) | \( \mathrm{C}_n\mathrm{H}_{2n+2} \) |
| Boiling condition in steam distillation | \( p = p_1 + p_2 \) |
| Retardation factor in chromatographic separation | \( R_f = \frac{x}{y} \) |
| Combustion of a C–H compound in C/H estimation | \( \mathrm{C}_x\mathrm{H}_y + \left(x+\frac{y}{4}\right)\mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + \frac{y}{2}\mathrm{H}_2\mathrm{O} \) |
| Percentage of carbon from combustion data | \( \%\mathrm{C} = \frac{12 \times m_2 \times 100}{44 \times m} \) |
| Percentage of hydrogen from combustion data | \( \%\mathrm{H} = \frac{2 \times m_1 \times 100}{18 \times m} \) |
| Volume of nitrogen at STP in Dumas method (corrects the measured volume) | \( V = \frac{P_1 V_1 \times 273}{760 \times T_1} \) |
| Percentage of nitrogen – Dumas method | \( \%\mathrm{N} = \frac{28 \times V \times 100}{22400 \times m} \) |
| Percentage of nitrogen – Kjeldahl’s method | \( \%\mathrm{N} = \frac{1.4 \times M \times 2\left(V-\frac{V_1}{2}\right)}{m} \) |
| Percentage of halogen – Carius method | \( \%\mathrm{X} = \frac{\text{atomic mass of } X \times m_1 \times 100}{\text{molar mass of } \mathrm{AgX} \times m} \) |
| Percentage of sulphur | \( \%\mathrm{S} = \frac{32 \times m_1 \times 100}{233 \times m} \) |
| Percentage of phosphorus (as ammonium phosphomolybdate) | \( \%\mathrm{P} = \frac{31 \times m_1 \times 100}{1877 \times m} \) |
| Percentage of phosphorus (as \( \mathrm{Mg_2P_2O_7} \)) | \( \%\mathrm{P} = \frac{62 \times m_1 \times 100}{222 \times m} \) |
| Percentage of oxygen (direct method) | \( \%\mathrm{O} = \frac{32 \times m_1 \times 100}{88 \times m} \) |
| Percentage of oxygen (by difference) | \( \%\mathrm{O} = 100 – \text{(sum of all other element percentages)} \) |
| Confirming nitrogen (Prussian blue test) | \( 3[\mathrm{Fe(CN)_6}]^{4-} + 4\mathrm{Fe}^{3+} \rightarrow \mathrm{Fe_4[Fe(CN)_6]_3} \) |
| Confirming sulphur (lead acetate test) | \( \mathrm{S}^{2-} + \mathrm{Pb}^{2+} \rightarrow \mathrm{PbS}\downarrow \ \text{(black)} \) |
| Confirming halogens (silver nitrate test) | \( \mathrm{X}^- + \mathrm{Ag}^+ \rightarrow \mathrm{AgX}\downarrow \) |
All Formulas, Grouped by Topic
All formulas below follow the section order of NCERT Chapter 8. Every percentage formula works the same way: find the mass of the element inside the weighed product, then express it as a percentage of the sample.
Anchor facts used again and again: 233 g of \( \mathrm{BaSO_4} \) contains 32 g of sulphur, 188 g of \( \mathrm{AgBr} \) contains 80 g of bromine, and 22400 mL of \( \mathrm{N_2} \) at STP weighs 28 g.
Structure of Carbon Compounds: Bond Counts and Shapes
\[ \mathrm{C-C} = 1\sigma;\qquad \mathrm{C=C} = 1\sigma + 1\pi;\qquad \mathrm{C\equiv C} = 1\sigma + 2\pi \]
A single bond is one \( \sigma \) bond; a double bond is one \( \sigma \) plus one \( \pi \); a triple bond is one \( \sigma \) plus two \( \pi \) bonds. To count a whole molecule, allocate one \( \sigma \) to every bond first, then collect the \( \pi \) bonds from each double and triple bond (NCERT, p. 258).
