Chemistry Lecture Slide Week 2
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Chemistry Lecture Slide Week 2

Chemistry Lecture Slide Week 2

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  • K 3 Fe(CN) 6 = potassium ferricyanide
  • H2O2 – HYDROGEN PEROXIDE
  • HCO 3 - - bicarbonate ion
  • Mg – 24.31 atomic mass
  • Using ion-electron method, Cr 2 O 7 2- dichromate ion
  • Yield always less because many reaction are reversible, some reaction are complex – product react among themselves or with the raectant

Chemistry Lecture Slide Week 2 Presentation Transcript

  • 1. LECTURE 2
  • 2. Contents:• Empirical & Molecular formula• Oxidation Number• Chemical Reaction• Stoichiometry calculation• Limiting Reagent
  • 3. • Formula kimia bagi sebatian ditulis menggunakan simbolbagi menunjukkan semua unsur dalam sebatianmengikut nisbah tertentu.• Formula empirikFormula empirik merupakan formula paling ringkas yangmenunjukkan nisbah terkecil bilangan atom bagi semuaunsur yang wujud dalam satu sebatian.• Contoh: CH2O menunjukkan nisbah unsur C, H dan Odalam sebatian ialah 1:2:1.• Formula empirik tidak menunjukkan bilangan sebenartidak menunjukkan bilangan sebenaratom-atom yang bergabung dalam suatu molekul.
  • 4. • Formula molekul ialah formula yang menunjukkanbilangan sebenar setiap atom yang wujud dalam satumolekul.• Terdapat yang mempunyai formula empirik yang samatetapi mempunyai formula molekul yang berbeza.• Perkaitan antara formula empirik dan formula molekuldiringkaskan seperti berikut:yang mana n ialah integer
  • 5. Apabila dianalisis, suatu sebatian didapati mengandungi25.5% Mg dan 74.5% Cl mengikut jisim. Apakah formulaempirik sebatian itu?Andaikan jisim sampel sebatian = 100 gJisim Mg = 25.5 g dan Cl = 74.5 gUnsur Mg ClJisim (g) 25.5 74.5Bil mol atom 25.5/24= 1.06374.5/35.5= 2.098Nisbahterkecil1.063/1.063= 12.098/1.063= 2Formula empirik sebatian = MgClMgCl22
  • 6. Pembakaran 0.202 g suatu sampel bahan organik yangmengandungi unsur-unsur C, H dan O menghasilkan 0.361 gkarbon dioksida dan 0.147 g air. Jika jisim molekul relatifsebatian ialah 148, tentukan formula molekulnya.% O dalam sampel = 100 – (48.51 + 8.07)= 43.42%= 43.42%Jisim C dalam CO2 = 1244X 0.361 = 0.098 g% C dalam sampel = 0.0980.202X 100 = 48.51%Jisim H dalam H2O = 218X 0.147 = 0.0163 g% H dalam sampel = 0.01630.202X 100 = 8.07%
  • 7. Pembakaran 0.202 g suatu sampel bahan organik yangmengandungi unsur-unsur C, H dan O menghasilkan 0.361 gkarbon dioksida dan 0.147 g air. Jika jisim molekul relatifsebatian ialah 148, tentukan formula molekulnya.Jisim formula molekul = n x jisim formula empirik148 = n x [ (3x12) + (6x1) + (2x16) ]n = 2Formula molekul sebatianFormula molekul sebatian = 2 (C3H6O2)= CC66HH1212OO44Unsur C H OJisim (g) 48.51 8.07 43.42Bil mol atom 48.51/12= 4.048.07/1= 8.0743.42/16= 2.71Nisbahterkecil4.04/2.71= 1.5 @ 38.07/2.71= 3 @ 62.71/2.71= 1 @ 2Formula empirik sebatian = CC33HH66OO22
  • 8. • Amaun bahan yang wujud dalam bentuk gas biasanyadiukur dalam unit isipadu.• 1 mol gas pada suhu dan tekanan piawai, s.t.p.(T=273.15 K, P = 1 atm) menempati isipadu 22.4 dm3dan dinamakan isipadu molarisipadu molar.
  • 9. • Larutan adalah campuran bahan larut dalam sejumlah pelarut.Kepekatan larutan boleh dinyatakan dalam sebutan:a.a. KemolaranKemolaran – bilangan mol bahan larut yang terlarut dalam 1dm3(1000 cm3) larutan (unit = mol dm-3).b.b. KemolalanKemolalan – bilangan mol bahan larut yang terlarut dalam 1kg pelarut (unit = mol kg-1).
