Qualitative Detection of Nitrogen, Sulphur, Phosphorus, and Halogens

In organic chemistry, identifying the presence of specific elements like Nitrogen (N), Sulphur (S), Phosphorus (P), and Halogens (X) within a compound is a crucial step in determining its structure and properties. This process is known as qualitative analysis. These tests are typically performed on the fused mass obtained from heating the organic compound with sodium metal. This process, called the Lassaigne's test or sodium fusion test, converts the covalent compounds of these elements into their corresponding ionic sodium salts, which are soluble in water and can then be tested.

Sodium Fusion Test (Lassaigne's Test)

Princ: Organic compounds containing N, S, P, and halogens are heated strongly with sodium metal in a fusion tube. The elements present get converted into ionic sodium salts (e.g., NaCN, Na₂S, Na₃PO₄, NaX). These salts are then extracted with distilled water to form the sodium extract (also known as the 'Lassaigne's extract' or 'boiling extract'). This aqueous extract is then used for specific tests.

Procedure for Sodium Extract Preparation:

• Take a small piece of clean sodium metal and heat it gently in a clean fusion tube until it melts.
• Add a small amount of the organic compound to the molten sodium. Heat the mixture strongly.
• Once the heating is complete, plunge the hot fusion tube into about 10-15 mL of distilled water in a porcelain dish. The tube will break, and the molten mass will dissolve in water.
• Boil the mixture for a few minutes and then filter it while hot. The clear filtrate is the sodium extract.
• If the extract is cloudy, it might contain unreacted sodium or carbon. Boil it again and filter.

Detection of Nitrogen

Test: Prussian Blue Test

Procedure: Take about 2 mL of the sodium extract in a test tube. Add a few drops of ferrous sulphate (FeSO₄) solution and boil. Acidify the solution with a few drops of concentrated sulphuric acid (H₂SO₄). A blue or greenish blue precipitate indicates the presence of nitrogen.

Chemical Reaction:

If nitrogen is present, sodium cyanide (NaCN) is formed in the sodium extract.

FeSO₄ + 2NaCN → Fe(CN)₂ + Na₂SO₄

The ferrous cyanide (Fe(CN)₂) reacts with excess NaCN to form sodium ferrocyanide (Na₄[Fe(CN)₆]).

Fe(CN)₂ + 4NaCN → Na₄[Fe(CN)₆]

When acidified with H₂SO₄, sodium ferrocyanide reacts with ferric ions (Fe³⁺), which are usually present as impurity in FeSO₄ or formed by oxidation of Fe²⁺, to form ferric ferrocyanide, commonly known as Prussian blue.

3Na₄[Fe(CN)₆] + 4Fe³⁺ → Fe₄[Fe(CN)₆]₃ (Prussian blue) + 12Na⁺

If the precipitate is greenish, it indicates the formation of ferrous ferrocyanide (Fe₂[Fe(CN)₆]).

Detection of Sulphur

Test 1: Sodium Nitroprusside Test

Procedure: Take about 2 mL of the sodium extract in a test tube. Add a few drops of sodium nitroprusside (Na₂[Fe(CN)₅NO]) solution. A deep violet or purple coloration indicates the presence of sulphur.

Chemical Reaction:

If sulphur is present, sodium sulphide (Na₂S) is formed in the sodium extract.

Na₂S + Na₂[Fe(CN)₅NO] → Na₄[Fe(CN)₅NOS] (Sodium nitrosyl-sulphur ferrocyanide)

This complex gives a deep violet colour.

Test 2: Lead Acetate Test

Procedure: Take about 2 mL of the sodium extract in a test tube. Add a few drops of lead acetate (Pb(CH₃COO)₂) solution. Heat gently. Formation of a black or dark brown precipitate of lead sulphide (PbS) indicates the presence of sulphur.

Chemical Reaction:

Na₂S + Pb(CH₃COO)₂ → PbS↓ + 2CH₃COONa

Detection of Nitrogen and Sulphur Together

If both nitrogen and sulphur are present in the organic compound, sodium thiocyanate (NaSCN) is formed during sodium fusion.

NaCN + S → NaSCN

Test: Ferric Chloride Test

Procedure: Take about 2 mL of the sodium extract in a test tube. Add a few drops of freshly prepared ferric chloride (FeCl₃) solution. The formation of a blood-red colour indicates the presence of both nitrogen and sulphur.

