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Message: <blockquote data-quote="ashishsinha" data-source="post: 2118161" data-attributes="member: 755260">carbon--percentage = 74.03atomic mass = 12number of moles = 6.17hydrogen--percentage = 8.70atomic mass = 1number of moles = 8.70/1 = 8.70nitrogen--percentage  = 17.27atomic mass = 14number of moles  = 1.23simples molar ratio of carbon = 6.17/1.23 = 5.01 or 5simplest molar ratio of hydrogen = 8.70/1.23 = 7.07 or 7simples molar ratio of nitrogen = 1.23/1.23 = 1so empirical formula is C5H7Nempirical mass = 12 X 5 + 7 X 1 + 14 = 81 g/molemolecular mass of Nicotine = 162 g/molemolecular mass / empirical mass = 162/81 = 2so molecular formula = (C5H7N)2 = C10H14N2see this link--http://chemistry.about.com/od/workedchemistryproblems/a/empirical.htm</blockquote>