EUDIOMETRY (GAS ANALYSIS) :- APPLICATION OF POAC ON EUDIOMETRY (gas analysis) NUMERICALS and HOW TO FIND OUT MOLECULAR FORMULA OF ANY HYDROCARBON USING "POAC"

Eudiometry:- Eudiometry means gas analysis. It means in this type of numericals either you have to find out vol. of GAS From the data given in question or you have to find out any other thing like molecular formula of hydrocarbon ,no. of mole , mass of any other reactant and vol. of GAS is given. Specific gases are absorbed in specific reagents as given below

Reactants                                                  Absorbed gases

KOH solution                                             CO₂ & SO₂

Alkaline Pyrogallol                                     O₂

Solution of ammonical cuprous chloride    CO

And if the mixture is cooled, H₂O vapour changes to liquid.

Hence, when temperature is above 100 ⁰C (373 K) then only volume of H₂O is considered.

Eudiometry is mainly based on Avogadro's law,

     A _(g) +       B _(g) →  C_(g)  +   D_(g)

 a volume   b volume   c volume   d volume

 a moles    b moles     c moles    d moles

Q.1 Three moles of Na₂CO₃ is reacted with 6 moles of HCl solution. Find out 

the volume of CO₂ gas produced at STP

Sol.  Na₂CO₃→ NaCl + CO₂

Apply POAC on C atom

No. of moles of C atoms in Na₂CO₃ = no. of moles of C atoms in CO₂

1* no. of moles of Na₂CO₃ = 1* no. of moles of CO₂

So no. of moles of Na₂CO₃ = no. of moles of CO₂ = 3 moles (given)

Vol. of CO₂ = 3 * 22.4 litre = 67.2 litre

  v HOW TO FIND OUT MOLECULAR FORMULA OF ANY HYDROCARBON USING POAC:-

Q. 10 ml. of a gaseous hydrocarbon was burnt completely in 80 ml. of O₂ 

at   NTP. The remaining gas occupied 70 ml. at NTP. This vol. became 50

ml on   treatment with KOH solution. What is the formula of hydrocarbon?

Solution: -  C_x H_y + O₂ → CO₂+ H₂O

 Vol. of ( CO₂ + unreacted O₂) = 70 ml.

Because CO₂  can be absorbed by KOH .  So when this 70 ml. mix. Is treated it 

becames 50 ml it means vol. of CO₂ in the mixture was 20 ml. which is absorbed

 by KOH and remaining 50 ml. was unreacted O₂  ( O₂  can NOT absorbed by 

KOH)

So vol. of CO₂ = 20 ml.

Vol. of O₂ reacted = 80 -50 = 30 ml.

Vol. of hydrocarbon = 10 ml.

Apply POAC on C atoms

X* no. of mole of C_x H_y = 1* no. of mole of CO₂   -----------------------( i  ) 

X* 10 = 20

So X = 20/10= 2  

Apply POAC on H atoms

Y* no. of moles of C_x H_y = 2* no. of mole of H₂O -----------------( ii )

Apply POAC on O atoms

2* no. of moles of O₂ = 1* no. of mole of H₂O+2* no. of mole of CO₂

From eq. (ii)

2* no. of moles of O₂ = y/2* no. of mole of C_x H_y + 2* no. of mole of CO₂

2*30 = Y/2*10 + 2 *20

20 = 10Y/2

Y = 4

Put the value of X and Y in C_x H_y

The molecular formula of hydrocarbon is C₂H₄

_{TRY TO SOLVE FOLLOWING QUESTION}

Q.2   7.5 ml of hydrocarbon gas was exploded with excess of oxygen. On

 cooling it was found to have undergone a contraction of 15 ml. if vapour 

density of hydrocarbon is 14. Determine its molecular formula.

Solution ( HINT ) :- molecular wt.= 2* vapour density

   Answer:- C₂H₄

Q.3  7.5 ml of hydrocarbon gas was exploded with 36 ml.of oxygen. The 

vol. of gases on cooling was found to be 28.5 ml., 15 ml. of which was 

absorbed by KOH and the rest was absorbed in a solution of alkaline 

pyrogallol. If all vol. are measured under the same conditions, determine 

the molecular formula of hydrocarbon.

       Answer:- C₂H₄