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Section 3.5

Empirical and Molecular Formulas. Section 3.5. Empirical Formulas. Tells us relative number of atoms of each element it contains Example: H 2 O: 2 atoms of H per 1 atom of O ALSO: H 2 O: 2 mol of H per 1 mol of O Mole concept allows us to calculate the empirical formula. Example.

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Section 3.5

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  1. Empirical and Molecular Formulas Section 3.5

  2. 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

  3. 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

  4. 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)

  5. 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

  6. 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

  7. Homework • 3.44- 3.50 even only on page 113-114

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