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The volumetric analysis of gaseous reaction using eudiometric tube called Eudiometry or “Volume analysis of gas”
In eudiometric tube all the measurement of volume is done at constant pressure and temperature and given gaseous reaction at least two components are gases.
The edudiometric relationship amongst gases, when they react with one another, is governed by two laws, they are illustrated as, Gay-Lussac law and Avogadro’s law.
(1) Gay-Lussac Law:
According to Gay Lussac’s law, the volumes of gaseous reactants reacted and the volumes of gaseous products formed, are always bear a simple ratio at same temperature and pressure.
(2) Avogadro’s Law:
According Avogadro law A samples of different gases which contain the same number of molecules occupy the same volume at the same temperature and pressure. This law is also known as Avogadro’s hypothesis and given by Amadeo Avogadro in 1812.
Gaseous reactions 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 Eudiometric tube and hence the name Eudiometry also used for “Volume analysis of Gas”.
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 oxidized.
The various reagents used for absorbing different gases are
O3 -- Turpentine oil
O2 - Alkaline pyrogallol
NO - FeSO4 solution
Cl2, SO2, CO2 -- Alkali solution NaOH, KOH, etc.
NH3, HCl - Water
H2O, - CuSO4, CaCl2
H2O(g) – Con H2SO4
CO – Ammonical Cuprus chloride solution (NH4OH+ Cu2Cl2 )
(1)  In Eudiometric tube Nitrogen does not react.
(2) 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.
(3) The excess of oxygen left after the combustion is also determined by difference if other gases formed during combustion have already been determined.

APPLICATION OF EUDIOMETRY: Data collected from eudiometric tube experiment a number of useful conclusions regarding gaseous reactions can be drawn.
(1) Molecular formulae of Gaseous Hydrocarbons.
(2) The composition of Gaseous mixtures.
(3) Molecular formulae of Gases.

(4) Volume-volume relationship Gaseous reactions.

(1) Determination of Molecular formula of Hydrocarbon using Eudiometry: 

A known amount of hydrocarbon is taken into a Eudiometry tube. O2 gas is then inserted to cause complete combustion of hydrocarbon & the reaction mixture is cooled back to the original room temperature. This gives 1st volume contraction V1C the 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.
Step 2: Write down Volume of components before reaction V(HC), V(O2).

Step 3: Write down Volume of components after the reaction using Gay Lussac Law, (After identifying limiting reagent)
Step 4:  Using Given Data
This will give the value of “y’
V11C = Due to change in volume because of absorption of CO2 in alc KOH
V11C = xV (Hydrocarbon)
Hence, both x and y can be calculated.
ILLUSTRATIVE EXAMPLE: (1) 10 ml of nitrogen and 40 ml of Hydrogen reacts to produce NH3 gas fined  out the final volume and volume contraction If :
(1) Reaction is 100%  completed 
(2) Reaction is 50% complete 
ILLUSTRATIVE EXAMPLE: (2) 50 ml of C3H8 is mixed with 300 ml of Oxygen for complete combustion find out the final volume and volume contraction.

ILLUSTRATIVE EXAMPLE: (3) 10 ml of Hydrogen on complete combustion gives 30 ml of CO2 for this 40 ml of O2 is required fined out the formula of Hydrocarbon.
ILLUSTRATIVE EXAMPLE: (4) 10 ml of gaseous Hydrocarbon burns Completely in 80 ml of Oxygen , the remaining gases occupy 70 ml of volume , this volume become 50 ml and on treatment with KOH find out the formula of Hydrocarbon .
ILLUSTRATIVE EXAMPLE: (5) A mixture of ethane (C2H6) and ethene (C2H4) occupies 40 litre at 1.00 atm and 400 K , the mixture reacts completely with 130 gm of O2 to produce CO2 and H2O . Assuming ideal gas behaviour , calculate the mole fraction of ethane (C2H6) and ethene (C2H4) in mixture (IIT 1915)
ILLUSTRATIE EXAMPLE: (6) A mixture of ethane (C2H6) and ethene (C2H4) occupies 35.5 litre at 1.00 bar and 405 K , this mixture reacts completely with 110.3 gm of O2 to produce CO2 and H2O.what was the compostion of original mixture.(Assuming ideal gas behaviour )
ILLUSTRATIVE EXAMPLE:(7) 1ml of gaseous aliphatic compound cnH3nOm is completely  burnt in an excess of O2 and cooled to room temperature . The contraction in volume is .
ILLUSTRATIVE EXAMPLE:(8) 1ml of gaseous aliphatic compound cnH3nOm is completely  burnt in an excess of O2 . The reacted number of moles of  oxygen is ?
ILLUSTRATIVE EXAMPLE: (9) A mixture of ethane (C2H6) and ethene (C2H4) occupies 42 litre at 1.00 atm and 500 K , this mixture reacts completely with 10/3 moles of O2 to produce CO2 and H2O. The mole fraction of ethane and ethene in the original mixture are respectively .(Assuming ideal gas behaviour )
(Given The = 0.082 L atm per K per mole )


