Mole Concept 5.319% = = g sample =
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1 Mole - a counting system Avogadro s number = Mole Concept Chemical calculation involving mass: Empirical formula: The simplest formula that shows the relative numbers of the different kinds of atoms in a molecule. Ex. Combustion of 8.3 g of alkyl halide gave 9.6 g of carbon dioxide and 3.94 g of water. Analysis of 5.3 g of the halide gave g of AgCl. Find the empirical formula. In 8.3 g of the sample: mass of carbon = =.64 g 44 mass of hydrogen = 394. = 4377 g 18 In 5.3 g of the sample: mass of chlorine = 35.5 = g Element C H Cl Mass percentage % = 100% = 100% = % 5.319% 6.73% Molar mass/ g mol Number of mole in =.657 = = g sample Ratio of mole = = = 1 Simple mole ratio 1.503x x 6 1x = Its empirical formula is C 3 H 6 Cl. Molecular formula: The formula that describes the total number of atoms of each element in molecule. Ex. If the molecular mass of the alkyl halide is 113, find the molecular formula of the halide. Let (C 3 H 6 Cl ) n be the molecular formula. Molecular mass = ( )n = n = 113 n = 1 Its molecular formula is C 3 H 6 Cl. Chemical calculation involving gas volumes: Ex. A student worked out the densities of different gases at 0 C and 1 atmosphere. (a) Complete the following table. (b) Plot a graph of molar mass against density by using the following data. (c) Find the slope of the curve formed. What is the physical meaning of this value? Gas Molar mass / g mol -1 Density at s.t.p. / g dm -3 Molar volume / dm 3 mol -1 hydrogen 089 carbon monoxide 1.5 chlorine 3.0 hydrogen chloride 1.64 sulphur dioxide.87 argon 1.79 Mole Concept / Page 1
2 Avogadro s Law: (a) At s.t.p., one mole of an ideal gas occupies.4 dm 3, or (b) Equal volume of any gas(ideal gas) at the same temperature and pressure contains the same number of molecules. Expression: n Ex. 83 cm 3 of a gaseous compound of carbon, hydrogen and oxygen required 498 cm 3 of oxygen for complete combustion. The volume of carbon dioxide and steam formed were found to be 33 cm 3 and 415 cm 3 respectively. All gas volume are measured under the same condition at 110 C. Find the molecular formula of the compound. Let C x H y O z be the molecular formula of the compound. C x H y O z (g) + 1 (x+ y -z) O (g) x CO (g) + y H O (g) Molecule C x H y O z O CO H O olume reacted / cm olume ratio = = = = 5 Mole ratio from equation 1 Q x = 4 y = 5 x + y 4 - z = 6 y = 10 z = 1 The molecular formula of the compound is C 4 H 10 O. x + y 4 - z Chemical calculation involving concentration: Standard solution a solution with known concentration Primary standard solution A standard solution is prepared by weighing a certain mass of a solute and then diluting to a certain concentration by using a volumetric flask. Secondary standard solution The concentration of a standard solution is standardized by titration. End point the point at which indicator changes colour x y Mole Concept / Page
3 Equivalence point the point at which the reaction (neutralization) is complete Two important primary standards for neutralization: oxalic acid--water (ethanedioic acid), H C O 4 H O; anhydrous sodium carbonate, Na CO 3 Indicators for neutralization: methyl orange (yellow red) phenolphthalein (pink colourless) Some important primary standards for redox titration: sodium thiosulphate, Na S O 3 ; potassium bromate (), KBrO 3 ; oxalic acid--water (ethanedioic acid), H C O 4 H O Indicator for redox titration: starch for detection of iodine (colourless dark blue) Ex g of calcium carbonate was dissolved in 40 cm 3 of a hydrochloric acid solution. The excess acid required 5.0 cm 3 of a certain sodium hydroxide solution for neutralization. 