Eudiometry — Volume-Volume Analysis
Volume-volume analysis of gas reactions. For gases at the same temperature and pressure, volume ratios match mole ratios — so you can predict product volumes directly from a balanced equation, no moles required.
Volume–Volume Analysis (Eudiometry)
For reactions that take place entirely in the gas phase, you can measure quantities by volume instead of mass. The technique is called eudiometry (from the Greek eudios, “fair weather” — historically used to measure atmospheric oxygen).
Vol–vol analysis is built on Avogadro's law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. So the volume ratios in a gaseous reaction are equal to the mole ratios — same as the stoichiometric coefficients of the balanced equation.
The experimental setup. Gaseous reactions are run in a sealed glass tube called a eudiometer tube, fitted with platinum electrodes for spark ignition. After the reaction, the individual product gases are identified and measured by passing the mixture through a series of selective absorbing reagents:
| Gas absorbed | Reagent used |
|---|---|
| , | KOH solution |
| alkaline pyrogallol | |
| ammoniacal cuprous chloride |
Water vapour produced during a reaction condenses on cooling — its volume is usually ignored when comparing gas volumes, but its moles must be counted when applying mole-based formulas.
The drop in total volume after each absorption step tells you how much of that specific gas was present. By chaining absorption steps, you can deduce the formula of an unknown gaseous reactant — a classic JEE-style problem.
The general hydrocarbon combustion equation is the workhorse of eudiometry problems. Any hydrocarbon burns in oxygen as:
So by Avogadro's law, the volume ratios of the gas-phase substances follow the coefficients directly:
(Water condenses on cooling, so its volume drops out — but if you're applying mole-based formulas, the moles of water do count.)
Air composition. When a problem asks for the volume of air rather than pure oxygen, remember that atmospheric air is approximately 21% O₂ by volume (the rest is mostly N₂, which is inert in combustion). So:
Quick recap of selective absorbents (used to identify which product gas is which in the cooled mixture):
| Absorbent | Gas it removes |
|---|---|
| KOH solution | and |
| Alkaline pyrogallol | |
| Ammoniacal |
Standard recipe for eudiometry problems:
- Write the general combustion equation symbolically: .
- Identify each gas volume in the problem (hydrocarbon used, O₂ consumed or left over, CO₂ formed, etc.). For absorption problems, the volume drop after each absorbent step tells you how much of that gas was present.
- Use volume ratios to set up linear equations in and .
- Solve, then read off the molecular formula.
Most JEE eudiometry questions are variations on this routine.
Loading simulator…
What volume of oxygen will be required for the complete combustion of 18.2 L of propane () at NTP?
20 mL of was mixed with 50 mL of and the mixture was exploded. On cooling, the resulting mixture was shaken with KOH solution. Find the volume of gas left.
What volume of air (containing 21% oxygen by volume) is required to completely burn 1 kg of carbon, assumed to be 100% combustible?
(Atomic mass: C = 12; molar volume at NTP = 22.4 L)
At 300 K and 1 atm, 10 mL of a hydrocarbon required 55 mL of O₂ for complete combustion, and 40 mL of CO₂ was formed. The formula of the hydrocarbon is:
(a) (b) (c) (d)
A mixture of methane and ethylene in the ratio by volume occupies 30 mL. On complete combustion, the mixture yields 40 mL of CO₂.
What volume of CO₂ would have been obtained if the ratio were instead?
(a) 50 mL (b) 30 mL (c) 40 mL (d) 60 mL
7.5 mL of a gaseous hydrocarbon was exploded with 36 mL of oxygen. The total volume of gases on cooling was 28.5 mL. Of that, 15 mL was absorbed by KOH and the rest was absorbed in a solution of alkaline pyrogallol.
If all volumes are measured under the same conditions, deduce the formula of the hydrocarbon.
Volume–Volume Analysis (Eudiometry)
For reactions that take place entirely in the gas phase, you can measure quantities by volume instead of mass. The technique is called eudiometry (from the Greek eudios, “fair weather” — historically used to measure atmospheric oxygen).
Vol–vol analysis is built on Avogadro's law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. So the volume ratios in a gaseous reaction are equal to the mole ratios — same as the stoichiometric coefficients of the balanced equation.
The experimental setup. Gaseous reactions are run in a sealed glass tube called a eudiometer tube, fitted with platinum electrodes for spark ignition. After the reaction, the individual product gases are identified and measured by passing the mixture through a series of selective absorbing reagents:
| Gas absorbed | Reagent used |
|---|---|
| , | KOH solution |
| alkaline pyrogallol | |
| ammoniacal cuprous chloride |
Water vapour produced during a reaction condenses on cooling — its volume is usually ignored when comparing gas volumes, but its moles must be counted when applying mole-based formulas.
The drop in total volume after each absorption step tells you how much of that specific gas was present. By chaining absorption steps, you can deduce the formula of an unknown gaseous reactant — a classic JEE-style problem.
The general hydrocarbon combustion equation is the workhorse of eudiometry problems. Any hydrocarbon burns in oxygen as:
So by Avogadro's law, the volume ratios of the gas-phase substances follow the coefficients directly:
(Water condenses on cooling, so its volume drops out — but if you're applying mole-based formulas, the moles of water do count.)
Air composition. When a problem asks for the volume of air rather than pure oxygen, remember that atmospheric air is approximately 21% O₂ by volume (the rest is mostly N₂, which is inert in combustion). So:
Quick recap of selective absorbents (used to identify which product gas is which in the cooled mixture):
| Absorbent | Gas it removes |
|---|---|
| KOH solution | and |
| Alkaline pyrogallol | |
| Ammoniacal |
Standard recipe for eudiometry problems:
- Write the general combustion equation symbolically: .
- Identify each gas volume in the problem (hydrocarbon used, O₂ consumed or left over, CO₂ formed, etc.). For absorption problems, the volume drop after each absorbent step tells you how much of that gas was present.
- Use volume ratios to set up linear equations in and .
- Solve, then read off the molecular formula.
Most JEE eudiometry questions are variations on this routine.
Loading simulator…
What volume of oxygen will be required for the complete combustion of 18.2 L of propane () at NTP?
20 mL of was mixed with 50 mL of and the mixture was exploded. On cooling, the resulting mixture was shaken with KOH solution. Find the volume of gas left.
What volume of air (containing 21% oxygen by volume) is required to completely burn 1 kg of carbon, assumed to be 100% combustible?
(Atomic mass: C = 12; molar volume at NTP = 22.4 L)
At 300 K and 1 atm, 10 mL of a hydrocarbon required 55 mL of O₂ for complete combustion, and 40 mL of CO₂ was formed. The formula of the hydrocarbon is:
(a) (b) (c) (d)
A mixture of methane and ethylene in the ratio by volume occupies 30 mL. On complete combustion, the mixture yields 40 mL of CO₂.
What volume of CO₂ would have been obtained if the ratio were instead?
(a) 50 mL (b) 30 mL (c) 40 mL (d) 60 mL
7.5 mL of a gaseous hydrocarbon was exploded with 36 mL of oxygen. The total volume of gases on cooling was 28.5 mL. Of that, 15 mL was absorbed by KOH and the rest was absorbed in a solution of alkaline pyrogallol.
If all volumes are measured under the same conditions, deduce the formula of the hydrocarbon.