Analysis of mass of reactants and products of given chemical reaction is known as Gravimetric analysis.
Two important method deal with the trapping and weighing of product in solid and gaseous state
(1) PRECIPITATATION METHOD: This method uses for metallic elements in their ionic form react with negative counter ions to produced stable (Insoluble) precipitate.
(I) Silver ppt with halides ions
(II) Calcium ppt with Oxalates
(III) Barium ppt with Sulphates
(2) VOATILIZATION METHOD: This method generally uses for analysis of Bicarbonate from the mixture of Carbonate and Bicarbonate.

ILLUSTRATIVE EXAMPLE (1): A mixture of FeO and Fe2O4 when heated in air to constant weight gain 5% in its weight, The % of FeO in sample is?
ILLUSTRATIVE EXAMPLE (2): A mixture of FeO and Fe2O3 is produced by 1 mole of Fe reacts Completely with 0.65 moles of O2. Find out the mole’s ratio of ferrous oxide to ferric oxide.?
8 gm mixture of sodium carbonate and sodium bicarbonate are heated until a constant mass equal to 6 gm is obtained calculate % of sodium bicarbonate in the original mixture ?.
1 litre of a mixture of CO and CO2 is passed through a tube containing red hot charcoal. The total volume of the mixture become 1.5 litre. Calculate Volume % of CO in the original mixture.? 
1-gram mixture of cuprous and cupric oxide was reduced to 0.839 gm metallic cupper. What is the weight of cuprous oxide in given sample.?  (Cu= 63.5, O= 16)
A mixture of pure AgCl and AgBr contains 60.94 % silver by mass, what is percentage of AgCl in sample.? (Ag= 108, Br= 80 Cl = 35.5)
1.48 gm mixture of CaCO3 and MgCO3 was heated to a constant weight till 0.96 gm residue formed, % of MgCO3 in the sample was?
ILLUSTRATIVE EXAMPLE (8): When 40 gm of 200% solution by weight was cooled, 50 gm of solute precipitated. The percentage of concentration of remaining solution is.
ILLUSTRATIVE EXAMPLE (9): 18.4 gm mixture of CaCO3 and MgCO3 was heated to produce a constant volume of carbon dioxide is 4.48 litre the calculate amount of MgCO3 and CaCO3 in mixture?
ILLUSTRATIVE EXAMPLE (10): 31.3 gm NaCl mixture of NaBr and NaCl treated with H2SO4, 4.48 gm of Na2SO4 is produced. Then Calculate the amount of NaCl and NaBr in mixture.

Answers: (1) 20.25 % (2) 4/3 (3) .......  (4) 50 %  (5) 0.45 grams (6) 19.64 % (7) 45.66 % (8) 8.57 % (9) CaCo3 = 10 gram, (10) NaCl=13.1grams and NaBr= 18.28 grams


The reagent which is consumed completely in the reaction and decides the amount of product is known as limiting reagent and left over reagent is known as excess reagent.
(1) Divide  the given moles of each reagent by its Stoichiometric coefficient.
(2) The reagent for which the division comes out minimum is called Limiting reagent.
(1) The amount of product is decided by Limiting reagent .
(2) Amount of excess reagent left over and participate in the reaction is decided by Limiting reagent.
ILLUSTRATIVE EXAMPLE (1) : The Chemical reaction 
8 moles of P and 5 moles of Q will produce how many moles of  R ?
ILLUSTRATIVE EXAMPLE (2): The Chemical reaction
Carried out by taking 24 gm of Carbon (2 moles ) and 96 gm (3 moles ) of oxygen (O2) find 
(1) Limiting reagent
(2) How many moles of CO formed
(3) percentage (%) of  reagent left over .

ILLUSTRATIVE EXAMPLE (3): 10 gm each of Lithium and Oxygen react to produce Li2O find out?
(1) Find the limiting reagent
(2) Mass of Li2O produced in reaction
(3) % of excess reagent consumed
(4) % of excess reagent undergoes in reaction.

ILLUSTRATIVE EXAMPLE (4): Equal mass of Zn (65) and S (32) reacts to form ZnS, Calculate fraction of excess reagent left unreacted.
Above reaction is carried out by taking 6 moles of each X and Y respectively then find % excess reagent react.
SOLUTION: Answer Key 25% unreacted  and 75% reacted 

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