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Understanding Molecular Formulas: Steps and Examples for Acetic Acid, Formaldehyde, and Glucose

This document explains the concept of molecular formulas, which specify the number of atoms for each element in a molecule. It includes steps to determine the molecular formula, starting from the empirical formula and its molar mass. Two examples illustrate the calculations for compounds such as acetic acid and glucose, detailing the percentages of carbon, hydrogen, and oxygen, as well as the molar mass and the derivation of empirical and molecular formulas. This provides clarity on how to effectively determine the molecular structures of various compounds.

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Understanding Molecular Formulas: Steps and Examples for Acetic Acid, Formaldehyde, and Glucose

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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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