Standard/ Class/ Grade XI ChemistryChapter 1 Basic ConceptsGurudatta K Wagh, gkwagh@gmail.com

Percentage Composition, Empirical and Molecular FormulaEthanol C2H5OH

Percent compositionCarbon % = 24/ 26 X 100 = 52.17 %Oxygen % = 16/ 46 X 100 = 34.78 %Hydrogen % = 6/ 46 X 100 = 13.05 %

Ethanol 100 g = C 52.17 g + O 34.78 g + H 13.05 g

Element Mole Atomic mass Mass (g)Hydrogen 6 1 6Carbon 2 12 24Oxygen 1 16 16

Ethanol C2H5OH 46

Number of moles and ratio

C number of moles = 52.17/ 12 = 4.3475O number of moles = 34.78/ 16 = 2.1735H number of moles = 13.05/ 1 = 13.05

Ratio of number of moles O : C : H = 2.1735 : 4.3475 : 13.052.1735/ 2.1735 : 4.3475/ 2.1735 : 13.05/ 2.1735 = 1 : 2 : 6

Empirical formula Molecular formulaIndicates the relative number of atoms in the simplest ratio

Indicates the actual number of constituent atoms in a molecule ratio

Ethanol C2H6O Ethanol C2H5OH or CH3CH2OH

Ratio r molecular mass/ empirical mass = 46/ 46 = 1Molecular mass = Empirical mass X r = 46 X 1 = 46

Ratio r molecular mass/ empirical mass = 46/ 46 = 1Molecular mass = Empirical mass X r = 46 X 1 = 46

ImportantThe sum of the percentages of the constituent elements must be hundredThe number of moles of each constituent in 100 g is obtained by dividing percentage of each element by its atomic massRatio of number of moles of constituent elements is determined and converted into smallest simple whole numberIf the ratio is a fraction, it is multiplied by a convenient integer to obtain a whole number ratioWhole numbers are written as the subscripts to the right side of the respective elementMolecular mass = Empirical mass X r