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Molecular Formula

Molecular Formula. Acetic Acid. Formaldehyde. Glucose. CH 2 O. 30.03 g/ mol. (30.03 x 1) = 30.03 g/ mol. CH 2 O. 30.03 g/ mol. (30.03 x 2) = 60.06 g/ mol. CH 2 O. 30.03 g/ mol. (30.03 x 6) = 180.18 g/ mol. Definition.

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Molecular Formula

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  1. Molecular Formula

  2. Acetic Acid Formaldehyde Glucose

  3. CH2O 30.03 g/mol (30.03 x 1) = 30.03 g/mol CH2O 30.03 g/mol (30.03 x 2) = 60.06 g/mol CH2O 30.03 g/mol (30.03 x 6) = 180.18 g/mol

  4. Definition Molecular Formula: formula that specifies the actual number of atoms of each element in one molecule or formula unit of the substance. It is a multiple of the empirical formula

  5. Steps • Find the empirical formula • Calculate the molar mass of the empirical formula • Divide the molar mass given, by the molar mass of the empirical formula to get the multiplier • Multiply empirical formula subscripts by the multiplier to get the molecular formula

  6. Example 1 A compound composed of 40.68% carbon, 5.08% hydrogen and 54.24% oxygen has a molar mass of 118.1 g/mol. Determine the empirical and molecular formula. E.F. = C2H3O2 40.68 g C 1 mol C 12.01 g C 5.08 g H 1 mol H 1.01 g H 54.24 g O 1 mol O 16.00 g O Molar Mass: C: 12.01 ×2 H: 1.01 ×3 + O: 16.00 ×2 59.05g = 5.03 mol H 3.387 = 3.387 mol C 3.387 = 3.390 mol O 3.387 = 1.49 molH ≈ 1.5 ×2 = 3 = 1.00 mol C ×2 = 2 = 1.00 mol O ×2 = 2 118.1 g 59.05g = 2 M.F. = C4H6O4

  7. Example 2 A compound has a molar mass of 462.8 g/mol and contains 77.87% C, 11.76% H, and 10.37% O. Determine the empirical and molecular formulas. E.F. = C10H18O 77.87g C 1 mol C 12.01 g C 11.76g H1 mol H 1.01 g H 10.37g O 1 mol O 16.00 g O Molar Mass: C: 12.01 ×10 H: 1.01 ×18 + O: 16.00 154.28 g = 11.64mol H 0.6481 = 6.484mol C 0.6481 = 0.6481mol O 0.6481 = 17.99molH ≈ 18 = 10 molC = 1.00 mol O 462.8g 154.28 g = 3 M.F. = C30H54O3

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