Chemistry

Qualitative and Quantitative Analysis

Chemistry·Revision Notes

Estimation of Carbon, Hydrogen, Nitrogen, Sulphur, Phosphorus — Revision Notes

NEET UG
Version 1Updated 22 Mar 2026

⚡ 30-Second Revision

  • Carbon & Hydrogen (Liebig's):Burn compound \rightarrow CO2_2 (absorb in KOH), H2_2O (absorb in CaCl2_2). \n %C=1244×mass of CO2mass of compound×100\text{\%C} = \frac{12}{44} \times \frac{\text{mass of CO}_2}{\text{mass of compound}} \times 100 \n %H=218×mass of H2Omass of compound×100\text{\%H} = \frac{2}{18} \times \frac{\text{mass of H}_2\text{O}}{\text{mass of compound}} \times 100 \n- Nitrogen (Dumas): Burn compound with CuO \rightarrow N2_2 gas. Collect N2_2 over KOH. Correct volume to STP. \n VSTP=V×273T×PPaq760\text{V}_{STP} = V \times \frac{273}{T} \times \frac{P - P_{aq}}{760} \n %N=2822400×VSTPmass of compound×100\text{\%N} = \frac{28}{22400} \times \frac{\text{V}_{STP}}{\text{mass of compound}} \times 100 \n- Nitrogen (Kjeldahl): Digest with H2_2SO4_4 \rightarrow (NH4_4)2_2SO4_4. Distill NH3_3 into std. acid. Back titrate. \n %N=(Vacid×Nacid)(Vbase×Nbase)1000×mass of compound×14×100\text{\%N} = \frac{(\text{V}_{acid} \times \text{N}_{acid}) - (\text{V}_{base} \times \text{N}_{base})}{1000 \times \text{mass of compound}} \times 14 \times 100 \n *Limitations:* Not for nitro, azo, or ring N. \n- Sulphur (Carius): Heat with fuming HNO3_3 \rightarrow H2_2SO4_4. Precipitate as BaSO4_4. \n %S=32233×mass of BaSO4mass of compound×100\text{\%S} = \frac{32}{233} \times \frac{\text{mass of BaSO}_4}{\text{mass of compound}} \times 100 \n- Phosphorus (Carius): Heat with fuming HNO3_3 \rightarrow H3_3PO4_4. Precipitate as MgNH4_4PO4_4, ignite to Mg2_2P2_2O7_7. \n %P=62222×mass of Mg2P2O7mass of compound×100\text{\%P} = \frac{62}{222} \times \frac{\text{mass of Mg}_2\text{P}_2\text{O}_7}{\text{mass of compound}} \times 100

2-Minute Revision

Quantitative estimation of elements is crucial for determining the composition of organic compounds. For carbon and hydrogen, Liebig's method involves complete combustion of a known mass of the compound to form CO2_2 and H2_2O.

These products are absorbed by specific reagents (anhydrous CaCl2_2 for H2_2O, KOH for CO2_2), and their increased masses are used to calculate the percentages of C and H. \n\nNitrogen can be estimated by two main methods: Dumas and Kjeldahl.

The Dumas method is universal; it converts all nitrogen to N2_2 gas, whose volume is measured and corrected to STP for calculation. Kjeldahl's method is for compounds where nitrogen can be converted to ammonium sulphate; it involves digestion, distillation of ammonia, and titration.

Remember, Kjeldahl's is not for nitro, azo, or ring nitrogen compounds. \n\nSulphur and phosphorus are typically estimated by the Carius method. The compound is heated with fuming nitric acid in a sealed tube.

Sulphur is oxidized to H2_2SO4_4, which is then precipitated and weighed as BaSO4_4. Phosphorus is oxidized to H3_3PO4_4, which is eventually weighed as magnesium pyrophosphate (Mg2_2P2_2O7_7).

Mastering the specific formulas and their stoichiometric factors for each element is key for NEET.

