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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.

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Chapter 3

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  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.

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