 # Stoichiometry and Empirical and Molecular Formulae and Percentage Composition

Chemical formula : (Percentage composition)

• \tt Mass \ percent = \frac{mass \ of \ element \ in \ a \ compound}{molar \ mass \ of \ that \ compound}\times 100
• \tt \% \ by \ mass \ \Rightarrow \left(w/w\right) = \frac{mass\ of \ solute}{mass \ of \ solution} \times 100
• \tt \% \ by \ volume \ \Rightarrow \left(w/v\right) = \frac{mass\ of \ solute}{volume \ of \ solution} \times 100
• \tt ppt \left(parts \ per \ thousand\right) = \frac{Parts \ of \ solute}{Parts \ of \ solution} \times 10^{3}
• \tt ppm \left(parts \ per \ million\right) = \frac{Parts \ of \ solute}{Parts \ of \ solution} \times 10^{6}
• \tt ppb \left(parts \ per \ billion\right) = \frac{Parts \ of \ solute}{Parts \ of \ solution} \times 10^{9}
• Oxidation is addition of ' O ' or removal of ' H ' from a compound.
\tt 4Na + O_{2} \rightarrow 2Na_{2}O
• Reduction is addition of ' H ' or removal of ' O ' from a compound.
\tt CuO + H_{2} \rightarrow Cu + H_{2}O

Oxidation Number :
The number of electrons donated or accepted by an atom during the formation of molecule, represents its oxidation number.
eg : FeCl3 molecule consists of one Fe+3 and three Cl-
Valency and Oxidation Number :
Valency represents combining capacity of an element. Oxidation number of an atom represents charge present on it. It is +ve (or) −ve.
Balancing equation in acid medium :
Add H2O → for ' O ' balanced atom
Add H+ → for ' H ' balanced atom .
Balancing equation in basic medium :
Add H2O → for ' O ' balanced atom
Add H2O → for ' H ' balanced atom and add same number of OH ions on opposite side .

### Part2: View the Topic in this Video from 0:12 to 13:20

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1. \tt Percentage \ of \ element = \frac{Mass \ of \ element}{Molecular \ mass}\times 100 = \frac{B}{M} \times100
M = mass of compound
B = mass of element

2. Molecular formula = n × empirical formula

3. \tt n = \frac{molecular \ weight \ of \ compound}{empirical \ formula \ weight \ of \ compound}