Chemical Reaction:

3NaSCN + FeCl₃ → Fe(SCN)₃ (Ferric thiocyanate) + 3NaCl

Ferric thiocyanate is blood-red in colour.

Detection of Halogens (Chlorine, Bromine, Iodine)

Test: Silver Nitrate Test

Procedure: Take about 2 mL of the sodium extract in a test tube. Acidify it with dilute nitric acid (HNO₃). This step is crucial to decompose any sodium cyanide (NaCN) or sodium sulphide (Na₂S) that might interfere with the test.

NaCN + HNO₃ → NaNO₃ + HCN

Na₂S + 2HNO₃ → 2NaNO₃ + H₂S

After acidification, add a few drops of silver nitrate (AgNO₃) solution. The formation of a precipitate of silver halide (AgX) indicates the presence of halogens.

Chemical Reaction:

NaX + AgNO₃ → AgX↓ + NaNO₃ (where X = Cl, Br, I)

The nature of the precipitate helps to identify the specific halogen:

• Silver Chloride (AgCl): White precipitate, soluble in ammonium hydroxide (NH₄OH).
• Silver Bromide (AgBr): Pale yellow precipitate, sparingly soluble in ammonium hydroxide.
• Silver Iodide (AgI): Yellow precipitate, insoluble in ammonium hydroxide.

Further Tests:

• For Chlorine: If a white precipitate is formed, add excess ammonium hydroxide. If it dissolves, chlorine is present.
• For Bromine: If a pale yellow precipitate is formed, add excess ammonium hydroxide. If it dissolves slightly, bromine is present.
• For Iodine: If a yellow precipitate is formed, add excess ammonium hydroxide. If it does not dissolve, iodine is present.

Distinction between AgCl and AgBr:

To distinguish between AgCl and AgBr, the precipitate is dissolved in dilute NH₄OH. Then, carbon disulphide (CS₂) is added, followed by a few drops of chlorine water. If AgCl is present, the CS₂ layer remains colourless. If AgBr is present, the CS₂ layer turns yellow.

Detection of Phosphorus

Test: Ammonium Molybdate Test

Procedure: Take about 2 mL of the sodium extract in a test tube. Acidify it with dilute nitric acid (HNO₃). Add ammonium molybdate ((NH₄)₂MoO₄) solution and heat gently. A yellow precipitate or yellow coloration indicates the presence of phosphorus.

Chemical Reaction:

If phosphorus is present, sodium phosphate (Na₃PO₄) is formed in the sodium extract.

Na₃PO₄ + 12(NH₄)₂MoO₄ + 21HNO₃ → (NH₄)₃PO₄·12MoO₃↓ (Ammonium phosphomolybdate) + 21NH₄NO₃ + 12H₂O

Ammonium phosphomolybdate is a yellow precipitate.

Quantitative Analysis of Organic Compounds

Quantitative analysis deals with the determination of the percentage composition of elements in an organic compound. This involves measuring the amount of each element present. Some common methods include:

Estimation of Carbon and Hydrogen (Liebig's Method)

Princ: A known mass of the organic compound is burnt completely in a stream of oxygen. Carbon dioxide (CO₂) and water (H₂O) are produced. These are absorbed in specific absorbents and their masses are determined.

Apparatus: A combustion tube containing the organic compound, heated strongly in a furnace. The products of combustion are passed through a series of tubes.

• The first set of tubes contains anhydrous calcium chloride (CaCl₂) or magnesium perchlorate (Mg(ClO₄)₂) to absorb the water formed.
• The second set of tubes contains potassium hydroxide (KOH) solution to absorb the carbon dioxide formed.

Procedure:

1. A known mass (say, 'm' grams) of the organic compound is taken.
2. The apparatus is set up with pre-weighed absorption tubes (U-tubes) for H₂O and CO₂. Let the initial weight of the H₂O absorption tube be W₁ and the final weight be W₂. Let the initial weight of the CO₂ absorption tube be C₁ and the final weight be C₂.
3. The compound is burnt completely in a current of dry oxygen.
4. The increase in weight of the first absorption tube (W₂ - W₁) gives the mass of water produced.
5. The increase in weight of the second absorption tube (C₂ - C₁) gives the mass of carbon dioxide produced.

