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

     

   

EVERY SUCCESS STORY IS ALSO A STORY OF GREAT FAILURE.

A man failed in business at the age of 21; was defeated in a legislative race at the age 22; failed again in business at age 24; his wife died when he was age 26; had nervous breakdown at age 27; lost a congressional race at age 32; lost a senatorial race at age 45; failed in an effort to become vice- president at age 47; lost a senatorial race again at age 49; and was elected president of united state at age 52. This man was Abraham Lincoin.

If we study the history, we will find that all stories of success are also stories of great failures. But people don’t see the failures. They only see the end result and think that person got lucky: “he must have been at the right place at the right time.”

"A SMOOTH SEA NEVER MADE A SKILLFUL MARINER. EVERYTHING IS DIFFICULT BEFORE IT BECOME EASY. WE CAN NOT RUNAWAY FROM OUR PROBLEMS. ONLY LOSERS QUIT AND GIVE UP."

One day a partially deaf four year old child came to home with a note in his pocket from his teacher , “ your tommy is too stupid too learn, get him out of school.” His mother read the note and answered, “ my tommy is not too stupid to learn,I will teach him myself.” And that tommy grew up and become the great scientist Thomas Edison. Thomas Edison had only three months of formal schooling.

LEARN TO LIKE THE THINGS THAT NEED TO BE DONE:-

"START BY DOING WHAT IS NECESSARY, THEN WHAT IS POSSIBLE; SUDDENLY YOU ARE DOING THE IMPOSSIBLE."

“LEARNING IS LOT LIKE EATING. IT IS NOT HOW MUCH YOU EAT THAT MATTER, WHAT MATTER HOW MUCH YOU DIGEST.”

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SOLVED NUMERICALS ON MOLE CONCEPT AND POAC :- BOOST UP YOUR CONCEPT ABOUT "APPLICATION OF POAC AND MOLE CONCEPT"

Q.1 what amount of CaO will be produced by 1 gm. of Ca?

This numerical we can solve by two methods

 1.     Mole concept  2. POAC method

Advantage of POAC over Mole concept is that we need not to write down the stoichiometry of reaction . means in POAC method NEED NOT to  BALANCE THE CHEMICAL REACTIONS

By mole concept

First wite down the BALANCE CHEMICAL REACTION

2Ca + O₂ → 2CaO

It means when 2mole of ca ( 2 *40=80 gm) react with 1 mole of oxygen it produce 2 

mole of CaO (2*56= 112 gm).

80 gm Ca produce = 112 gm Cao

So 1 gm Ca will produce = 112/ 80 = 1.4 gm CaO (unitary method)

By POAC (ADAVANTAGE OVER mole concept NO NEED TO BALANCE EQUATION)

Ca + O₂ → CaO

 Apply POAC on Ca

Total no. of moles of Ca atoms in reactant side = Total no. of moles of Ca atoms in product side

1* no. of moles of Ca = 1* no. of moles of CaO

 No. of moles of Ca /  no. of moles of CaO = 1

So mole of CaO produced in reaction would be equal to mole of Ca  used in chemical reaction

because we are using 1 gm Ca means ( no. of mole =mass/ atomic wt ; here atomic wt

 of Ca = 40 ; so  no. moles of Ca =1/40 =0.025 moles)  0.025  moles of Ca ,the total moles of Cao would be same means 0.025 moles of Cao means (0.025 * 56 = 

1.4) 1.4 gm Cao . Here m.wt of CaO =56 gm

POAC method is a good method to solve these type of numerical because you can  apply POAC concept directly without balancing of equation. So u can save your time  in competitive exams. Even in starting students may feel to apply POAC something difficult but actually it is better than mole concept and time saving method. So students  have to be practice to solve this type of numerical by POAC method. Here some numericals are solved by POAC method to clarify your concept about POAC .observe these numericals to improve your understanding about POAC.

Q.2 A sample of  KClO₃ on decomposition yielded 448 ml.of oxygen gas at N.T.P. calculate