0 cm 3 of this sodium hydroxide neutralized 16.8 cm 3 of the original acid. If the concentration of the sodium hydroxide solution is 800 M, find the purity of the calcium carbonate. NaOH (aq) + HCl (aq) NaCl (aq) + H O (l) The number of mole of NaOH reacted with 16.8 cm 3 HCl = 0 8 = 016 mol The number of mole of HCl reacted with 0 cm 3 NaOH = 016 mol The original [HCl] = = 954 M 0168 The total number of mole of HCl used = = mol The number of mole of 5 cm 3 NaOH = 8 05 = 0 mol The number of mole of HCl reacted with CaCO 3 = = mol CaCO 3 (s) + HCl (aq) CaCl (aq) + H O (l) + CO (g) The number of mole of pure CaCO 3 used = = mol The mass of pure CaCO 3 used = = 9048 g The purity of CaCO 3 = % = 63% 15. Ideal gas equation: Charle s Law : 1 P (At constant temperature and number of mole) Boyle s Law : T (At constant pressure and number of mole) Avogadro s Law : n (At constant pressure and temperature) Combining the above three laws : 1 P (T)(n) =R 1 P Tn Ideal gas law : P = nrt or PM = ρrt where ρ is the density of the gas, M is the molar mass of the gas, and R is the gas constant. R = J K -1 mol -1 R = dm 3 atm K -1 mol -1 Ex. Hydrogen gas in a cylinder of volume 15 dm 3 has a pressure of 00 kpa at 5 C. What is the mass of the hydrogen in the tank? P = Pa = 15 dm 3 = 015 m 3 T = 98 K The number of mole of hydrogen = P = = 1.11 mol RT The mass of hydrogen = 1.11 =.4 g Mole Concept / Page 3
4 Ex. What is the density of helium at 500 C and 100 mm Hg pressure? T = 773 K P = atm Molar mass of He M = 4 g mol -1 The density ρ at that condition = PM RT = = g dm -3 Dalton s law of partial pressures: The total pressure of a mixture of gases is equal to the sum of the partial pressures of the gases in the mixture. Partial pressure: The gas would exert the same pressure as it would if it alone occupied the same volume, at the specified temperature. For ideal gases: n t = n A + n B + n C + Multiplying through by RT : n RT t = n RT A Q P = nrt P t = P A + P B + P C + ( ) : ( 1) + n RT B + n RT C + (1st law) Q P t = n RT t (1) P A = n RT A () PA = n A Pt nt P A = P t X A where X A is the mole fraction of the gas A. n A X A = (nd law) n + n + n + A B C Ex. 10 g of hydrogen and 64 g of oxygen are contained in a 10 dm 3 flask at 00 C. Calculate the total pressure of the mixture. If a spark ignites the mixture, what will be the final pressure? n H = 10 = 5 mol n O = 64 = mol 3 P t = ( 5 + ) = 7.17 atm 10 The reaction occurring on sparking is H (g) + O (g) H O (g) Since moles of oxygen and 5 moles of hydrogen were initially present, the net result of the reaction is that 4 moles of steam are formed, leaving 1 mole of hydrogen unreacted. After sparking: n H = 1 mol n H O = 4 mol P t = ( ) = atm 10 Electrolysis: Faraday s first law of electrolysis: The mass of a substance discharged at an electrode during electrolysis is directly proportional to the quantity of electricity. Mole Concept / Page 4
5 m Q Q = It where Q is the quantity of electricity in Coulomb, I is current in ampere, t is time in second. Charge of an electron = Coulomb Charge of one mole of electrons = e = C = C = 1 Faraday mass/g Q/Coul Ex. A solution of potassium sulphate is electrolysed using graphite electrodes. A current of 5 A is passed through the electrolyte for 1930 s. Calculate (a) the quantity of electricity passed in terms of Faraday. (b) the volume of gas formed at the cathode at s.t.p. (c) the mass of gas collected at the anode. Overall reaction: H O (l) H (g) + O (g) (a) Q I = 5 A t = 1930 s Q = 5x1930 = 965 C = 01 F (b) H O (aq) + e - H (g) + OH - (aq) Q moles of electrons produce 1 mole of hydrogen. Number of mole of hydrogen formed = = 005 mol olume of hydrogen formed at s.t.p. = dm 3 = 11 dm 3 (c) H O (aq) O (g) + 4H + (aq) + 4 e - Q 4 moles of electrons produces 1 mole of oxygen. Number of mole of oxygen formed at s.t.p. = = 005 mol mass of oxygen formed = = 08 g 4 Faraday s second law of electricity: The number of moles of different substances produced by the same electricity form simple whole number ratio. Consider 1 mole of electrons pass through the following circuit. Pt Pt Ag Ag C C voltammeter A voltammeter B voltammeter C H SO 4 (aq) AgNO 3 (aq) Conc. NaCl (aq) For voltammeter A: Anode: H O (l) O (g) + 4 H + (aq) + 4 e - 1 mole of O is produced. 4 Cathode: H + (aq) + e - H (g) For voltammeter B: 1 mole of H is produced. Mole Concept / Page 5
6 Anode: Ag (aq) Ag + (aq) + e - 1mole of Ag + is produced. Cathode: Ag + (aq) + e - Ag (aq) 1mole of Ag is produced. For voltammeter C: Anode: Cl - (aq) Cl (g) + e - 1 is produced. Cathode: H + (aq) + e - H (g) Ex. Consider the following circuit: A 1 mole of H is produced. Mg Cu Cu Cu MgSO 4 (aq) CuSO 4 (aq) Left-hand side is an. Anode: CuSO 4 (aq) Right-hand side is an. Anode: Cathode: Cathode: Overall reaction: Overall reaction: Changes in solution: Changes in solution: Changes at electrodes: Changes at electrodes: Mole Concept / Page 6
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