# Std XI-Chem-Ch1-Concepts-Percent-composition-Molecular-formula

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Published on December 15, 2016

Author: gkwagh

Source: slideshare.net

1. Standard/ Class/ Grade XI Chemistry Chapter 1 Basic Concepts Gurudatta K Wagh, gkwagh@gmail.com Percentage Composition, Empirical and Molecular Formula Ethanol C2H5OH Percent composition Carbon % = 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 6 Carbon 2 12 24 Oxygen 1 16 16 Ethanol C2H5OH 46

2. Number of moles and ratio C number of moles = 52.17/ 12 = 4.3475 O number of moles = 34.78/ 16 = 2.1735 H number of moles = 13.05/ 1 = 13.05 Ratio of number of moles O : C : H = 2.1735 : 4.3475 : 13.05 2.1735/ 2.1735 : 4.3475/ 2.1735 : 13.05/ 2.1735 = 1 : 2 : 6 Empirical formula Molecular formula Indicates 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 = 1 Molecular mass = Empirical mass X r = 46 X 1 = 46

3. Ratio r molecular mass/ empirical mass = 46/ 46 = 1 Molecular mass = Empirical mass X r = 46 X 1 = 46 Important The sum of the percentages of the constituent elements must be hundred The number of moles of each constituent in 100 g is obtained by dividing percentage of each element by its atomic mass Ratio of number of moles of constituent elements is determined and converted into smallest simple whole number If the ratio is a fraction, it is multiplied by a convenient integer to obtain a whole number ratio Whole numbers are written as the subscripts to the right side of the respective element Molecular mass = Empirical mass X r