  • 10. Hitungkan kemolaran larutan yang disediakan denganmelarutkan 3.42 g sukrosa (C12H22O11) dalam 500 cm3air.Bilangan mol sukrosa = 3.42 = 0.01 mol342Mr C12H22O11= (12 x 12) + (1 x 22) + (16 x 11)= 342Kemolaran = 0.01500/1000= 0.02 mol/dm3Hitungkan kemolalan larutan yang mengandungi 6.50 g etilenaglikol (C2H6O2) dalam 200 g air.Kemolalan C2H6O2 = 0.105200/1000= 0.524 mol/kgBilangan mol C2H6O2 = 6.52 (12) + 1 (6) + 2 (16)= 0.105 mol
  • 11. How to prepare 1 dm3of 0.1 M CuSO4 starting with 2 M CuSO4 ?50 g of benzene (C6H6) is dissolved in 80 g of toluene (C6H5CH3).Calculate the molality of the solution.n C6H6 = 5012 (6) + 1 (6)= 0.640 molUsing M1V1 = M2V2= 0.1 M x 1 dm32 M= 0.05 dm3= 50 cm3V1 = M2V2M1Molality C6H5CH3 = =n C6H680/1000 kg= 8.01 mol/kg0.640 mol80/1000
  • 12. c.c. Pecahan mol (X)Pecahan mol (X) Bilangan mol bahan larut dibahagikan dengan jumlahbilangan mol semua komponen yang terdapat dalamlarutan. Contoh: Larutan yang mengandungi sebatian A, B dan C.Pecahan mol bagi sebatian A dalam larutan ialah:nA, nB dan nC adalah bilangan mol A, B dan C masing-masing Pecahan mol tidak mempunyai unit dan jumlah pecahanmol semua komponen bernilai 1 iaitu:
  • 13. d. Peratusan mengikut jisim atau isipadud. Peratusan mengikut jisim atau isipadu Kepekatan larutan dalam sebutan peratusan bolehdinyatakan sebagai jisim bahan larut/jisim larutan (w/w)%atau isipadu bahan larut/isipadu larutan (V/V)% Contohnya asid sulfurik, H2SO4 berkepekatan 36%mengikut jisim mengandungi 36 g H2SO4 dalam 64 g air.Bagi larutan 5% NaCl mengikut isipadu, terdapat 5 g NaCldalam 100 cm3larutan.
  • 14. • In solution prepared by dissolving 24 g of NaCl in 152 g of water,• 20% w/w HCl → 20 g of HCl dissolved in 80 g of water.Calculate the percentage by mass of NaBr in an aqueous solutingcontaining 8 g of NaBr in 80 g of water.% w/w = 8(8 + 80)x 100%% w/w NaCl = 24 g x 100%24 g + 152 g= 14%= 9.09%
  • 15. e. ppme. ppm Kepekatan larutan yang sangat cair dinyatakandalam bahagian persejuta (ppm).Calculate the volume of K3Fe(CN)6 used to prepare a 125 ml 85%v/v solution.% v/v = Volume of K3(CN)6(125)x 100%= 106.25 mlVolume of K3(CN)6 = 85 x 125100
  • 16. The charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.The charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.1.1. Free elementsFree elements (uncombined state) have an oxidationnumber of zero.Na, Be, K, Pb, H2, O2, P4 = 02. In monatomic ionsmonatomic ions, the oxidation number is equal to thecharge on the ion.Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -23. The oxidation number of oxygen is usually –2. In H2O2 andO22-it is –1.
  • 17. 4. The oxidation number of hydrogen is +1 except when it isbonded to metals in binary compounds. In these cases, itsoxidation number is –1 (hydride eg: H2S).6. The sum of the oxidation numbers of all the atoms in amolecule or ion is equal to the charge on the molecule orion.5. Group IA metals are +1, IIA metals are +2 and fluorine isalways –1.HCO3-O = -2 H = +13x(-2) + 1 + ? = -1C = +4Oxidation numbers of C inHCO3-?
