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.