Dumas Method: Percentage of Nitrogen Numericals
Quantitative elemental analysis is an essential topic in Class 11 organic chemistry and competitive entrance examinations such as JEE Main, NEET, and CUET-UG. Among the techniques used to determine nitrogen in organic compounds, the Dumas method is the most reliable and universal. Unlike Kjeldahl's method, which fails for nitro compounds, azo dyes, and nitrogen-containing heterocyclic rings like pyridine, the Dumas method works accurately for almost all organic substances. Mastering its numerical steps helps students secure full marks in school tests and competitive exams.
Principle of the Dumas Method
In the Dumas method, a weighed sample of the organic compound is mixed with dry copper(II) oxide (cupric oxide, CuO) and heated strongly in a combustion tube in an atmosphere of pure carbon dioxide gas. Carbon oxidises to carbon dioxide, hydrogen oxidises to water vapour, and nitrogen is liberated as dinitrogen gas (N₂) along with traces of nitrogen oxides.
To convert all nitrogen oxides completely into dinitrogen, the gas stream passes through a heated copper gauze. The resulting gases are carried by carbon dioxide into a graduated nitrometer containing concentrated potassium hydroxide (KOH) solution. The potassium hydroxide absorbs all carbon dioxide gas by forming potassium carbonate. Water vapour condenses into liquid. Pure, insoluble nitrogen gas collects in the upper graduated portion of the nitrometer above the liquid surface.
Pressure Correction and Molar Volume at STP
The volume of nitrogen is recorded at room temperature and laboratory atmospheric pressure. Because nitrogen is collected over an aqueous potassium hydroxide solution, it is moist and contains water vapour. The observed atmospheric pressure equals the partial pressure of dry nitrogen plus the aqueous tension (water vapour pressure) at that temperature.
To find the pressure of dry nitrogen, subtract aqueous tension from the total barometric pressure:
P₁ = P_observed - Aqueous Tension (p)
The dry volume is then reduced to standard temperature and pressure (STP) using the combined gas law:
(P₁ × V₁) / T₁ = (P₀ × V₀) / T₀
Two standards appear in chemistry textbooks:
- Traditional NCERT / School standard: T₀ = 273 K, P₀ = 760 mm Hg (1 atm). One mole of gas occupies 22.4 L (22400 mL) at STP.
- Modern IUPAC standard: T₀ = 273.15 K, P₀ = 1 bar ≈ 750.06 mm Hg. One mole of gas occupies 22.7 L (22700 mL) at STP.
Most Indian board and entrance examinations follow the 22.4 L (22400 mL at 760 mm Hg) convention. Both values are provided here for completeness.
Formulas for Dumas Method:
Dry nitrogen pressure: P₁ = P_barometric - p_aqueous_tension (in mm Hg)
Volume of dry N₂ at STP: V₀ = (P₁ × V₁ × 273) / (760 × T₁) (where V₁ is in mL, T₁ in K)
Percentage of nitrogen (% N) using traditional STP (22400 mL/mol):
% N = [ (28 × V₀) / (22400 × m) ] × 100 = V₀ / (8 × m)
(where m is the mass of the organic compound in grams, and V₀ is in mL).
Percentage of nitrogen (% N) using modern IUPAC STP (22700 mL/mol):
% N = [ (28.014 × V₀_bar) / (22700 × m) ] × 100
Worked Numerical Examples
Question: In a Dumas estimation of nitrogen, 0.250 g of an organic compound yielded 32.5 mL of moist nitrogen gas at 300 K and 755 mm Hg. The aqueous tension at 300 K is 15 mm Hg. Calculate the percentage of nitrogen in the compound using standard molar volume 22400 mL at 760 mm Hg and 273 K.
Step 1: Find the pressure of dry nitrogen gas.
P₁ = 755 mm Hg - 15 mm Hg = 740 mm Hg
Step 2: Convert nitrogen volume to STP.