  • 18. Determine the oxidation number of the following underlinedelements:a. Cu2O b. SO42-c. ICl3 d. OF2a. Oxidation no of O = - 2Let oxidation no of Cu be x2x - 2 = 0x = +1b. Oxidation no of O = -2Let oxidation no of S be yy - 8 = - 2y = +6c. Cl more electronegative than IOxidation no of Cl = - 1Let oxidation no of I be qq - 3 = 0q = +3d. F more electronegative than OOxidation no of F = - 1Let oxidation no of O be rr - 2 = 0r = +2
  • 19. • Persamaan kimia menggunakan simbol dan formula bagimenggambarkan secara ringkas perubahan yang berlakudalam suatu tindak balas kimia.• Tindak balas am: A dan B adalah reaktan manakala C dan D adalah hasiltindak balas. Tanda ‘+’ = memisahkan setiap reaktan dan hasil tindakbalas. Anak panah = bermakna ‘untuk menghasilkan’
  • 20. Example: 3 ways of representing the reaction of H2 with O2 to formH2O
  • 21. 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO2 moles Mg + 1 mole O2 makes 2 moles MgO48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgOIS NOT2 grams Mg + 1 gram O2 makes 2 g MgO
  • 22. 1. Write the correct formula(s) for the reactants on the leftside and the correct formula(s) for the product(s) on theright side of the equation.Ethane reacts with oxygen to form carbon dioxide and waterC2H6 + O2 CO2 + H2O2. Change the numbers in front of the formulas (coefficients)to make the number of atoms of each element the sameon both sides of the equation. Do not change thesubscripts.2C2H6NOT C4H12
  • 23. C2H6 + O2 CO2 + H2O start with C or H but not O2 carbonon left1 carbonon rightmultiply CO2 by 2C2H6 + O2 2CO2 + H2O6 hydrogenon left2 hydrogenon rightmultiply H2O by 33. Start by balancing those elements that appear in only onereactant and one product.
  • 24. 4. Balance those elements that appear in two or morereactants or products.multiply O2 by722 oxygenon left4 oxygen(2x2)C2H6 + O2 2CO2 + 3H2O+ 3 oxygen(3x1)= 7 oxygenon rightC2H6 + O2 2CO2 + 3H2O72remove fractionmultiply both sides by 22C2H6 + 7O2 4CO2 + 6H2O2C2H6 + 7O2 4CO2 + 6H2O
  • 25. 5. Check to make sure that you have the same number ofeach type of atom on both sides of the equation.2C2H6 + 7O2 4CO2 + 6H2OReactants Products4 C12 H14 O4 C12 H14 O4 C (2 x 2) 4 C12 H (2 x 6) 12 H (6 x 2)14 O (7 x 2) 14 O (4 x 2 + 6)
  • 26. Ion-electron methodIon-electron method• The ion-electron method is mainly used forbalancing the redox (reduction-oxidation)equations.• Reduction is a process whereby the oxidationnumber of an atom decreases.• Conversely, oxidation involves an increase in theoxidation number.
  • 27. 2Mg 2Mg2++ 4e-O2 + 4e-2O2-2Mg (s) + O2 (g) 2MgO (s)0 0 2+ 2-Oxidation half-reaction- Lose e-.- Also a reducing agent by giving up itselectrons causing the other substance tobe reduced.Reduction half-reaction- Gain e-.- Also a oxidizing agent removeselectrons from another substance byacquiring them itself; thus the agent isitself reduced.
  • 28. Example: Balance the equation showing oxidation of Fe2+to Fe3+byCr2O72-in acid solution?Using ion-electron method (reaction divided into two half reactions)1. Write the unbalanced equation for the reaction ion ionicform. Fe2++ Cr2O72-Fe3++ Cr3+2. Separate the equation into two half-reactions.(reducing agent) Oxidation:Cr2O72-Cr3++6 +3(oxidizing agent) Reduction:Fe2+Fe3++2 +33. Balance the atoms other than O and H in each half-reaction.Cr2O72-2Cr3+
  • 29. 4. For reactions in acid, add H2O to balance O atoms and H+tobalance H atoms.Cr2O72-2Cr3++ 7H2O14H++ Cr2O72-2Cr3++ 7H2O6e-+ 14H++ Cr2O72-2Cr3++ 7H2O6. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriatecoefficients.6Fe2+6Fe3++ 6e-5. Add electrons to one side of each half-reaction to balance thecharges on the half-reaction.6e-+ 14H++ Cr2O72-2Cr3++ 7H2OFe2+Fe3++ 1e-
  • 30. 7. Add the two half-reactions together and balance the finalequation by inspection. The number of electrons on both sidesThe number of electrons on both sidesmust cancel.must cancel.6e-+ 14H++ Cr2O72-2Cr3++ 7H2O6Fe2+6Fe3++ 6e-Oxidation:Reduction:14H++ Cr2O72-+ 6Fe2+6Fe3++ 2Cr3++ 7H2O8. Verify that the number of atoms and the charges are balanced.14x1 – 2 + 6x2 = 24 = 6x3 + 2x3
  • 31. Stoichiometrically Equivalent Molar ratiosStoichiometrically Equivalent Molar ratios• Stoichiometry is the quantitative study of reactants andproducts in a chemical reaction.• Use moles to calculate the amount of product formed ina reaction – mole methodmole method.• Therefore, the stoichiometric coefficients in a chemicalequation can be interpreted as the number of moles ofeach substance.
