1. Unit 3 the mole, stoichiometry

4. yesterday... You translated the reaction equation (in moles) into practical lab measurements (in grams). C 8 H 18 +O2 →O2 →CO 2 +H2OH2O8912.51 2251618 2 mol C 8 H 18 114.2302 g C 8 H 18 1 mole C 8 H 18 = 228.4604 g C 8 H 18 25 mol O 2 31.9988 g O 2 1 mole O 2 = 799.97 g O 2 BUT we might want to use different amounts of fuel. (Half a pound is quite a bit of gasoline.)

5. To do this, we need one more type of conversion factor: from moles to moles. These conversion factors are based on the coefficients in the balanced reaction equation. C 8 H 18 +O2 →O2 →CO 2 +H2OH2O2251618 2 mole C 8 H 18 25 mole O 2 examples: This means that 25 moles of O 2 are consumed by every 2 moles of gasoline (C 8 H 18 ) that burn. Like all other conversion factors you have used, it is equally true when inverted. 2 mole C 8 H 18 25 mole O 2

6. You can make many more mole-mole conversion factors from this equation. C 8 H 18 +O2 →O2 →CO 2 +H2OH2O2251618 2 mole C 8 H 18 16 mole CO 2 more examples: Every 2 moles of gasoline (C 8 H 18 ) that burn make 16 moles of CO 2. 2 mole C 8 H 18 16 mole CO 2 …and the inverse… …and still more… 25 mole O 2 16 mole CO 2 25 mole O 2 16 mole CO 2 2 mole C 8 H 18 18 mole H 2 O 2 mole C 8 H 18 18 mole H 2 O

7. Now use all of these together: gram→mole, mole→mole, and mole→gram. Example: How much oxygen is required to burn 0.75 g of gasoline? C 8 H 18 +O2 →O2 →CO 2 +H2OH2O2251618 0.75 g C 8 H 18 114.2302 g C 8 H 18 1 mole C 8 H 18 2 mole C 8 H 18 25 mole O 2 = 2.626… g O 2 31.9988 g O 2 1 mole O 2 (0.75)(25)(31.9988) g O 2 (114.2302)(2) = ≈ 2.6 g O 2

8. Notice the pathway you followed: 0.75 g C 8 H 18 114.2302 g C 8 H 18 1 mole C 8 H 18 2 mole C 8 H 18 25 mole O 2 31.9988 g O 2 1 mole O 2 grams of A moles of A moles of B grams of B Notice the source of each conversion… molar mass molar mass balanced equation molar mass molar mass

9. Try to set this one up yourself. Example: How much carbon dioxide is made when burning 0.75 g of gasoline? C 8 H 18 +O2 →O2 →CO 2 +H2OH2O2251618 0.75 g C 8 H 18 114.2302 g C 8 H 18 1 mole C 8 H 18 2 mole C 8 H 18 16 mole CO 2 = 2.3116… g CO 2 44.0098 g CO 2 1 mole CO 2 (0.75)(16)(44.0098) g CO 2 (114.2302)(2) = ≈ 2.3 g CO 2 HW: unit 3 #11helpful reading: pp. 302-307

10. finishing Lab 3.1 put on goggles, get key get your beaker from oven measure beaker mass with silver residue scrape silver into “retirement fund” beaker wash your beaker, put things away begin lab calculations and/or homework ask questions about calcs tomorrow HW: unit 3 #11helpful reading: pp. 302-307