Calculations:

• Mass of water produced = (W₂ - W₁) g
• Mass of CO₂ produced = (C₂ - C₁) g
• From the mass of CO₂, the mass of carbon can be calculated. Molar mass of CO₂ = 44 g/mol. Molar mass of C = 12 g/mol.
• Mass of Carbon (C) = (12 / 44) × (Mass of CO₂) g
• Percentage of Carbon (% C) = (Mass of Carbon / Mass of organic compound) × 100
• % C = [(12 / 44) × (C₂ - C₁) / m] × 100
• From the mass of water, the mass of hydrogen can be calculated. Molar mass of H₂O = 18 g/mol. Molar mass of H = 2 g/mol.
• Mass of Hydrogen (H) = (2 / 18) × (Mass of H₂O) g
• Percentage of Hydrogen (% H) = (Mass of Hydrogen / Mass of organic compound) × 100
• % H = [(2 / 18) × (W₂ - W₁) / m] × 100

Estimation of Nitrogen (Dumas' Method)

Princ: A known mass of the organic compound is heated with copper oxide (CuO) in a combustion tube. Nitrogen present in the compound is converted into nitrogen gas (N₂). Other combustible elements are oxidized to CO₂ and H₂O.

Apparatus: Similar combustion setup as Liebig's method, but the evolved gases are passed through a heated copper spiral to reduce any oxides of nitrogen to N₂. The mixture of gases (N₂, CO₂, H₂O vapour) is then passed through a solution of KOH. KOH absorbs CO₂ and H₂O vapour (if not already absorbed). The remaining gas is collected over mercury or water, and its volume is measured.

Procedure:

1. A known mass (say, 'm' grams) of the organic compound is mixed with excess copper oxide.
2. The mixture is heated strongly in a combustion tube.
3. The evolved gases (N₂, CO₂, H₂O vapour) are passed over heated copper, then through KOH solution, and finally collected over mercury.
4. The volume of the nitrogen gas collected is measured at a specific temperature and pressure.

Calculations:

• Let the mass of the organic compound be 'm' grams.
• Let the volume of nitrogen gas collected be 'V' mL at STP.
• From the ideal gas equation, 1 mole of any gas occupies 22.4 L (or 22400 mL) at STP and has a mass equal to its molar mass.
• Molar mass of N₂ = 28 g/mol.
• Mass of nitrogen collected = (Volume of N₂ at STP / 22400 mL) × 28 g
• Mass of N₂ = (V / 22400) × 28 g
• Percentage of Nitrogen (% N) = (Mass of Nitrogen / Mass of organic compound) × 100
• % N = [(V / 22400) × 28 / m] × 100

Note: If the volume of nitrogen is collected at room temperature (T K) and pressure (P atm), it needs to be converted to STP conditions (273.15 K and 1 atm).

Volume at STP (V_STP) = V × (P / 1) × (273.15 / T)

Then, use V_STP in the percentage calculation.

Estimation of Halogens (Carius Method)

Princ: A known mass of the organic compound containing a halogen is heated with fuming nitric acid (HNO₃) in a sealed Carius tube. This oxidizes the halogen to the corresponding halogen acid (HX), which is then precipitated as silver halide (AgX) by adding silver nitrate (AgNO₃) solution.

Procedure:

1. A known mass (say, 'm' grams) of the organic compound is taken in a Carius tube.
2. To this, about 3-5 mL of fuming nitric acid and a few crystals of silver nitrate are added.
3. The tube is sealed and heated in a furnace at about 250-300 °C for several hours.
4. After cooling, the tube is broken open, and the contents are transferred to a beaker.
5. If the compound contains chlorine or bromine, the precipitate of AgCl or AgBr is filtered, washed, dried, and weighed.
6. If the compound contains iodine, the precipitate of AgI is filtered, washed, dried, and weighed.

