Section 3 5 Empirical and Molecular Formulas Empirical
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Section 3. 5 Empirical and Molecular Formulas
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 • 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 = Hg. Cl 2 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 = C 3 H 4 O 3 • 1 st, 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 = C 9 H 12 O 9
Homework • 3. 44 - 3. 50 even only on page 113 -114