| Hybridisation of carbon | Example from the chapter | Shape |
|---|---|---|
| \( sp^3 \) | \( \mathrm{CH_4} \), \( \mathrm{CH_3Cl} \), \( \mathrm{CH_3F} \) | tetrahedral |
| \( sp^2 \) | \( \mathrm{H_2C=O} \), carbonyl carbon of \( (\mathrm{CH_3})_2\mathrm{CO} \) | trigonal planar |
| \( sp \) | \( \mathrm{HC\equiv N} \), nitrile carbon of \( \mathrm{CH_3CN} \) | linear |
\[ \text{s-character: } sp\ (50\%) \gt sp^2 \gt sp^3 \]
The more \( s \)-character a hybrid orbital has, the closer its electrons sit to the nucleus, so the bond is shorter and stronger and the carbon is more electronegative (NCERT, p. 258).
The \( \pi \)-electron cloud lies above and below the bond plane and is easily reached by attacking reagents, which is why multiple bonds are the reactive centres of a molecule. Rotation about a \( \mathrm{C=C} \) double bond is restricted because it would destroy the sideways \( p \)-orbital overlap (NCERT, p. 258).
Homologous Series
Members of a homologous series share the same functional group, can be written with one general molecular formula, and successive members differ by a \( \mathrm{CH_2} \) unit (NCERT, p. 264).
\[ \mathrm{C}_n\mathrm{H}_{2n+2} \quad (n \ge 1)\ \text{for alkanes, read from Table 8.2} \]
Check the pattern yourself: every alkane listed in Table 8.2, from methane \( \mathrm{CH_4} \) to triacontane \( \mathrm{C_{30}H_{62}} \), fits \( \mathrm{C}_n\mathrm{H}_{2n+2} \) (NCERT, p. 264).
Purification Techniques: Steam Distillation and Chromatography
A liquid boils when its vapour pressure equals the external pressure. In steam distillation the organic liquid and water vaporise together, so the mixture boils when the sum of their vapour pressures reaches atmospheric pressure (NCERT, pp. 281-282).
\[ p = p_1 + p_2 \]
Because \( p_1 \) alone is less than \( p \), the organic liquid vaporises below its normal boiling point; a water-organic mixture boils close to, but below, 373 K. Use this method for steam-volatile compounds that are immiscible with water, such as aniline (NCERT, pp. 281-282).
\[ R_f = \frac{\text{Distance moved by the substance from base line }(x)}{\text{Distance moved by the solvent from base line }(y)} \]
The retardation factor compares how far a component has moved with how far the solvent front has moved, both measured from the base line. \( R_f \) is constant for a given compound under fixed conditions, so it identifies components and helps check purity (NCERT, p. 284).
Qualitative Analysis: Lassaigne’s Test
In Lassaigne’s test, the elements are first converted into ionic sodium salts by fusing the compound with sodium metal (NCERT, p. 285).
\[ \mathrm{Na + C + N \xrightarrow{\Delta} NaCN;\qquad 2Na + S \xrightarrow{\Delta} Na_2S;\qquad Na + X \xrightarrow{\Delta} NaX}\ (X = \mathrm{Cl, Br, I}) \]
| Element | Fusion product | Reagent used | Positive result |
|---|---|---|---|
| Nitrogen (N) | \( \mathrm{NaCN} \) | FeSO₄, then acidify with conc. \( \mathrm{H_2SO_4} \) | Prussian blue colour |
| Sulphur (S) | \( \mathrm{Na_2S} \) | lead acetate; or sodium nitroprusside | black \( \mathrm{PbS} \) precipitate; violet colour |
| Halogens (Cl, Br, I) | \( \mathrm{NaX} \) | \( \mathrm{AgNO_3} \) after acidifying with \( \mathrm{HNO_3} \) | white \( \mathrm{AgCl} \) (soluble in \( \mathrm{NH_4OH} \)), yellowish \( \mathrm{AgBr} \) (sparingly soluble), yellow \( \mathrm{AgI} \) (insoluble) |
| Phosphorus (P) | phosphate (after heating with \( \mathrm{Na_2O_2} \)) | ammonium molybdate + \( \mathrm{HNO_3} \) | yellow colour / precipitate |
\[ 6\mathrm{CN^- + Fe^{2+} \rightarrow [Fe(CN)_6]^{4-};\qquad 3[Fe(CN)_6]^{4-} + 4Fe^{3+} \rightarrow Fe_4[Fe(CN)_6]_3 \cdot xH_2O}\ \text{(Prussian blue)} \]
\[ \mathrm{S^{2-} + Pb^{2+} \rightarrow PbS\downarrow}\ \text{(black);}\qquad \mathrm{X^- + Ag^+ \rightarrow AgX\downarrow} \]
If nitrogen and sulphur are both present, fusion gives sodium thiocyanate \( (\mathrm{NaSCN}) \) instead of sodium cyanide. The extract then gives a blood-red colour with \( \mathrm{Fe^{3+}} \) and no Prussian blue; fusing with excess sodium decomposes \( \mathrm{NaSCN} \) to \( \mathrm{NaCN} \) and \( \mathrm{Na_2S} \), so the usual tests work (NCERT, p. 285).