i. weight of oxygen produced

ii. weight of KClO₃ originally taken, and

iii. wt. of KCl produced

solution:-

i. Calculation of  Wt. of  O₂

  VOL. of O₂ = 448 ml = 0.448 litre

  Mole of  O₂ = 0.448/22.4 = 0.02 moles means

  Mass = no. of mole *  m.wt = 0.02 * 32 = 0.064 gm

ii.Calculation of  Wt of KClO₃

KClO₃  → KCl + O₂
Apply POAC on O atom

3* no. of moles of KClO₃ = 2* no. of moles of O₂

No. of moles of KClO₃ / No. of moles of O₂ = 2/3

No. of moles of KClO₃ = 2/3 * no. of moles O₂

No. of moles of KClO₃ = 2/3 * 0.02 = 0.0133 moles

Mass of  KClO₃ = 0.0133 * 122.5 = 1.633 gm  

Here 122.5 is the m.wt of  KClO₃

iii.Calculation of wt. of  KCl produced 

Apply POAC on K atoms

1* no. of mole of KClO₃ = 1* no. of mole of KCl

So no. of mole of  KCl = NO. of mole of KClO₃ = 0.0133 moles

Mass of KCl = 0.0133 * 74.5 = 0.99 gm

Here 74.5 is the m.wt. of  KCl 

Q.3 A mixture of  KBr and NaBr weighing 0.560 gm was treated with aq. Ag+ ion and all bromide ion was recovered as 0.970 gm of pure AgBr. What was the friction by weight of KBr in the sample?

Solution :- ( KBr   +   NaBr)   + Ag⁺ → AgBr

              (x gm      0.560 –x gm)                                0.970 gm

Because no. of moles of Br atoms are conserved in reaction so

Apply POAC on Br atom

Moles of Br in KBr + moles of Br in NaBr = moles of Br in AgBr 

1* no.of mole of KBr + 1*no. of mole of NaBr =   1* no. of mole of AgBr

Mass of KBr/ m.wt of KBr + Mass of NaBr / M.wt of NaBr  = Mass of AgBr/ M.wt  of AgBr

x/ 119 + 0.560- x / 103 = 0.97/188

x = 0.1332 gm

So KBr in the mixture is 0.1332 gm

Fraction of KBr  in mixture = 0.1332/ 0.560 = 0.2378

Q.4 27.6 gm of K₂CO₃  was treated by a series of reagents so as to convert all of its carbon to K₂Zn₃[ Fe(CN)₆]₂. Calculate the weight of the product.

Solution: - K₂ CO₃   → → → → K₂Zn₃[ Fe(CN)₆]₂

No. of moles of C atoms are conserved in the reaction

Apply POAC on C atoms

Moles of C atoms in K₂ CO₃   = Moles of C atoms in K₂Zn₃[ Fe(CN)₆]₂

1* no. of moles of  K₂ CO₃   =  12* no. of moles of  K₂Zn₃[ Fe(CN)₆]₂

Now you can calculate mass of K₂Zn₃[ Fe(CN)₆]₂

Hint:-

1* mass of K₂ CO₃  / m.wt  of K₂ CO₃  = 12* mass of  K₂Zn₃[ Fe(CN)₆]₂ /m.wt of K₂Zn₃[ Fe(CN)₆]

m.wt of  K₂Zn₃[ Fe(CN)₆]₂ = 698

m.wt of  K₂ CO₃   = 138

answer  mass of K₂Zn₃[ Fe(CN)₆]₂  =   11.6 gm

Mole concept:

The mole is defined as the amount of a substance containing as many atoms, molecules, ions, electrons or other elementry entities as there are carbon atoms in exactly 12 g of ¹²C.

N_A = 6.022 X 10²³

The following are the definitions of 'mole' represented in the form of equations:

(1) no. of moles of molecules/atoms =        weight in g     

                                                             molecular/atomic wt

(2) no. of moles of gases =                volume at NTP            

                                            standard molar volume (i.e. 22.4 L)

(3) no. of moles of entities =                  no. of entities                

                                           Avagadro constant (i.e. 6.022 X 10²³)

(4) no. of moles of solute = molarity X volume of solution in L

    no. of millimoles = molarity X volume of solution in mL

(5) For a compound M_xN_y, x moles of N = y moles of M

 Note: 1 mole is a fixed no. of particles but not a fixed weight.

Principle of Atom Conservation (POAC):

This principle states that the moles of atoms of an element are conserved throughout the reaction.

This means, moles of an element in reactants = moles of the element in products.

eg. CO + O₂ -----> CO₂

In this moles of C in CO = moles of C in CO₂

This means, moles of CO = moles of CO₂.

Similarly, for O,

moles of O in CO + moles of O in O₂ = moles of CO₂

Thus, moles of CO + 2(moles of O₂) = 2(moles of CO₂)

Advantages of Mole Method over other Methods:

(1) Balancing of chemical equations is not required in the majority of problems as the method of balancing the chemical equation is based on the principle of conservation of atom conservation.

(2) Number of reactions and their sequences, leading from reactants to products, need not be given.

