Eudiometry Class XI

Eudiometry or “gas analysis” involves a calculation based on gaseous reactions in which at least two components are gases & their amount is given in terms of volumes measured at same pressure & Temperature.
The relationship amongst gases, when they react with one another, is governed by two laws, namely Gay-Lussac law and Avogadro’s law.
Gay-Lussac Law
According to Gay – Lussac’s law, the volumes of gaseous reactants reacted and the volumes of gaseous products formed. all measured at the same temperature and pressure. Bear a simple ratio.

Avogadro’s Law
In 1812, Amadeo Avogadro stated that samples of different gases which contain the same number of molecules (any complexity, size, shape) occupy the same volume at the same temperature and pressure. It follows from Avogadro’s hypothesis that Vn (when T and P are constant).
Gaseous reactions for investigation purposes are studied in a closed graduated tube open at one end and the other closed end of which is provided with platinum terminals for the passage of electricity through the mixture of gases, Such a tube is known as Eudiometer tube and hence the name Eudiometry also used for Gas analysis.
During Gas analysis, the Eudiometer tube filled with mercury is inverted over a trough containing mercury. A known volume of the gas or gaseous mixture to be studied is next introduced, which displaces an equivalent amount of mercury. Next, a known excess of oxygen is introduced and the electric spark is passed, whereby the combustible material gets oxidised. The volumes of carbon dioxide, water vapour or other gaseous products of combustion are next determined by absorbing them in suitable reagents. For example, the volume of CO2 is determined by absorption in KOH solution and that of an excess of oxygen in an alkaline solution of pyrogallol. Water vapour produced during the reaction can be determined by noting contraction in volume caused due to cooling, as by cooling the steam formed during combustion forms liquid (water) which occupies a negligible volume as compared to the volumes of the gases considered. The excess of oxygen left after the combustion is also determined by difference if other gases formed during combustion have already been determined. From the data thus collected a number of useful conclusions regarding reactions amongst gases can be drawn.
(a) Volume-volume relationship amongst Gases or simple Gaseous reactions.
(b) The composition of Gaseous mixtures.
(c) Molecular formulae of Gases.
(d) Molecular formulae of Gaseous Hydrocarbons.
The various reagents used for absorbing different gases are
O3turpentine oil
O2 alkaline pyrogallol
NO FaSO4 solution
Cl2, alkali solution (NaOH, KOH, etc.)
NH3 Water
H2O CuSO4, CaCl2
CO Ammonical Cu2Cl2
General Assumptions: In all problems, it is assumed that the sparking occurs at room temperature. This implies that water formed would be in liquid state and that nitrogen gas is inert toward oxidation.
Also, if a gas is added externally, it will be assumed to be in excess unless otherwise stated.
Determination of Molecular formula of Hydrocarbon using Eudiometry

A known amount of hydrocarbon is taken into a Eudiometry tube. O2gas is then inserted to cause complete combustion of hydrocarbon & the reaction mixture is cooled back tot he original room temperature. This gives 1st volume contraction V1CThe resultant gaseous mixture is then passed through alc. KOH which gives second volume contraction, V11C. These data can help to calculate the molecular formula of the hydrocarbon as explained below.
Step 1: Write down the balanced chemical reaction along with their states.

















Illustration 18: A gaseous hydrocarbon requires 6 times its own volume of O2 for complete oxidation and produces 4 times its volume of CO2. What is its formula?





Again x = 4 since evolved CO2 is 4 times that of hydrocarbon
∴ 16 + y = 24 or y = 8  ∴ formula of hydrocarbon C4H8
Illustration 19: 0.378 gm of an organic acid gave on combustion 0.264 g of carbon dioxide and 0.162 g of water vapours. Calculate the percentage of C of H.
Solution : Mass of  orgnic compound = 0.378 g
Mass of CO2 formed = 0.264 g
Mass of H2O formed = 0.162 g





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