1 / 7

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.

Download Presentation

Molecular Formula

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related