Quantitative Analysis: Carbon and Hydrogen
A known mass of the compound is burnt in excess oxygen and copper(II) oxide. Carbon is oxidised to \( \mathrm{CO_2} \) and hydrogen to \( \mathrm{H_2O} \); the \( \mathrm{CO_2} \) is absorbed in KOH solution and the water in anhydrous \( \mathrm{CaCl_2} \), and the two U-tubes are weighed (NCERT, pp. 286-287).
\[ \mathrm{C}_x\mathrm{H}_y + \left(x + \frac{y}{4}\right)\mathrm{O_2} \rightarrow x\mathrm{CO_2} + \frac{y}{2}\mathrm{H_2O} \]
\[ \%\mathrm{C} = \frac{12 \times m_2 \times 100}{44 \times m};\qquad \%\mathrm{H} = \frac{2 \times m_1 \times 100}{18 \times m} \]
The factor 12/44 is the fraction of carbon in \( \mathrm{CO_2} \), and 2/18 is the fraction of hydrogen in \( \mathrm{H_2O} \) (NCERT, pp. 286-287).
Quantitative Analysis: Nitrogen – Dumas and Kjeldahl
In Dumas method, the compound is heated with copper oxide and the nitrogen gas is collected over KOH solution, which absorbs \( \mathrm{CO_2} \). The measured volume is corrected to STP before the percentage is calculated (NCERT, pp. 287-288).
\[ \mathrm{C}_x\mathrm{H}_y\mathrm{N}_z + \left(2x + \frac{y}{2}\right)\mathrm{CuO} \rightarrow x\mathrm{CO_2} + \frac{y}{2}\mathrm{H_2O} + \frac{z}{2}\mathrm{N_2} + \left(2x + \frac{y}{2}\right)\mathrm{Cu} \]
\[ V = \frac{P_1 V_1 \times 273}{760 \times T_1};\qquad P_1 = \text{atmospheric pressure} – \text{aqueous tension} \]
\[ \%\mathrm{N} = \frac{28 \times V \times 100}{22400 \times m} \]
The molar volume of any gas at STP is 22400 mL, and 22400 mL of \( \mathrm{N_2} \) weighs 28 g; both numbers in the formula come from this fact.
In Kjeldahl’s method, nitrogen is converted to ammonium sulphate, then to ammonia, which is absorbed in a known volume of standard \( \mathrm{H_2SO_4} \). The unused acid is found by back-titration with NaOH (NCERT, p. 289).
\[ \text{Compound} \xrightarrow[\text{conc. } \mathrm{H_2SO_4}]{\Delta} (\mathrm{NH_4})_2\mathrm{SO_4} \xrightarrow[\Delta]{\text{excess } \mathrm{NaOH}} 2\mathrm{NH_3} \xrightarrow{\mathrm{H_2SO_4}} (\mathrm{NH_4})_2\mathrm{SO_4} \]
\[ \%\mathrm{N} = \frac{1.4 \times M \times 2\left(V – \frac{V_1}{2}\right)}{m} \]
Kjeldahl’s method is not applicable to compounds containing nitrogen in nitro or azo groups, or in a ring (e.g. pyridine), because such nitrogen does not form ammonium sulphate; use Dumas method for those (NCERT, p. 289).
