 # Oxidation Number and its Applications

Calculation of % composition :
\tt \% \ composition = \frac{wt. of \ element \ in \ the \ compound}{M.wt \ of \ compound}\times 100
NaOH = M.wt = 40g
\% \ O = \frac{16}{40}\times 100 = 40\%
\% \ H = \frac{1}{40}\times 100 = 2.5\%
\% \ Na = \frac{23}{40}\times 100 = 57.5\%

Empirical and molecular calculations :
E.F : Relative no. of atoms [indicates] or simplest ratio of atoms.
M.F : It indicates actual atoms in different elements.

ex :

 Hydrogen M.F E.F Peroxide H2O2 HO Acetylene C2H2 CH Benzene C6H6 CH

Relation between empirical formula and molecular formula
M.F = n × E.F
n = \frac{M.F\ wt}{E.F\ wt}
M.wt = 2 × vapour density
ex : E.F = CH2Cl
n = \frac{M.\ wt}{E.\ wt} = \frac{98.96}{49.5} = 2
M.F = n × E.F ⇒ 2 × CH2Cl → C2H4Cl2
O.C = organic compound
\% \ C = \frac{12}{24} \times \frac{W_{{CO_{2}}}}{W.O.C} \times 100
\% \ H = \frac{2}{18} \times \frac{W_{{H_{2}O}}}{W.O.C} \times 100
\% \ N = \frac{28 \times V_{ml}}{22400} \times \frac{100}{W.O.C}
\% \ N = \frac{ V_{ml}}{8 \ W.O.C}

Equivalent weight :
E.wt of element = \tt \frac{ A.wt}{Valency}

 Element E.wt Al \frac{27}{3} = 9 Mg \frac{24}{2} = 12 Na \frac{23}{1} = 23 Ca \frac{40}{2} = 20 O \frac{16}{2} = 8

Eq.wt of ions = \tt \frac{Gram \ ion \ wt}{Valency(X)}
Eq.wt of acid = \tt \frac{M.wt}{Basicity}
Equivalent of base = \tt \frac{M.wt}{Acidity(a)}

Equivalent of salt = \tt \frac{M.wt}{No. of \ +ve \ charge \ of \ cation \ (or) \ No. of \ −ve \ charge \ of \ anion}
Equivalent of OA = \tt \frac{M.wt}{No. of \ e's \ gained}
Equivalent of RA = \tt \frac{M.wt}{No. of \ e's \ lost}

### Part3: View the Topic in this Video from 0:40 to 1:05:40

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