V₁ = 32.5 mL, T₁ = 300 K, P₀ = 760 mm Hg, T₀ = 273 K
V₀ = (740 × 32.5 × 273) / (760 × 300)
V₀ = 6,565,650 / 228,000 = 28.80 mL
Step 3: Calculate the percentage of nitrogen.
Mass of N₂ = (28 × 28.80) / 22400 = 0.0360 g
% N = (0.0360 g / 0.250 g) × 100 = 14.40%
Shortcut formula: % N = V₀ / (8 × m) = 28.80 / (8 × 0.250) = 14.40%
Route 2 verification (Ideal gas equation n = PV/RT):
P = 740/760 = 0.9737 atm, V = 0.0325 L, R = 0.0821 L atm/(mol K), T = 300 K
n = (0.9737 × 0.0325) / (0.0821 × 300) = 0.001285 mol
Mass of N = 0.001285 × 28.01 = 0.03600 g
% N = (0.03600 / 0.250) × 100 = 14.40%
Answer: 14.40%
Question: An organic substance weighing 0.184 g was analysed by Dumas method. The volume of nitrogen collected over aqueous KOH at 288 K and 748 mm Hg was 30.0 mL. Aqueous tension at 288 K is 12 mm Hg. Find the percentage of nitrogen in the substance using molar volume 22400 mL at 760 mm Hg and 273 K.
Step 1: Calculate dry gas pressure.
P₁ = 748 mm Hg - 12 mm Hg = 736 mm Hg
Step 2: Calculate volume at STP.
V₁ = 30.0 mL, T₁ = 288 K
V₀ = (736 × 30.0 × 273) / (760 × 288)
V₀ = 6,027,840 / 218,880 = 27.54 mL
Step 3: Calculate percentage of nitrogen.
% N = V₀ / (8 × m) = 27.54 / (8 × 0.184) = 27.54 / 1.472 = 18.71%
Route 2 verification:
n = (736/760 × 0.0300) / (0.0821 × 288) = 0.02905 / 23.645 = 0.001229 mol
Mass of N = 0.001229 × 28.01 = 0.03442 g
% N = (0.03442 / 0.184) × 100 = 18.71%
Answer: 18.71%
Common Mistakes Students Make
- Ignoring aqueous tension: Using total observed pressure rather than subtracting water vapour pressure overestimates dry nitrogen volume and produces an incorrectly high percentage.
- Using temperature in Celsius: The gas equation requires absolute temperature in Kelvin. Using degrees Celsius directly ruins the calculation.
- Confusing N and N₂ molar mass: Nitrogen gas is diatomic N₂ (28.01 g/mol). When calculating moles from 22400 mL, 1 mole equals 28 g of nitrogen.
- Units inconsistency: Mixing volume in litres with 22400 mL without converting creates large power-of-ten errors. Always match millilitres to 22400 mL.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CBSE Class 11, JEE Main, NEET, State Engineering and Medical Entrances, CUET-UG
- Topic: Quantitative Elemental Analysis of Organic Compounds (Dumas Method)
- Difficulty: Medium
Exam Relevance and Testing Patterns
The Dumas method is tested in theoretical and numerical formats across school boards and competitive entrance exams.
| Exam | Question Style | Core Concepts Tested |
|---|---|---|
| CBSE Class 11 & State Boards | Structured descriptive problems and apparatus questions | Chemical reactions with CuO, nitrometer role of KOH, aqueous tension deduction, and step-by-step STP conversions |
| JEE Main | Numerical value questions and multiple-choice items | Multi-step gas law reduction, high numerical precision, and distinction between Dumas and Kjeldahl applicability |
| NEET | Direct calculation MCQs | Rapid application of % N = V₀ / (8m), identifying nitrogen-containing functional groups suitable for Dumas analysis |
| CUET-UG | Conceptual and calculation questions | Principles of combustion analysis, gas laws, and molar volume relationships |
Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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