Calculations:

• Let the mass of the organic compound be 'm' grams.
• Let the mass of the silver halide (AgX) precipitate be 'w' grams.
• The atomic mass of Ag is 108.
• For Chlorine (X = Cl): Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol.
• For Bromine (X = Br): Molar mass of AgBr = 108 + 80 = 188 g/mol.
• For Iodine (X = I): Molar mass of AgI = 108 + 127 = 235 g/mol.
• The mass of halogen (X) in 'w' grams of AgX is calculated as:
• Mass of X = (Atomic mass of X / Molar mass of AgX) × w
• Percentage of Halogen (% X) = (Mass of Halogen / Mass of organic compound) × 100
• For Chlorine: % Cl = [(35.5 / 143.5) × w / m] × 100
• For Bromine: % Br = [(80 / 188) × w / m] × 100
• For Iodine: % I = [(127 / 235) × w / m] × 100

Estimation of Sulphur (Carius Method)

Princ: Similar to the estimation of halogens, a known mass of the organic compound containing sulphur is heated with fuming nitric acid in a sealed Carius tube. Sulphur is oxidized to sulphuric acid (H₂SO₄). Sulphuric acid is then precipitated as barium sulphate (BaSO₄) by adding barium chloride (BaCl₂) solution.

Procedure:

1. A known mass (say, 'm' grams) of the organic compound is taken in a Carius tube.
2. To this, about 3-5 mL of fuming nitric acid is added.
3. The tube is sealed and heated in a furnace at about 250-300 °C for several hours.
4. After cooling, the tube is broken open. The contents are diluted with water, and the excess nitric acid is neutralized.
5. Barium chloride (BaCl₂) solution is added, and the mixture is heated. A precipitate of barium sulphate (BaSO₄) is formed.
6. The precipitate is filtered, washed, dried, and weighed.

Calculations:

• Let the mass of the organic compound be 'm' grams.
• Let the mass of the barium sulphate (BaSO₄) precipitate be 'w' grams.
• Molar mass of S = 32 g/mol.
• Molar mass of BaSO₄ = 137 (Ba) + 32 (S) + 4 × 16 (O) = 137 + 32 + 64 = 233 g/mol.
• The mass of sulphur (S) in 'w' grams of BaSO₄ is calculated as:
• Mass of S = (Atomic mass of S / Molar mass of BaSO₄) × w
• Mass of S = (32 / 233) × w grams
• Percentage of Sulphur (% S) = (Mass of Sulphur / Mass of organic compound) × 100
• % S = [(32 / 233) × w / m] × 100

Estimation of Phosphorus (Carius Method)

Princ: A known mass of the organic compound containing phosphorus is heated with nitric acid and sodium carbonate. The phosphorus is oxidized to phosphoric acid (H₃PO₄). This phosphoric acid is then precipitated as ammonium phosphomolybdate ((NH₄)₃PO₄·12MoO₃) by adding ammonium molybdate solution.

Procedure:

1. A known mass (say, 'm' grams) of the organic compound is mixed with sodium carbonate and concentrated nitric acid.
2. The mixture is heated strongly.
3. The resulting solution is treated with ammonium molybdate solution and heated.
4. A yellow precipitate of ammonium phosphomolybdate is formed.
5. This precipitate is filtered, washed, dried, and weighed.

Calculations:

• Let the mass of the organic compound be 'm' grams.
• Let the mass of the ammonium phosphomolybdate precipitate be 'w' grams.
• Molar mass of P = 31 g/mol.
• Molar mass of Ammonium Phosphomolybdate ((NH₄)₃PO₄·12MoO₃) = 3 × 18 (NH₄) + 31 (P) + 4 × 16 (O) + 12 × (96 (MoO₃))
• Molar mass = 54 + 31 + 64 + 1152 = 1301 g/mol.
• The mass of phosphorus (P) in 'w' grams of ammonium phosphomolybdate is calculated as:
• Mass of P = (Atomic mass of P / Molar mass of Ammonium Phosphomolybdate) × w
• Mass of P = (31 / 1301) × w grams
• Percentage of Phosphorus (% P) = (Mass of Phosphorus / Mass of organic compound) × 100
• % P = [(31 / 1301) × w / m] × 100

Empirical and Molecular Formula Calculations

The empirical formula of a compound represents the simplest whole-number ratio of atoms of each element present in one molecule of the compound. The molecular formula represents the actual number of atoms of each element in one molecule.

Empirical Formula

Steps to determine the Empirical Formula:

1. Find the percentage composition of each element in the compound. This is usually given or calculated from quantitative analysis data.
2. Convert the percentage composition into grams by assuming a total mass of 100 g for the compound. For example, if a compound has 40% Carbon, 6.67% Hydrogen, and 53.33% Oxygen, assume you have 100 g of the compound, which means you have 40 g of C, 6.67 g of H, and 53.33 g of O.
3. Convert the mass of each element into moles by dividing the mass by its atomic mass.
4. Divide the molar ratio by the smallest molar ratio obtained. This gives the relative number of atoms.
5. If the ratios are not whole numbers, multiply them by the smallest integer that will convert them all to whole numbers. This gives the empirical formula.

