Kjeldahl's Method: Percentage of Nitrogen Calculations
Quantitative analysis of organic compounds is a vital topic in chemistry examinations. Among analytical procedures, Kjeldahl's method is the standard laboratory technique for determining nitrogen percentage in organic samples, fertilizers, and food products. In competitive examinations, numerical questions on Kjeldahl's method appear regularly, testing acid-base back-titration concepts, chemical equivalence, and volumetric calculations.
Principle of Kjeldahl's Method
Kjeldahl's method consists of three successive laboratory steps:
- Digestion: The organic sample is heated with concentrated H2SO4 in a Kjeldahl flask. K2SO4 raises the boiling point, and CuSO4 acts as a catalyst. Organic nitrogen converts into ammonium sulfate:
Organic Nitrogen + conc. H2SO4 → (NH4)2SO4 - Distillation: The digested mixture is heated with excess NaOH to liberate ammonia gas:
(NH4)2SO4 + 2 NaOH → Na2SO4 + 2 NH3 + 2 H2O - Absorption and Titration: Evolved ammonia is absorbed into a known excess volume of standard acid:
2 NH3 + H2SO4 → (NH4)2SO4
Unreacted residual acid is back-titrated against standard NaOH using methyl orange indicator.
Derivation of the Nitrogen Percentage Formula
Let the organic sample mass be W grams. Evolved ammonia is absorbed into V1 mL of standard acid of normality N1. Unreacted acid requires V2 mL of standard NaOH of normality N2 for neutralisation.
- Initial acid milliequivalents = N1 × V1
- Unreacted acid milliequivalents = N2 × V2
- Acid milliequivalents reacting with ammonia = (N1 × V1) - (N2 × V2)
Because one equivalent of ammonia neutralises one equivalent of acid and contains one mole of nitrogen. By IUPAC standards, nitrogen has atomic mass 14.007 g/mol (commonly 14.0 g/mol). Therefore, 1 milliequivalent of acid consumed equals 0.014 g of nitrogen. Dividing by sample mass W and multiplying by 100 gives the master formula:
Master Formula for Kjeldahl's Nitrogen Estimation:
% N = [1.4 × (N1 × V1 - N2 × V2)] / W
When volume V mL of standard acid of normality N neutralises ammonia directly:
% N = [1.4 × N × V] / W
Where:
- % N = mass percentage of nitrogen (% w/w).
- N1, V1 = normality (N) and volume (mL) of initial standard acid.
- N2, V2 = normality (N) and volume (mL) of standard NaOH in back-titration.
- W = mass of organic compound taken (g).
- 1.4 = constant equal to (14 g/mol × 100) / 1000 mL.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: JEE Main, NEET, CUET-UG, State Board Examinations
- Topic: Quantitative Organic Analysis (Kjeldahl's Method)
- Difficulty: Medium
Step-by-Step Worked Examples
Question: In an estimation of nitrogen by Kjeldahl's method, 0.500 g of an organic compound was digested. The evolved ammonia was absorbed into 50.0 mL of 0.100 N H2SO4. The unreacted acid required 30.0 mL of 0.100 N NaOH for neutralisation. Calculate the percentage of nitrogen in the compound.
Solution (Route 1 - Equivalence Method):
Step 1: Identify values: W = 0.500 g, N1 = 0.100 N, V1 = 50.0 mL, N2 = 0.100 N, V2 = 30.0 mL.
Step 2: Initial acid meq = 0.100 × 50.0 = 5.00 meq.
Step 3: Unreacted acid meq = 0.100 × 30.0 = 3.00 meq.
Step 4: Acid meq consumed by ammonia = 5.00 - 3.00 = 2.00 meq.
Step 5: Apply formula: % N = [1.4 × 2.00] / 0.500 = 2.80 / 0.500 = 5.60%.
Verification (Route 2 - Mole and Mass Method):
2.00 meq acid = 2.00 mmol N = 0.00200 mol × 14.0 g/mol = 0.0280 g N.
% N = (0.0280 g / 0.500 g) × 100% = 5.60%.
Both routes give 5.60%.
