Presentation is loading. Please wait.

Presentation is loading. Please wait.

Atoms, Molecules, and Stoichiometry Recap of Lecture 5.

Similar presentations


Presentation on theme: "Atoms, Molecules, and Stoichiometry Recap of Lecture 5."— Presentation transcript:

1 Atoms, Molecules, and Stoichiometry Recap of Lecture 5

2 YISHUN JC Recap 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 Avogadros law to compare volumes

3 YISHUN JC 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 CO 2 = M C M CO2 x m CO x = = g Combustion Analysis Mass of C Mass of CO 2 MCMC M CO2 =

4 YISHUN JC 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 CO 2 = M C M CO2 x m CO x = = g Combustion Analysis Mass of C Mass of CO 2 MCMC M CO2 = x

5 YISHUN JC 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? Combustion Analysis Mass of H in 0.405g H 2 O = 2 x M H M H2O x m H2O = 2 x x = g Mass of oxygen in the compound = – – = g

6 YISHUN JC CHO Mass ratio = = 0.045= Mole ratio = 1= 1.320= 1 Simplest whole number ratio Compounds empirical formula is C 3 H 4 O 3 Combustion Analysis

7 YISHUN JC Avogadros 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 10 (y/2) cm 3 10x cm 3 10(x + y/4) cm 3 10 cm 3 y/2 cm 3 x cm 3 (x + y/4) cm 3 1 cm 3 y/2 molx mol(x + y/4) mol1 mol H2OH2OCO 2 O2O2 CXHYCXHY Mole ratio Volume ratio C X H Y + (x + y/4) O 2 x CO 2 + y/2 H 2 O Combustion Analysis

8 YISHUN JC Example 1 When 10 cm 3 of a gaseous hydrocarbon was completely burnt in excess oxygen, 20 cm 3 of carbon dioxide and 30 cm 3 of steam was produced (all gas volumes measured under the same conditions). Calculate the molecular formula of the hydrocarbon. C X H Y + (x + y/4) O 2 x CO 2 + y/2 H 2 O By Avogadros law, n CO2 n CxHy V CO2 V CxHy = = x = 2 Combustion Analysis x1x1 =

9 YISHUN JC Example 1 When 10 cm 3 of a gaseous hydrocarbon was completely burnt in excess oxygen, 20 cm 3 of carbon dioxide and 30 cm 3 of steam was produced (all gas volumes measured under the same conditions). Calculate the molecular formula of the hydrocarbon. C X H Y + (x + y/4) O 2 x CO 2 + y/2 H 2 O By Avogadros law, n H2O n CxHy V H2O V CxHy = y/2 1 = y = 6 Molecular formula is C 2 H 6 Combustion Analysis =

10 YISHUN JC Points to note: 1.The volume of CO 2 may be given as a decrease in volume when the residual gases are passed through NaOH or other alkali 2.The volume of H 2 O may be given as a.Decrease in volume when the residual gases are passed through anhydrous CaCl 2 or conc. H 2 SO 4 b.Decrease in volume when the residual gases are cooled to below 100 o C at atm pressure 3.Oxygen is usually added in excess and not in stoichiometric amount, so there will be excess oxygen as residual gas 4.Contraction = volume of reactants – volume of products 5.Expansion = volume of products – volume of reactants 19 Combustion Analysis

11 YISHUN JC Example 2 In an experiment, 10.0 cm 3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm 3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm 3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon. C X H Y + (x + y/4) O 2 x CO 2 + y/2 H 2 O liquid state

12 YISHUN JC Example 2 In an experiment, 10.0 cm 3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm 3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm 3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon. Volume of CO 2 gas = 40.0 cm 3 [absorbed by aq NaOH] By Avogadros Law, n CO2 n CxHy V CO2 V CxHy = x1x = x = 4 C X H Y + (x + y/4) O 2 x CO 2 + y/2 H 2 O liquid state

13 YISHUN JC Example 2 In an experiment, 10.0 cm 3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm 3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm 3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon. C X H Y + (x + y/4) O 2 x CO 2 + y/2 H 2 O y/2 molx mol(x + y/4) mol1 mol 0 cm 3 10x cm 3 10(x + y/4) cm 3 10 cm 3 H2OH2OCO 2 O2O2 CXHYCXHY Mole ratio Volume ratio

14 YISHUN JC Example 2 In an experiment, 10.0 cm 3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm 3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm 3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon. C X H Y + (x + y/4) O 2 x CO 2 + y/2 H 2 O y/2 molx mol(x + y/4) mol1 mol 0 cm 3 40 cm 3 10(x + y/4) cm 3 10 cm 3 H2OH2OCO 2 O2O2 CXHYCXHY Mole ratio Volume ratio

15 YISHUN JC Example 2 In an experiment, 10.0 cm 3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm 3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm 3. 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 O 2 } – { 40 + excess O 2 } 65= 10(x+y/4) Since x = 4, solvingy = 10 Hence the empirical formula of the hydrocarbon is C 4 H 10 C X H Y + (x + y/4) O 2 x CO 2 + y/2 H 2 O

16 YISHUN JC Example 3 In an experiment, 10.0 cm 3 of a gaseous hydrocarbon was exploded with cm 3 of oxygen. The volume of the resulting gas mixture was 80.0 cm 3, which decreased to 40.0 cm 3 when passed through aqueous sodium hydroxide. All volumes were measured at room temperature and pressure. What is the formula of the hydrocarbon? Volume of CO 2 produced = 80.0 – 40.0 = 40.0 cm 3 C X H Y + (x + y/4) O 2 x CO 2 + y/2 H 2 O n CO2 n CxHy V CO2 V CxHy = x1x = x = 4 Volume of resulting gas mixture = 80.0 cm 3 = volume of CO 2 + volume of unreacted O 2 Hence volume of unreacted O 2 = 80.0 – 40.0 = 40.0 cm 3 20

17 YISHUN JC Example 3 In an experiment, 10.0 cm 3 of a gaseous hydrocarbon was exploded with cm 3 of oxygen. The volume of the resulting gas mixture was 80.0 cm 3, which decreased to 40.0 cm 3 when passed through aqueous sodium hydroxide. All volumes were measured at room temperature and pressure. What is the formula of the hydrocarbon? C X H Y + (x + y/4) O 2 x CO 2 + y/2 H 2 O Therefore volume of O 2 used up in the reaction = – 40.0 = 60.0 cm 3 n O2 n CxHy V O2 V CxHy = x + y/ = y = 8 The formula for the hydrocarbon is C 4 H 8

18 YISHUN JC 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!)


Download ppt "Atoms, Molecules, and Stoichiometry Recap of Lecture 5."

Similar presentations


Ads by Google