Empirical and Molecular Formulas Section 3.5 Empirical and Molecular Formulas

Empirical Formulas Tells us relative number of atoms of each element it contains Example: H2O: 2 atoms of H per 1 atom of O ALSO: H2O: 2 mol of H per 1 mol of O Mole concept allows us to calculate the empirical formula

Example Compound is 73.9% Hg and 26.1% Cl. Find empirical formula. Assume 100 grams 73.9 g Hg x (1 mol Hg) = 0.368 mol Hg 200.6 g Hg 26.1 g Cl x (1 mol Cl) = 0.735 mol Cl 35.5 g Cl mol Cl = 0.735 mol Cl = 1.99 mol Cl = 2 = HgCl2 mol Hg 0.368 mol Hg = 1 mol 1

Sample Exercise 3.13 p. 97 Ascorbic acid: 40.92% C, 4.58% H, and 54.50 %O Always divide the larger numbers (C and H) of moles by the smallest number of moles (O)

Molecular Formula Have to find Empirical Formula first!!! Whole number multiple = molecular weight empirical formula weight Multiple each subscript of the empirical formula by the whole number multiple

Back to Sample Exercise 3.13 Empirical formula = C3H4O3 1st, find empirical formula weight: 3(12.0) + 4(1.0) + 3(16.0) = 88.0 amu Given: experimentally determined molecular weight = 264 amu W.N.M.= molecular weight = 264 = 3 empirical formula weight 88 Molecular Formula = C9H12O9

Homework 3.44- 3.50 even only on page 113-114