Quantitative Analysis: Halogens, Sulphur, Phosphorus and Oxygen
In Carius method, the compound is heated with fuming nitric acid and silver nitrate; the halogen forms \( \mathrm{AgX} \), which is filtered, washed, dried and weighed (NCERT, p. 290).
\[ \%\mathrm{X} = \frac{\text{atomic mass of } X \times m_1 \times 100}{\text{molar mass of } \mathrm{AgX} \times m} \]
Sulphur is oxidised to sulphuric acid and precipitated as \( \mathrm{BaSO_4} \); 233 g of \( \mathrm{BaSO_4} \) contains 32 g of sulphur (NCERT, p. 290).
\[ \%\mathrm{S} = \frac{32 \times m_1 \times 100}{233 \times m} \]
Phosphorus is oxidised to phosphoric acid and weighed either as ammonium phosphomolybdate \( (\mathrm{NH_4})_3\mathrm{PO_4} \cdot 12\mathrm{MoO_3} \) (molar mass 1877 u) or as \( \mathrm{Mg_2P_2O_7} \) (molar mass 222 u, which contains two phosphorus atoms) (NCERT, p. 291).
\[ \%\mathrm{P} = \frac{31 \times m_1 \times 100}{1877 \times m}\quad \text{(as ammonium phosphomolybdate)} \]
\[ \%\mathrm{P} = \frac{62 \times m_1 \times 100}{222 \times m}\quad \text{(as } \mathrm{Mg_2P_2O_7} \text{)} \]
Oxygen is usually found by difference. In the direct method, oxygen from the decomposed compound is converted to \( \mathrm{CO} \) over red-hot coke, then to \( \mathrm{CO_2} \) by iodine pentoxide \( (\mathrm{I_2O_5}) \); each mole of oxygen liberated ultimately gives two moles of \( \mathrm{CO_2} \), so 32 g of \( \mathrm{O_2} \) corresponds to 88 g of \( \mathrm{CO_2} \) (NCERT, p. 291).
\[ \%\mathrm{O} = \frac{32 \times m_1 \times 100}{88 \times m}\quad \text{(direct method)} \]
\[ \%\mathrm{O} = 100 – \left(\%\mathrm{C} + \%\mathrm{H} + \%\mathrm{N} + \%\mathrm{X} + \%\mathrm{S} + \%\mathrm{P}\right)\quad \text{(by difference)} \]
What Each Symbol Means
One warning before you substitute: the textbook reuses the symbol \( V \) for two different volumes. In Dumas method \( V \) is the STP volume of nitrogen; in Kjeldahl’s method \( V \) is the volume of \( \mathrm{H_2SO_4} \) taken. Always read which method the question uses.
| Symbol | What it means | Unit / nature |
|---|---|---|
| \( m \) | mass of the organic compound taken | \( \text{g} \) |
| \( m_1 \) | mass of water formed (C/H estimation); mass of \( \mathrm{AgX} \), \( \mathrm{BaSO_4} \), ammonium phosphomolybdate or \( \mathrm{Mg_2P_2O_7} \) (element estimations); mass of \( \mathrm{CO_2} \) (direct oxygen method) | \( \text{g} \) |
| \( m_2 \) | mass of \( \mathrm{CO_2} \) formed (C/H estimation) | \( \text{g} \) |
| \( V_1 \) | measured volume of nitrogen gas collected (Dumas); volume of NaOH used in back-titration (Kjeldahl) | \( \text{mL} \) |
| \( V \) | volume of nitrogen corrected to STP (Dumas); volume of \( \mathrm{H_2SO_4} \) taken (Kjeldahl) | \( \text{mL} \) |
| \( P_1 \) | pressure of the nitrogen gas = atmospheric pressure − aqueous tension | \( \text{mm Hg} \) |
| \( T_1 \) | temperature at which the nitrogen gas is collected | \( \text{K} \) |
| \( M \) | molarity of the \( \mathrm{H_2SO_4} \) (and the NaOH used to titrate the excess) | \( \text{mol L}^{-1} \) |
| \( x \) | distance moved by the substance spot from the base line | \( \text{cm} \) |