Note: Whenever balanced chemical equation is given, mole method is very useful.

eg. 2KClO₃ -----> 2KCl + 3O₂

Thus, 2(moles of KClO₃) = 2(moles of KCl)

     3(moles of KClO₃) = 2(moles of O₂)

     3(moles of KCl) = 2(moles of O₂)

Beware of Half –Truth or Misrepresentation of Truth

There was a sailor who worked on the same boat for three years. One night he got drunk this was the first time it had ever happened. The captain recorded it in the log, “The sailor was drunk tonight. “ The sailor read it, and he knew this comment would affect his career, so he went to captain, apologized and asked the captain to add that it only happened once in three years because that was complete truth. The captain refused and said, “What I have written in the log is the truth.”

         The next day it was the sailor’s turn to fill in the log. He wrote, “The captain was sober tonight.” The captain read the comment and asked the sailor to change or add to it explaining the complete truth because this implied that captain was drunk every other night. The sailor told the captain that what he written in the log was the truth.

Both statements were truth but they conveyed misleading messages.
So always beware of HALF -TRUTH.

  There was a man who made his living selling balloons at a fair. He had balloons of many different colors, including red, yellow, blue, green. Whenever business was slow, he would release a helium filled balloon into the air. When the children saw the balloon go up, they all wanted one. They would come up to him, buy a balloon and his sales would go up. All day, he continued to release a balloon whenever his sales down. One day the balloon man felt someone tugging at his jacket. He turned around and a little boy asked, " if you release a black balloon, would that also fly?” moved by the boy concern, the man replied gently, “Son, it is not the color of the balloon, it is what’s inside that makes it go up.”  

The same principle applies to our lives: it’s what’s inside us that counts

LATEST NEET UPDATE :- NEET ( National Eligibility Entrance Test) for admission to medical & dental colleges is CANCELLED by supreme court.

 Supreme Court (SC) on Thursday, July 18  issued orders to the effect of cancelling National Eligibility Entrance Test (NEET) for admissions to medical and dental courses, post graduation courses in government and private colleges.

It is the most significant SC judgment, denying the right of Indian Medical Council (IMC) to hold uniform entrance test across the nation to seek admissions in both government and private medical colleges. 

The verdict will not hinder the admissions this year who are qualified in NEET, said SC chief judge Altamus Kabir and advocate Vikramjit Sen.

SOLID STATE "BRAVAIS LATTICE"

The three Bravais lattices which form cubic crystal systems are:

Cubic Bravais lattices

Name	Primitive cubic	Body-centered cubic	Face-centered cubic
Pearson symbol	cP	cI	cF
Unit cell

The primitive cubic system (cP) consists of one lattice point on each corner of the cube. Each atom at a lattice point is then shared equally between eight adjacent cubes, and the unit cell therefore contains in total one atom (¹⁄₈ × 8).
The body-centered cubic system (cI) has one lattice point in the center of the unit cell in addition to the eight corner points. It has a net total of 2 lattice points per unit cell (¹⁄₈ × 8 + 1).
The face-centered cubic system (cF) has lattice points on the faces of the cube, that each gives exactly one half contribution, in addition to the corner lattice points, giving a total of 4 lattice points per unit cell (¹⁄₈ × 8 from the corners plus ¹⁄₂ × 6 from the faces).

The 7 lattice systems The 14 Bravais lattices Triclinic P Monoclinic P C Orthorhombic P C I F Tetragonal P I Rhombohedral P Hexagonal P Cubic P (pcc) I (bcc) F (fcc)

Lattice System	Possible Variations	Axial Distances (edge lengths)	Axial Angles	Examples
Cubic	Primitive, Body-centred, Face-centred	a = b = c	α = β = γ = 90°	NaCl, Zinc Blende, Cu
Tetragonal	Primitive, Body-centred	a = b ≠ c	α = β = γ = 90°	White tin, SnO2, TiO2, CaSO4
Orthorhombic	Primitive, Body-centred, Face-centred, Base-centred	a ≠ b ≠ c	α = β = γ = 90°	Rhombic sulphur, KNO3, BaSO4
Hexagonal	Primitive	a = b ≠ c	α = β = 90°, γ = 120°	Graphite, ZnO, CdS
Rhombohedral	Primitive	a = b = c	α = β = γ ≠ 90°	Calcite (CaCO3, Cinnabar (HgS)
Monoclinic	Primitive, Base-centred	a ≠ b ≠ c	α = γ = 90°, β ≠ 90°	Monoclinic sulphur, Na2SO4.10H2O
Triclinic	Primitive	a ≠ b ≠ c	α ≠ β ≠ γ ≠ 90°	K2Cr2O7, CuSO4.5H2O, H3BO3