Answer: 5.60%
Question: A 0.280 g sample of an organic compound was analysed by Kjeldahl's method. Ammonia evolved was trapped in 20.0 mL of 0.100 M H2SO4. The excess acid required 15.0 mL of 0.100 M NaOH for neutralisation. Determine the percentage of nitrogen in the sample.
Solution (Route 1 - Conversion to Normality):
Step 1: Convert molarity to normality. H2SO4 has n = 2: N1 = 0.100 M × 2 = 0.200 N. NaOH has n = 1: N2 = 0.100 M × 1 = 0.100 N.
Step 2: Initial acid meq = 0.200 N × 20.0 mL = 4.00 meq.
Step 3: Unreacted acid meq = 0.100 N × 15.0 mL = 1.50 meq.
Step 4: Net meq consumed = 4.00 - 1.50 = 2.50 meq.
Step 5: Compute percentage: % N = (1.4 × 2.50) / 0.280 = 3.50 / 0.280 = 12.50%.
Verification (Route 2 - Millimole Balance):
Initial mmol H+ = 20.0 mL × 0.100 M × 2 = 4.00 mmol. Unreacted mmol OH- = 15.0 mL × 0.100 M = 1.50 mmol.
H+ consumed by NH3 = 4.00 - 1.50 = 2.50 mmol N.
Mass of N = 0.00250 mol × 14.0 g/mol = 0.0350 g.
% N = (0.0350 g / 0.280 g) × 100% = 12.50%.
Both methods give 12.50%.
Answer: 12.50%
Question: In a Kjeldahl determination, 0.400 g of an organic substance released ammonia that was absorbed into excess acid. The distillate required 25.0 mL of 0.200 N HCl for complete neutralisation. Calculate the percentage of nitrogen.
Solution (Route 1 - Direct Equation):
Step 1: Identify values: W = 0.400 g, N = 0.200 N, V = 25.0 mL.
Step 2: Apply formula: % N = (1.4 × N × V) / W = (1.4 × 0.200 × 25.0) / 0.400 = 7.00 / 0.400 = 17.50%.
Verification (Route 2 - Stoichiometric Mass Route):
Equivalents of HCl = 0.200 N × 0.0250 L = 0.00500 eq = 0.00500 mol HCl.
1 mol HCl neutralises 1 mol NH3 (1 mol N): Mass of N = 0.00500 mol × 14.0 g/mol = 0.0700 g.
% N = (0.0700 g / 0.400 g) × 100% = 17.50%.
Both routes give 17.50%.
Answer: 17.50%
Common Mistakes Students Make
- Confusing Molarity with Normality: H2SO4 has n = 2. Forgetting to multiply molarity by 2 incorrectly halves the nitrogen percentage.
- Adding instead of subtracting in back-titration: Base milliequivalents represent unreacted acid and must be subtracted from initial acid milliequivalents.
- Applying Kjeldahl's method to nitro and azo compounds: Kjeldahl's method fails for nitro (-NO2), azo (-N=N-), and pyridine ring nitrogen because their nitrogen is not converted into ammonium sulfate.
- Entering sample mass in milligrams: The factor 1.4 requires sample mass in grams. A 500 mg sample must be converted to 0.500 g.
Exam Relevance and Question Patterns
| Examination | Question Pattern | Preparation Strategy |
|---|---|---|
| School Board Examinations | Theory questions on digestion and distillation reactions, plus standard back-titration numericals. | Write out balanced chemical equations for ammonium sulfate formation and ammonia release. |
| JEE Main & Advanced | Multi-step numericals with dilution of digest, aliquot sampling, or mixing molarity and normality. | If a diluted digest is sampled, multiply by the dilution factor before calculating % N. |
| NEET | Direct formula numericals and questions on which compounds fail the Kjeldahl test. | Remember that pyridine, nitrobenzene, and azobenzene give negative results with Kjeldahl's method. |
| CUET-UG | Objective questions on percentage calculation, acid-base equivalence, and indicator selection. | Check whether the absorbing acid is monoprotic (HCl) or diprotic (H2SO4) before calculating equivalents. |
Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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