5-Minute Revision

For NEET, a solid grasp of elemental estimation methods is non-negotiable. Start with Carbon and Hydrogen (Liebig's Method). The principle is combustion: C \rightarrow CO2_2, H \rightarrow H2_2O.

Remember the absorbents: anhydrous CaCl2_2 for H2_2O and KOH for CO2_2. The key formulas are: \n%C=1244×mass of CO2mass of compound×100\text{\%C} = \frac{12}{44} \times \frac{\text{mass of CO}_2}{\text{mass of compound}} \times 100 and %H=218×mass of H2Omass of compound×100\text{\%H} = \frac{2}{18} \times \frac{\text{mass of H}_2\text{O}}{\text{mass of compound}} \times 100.

\n\nNext, Nitrogen estimation has two methods. Dumas Method is universal. Nitrogen is converted to N2_2 gas, collected over KOH (which absorbs CO2_2), and its volume is measured. Crucially, convert this volume to STP using VSTP=V×273T×PPaq760V_{STP} = V \times \frac{273}{T} \times \frac{P - P_{aq}}{760}.

Then, %N=2822400×VSTPmass of compound×100\text{\%N} = \frac{28}{22400} \times \frac{\text{V}_{STP}}{\text{mass of compound}} \times 100. \nKjeldahl's Method is for specific nitrogen compounds (not nitro, azo, or ring N). It involves digestion with conc.

H2_2SO4_4 to form (NH4_4)2_2SO4_4, followed by distillation of NH3_3 into a known excess of standard acid, and then back-titration. The formula is %N=(Vacid×Nacid)(Vbase×Nbase)1000×mass of compound×14×100\text{\%N} = \frac{(\text{V}_{acid} \times \text{N}_{acid}) - (\text{V}_{base} \times \text{N}_{base})}{1000 \times \text{mass of compound}} \times 14 \times 100.

Pay attention to milliequivalents. \n\nFor Sulphur and Phosphorus (Carius Method), the compound is heated with fuming HNO3_3. For sulphur, it forms H2_2SO4_4, which is precipitated as BaSO4_4.

\n%S=32233×mass of BaSO4mass of compound×100\text{\%S} = \frac{32}{233} \times \frac{\text{mass of BaSO}_4}{\text{mass of compound}} \times 100. \nFor phosphorus, it forms H3_3PO4_4, which is eventually weighed as Mg2_2P2_2O7_7. \n%P=62222×mass of Mg2P2O7mass of compound×100\text{\%P} = \frac{62}{222} \times \frac{\text{mass of Mg}_2\text{P}_2\text{O}_7}{\text{mass of compound}} \times 100.

\n\nWorked Example (Kjeldahl): 0.2 g compound, NH3_3 absorbed in 20 mL 0.1 N H2_2SO4_4. Excess acid needed 10 mL 0.1 N NaOH. \nMilliequivalents of H2_2SO4_4 reacted with NH3_3 = $(20 \times 0.

1) - (10 \times 0.1) = 2 - 1 = 1.\nMassofN=. \nMass of N =1 \times 14, ext{mg} = 0.014, ext{g}.\n. \n\text{\%N} = \frac{0.014}{0.

