1 / 26

Atoms, Molecules, and Stoichiometry

Atoms, Molecules, and Stoichiometry. Recap. n Al =. n Cl2 =. m Al M Al. m Cl2 M Cl2. 27.0g 27.0gmol -1. 100g 71.0gmol -1. =. =. 14. Limiting Reagent. Example 1 What is the maximum amount of AlCl 3 that can produced from 27.0g of Al and 100g of Cl 2 by the following reaction?

venice
Download Presentation

Atoms, Molecules, and Stoichiometry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Atoms, Molecules, and Stoichiometry Recap

  2. nAl = nCl2 = mAl MAl mCl2 MCl2 27.0g 27.0gmol-1 100g 71.0gmol-1 = = 14 Limiting Reagent Example 1 What is the maximum amount of AlCl3 that can produced from 27.0g of Al and 100g of Cl2 by the following reaction? 2 Al + 3 Cl2 2 AlCl3 = 1.41 mol = 1.0 mol From the balanced equation: 2 mol Al  3 mol Cl2 Since 1.0 mol of Al would require 1.5 mol of Cl2 Hence Cl2 is the limiting reagent

  3. 2 3 = nAlCl3 nCl2 Limiting Reagent From balanced equation: nAlCl3 = 2/3 x nCl2 = 2/3 x 1.41 = 0.94 mol The maximum amount of AlCl3 produced is 0.94 mol.

  4. Actual Yield Theoretical Yield Percentage Yield = X 100% 15 Percentage Yield The mass of the product formed in a chemical reaction is called the yield. If the yield is calculated based on the chemical equation and the amount of reactants present, it is called theoretical yield. The amount of product actually obtained in a chemical reaction that is really carried out, is called the actual yield.

  5. Example 1 15.0g of barium chloride was added to 10.0g of iron (III) sulphate and 15.6g of barium sulphate was precipitated. Calculate the percentage yield of barium sulphate. mBaCl2 MBaCl2 15.0g (137.3 + 2 x 35.5) gmol-1 = nBaCl2 = = 0.0720 mol 16 Percentage Yield 3 BaCl2 + Fe2(SO4)3 3 BaSO4 + 2 FeCl3

  6. Example 1 15.0g of barium chloride was added to 10.0g of iron (III) sulphate and 15.6g of barium sulphate was precipitated. Calculate the percentage yield of barium sulphate. mFe2(SO4)2 MFe2(SO4)3 10.0g [2 x 55.8 + 3(32.1 + 4 x 16.0)] gmol-1 = = 0.0250 mol Percentage Yield 3 BaCl2 + Fe2(SO4)3 3 BaSO4 + 2 FeCl3 nFe2(SO4)3 =

  7. Example 1 15.0g of barium chloride was added to 10.0g of iron (III) sulphate and 15.6g of barium sulphate was precipitated. Calculate the percentage yield of barium sulphate. Percentage Yield 3 BaCl2 + Fe2(SO4)3 3 BaSO4 + 2 FeCl3 From the balanced equation: 3 mol BaCl2 : 1 mol Fe2(SO4)3 Hence 0.0720 mol of BaCl2 will require 0.0240 mol of Fe2(SO4)3 Since 0.0250 mol of Fe2(SO4)3 is provided, there is excess And thus BaCl2 is the limiting reagent

  8. Example 1 15.0g of barium chloride was added to 10.0g of iron (III) sulphate and 15.6g of barium sulphate was precipitated. Calculate the percentage yield of barium sulphate. Percentage Yield 3 BaCl2 + Fe2(SO4)3 3 BaSO4 + 2 FeCl3 nBaSO4 = nBaCl2 = 0.0720 mol mBaSO4 = nBaSO4 x MBaSO4 = 0.0720mol x (137.3+32.1+64.0)gmol-1 = 16.8g (theoretical)

  9. Actual Yield Theoretical Yield Percentage Yield = X 100% Example 1 15.0g of barium chloride was added to 10.0g of iron (III) sulphate and 15.6g of barium sulphate was precipitated. Calculate the percentage yield of barium sulphate. = 15.6g 16.8g x 100% Percentage Yield 3 BaCl2 + Fe2(SO4)3 3 BaSO4 + 2 FeCl3 = 92.8%

  10. Atoms, Molecules, and Stoichiometry Combustion Analysis

  11. 17 Combustion analysis apparatus • To absorb water : anhydrous cobalt chloride / concentrated H2SO4 • To absorb CO2 : alkaline solutions e.g. aq NaOH Note: CO2 is an acidic oxide

  12. = 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 = In 44g of CO2 , there is 12g of C ; In 1.5g of CO2 , there is ? of C

  13. 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

  14. Combustion Analysis CXHYOZ + ? O2 ? CO2 + ? H2O By conservation of atoms/mass, C in CO2 comes from C in compound H in H2O comes from H in compound Mass of C + mass of H + mass of O = mass of compound 0.409 g + 0.045 g + mass of O = 1.00g Mass of oxygen in the compound = 1.000 – 0.409 – 0.045 = 0.546 g

  15. 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

  16. 18 Combustion Analysis When the unknown organic compound is a gaseous hydrocarbon, the quantities are commonly expressed in terms of gaseous volume under specified conditions. A hydrocarbon is an organic compound containing only C and H as constituent elements. Expressed as CXHY. Useful combustion equation for hydrocarbons only: CXHY + (x + y/4) O2 x CO2 + y/2 H2O

  17. 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 Applicable for all gaseous substances This means that, Mole ratio is equivalent to the Volume ratio CXHY + (x + y/4) O2 x CO2 + y/2 H2O

  18. 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

  19. 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

  20. 19 Combustion Analysis Points to note: • The volume of CO2 may not be given directly • 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

  21. 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

  22. 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

  23. 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

  24. 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

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

  26. End of Lecture 5 “As cowardly as a coward is, it is not safe to call a coward a coward” Anonymous

More Related