1 / 18

Chapter 3

Chapter 3. Molecules, Compounds, & Chemical Equations. Hydrogen, Oxygen, & Water. What is the main difference between a mixture and a compound?. Ball-and-stick model ***check pg 87 and A-7 (appendix IIA) for colors. Space-filling molecular model.

aoife
Download Presentation

Chapter 3

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. Chapter 3 Molecules, Compounds, & Chemical Equations

  2. Hydrogen, Oxygen, & Water

  3. What is the main difference between a mixture and a compound?

  4. Ball-and-stick model ***check pg 87 and A-7 (appendix IIA) for colors Space-filling molecular model

  5. Example: Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the empirical formula for methyl acetate, which has the following chemical analysis: 48.64% C, 8.16% H, and 43.20% O. Step 1: Determine the number of moles of each element in the compound. 48.64 g C 1 mol C = 4.050 mol C 12.01 g C 8.16 g H 1 mol H = 8.10 mol H 1.01 g H 43.20 g O 1 mol O = 2.70 mol O 16.00 g O

  6. Step 2: Calculate the simplest ratio of moles of the elements by dividing each number of moles by the smallest value in the mole ratio. 4.050 mol C = 1.5 mol C 2.70 8.10 mol H = 3 mol H 2.70 2.70 mol O = 1 mol O 2.70

  7. Step 3: Multiple the numbers of moles in the ratio by the smallest number that will produce a ratio of whole numbers. (This step is not always needed) 2 × 1.5 mol C = 3 mol C 2 ×3 mole H = 6 mol H 2 × 1 mole O = 2 mol O The empirical formula is C3H6O2

  8. Example: Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% carbon, 5.08% hydrogen, and 54.24% O and has a molecular mass of 118.1 g/mol. Determine the empirical and molecular formulas for succinic acid. 48.68 g C 1 mol C = 3.387 mol C 12.01 g C 5.08 g H 1 mol H = 5.04 mol H 1.01 g H 54.24 g O 1 mol O = 3.390 mol O 16.00 g O

  9. 3.387 mol C = 1 mol C 3.387 5.040 mol H = 1.5 mol H 3.387 3.390 mol O = 1 mol O 3.387 2 × 1 mol C = 2 mol C 2 × 1.5 mol H = 3 mol H 2 × 1 mol O = 2 mol O The empirical formula is C2H3O2

  10. Calculate the empirical formula mass using the molar mass of each element. C = 2(12.01) = 24.02 g C H = 3(1.01) = 3.024 g H O = 2(16.0) = 32.0 g O Molar mass C2H3O2 = 59.04 g/mol Divide the experimentally determined molar mass of succinic acid by the mass of the empirical formula to determine n. n = Molar mass of succinic acid = = 2.00 118.1 g/mol Molar mass of C2H3O2 59.04 g/mol

  11. Multiple the subscripts in the empirical formula by 2 to determine the actual subscripts in the molecular formula. (C2H3O2) (2) = C4H6O4 The molecular formula for succinic acid is C4H6O4

  12. Example: Upon combustion, a 0.8233 g sample of a compound containing only carbon, hydrogen, and oxygen produced 2.445 g CO2 and 0.6003 g H2O. Find the empirical formula of the compound. Step 1: Convert the masses of CO2 and H2O to moles. 2.445 g CO2 1 mol CO2 = 0.05556 mol CO2 44.01 g CO2 0.6003 g H2O 1 mol H2O = 0.03331 mol H2O 18.02 g H2O

  13. Step 2: Convert moles of CO2 and H2O to moles of C and H using ratios of C and H in compounds. 1 mol C 0.05556 mol CO2 = 0.05556 mol C 1 mol CO2 0.03331 mol H2O 2 mol H2O = 0.06662 mol H 1 mol H2O

  14. Step 3: If the compound contains an element other than C and H, find the mass of the other element by subtracting the sum of the masses of C and H from the total mass of the sample. Convert the mass of the other element to moles Mass C = 0.0556 mol C 12.01 g C = 0.6673 g C 1 mol C Mass H = 0.06662 mol H 1.01 g H = 0.06715 g H 1 mol H Mass O = 0.8233 g − (0.6673 g + 0.06715 g) = 0.0889 g O 0.0889 g O 1 mol O = 0.00556 mol O 16.00 g O

  15. Step 4: Calculate the simplest ratio of moles of the elements by dividing each number of moles by the smallest value in the mole ratio. 0.05556 mol C = 10 mol C Empirical Formula: C10H12O 0.00556 0.06662 mol H = 12 mol H 0.00556 0.00556 mol O = 1 mol C 0.00556 Step 5: (If needed) Multiple the numbers of moles in the ratio by the smallest number that will produce a ratio of whole numbers.

More Related