  • 32. • Example:N2(g) + 3H2(g) 2NH3(g)• The equation can be read as “ 1 mole of N2 gas combineswith 3 moles of H2 gas to form 2 moles of NH3 gas”.• In stoichiometric calculationstoichiometric calculation– 3 moles of H2 are stoichiometrically equivalent to 2 moles of NH3– 1 mole N2 are stoichiometrically equivalent to 2 mol NH3– 1 mole N2 are stoichiometrically equivalent to 3 mol H2• The conversion factorsconversion factors from this equivalence:1 molecule 3 molecule 2 molecule3 mol H22 mol NH32 mol NH33 mol H2and
  • 33. The general approach for solving stoichiometryproblems:i. Write a balanced equation for the reaction.ii. Convert the given amount of the reactant (gram@ other unit) to number of moles.iii. Use the mole ratio from the balanced equation tocalculate the number of moles of product formed.iv. Convert the moles of product to gram ( or otherunits) of product.
  • 34. • Example: 16.0 g of H2 react completely with N2 to formNH3. How many grams of NH3 will be formed?grams H2 moles of H2 moles of NH3 grams of NH316 g H21 mol H22.016 g H2x 2 mol NH33 mol H2x 17.03 g NH31 mol NH3xGrams of NHGrams of NH33 === 90.61 g NH= 90.61 g NH33N2 (g) + 3H2 (g) 2NH3(g)
  • 35. Methanol burns in air according to the equation2CH3OH + 3O2 2CO2 + 4H2OIf 209 g of methanol are used up in the combustion,what mass of water is produced?grams CH3OH moles CH3OH moles H2O grams H2Omolar massCH3OHcoefficientschemical equationmolar massH2O209 g CH3OH1 mol CH3OH32.0 g CH3OHx4 mol H2O2 mol CH3OHx18.0 g H2O1 mol H2Ox= 235 g H2O
  • 36. • Goal of reaction is to produce the maximum quantityfrom the starting material.• Consequently, some reactant will be left over at the endof reaction.• The reactant used up first in a reaction is called thelimiting reagentlimiting reagent - the maximum amount of productformed depends on how much of this reactant wasoriginally present.• Excess reagentsExcess reagents are the reactants present in quantitiesgreater than necessary to react with the quantity of thelimiting reagent.
  • 37. 6 green used up6 red left over
  • 38. Sebanyak 12 g zink dan 6.5 g sulfur bertindak balasmengikut persamaan:a.Apakah bahan yang merupakan reaktan penghad di dalamtindak balas?Daripada persamaan,Zn(p) + S(p) ZnS(p)Bilangan mol Zn = = 0.1836 mol1265.37Bilangan mol S = = 0.2031 mol6.5321 mol Zn bertindak balas dengan 1 mol SOleh itu 0.1836 mol Zn memerlukan 0.1836 mol S# Reaktan penghad ialah Zn# Reaktan penghad ialah Zn
  • 39. Sebanyak 12 g zink dan 6.5 g sulfur bertindak balasmengikut persamaan:b. Berapa gramkah ZnS yang terhasil?Daripada persamaan,Zn(p) + S(p) ZnS(p)Bilangan mol ZnS terbentuk = 0.1836 molJisim ZnS terbentuk = 0.1836 x 97.37= 17.88 gOleh kerana Zn ialah reaktan penghad maka amaun hasilditentukan oleh amaun Zn1 mol Zn menghasilkan 1 mol ZnS
  • 40. Sebanyak 12 g zink dan 6.5 g sulfur bertindak balasmengikut persamaan:c. Namakan bahan yang berlebihan dan kirakanbilangan gram bahan tersebut yang masih belum bertindakbalas.Bahan berlebihan ialah SZn(p) + S(p) ZnS(p)Bilangan mol S berlebihan = 0.2031 - 0.1836= 0.0195 molJisim Sulfur berlebihan = 0.0195 x 32= 0.624 g
  • 41. • Example: Urea is prepares by reacting ammonia with CO22NH3 (g) + CO2 (g) (NH2)2CO (aq) + H2O(l)637.2 g of NH3 are treated with 1142 g CO2. Which of the tworeactant is limiting reagent?grams NH3 moles of NH3 moles of (NH2)2CO637.2 g NH31 mol NH317.03 g NH3x 1 mol (NH2)2CO2 mol NH3xmoles of (NH2)2CO === 18.71 mol (NH= 18.71 mol (NH22)) 22COCO