| \( y \) | distance moved by the solvent front from the base line | \( \text{cm} \) |
| \( R_f \) | retardation factor | dimensionless |
| \( p \) | atmospheric pressure (steam distillation) | \( \text{mm Hg} \) |
| \( p_1 \) | vapour pressure of the organic liquid (steam distillation) | \( \text{mm Hg} \) |
| \( p_2 \) | vapour pressure of water (steam distillation) | \( \text{mm Hg} \) |
| \( \sigma, \ \pi \) | sigma bond and pi bond | bond type (count) |
Constants locked into the formulas:
| Constant | Value | Used in |
|---|---|---|
| \( M_{\mathrm{C}} : M_{\mathrm{CO_2}} \) | 12 : 44 u | \( \%\mathrm{C} \) |
| \( 2M_{\mathrm{H}} : M_{\mathrm{H_2O}} \) | 2 : 18 u | \( \%\mathrm{H} \) |
| \( M_{\mathrm{N_2}} : V_m \) (molar volume at STP) | 28 \( \text{g mol}^{-1} \) : 22400 \( \text{mL mol}^{-1} \) | Dumas \( \%\mathrm{N} \) |
| \( M_{\mathrm{S}} : M_{\mathrm{BaSO_4}} \) | 32 : 233 u | \( \%\mathrm{S} \) |
| \( M_{\mathrm{P}} : M_{(\mathrm{NH_4})_3\mathrm{PO_4} \cdot 12\mathrm{MoO_3}} \) | 31 : 1877 u | \( \%\mathrm{P} \) |
| \( 2M_{\mathrm{P}} : M_{\mathrm{Mg_2P_2O_7}} \) | 62 : 222 u | \( \%\mathrm{P} \) (as \( \mathrm{Mg_2P_2O_7} \)) |
| \( M_{\mathrm{O_2}} : 2M_{\mathrm{CO_2}} \) | 32 : 88 g | \( \%\mathrm{O} \) (direct) |
When to Use Each Formula
Reach for the right formula with this decision aid.
| Formula or method | Use it when… |
|---|---|
| \( \sigma \)/\( \pi \) counting rule | you are asked how many \( \sigma \) and \( \pi \) bonds a given structure has. |
| \( \mathrm{C}_n\mathrm{H}_{2n+2} \) | you need to check whether a formula fits the alkane family or predict the next homologue. |
| \( p = p_1 + p_2 \) | a water-immiscible, steam-volatile compound (e.g. aniline) is being separated by steam distillation. |
| \( R_f = x/y \) | you are comparing spots on a TLC plate or paper chromatogram to identify components. |
| \( \%\mathrm{C} \), \( \%\mathrm{H} \) | combustion gives the masses of \( \mathrm{CO_2} \) and \( \mathrm{H_2O} \) produced. |
| STP correction \( V \) | nitrogen in Dumas method is collected at room temperature and laboratory pressure — this step is always needed. |
| \( \%\mathrm{N} \) (Dumas) | a nitrogen gas volume is measured over KOH solution. |
| \( \%\mathrm{N} \) (Kjeldahl) | nitrogen is converted to ammonia (amines, amides); do not use for nitro, azo or ring nitrogen. |
| \( \%\mathrm{X} \) (Carius) | the halogen is weighed as \( \mathrm{AgX} \). |
| \( \%\mathrm{S} \) | sulphur is weighed as \( \mathrm{BaSO_4} \). |
| \( \%\mathrm{P} \) | phosphorus is weighed as ammonium phosphomolybdate or \( \mathrm{Mg_2P_2O_7} \). |
| \( \%\mathrm{O} \) | oxygen is to be reported — by difference unless the direct method is specified. |
Worked Examples
The worked examples below use original numbers and show the formula being chosen, the substitution, and the answer with its unit. Practise the same steps on the textbook’s questions, and find other chapter sheets in the chemistry formulas index.
Worked Example 1: Percentage of C, H and O from combustion data
Given: 0.320 g of an organic compound containing only C, H and O gave 0.704 g of \( \mathrm{CO_2} \) and 0.288 g of \( \mathrm{H_2O} \) on complete combustion.