Example: A compound contains C, H, and O. Its percentage composition is C = 40.0%, H = 6.67%, O = 53.33%. Find its empirical formula.

1. Assume 100 g of the compound. So, we have 40 g C, 6.67 g H, 53.33 g O.
2. Convert to moles:
   • Moles of C = 40 g / 12 g/mol = 3.33 mol
   • Moles of H = 6.67 g / 1 g/mol = 6.67 mol
   • Moles of O = 53.33 g / 16 g/mol = 3.33 mol
3. Divide by the smallest molar ratio (3.33):
   • C: 3.33 / 3.33 = 1
   • H: 6.67 / 3.33 ≈ 2
   • O: 3.33 / 3.33 = 1
4. The simplest whole-number ratio is C:H:O = 1:2:1.

Therefore, the empirical formula is CH₂O.

Molecular Formula

The molecular formula is always a whole-number multiple of the empirical formula.

Molecular Formula = (Empirical Formula)_n

Where 'n' is a positive integer.

The value of 'n' can be calculated using the molar mass (molecular weight) of the compound:

n = (Molar Mass of Compound) / (Empirical Formula Mass)

Steps to determine the Molecular Formula:

1. Determine the empirical formula of the compound using the steps described above.
2. Calculate the empirical formula mass.
3. Determine the molar mass (molecular weight) of the compound. This is usually given in the problem or can be determined using methods like Victor Meyer's method or from colligative properties.
4. Calculate 'n' using the formula: n = Molar Mass / Empirical Formula Mass.
5. Multiply the subscripts in the empirical formula by 'n' to get the molecular formula.

Example: The empirical formula of a compound is CH₂O. Its molar mass is 180 g/mol. Find its molecular formula.

1. Empirical Formula = CH₂O
2. Empirical Formula Mass = 12 + (2 × 1) + 16 = 30 g/mol
3. Molar Mass = 180 g/mol
4. Calculate 'n': n = Molar Mass / Empirical Formula Mass = 180 g/mol / 30 g/mol = 6
5. Molecular Formula = (CH₂O)₆ = C₆H₁₂O₆

Calculation involving percentage composition and Molar Mass

Sometimes, you are given the percentage composition of elements and asked to find both the empirical and molecular formulas without the molar mass. In such cases, you first find the empirical formula as usual. Then, you use the molar mass determination techniques to find the molecular weight of the compound.

Example: A compound on analysis gave the following percentage composition: C = 72.0%, H = 6.67%, and the rest is Oxygen. Its vapour density is 75. Find its molecular formula.

1. Percentage Composition:
   • % C = 72.0%
   • % H = 6.67%
   • % O = 100 - (72.0 + 6.67) = 100 - 78.67 = 21.33%
2. Convert to grams (assume 100 g): 72 g C, 6.67 g H, 21.33 g O.
3. Convert to moles:
   • Moles C = 72 / 12 = 6
   • Moles H = 6.67 / 1 = 6.67
   • Moles O = 21.33 / 16 = 1.33
4. Divide by the smallest molar ratio (1.33):
   • C: 6 / 1.33 ≈ 4.5
   • H: 6.67 / 1.33 ≈ 5
   • O: 1.33 / 1.33 = 1
5. Multiply to get whole numbers: Since C is 4.5, we multiply all by 2.
   • C: 4.5 × 2 = 9
   • H: 5 × 2 = 10
   • O: 1 × 2 = 2
6. Empirical Formula: C₉H₁₀O₂
7. Empirical Formula Mass: (9 × 12) + (10 × 1) + (2 × 16) = 108 + 10 + 32 = 150 g/mol.
8. Molar Mass: Vapour Density = 75. Molar Mass = 2 × Vapour Density = 2 × 75 = 150 g/mol.
9. Calculate 'n': n = Molar Mass / Empirical Formula Mass = 150 / 150 = 1.
10. Molecular Formula: (C₉H₁₀O₂)₁ = C₉H₁₀O₂.