Prelims Revision Notes

Elemental Estimation: Key Facts for NEET\n\n1. Carbon and Hydrogen (Liebig's Method):\n* Principle: Complete combustion of organic compound in excess O$_2$ over CuO.\n* Products: Carbon $\rightarrow$ CO$_2$, Hydrogen $\rightarrow$ H$_2$O.\n* Absorption: H$_2$O absorbed by anhydrous CaCl$_2$ (or Mg(ClO$_4$)$_2$). CO$_2$ absorbed by concentrated KOH solution.\n* Order of Absorption: H$_2$O first, then CO$_2$. (KOH absorbs both, CaCl$_2$ only H$_2$O).\n* Formulas:\n * $\text{\%C} = \frac{12}{44} \times \frac{\text{Mass of CO}_2}{\text{Mass of organic compound}} \times 100$\%\n * $\text{\%H} = \frac{2}{18} \times \frac{\text{Mass of H}_2\text{O}}{\text{Mass of organic compound}} \times 100$\%\n\n2. Nitrogen Estimation:\n A. Dumas Method:\n * Principle: Organic compound heated with CuO in CO$_2$ atmosphere. Nitrogen $\rightarrow$ N$_2$ gas.\n * Reduction: Oxides of nitrogen (if formed) reduced to N$_2$ by hot copper gauze.\n * Collection: N$_2$ collected over aqueous KOH solution (absorbs CO$_2$).\n * Volume Correction: Convert observed N$_2$ volume to STP using: \n $\text{V}_{STP} = V \times \frac{273}{T} \times \frac{P - P_{aq}}{760}$ (T in K, P in mmHg)\n * Formula: $\text{\%N} = \frac{28}{22400} \times \frac{\text{V}_{STP}}{\text{Mass of organic compound}} \times 100$\%\n * Applicability: Universal for all N-containing organic compounds.\n\n B. Kjeldahl's Method:\n * Principle: Nitrogen $\rightarrow$ (NH$_4$)$_2$SO$_4$ (digestion with conc. H$_2$SO$_4$ + catalyst). (NH$_4$)$_2$SO$_4$ $\rightarrow$ NH$_3$ (with NaOH). NH$_3$ absorbed in excess standard acid, then back-titrated.\n * Catalysts: CuSO$_4$ (accelerates reaction), K$_2$SO$_4$ (raises boiling point of H$_2$SO$_4$).\n * Limitations: NOT for compounds with N in nitro ($-$NO$_2$), azo ($-$N=N$-$), or heterocyclic rings (e.g., pyridine, quinoline) as they don't convert to (NH$_4$)$_2$SO$_4$ quantitatively.\n * Formula: $\text{\%N} = \frac{(\text{V}_{acid} \times \text{N}_{acid}) - (\text{V}_{base} \times \text{N}_{base})}{1000 \times \text{Mass of organic compound}} \times 14 \times 100$\%\n (Where 14 is atomic mass of N, 1000 converts mg to g)\n\n3. Sulphur (Carius Method):\n* Principle: Organic compound heated with fuming HNO$_3$ in sealed tube. Sulphur $\rightarrow$ H$_2$SO$_4$.\n* Precipitation: H$_2$SO$_4$ precipitated as BaSO$_4$ by adding BaCl$_2$.\n* Formula: $\text{\%S} = \frac{32}{233} \times \frac{\text{Mass of BaSO}_4}{\text{Mass of organic compound}} \times 100$\%\n (Atomic mass S = 32, Molecular mass BaSO$_4$ = 233)\n\n4. Phosphorus (Carius Method):\n* Principle: Organic compound heated with fuming HNO$_3$ in sealed tube. Phosphorus $\rightarrow$ H$_3$PO$_4$.\n* Precipitation: H$_3$PO$_4$ precipitated as MgNH$_4$PO$_4$ (using magnesia mixture), then ignited to Mg$_2$P$_2$O$_7$.\n* Formula: $\text{\%P} = \frac{62}{222} \times \frac{\text{Mass of Mg}_2\text{P}_2\text{O}_7}{\text{Mass of organic compound}} \times 100$\%\n (Mass of 2P = 62, Molecular mass Mg$_2$P$_2$O$_7$ = 222)\n\nGeneral Tips: Pay attention to atomic/molecular masses, unit conversions (mL to L, mg to g), and significant figures. Practice numerical problems extensively.

Vyyuha Quick Recall

To remember the elements and their estimation methods, think: 'C H N S P - L D K C C'\n\n* C H: Liebig's (Combustion)\n* N: Dumas (Gasometric), Kjeldahl (Titrimetric)\n* S: Carius (BaSO4_4 ppt)\n* P: Carius (Mg2_2P2_2O7_7 ppt)\n\nThis helps recall the primary method for each element. For Kjeldahl's limitation, remember: 'No Nitro, No Azo, No Ring N for Kjeldahl!'

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