Step 1 — choose the carbon formula: \( \%\mathrm{C} = \frac{12 \times m_2 \times 100}{44 \times m} \), with \( m = 0.320\ \text{g} \) (compound) and \( m_2 = 0.704\ \text{g} \) (\( \mathrm{CO_2} \)).
\[ \%\mathrm{C} = \frac{12 \times 0.704 \times 100}{44 \times 0.320} = 60.0\% \]
Step 2 — choose the hydrogen formula: \( \%\mathrm{H} = \frac{2 \times m_1 \times 100}{18 \times m} \), with \( m_1 = 0.288\ \text{g} \) (\( \mathrm{H_2O} \)).
\[ \%\mathrm{H} = \frac{2 \times 0.288 \times 100}{18 \times 0.320} = 10.0\% \]
Step 3 — oxygen by difference: only C, H and O are present, so \( \%\mathrm{O} = 100 – (\%\mathrm{C} + \%\mathrm{H}) \).
\[ \%\mathrm{O} = 100 – (60.0 + 10.0) = 30.0\% \]
Final answer: C = 60.0%, H = 10.0%, O = 30.0% by mass.
Worked Example 2: Percentage of nitrogen by Dumas method
Given: 0.400 g of an organic compound gave 60 mL of nitrogen collected over water at 300 K and 750 mm pressure.
Aqueous tension at 300 K = 15 mm.
Step 1 — correct the pressure: the gas is wet, so \( P_1 = 750 – 15 = 735\ \text{mm} \).
Step 2 — correct the volume to STP: \( V = \frac{P_1 V_1 \times 273}{760 \times T_1} \).
\[ V = \frac{735 \times 60 \times 273}{760 \times 300} = 52.8\ \text{mL} \]
Step 3 — apply the Dumas formula: \( \%\mathrm{N} = \frac{28 \times V \times 100}{22400 \times m} \), with \( m = 0.400\ \text{g} \) and \( V = 52.8\ \text{mL} \).
\[ \%\mathrm{N} = \frac{28 \times 52.8 \times 100}{22400 \times 0.400} = 16.5\% \]
Final answer: the compound contains 16.5% nitrogen by mass.
Worked Example 3: Percentage of nitrogen by Kjeldahl’s method
Given: 0.700 g of an amine gave ammonia which was absorbed in 20 mL of 1 M \( \mathrm{H_2SO_4} \).
The unused acid required 10 mL of 1 M NaOH for neutralisation.
Step 1 — find the unused acid: 10 mL of 1 M NaOH neutralises only 5 mL of 1 M \( \mathrm{H_2SO_4} \) (2 mol NaOH \( \equiv \) 1 mol \( \mathrm{H_2SO_4} \)), so unused acid = 20 − 5 = 15 mL.
Step 2 — apply the Kjeldahl formula: \( \%\mathrm{N} = \frac{1.4 \times M \times 2\left(V – \frac{V_1}{2}\right)}{m} \), with \( M = 1\ \text{mol L}^{-1} \), \( V = 20\ \text{mL} \), \( V_1 = 10\ \text{mL} \), \( m = 0.700\ \text{g} \).
\[ \%\mathrm{N} = \frac{1.4 \times 1 \times 2\left(20 – \frac{10}{2}\right)}{0.700} = \frac{1.4 \times 30}{0.700} = 60.0\% \]
Step 3 — check with the 14 g rule: the ammonia is equivalent to 2 × 15 = 30 mL of 1 M \( \mathrm{NH_3} \), which contains \( 14 \times 30/1000 = 0.42\ \text{g N} \); \( (0.42/0.700) \times 100 = 60\% \).
Final answer: the amine contains 60.0% nitrogen by mass.
Common Mistakes to Avoid
These are the errors students actually make when substituting into this chapter’s formulas.
