Victor Meyer's Method: Molar Mass of Volatile Liquids
Victor Meyer's method is a standard laboratory procedure for determining the molar mass of volatile organic liquids. In Class 11 chemistry, school board examinations, and competitive entrance tests like JEE Main, NEET, and CUET, numerical questions on this method test your understanding of ideal gas laws, Dalton's law of partial pressures, and the mole concept. Because displaced air is collected over water, students must correctly subtract aqueous tension, convert gas volumes to standard temperature and pressure (STP), and apply molar volume formulas accurately.
Experimental Setup and Operating Principle
The apparatus consists of an inner glass Victor Meyer tube with an elongated neck and a side delivery tube leading to a water trough, where a graduated eudiometer tube collects displaced air over water. The base of the inner tube contains an asbestos pad to cushion the fall of the sample. This inner tube is enclosed in an outer jacket containing a heating liquid whose boiling point is 20 to 30 K higher than that of the volatile compound.
When the jacket liquid boils steadily, the air inside reaches uniform temperature. A small glass bulb containing a weighed mass (w) of volatile liquid is dropped into the tube, vaporising instantly at the hot bottom. The sudden vapour expansion expels an identical volume of air through the delivery tube into the eudiometer. The displaced air collects over water at ambient room temperature and atmospheric pressure. Under identical conditions, the volume of air collected equals the volume that the vapour of the substance would occupy at room conditions.
Key Formulas for Victor Meyer Calculations
Because displaced air is collected over water, it is saturated with water vapour. According to Dalton's law of partial pressures, total atmospheric pressure is the sum of dry air pressure and aqueous tension:
1. Dry air pressure: Pdry = Ptotal - p
where Ptotal is total atmospheric pressure and p is aqueous tension at temperature T1 (in mm Hg, torr, or atm).
2. Reduction to STP (Combined Gas Law):
VSTP = (Pdry × V1 × TSTP) / (PSTP × T1)
where V1 is observed volume in mL, T1 is room temperature in Kelvin (T1 = °C + 273.15), and VSTP is dry air volume at standard conditions.
3. Molar mass calculation:
M = (w × Vm) / VSTP
where w is sample mass in grams, Vm is standard molar volume in mL/mol, and M is molar mass in g/mol.
Textbooks differ on the definition of standard temperature and pressure (STP):
1. Traditional standard: 0 °C (273.15 K) and 1 atm (760 mm Hg), where molar volume is 22.4 L/mol (22,400 mL/mol).
2. Modern IUPAC standard: 0 °C (273.15 K) and 1 bar (750.06 mm Hg), where molar volume is 22.71 L/mol (22,710 mL/mol).
School and competitive examinations predominantly use 22,400 mL/mol unless 1 bar is specified. Molar mass can also be computed directly from the ideal gas equation: M = (w × R × T1) / (Pdry × V1), where R = 0.082057 L·atm/(mol·K).
Worked Numerical Problems
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CBSE Class 11, State Boards, JEE Main, NEET, CUET-UG
- Topic: Victor Meyer's Method and Molar Mass Calculations
- Difficulty: Intermediate
Question: In a Victor Meyer determination, 0.202 g of a volatile organic liquid displaced 58.0 mL of moist air collected over water at 288 K and 755 mm Hg barometric pressure. The aqueous tension of water at 288 K is 13.0 mm Hg. Find the molar mass of the compound using the standard molar volume of 22,400 mL/mol at 760 mm Hg and 273 K.
Solution step by step:
Step 1: Calculate the pressure of dry displaced air.
Pdry = Ptotal - aqueous tension = 755 mm Hg - 13.0 mm Hg = 742.0 mm Hg
Step 2: Convert the gas volume to STP conditions (760 mm Hg and 273 K).
VSTP = (Pdry × V1 × 273) / (760 × 288) = (742.0 × 58.0 × 273) / (760 × 288) = 11,748,768 / 218,880 = 53.677 mL
Step 3: Calculate the molar mass.
M = (w × 22,400) / VSTP = (0.202 × 22,400) / 53.677 = 4524.8 / 53.677 = 84.3 g/mol
Verification by second route (Ideal Gas Law):
Pdry = 742.0 / 760 = 0.97632 atm
Moles n = (0.97632 atm × 0.0580 L) / (0.082057 × 288 K) = 0.056627 / 23.6324 = 0.002396 mol
M = w / n = 0.202 g / 0.002396 mol = 84.31 g/mol
Both calculation methods give 84.3 g/mol.
Answer: 84.3 g/mol
Question: A sample of 0.180 g of a volatile liquid displaced 54.0 mL of moist air in a Victor Meyer apparatus at 300 K and 750 mm Hg. The aqueous tension at 300 K is 26.7 mm Hg. Calculate the molar mass under traditional STP (22,400 mL/mol at 760 mm Hg, 273 K), and compare with modern IUPAC standard (22,710 mL/mol at 1 bar = 750.06 mm Hg, 273.15 K).
Solution step by step:
Step 1: Calculate dry air pressure.
Pdry = 750.0 mm Hg - 26.7 mm Hg = 723.3 mm Hg
Step 2: Convert dry air volume to traditional STP (760 mm Hg, 273 K).
VSTP = (723.3 × 54.0 × 273) / (760 × 300) = 10,662,817.8 / 228,000 = 46.767 mL
Step 3: Compute molar mass using 22,400 mL/mol.
M = (0.180 × 22,400) / 46.767 = 4032 / 46.767 = 86.2 g/mol
Verification by second route (Modern IUPAC standard at 1 bar = 750.06 mm Hg):
VIUPAC = (723.3 × 54.0 × 273.15) / (750.06 × 300) = 10,668,675.21 / 225,018 = 47.413 mL
M = (0.180 × 22,710) / 47.413 = 4087.8 / 47.413 = 86.2 g/mol
Both standard conventions give identical molar mass of 86.2 g/mol.
Answer: 86.2 g/mol
Common Errors in Victor Meyer Problems
1. Ignoring aqueous tension: Using total barometric pressure directly is the most frequent blunder. Because the air is moist, you must subtract aqueous tension to obtain dry air pressure. Neglecting it makes Pdry too large, giving a molar mass that is too low.
2. Using Celsius instead of Kelvin: Gas equations require thermodynamic absolute temperature. Always add 273.15 to convert Celsius values to Kelvin.
3. Mixing volume units: When using 22.4 L/mol, express gas volume in litres. When using 22,400 mL/mol, keep volume in millilitres.
4. Overlooking thermal dissociation: If a substance dissociates into smaller fragments inside the hot tube, the vapour volume increases, producing an artificially low molar mass.
Exam Relevance and Question Patterns
| Examination | Question Format | Key Skills Tested |
|---|---|---|
| CBSE and State Boards | Numerical and conceptual problems | Apparatus principle, aqueous tension deduction, and reduction to STP. |
| JEE Main | Single-correct MCQs and numerical questions | Rapid calculation of dry pressure, molar mass, and empirical formula links. |
| NEET | Formula-based multiple choice | Direct substitution into gas formulas and relating displaced volume to moles. |
| CUET-UG | Objective conceptual items | Differences between 22.4 L and 22.7 L standards and real vapour behavior. |
Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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