Atoms, Molecules, and Stoichiometry

1. Atoms, Molecules, and Stoichiometry Recap of Lecture 5

2. Actual Yield Theoretical Yield Percentage Yield = X 100% Recap Combustion Analysis Values given in terms of mass Values given in terms of volume Find the mass of each element  Find mole ratio and hence empirical formula Make use of Avogadro’s law to compare volumes

3. = 0.409 g = MC MCO2 12.0 44.0 x mCO2 x 1.500 Mass of C MC = Mass of CO2 MCO2 Combustion Analysis Example 1: Complete combustion of an unknown organic compound of mass 1.000g containing only carbon, hydrogen, and oxygen, gave 1.500g of carbon dioxide and 0.405g of water. What is the empirical formula of the organic compound? Using mass proportion: Mass of C in 1.500g CO2 =

4. = 0.409 g = MC MCO2 12.0 44.0 x mCO2 x 1.500 Combustion Analysis Example 1: Complete combustion of an unknown organic compound of mass 1.000g containing only carbon, hydrogen, and oxygen, gave 1.500g of carbon dioxide and 0.405g of water. What is the empirical formula of the organic compound? Using mass proportion: Mass of C in 1.500g CO2 = MC x Mass of CO2 Mass of C = MCO2

5. 2 x MH MH2O x mH2O 2 x 1.0 18.0 x 0.405 = Combustion Analysis Example 1: Complete combustion of an unknown organic compound of mass 1.000g containing only carbon, hydrogen, and oxygen, gave 1.500g of carbon dioxide and 0.405g of water. What is the empirical formula of the organic compound? Mass of H in 0.405g H2O = = 0.045 g Mass of oxygen in the compound = 1.000 – 0.409 – 0.045 = 0.546 g

6. 3 4 3 Combustion Analysis C H O Mass ratio 0.409 0.045 0.546 Mole ratio 0.40912.0 0.0451.0 0.54616.0 = 0.0341 = 0.045 = 0.0341 Simplest whole number ratio 0.03410.0341 0.0450.0341 0.03410.0341 = 1 = 1.320 = 1 Compound’s empirical formula is C3H4O3

7. CXHY O2 CO2 H2O 1 mol (x + y/4) mol x mol y/2 mol Mole ratio 1 cm3 (x + y/4) cm3 x cm3 y/2 cm3 Volume ratio 10 cm3 10(x + y/4) cm3 10x cm3 10 (y/2) cm3 Combustion Analysis Avogadro’s Law: no of moles of gas  volume of gas Provided that temperature and pressure are kept constant This means that, Mole ratio is equivalent to the Volume ratio CXHY + (x + y/4) O2 x CO2 + y/2 H2O

8. nCO2 nCxHy VCO2 VCxHy = x 1 20 10 = = Combustion Analysis Example 1 When 10 cm3 of a gaseous hydrocarbon was completely burnt in excess oxygen, 20 cm3 of carbon dioxide and 30 cm3 of steam was produced (all gas volumes measured under the same conditions). Calculate the molecular formula of the hydrocarbon. CXHY + (x + y/4) O2 x CO2 + y/2 H2O By Avogadro’s law,  x = 2

9. nH2O nCxHy VH2O VCxHy = y/2 1 30 10 = = Combustion Analysis Example 1 When 10 cm3 of a gaseous hydrocarbon was completely burnt in excess oxygen, 20 cm3 of carbon dioxide and 30 cm3 of steam was produced (all gas volumes measured under the same conditions). Calculate the molecular formula of the hydrocarbon. CXHY + (x + y/4) O2 x CO2 + y/2 H2O By Avogadro’s law,  y = 6 Molecular formula is C2H6

10. 19 Combustion Analysis Points to note: • The volume of CO2 may be given as a decrease in volume when the residual gases are passed through NaOH or other alkali • The volume of H2O may be given as • Decrease in volume when the residual gases are passed through anhydrous CaCl2 or conc. H2SO4 • Decrease in volume when the residual gases are cooled to below 100oC at atm pressure • Oxygen is usually added in excess and not in stoichiometric amount, so there will be excess oxygen as residual gas • Contraction = volume of reactants – volume of products • Expansion = volume of products – volume of reactants

11. liquid state CXHY + (x + y/4) O2 x CO2 + y/2 H2O Example 2 In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon.

12. liquid state nCO2 nCxHy VCO2 VCxHy  = x 1 40 10 = CXHY + (x + y/4) O2 x CO2 + y/2 H2O Example 2 In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon. Volume of CO2 gas = 40.0 cm3 [absorbed by aq NaOH] By Avogadro’s Law,  x = 4

13. CXHY O2 CO2 H2O 1 mol (x + y/4) mol x mol y/2 mol Mole ratio Volume ratio 10 cm3 10(x + y/4) cm3 10x cm3 0 cm3 CXHY + (x + y/4) O2 x CO2 + y/2 H2O Example 2 In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon.

14. CXHY O2 CO2 H2O 1 mol (x + y/4) mol x mol y/2 mol Mole ratio Volume ratio CXHY + (x + y/4) O2 x CO2 + y/2 H2O Example 2 In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon. 10 cm3 10(x + y/4) cm3 40 cm3 0 cm3

15. CXHY + (x + y/4) O2 x CO2 + y/2 H2O Example 2 In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon. 35.0 = Initial volume of gases – final volume of gases 35.0 = { 10 + 10( x + y/4 ) + excess O2 } – { 40 + excess O2 } 65 = 10(x+y/4) Since x = 4, solving y = 10 Hence the empirical formula of the hydrocarbon is C4H10

16. nCO2 nCxHy VCO2 VCxHy  = x 1 40 10 = 20 CXHY + (x + y/4) O2 x CO2 + y/2 H2O Example 3 In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with 100.0 cm3 of oxygen. The volume of the resulting gas mixture was 80.0 cm3, which decreased to 40.0 cm3 when passed through aqueous sodium hydroxide. All volumes were measured at room temperature and pressure. What is the formula of the hydrocarbon? Volume of CO2 produced = 80.0 – 40.0 = 40.0 cm3  x = 4 Volume of resulting gas mixture = 80.0 cm3 = volume of CO2 + volume of unreacted O2 Hence volume of unreacted O2 = 80.0 – 40.0 = 40.0 cm3

17. x + y/4 1 60 10 = nO2 nCxHy VO2 VCxHy = CXHY + (x + y/4) O2 x CO2 + y/2 H2O Example 3 In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with 100.0 cm3 of oxygen. The volume of the resulting gas mixture was 80.0 cm3, which decreased to 40.0 cm3 when passed through aqueous sodium hydroxide. All volumes were measured at room temperature and pressure. What is the formula of the hydrocarbon? Therefore volume of O2 used up in the reaction = 100.0 – 40.0 = 60.0 cm3  y = 8 The formula for the hydrocarbon is C4H8

18. What have I learnt? • Determine the formula of a hydrocarbon given the combustion analysis data • In terms of mass • In terms of volume

19. End of Lecture 6(End of part I) Have a prosperous and happy lunar new year(and do take care of your health!)