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| In Dumas method, using the collected pressure of nitrogen (750 mm) without subtracting aqueous tension. | When the gas is collected over water, always use \( P_1 = P_{\text{atm}} – \text{aqueous tension} \). | Your corrected pressure must be smaller than the atmospheric pressure. |
| Plugging the measured volume \( V_1 \) into the \( \%\mathrm{N} \) formula without converting to STP. | Convert first: \( V = \frac{P_1 V_1 \times 273}{760 \times T_1} \). The 22400 mL molar volume is valid only at STP. | The volume you substitute into \( \%\mathrm{N} \) must be the STP volume, not the lab volume. |
| In Kjeldahl’s method, treating \( V_1 \) mL of NaOH as the volume of unused acid directly. | One mole of \( \mathrm{H_2SO_4} \) needs two moles of NaOH, so \( V_1 \) mL of M NaOH = \( V_1/2 \) mL of M \( \mathrm{H_2SO_4} \). | Unused acid = \( V – V_1/2 \); ammonia-equivalent volume = \( 2(V – V_1/2) \). |
| Swapping \( m_1 \) and \( m_2 \) in the C and H estimation. | \( m_2 \) belongs to \( \mathrm{CO_2} \) and goes into \( \%\mathrm{C} \); \( m_1 \) belongs to \( \mathrm{H_2O} \) and goes into \( \%\mathrm{H} \). | Check that \( m_{\mathrm{C}} = (12/44) \times m_2 \) and \( m_{\mathrm{H}} = (2/18) \times m_1 \) are each smaller than the sample mass \( m \). |
| In Carius estimation, writing the molecular mass of the halogen (e.g. \( \mathrm{Br_2} = 160 \)) or forgetting the sample mass in the denominator. | The denominator is the molar mass of \( \mathrm{AgX} \) (e.g. \( \mathrm{AgBr} = 188 \) u) × the sample mass \( m \). | Mass of halogen in the precipitate = \( \frac{\text{atomic mass of } X}{M_{\mathrm{AgX}}} \times m_1 \), and it must be less than \( m_1 \). |
| Applying Kjeldahl’s method to a nitro or azo compound. | Kjeldahl’s method works only when nitrogen forms ammonium sulphate; nitro, azo and ring nitrogen (e.g. pyridine) do not — use Dumas method. | If the compound contains \( \mathrm{-NO_2} \) or \( \mathrm{-N=N-} \), do not trust a Kjeldahl percentage. |
Frequently Asked Questions
Which nitrogen method should I use — Dumas or Kjeldahl?
Dumas method works for any nitrogen-containing compound and measures the nitrogen gas volume. Kjeldahl’s method is quicker but only for compounds whose nitrogen converts to ammonium sulphate — amines and amides. For nitro compounds, azo compounds and ring nitrogen (pyridine), use Dumas method (NCERT, p. 289).
Why must the nitrogen volume be corrected to STP in Dumas method?
Because the 22400 mL molar volume applies only at STP (273 K and 760 mm). The gas collected in the laboratory is at room temperature, lab pressure and saturated with water vapour, so it must be converted with \( V = \frac{P_1 V_1 \times 273}{760 \times T_1} \) before the \( 28 \times V \times 100/(22400 \times m) \) formula is used (NCERT, pp. 287-288).
Why does the nitrogen test fail when sulphur is also present in the compound?
If nitrogen and sulphur are both present, fusion gives sodium thiocyanate \( (\mathrm{NaSCN}) \) instead of \( \mathrm{NaCN} \). The extract then gives a blood-red colour with \( \mathrm{Fe^{3+}} \) and no Prussian blue. Fusing with excess sodium decomposes \( \mathrm{NaSCN} \) to \( \mathrm{NaCN} \) and \( \mathrm{Na_2S} \), which give the usual tests (NCERT, p. 285).
Why is oxygen usually estimated by difference?
Direct oxygen estimation needs a special train: oxygen from the decomposed compound is turned into \( \mathrm{CO} \) over red-hot coke, then into \( \mathrm{CO_2} \) with iodine pentoxide. The routine route is simpler: subtract the sum of all other element percentages from 100. Use the direct formula only when the question specifies it (NCERT, p. 291).
Reference: NCERT Class 11 Chemistry textbook, chapter Organic Chemistry – Some Basic Principles and Techniques.
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