  • 42. • Example: Urea is prepares by reacting ammonia with CO22NH3 (g) + CO2 (g) (NH2)2CO (aq) + H2O(l)637.2 g of NH3 are treated with 1142 g CO2. Which of the tworeactant is limiting reagent?grams CO2 moles of CO2 moles of (NH2)2CO1142 g CO21 mol CO244.01 g CO2x 1 mol (NH2)2CO1 mol CO2xmoles of (NH2)2CO === 25.95 mol (NH= 25.95 mol (NH22)) 22COCOTherefore, NH3 must be the limiting reagent because it produces asmaller amount of (NH2)2CO
  • 43. • Example: Urea is prepares by reacting ammonia with CO22NH3 (g) + CO2 (g) (NH2)2CO (aq) + H2O(l)Calculate the mass of (NH(NH22))22COCO formed.18.71 mol (NH2)2CO 60.06 g (NH2)2CO1 mol (NH2)2COxMass of (NH2)2CO === 1124 g (NH= 1124 g (NH22)) 22COCOThe molar mass of (NH2)2CO is 60.06 g. Use the conversionconversionfactorfactor to convert from moles of (NH2)2CO to gram of(NH2)2CO.
  • 44. Do You Understand Limiting Reagents?In one process, 124 g of Al are reacted with 601 g of Fe2O32Al + Fe2O3 Al2O3 + 2FeCalculate the mass of Al2O3 formed.g Al mol Al mol Fe2O3 needed g Fe2O3 neededorg Fe2O3 mol Fe2O3 mol Al needed g Al needed124 g Al1 mol Al27.0 g Alx1 mol Fe2O32 mol Alx160. g Fe2O31 mol Fe2O3x = 367 g Fe2O3Start with 124 g Al need 367 g Fe2O3Have more Fe2O3 (601 g) so Al is limiting reagent
  • 45. Use limiting reagent (Al) to calculate amount of product thatcan be formed.g Al mol Al mol Al2O3 g Al2O3124 g Al1 mol Al27.0 g Alx1 mol Al2O32 mol Alx102. g Al2O31 mol Al2O3x= 234 g Al= 234 g Al22OO33Do You Understand Limiting Reagents?In one process, 124 g of Al are reacted with 601 g of Fe2O32Al + Fe2O3 Al2O3 + 2FeCalculate the mass of Al2O3 formed.
  • 46. • Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted (from calculationusing the balanced equation).• Actual Yield is the amount of product actually obtainedfrom a reaction (almost always less than the theoreticalyield).• Percent yield - to determine reaction efficiency, whichdescribe the proportion of the actual yield to thetheoretical yield.% Yield =Actual YieldTheoretical Yieldx 100
  • 47. • Example:TiCl4 (g) + 2Mg(l) Ti(s) + 2MgCl2 (l)3.54 x 107g of TiCl4 are reacted with 1.13 x 107g of Mg.Calculate the theoretical yield of Ti in grams.Calculate the theoretical yield of Ti in grams.g of TiCl4 Mol of TiCl4 Mol of TiFirst: to find limiting reagent, calculate the no. of moles of Tithat could produces if all the TiCl4 reacted= 1.87 x 105mol TiNext: calculate the no. of moles of Ti formed from 1.13 x 107g of Mg.g of Mg Mol of Mg Mol of Ti = 2.32 x 105mol Ti
  • 48. • Example:TiCl4 (g) + 2Mg(l) Ti(s) + 2MgCl2 (l)3.54 x 107g of TiCl4 are reacted with 1.13 x 107g of Mg.Calculate the theoretical yield of Ti in grams.Calculate the theoretical yield of Ti in grams.Therefore, TiCl4 is the limiting reagent because it produces asmaller amount of Ti= 8.95 x 106g Ti1.87 x 105mol Ti47.88 g Ti1 mol TixMass of Ti =Mass of Ti =
  • 49. • Example:TiCl4 (g) + 2Mg(l) Ti(s) + 2MgCl2 (l)3.54 x 107g of TiCl4 are reacted with 1.13 x 107g of Mg.Calculate the percent yield if 7.91 x 10Calculate the percent yield if 7.91 x 1066g of Ti are actuallyg of Ti are actuallyobtained.obtained.The percent yield is given by:% Yield = Actual YieldTheoretical Yieldx 100% Yield = 7.91 x 106g8.95 x 106gx 100= 88.4%