 Atoms, Molecules, and Stoichiometry

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Atoms, Molecules, and Stoichiometry
Recap of Lecture 5

Recap Combustion Analysis Actual Yield Theoretical Yield
Percentage Yield = Actual Yield Theoretical Yield X 100% 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

Combustion Analysis Example 1: Using mass proportion:
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 MCO2 x mCO2 12.0 44.0 x = = g Mass of C Mass of CO2 MC MCO2 =

Combustion Analysis Example 1: Using mass proportion:
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 MCO2 x mCO2 12.0 44.0 x = = g MC Mass of C = x Mass of CO2 MCO2

Combustion Analysis Example 1: 2 x MH MH2O Mass of H in 0.405g H2O =
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? 2 x MH MH2O x mH2O Mass of H in 0.405g H2O = = 2 x 1.0 18.0 x 0.405 = g Mass of oxygen in the compound = – – 0.045 = g

Compound’s empirical formula is C3H4O3
Combustion Analysis C H O Mass ratio Mole ratio 0.409  16.0 = = = Simplest whole number ratio 0.0341  0.0341 = 1 = = 1 Compound’s empirical formula is C3H4O3

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 CO y/2 H2O H2O CO2 O2 CXHY Mole ratio Volume ratio y/2 mol x mol (x + y/4) mol 1 mol y/2 cm3 x cm3 (x + y/4) cm3 1 cm3 10 (y/2) cm3 10x cm3 10(x + y/4) cm3 10 cm3

CXHY + (x + y/4) O2  x CO2 + y/2 H2O
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 CO y/2 H2O By Avogadro’s law, x 1 = nCO2 nCxHy VCO2 VCxHy = 20 10 =  x = 2

Combustion Analysis Example 1 CXHY + (x + y/4) O2  x CO2 + y/2 H2O
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 CO y/2 H2O By Avogadro’s law, y/2 1 = nH2O nCxHy VH2O VCxHy = 30 10 =  y = 6 Molecular formula is C2H6

Combustion Analysis 19 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

CXHY + (x + y/4) O2  x CO2 + y/2 H2O
liquid state CXHY + (x + y/4) O2  x CO 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.

CXHY + (x + y/4) O2  x CO2 + y/2 H2O
liquid state CXHY + (x + y/4) O2  x CO 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, nCO2 nCxHy VCO2 VCxHy = x 1 40 10 =  x = 4

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. H2O CO2 O2 CXHY Mole ratio Volume ratio y/2 mol x mol (x + y/4) mol 1 mol 0 cm3 10x cm3 10(x + y/4) cm3 10 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. H2O CO2 O2 CXHY Mole ratio Volume ratio y/2 mol x mol (x + y/4) mol 1 mol 10 cm3 10(x + y/4) cm3 40 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. = Initial volume of gases – final volume of gases 35.0 = { ( 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

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 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 nCO2 nCxHy VCO2 VCxHy = x 1 40 10 =  x = 4 20 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

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 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 = – 40.0 = 60.0 cm3 nO2 nCxHy VO2 VCxHy = x + y/4 1 60 10 = y = 8 The formula for the hydrocarbon is C4H8

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

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