Principle of atom conservation (PAOC):

We know that, in a chemical reaction, the atoms of each element remain conserved. If atoms are conserved, moles of atoms will also be conserved. This is known as the principle of atom conservation.

Total number of moles of an atoms of an element on reactant side = Total number of moles of atoms of an on product side.

Illustration of POAC: Take a chemical reaction as example.

(A) According to the principle of atom conservation (POAC) for K atoms:

    Total moles of K atoms in reactant = total mole of K atoms in products

or

 Moles of K atoms in KClO₃ = moles of K atoms in KCl.

Now, since 1 molecule of KClO₃ contains 1 atom of K, 1mole of KClO₃ contains 1 mole of K similarly, 1mole of KCl contains 1mole of K.

     Thus, mole of K atoms in KClO₃ = 1 × moles of KClO₃

        And mole of K atoms in KCl = 1 × moles of KCl

Hence            1 × moles of KClO₃ = 1 × moles of KCl

 (B) Similarly applying the principle of atom conservation for O atoms:

         Mole of O in KClO₃ = 3 ×moles of KClO₃

 And moles of O in O₂ = 2 × moles of O₂ 

      3 × moles of KClO₃ = 2 × moles of O₂

Illustrative example (1): Write the POAC equations for all the atoms in the following reaction.

Illustrative solution:

Applying POAC on (P) atoms:

                                                          P à H₃PO₄

Numbers of moles of (P) atoms in P₄ = Numbers of moles of (P) atoms in H₃PO₄

                                                         4x nP₄= 1x nH₃PO₄

Applying POAC on (H) atoms:

                                       1x nHNO3 = 3 x nH₃PO₄ + 2 x nH₂O

Applying POAC on (N) atoms:

                                       1 x nHNO₃ = 1 x nNO₂

Applying POAC on (O) atoms:

                                       3 x nHNO₃ = 4 x nH₃PO₄ +2 x nNO₂ + 1nH₂O

Related Questions:

Illustrative example (2) 27.6 gm K2CO3 (138) was treated by a series of reagents so as to convert all its carbon into K2Zn3[Fe(CN)6]2(MM=698) Calculate mass of K2Zn3[Fe(CN)6]2.

Illustrative example (3)0.32 gm mole of LiAlH4 in ether solution was placed in a flask and 74 gm (1 mole ) of t- butyl alcohol was added . The product is LiAlHC12H27O3. Calculate mass of product if all lithium atoms are converted.

Illustrative example (4) All the carbon atom present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How many gram of CO2 were obtained.

Illustrative example (5) For BaCl2.XH2O , 2.1 gm of Compound gives 2 gm of anhydrous BaSO4 (M.mass =233)upon treatment with H2SO4 then calculate value of X.

Illustrative example (6) In a determination of P an aqueous solution of NaH2PO4 is treated with a mixture of Ammonium and magnesium ions to precipitate magnesium Ammonium phosphate (Mg(NH4)PO4.6H2O. This is heated and decomposed to magnesium pyrophosphate. Which is weighed. A solution of NaH2PO4 yield 3.33 gm of Mg2P2O7. What is weighed of NaH2PO4 was present originally.

DEGREE OF HARDNESS OF HARD WATER:

(1)The amount of hardness causing substances (soluble salts of calcium or magnesium) in a certain volume of water measures the extent of hardness or degree of hardness.

(2) Hardness of water is always calculated in terms of calcium carbonate although this is never responsible for causing hardness of water because of its insoluble character.

(3) The reason for choosing CaCO₃ as the standard for calculating hardness of water is the ease in calculation as its molecular weight is exactly 100.Thus the amount of various hardness causing substances in terms of CaCO₃ can be calculated on the basis of the following relations.

Thus the various types of harnesses in a water sample may be calculated as below.

Temporary hardness =Hardness due to Ca(HCO₃)₂ + Hardness due to Mg(HCO₃)₂

Permanent hardness = Hardness due to CaCl₂ + due to CaSO₄ + due toMgCl₂ + due toMgSO₄

Degree of hardness is usually expressed as parts per million (ppm) and thus may be defined as the number of parts by weight of CaCO₃ (equivalent to calcium and magnesium salts) present in a million (10⁶) parts by weight of water.

From the above definition, we can say that;

ILLUSTRATIVE EXAMPLE (1): Determine the degree of hardness of a sample of water containing 30 ppm of MgSO₄.

SOLUTION: 1 MgSO₄ =1CaCO₃

                      120 ppm = 100 ppm

                      Hence 30 ppm = 25 ppm   Ans

EUDIOMETRY-VOLUME-VOLUME ANALYSIS OF GAS:

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.

ILLUSTRATION OF EDUDIOMETRY:
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.

REAGENT USE FOR ABSORPTION OF GASES IN EUDIOMETRIC TUBE:

The various reagents used for absorbing different gases are
O_{3 --} 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 )

SPECIAL NOTE:

(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 

SOLUTION:

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.

SOLUTION:

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 )