Question

2. 0.342g of a compound vield 0.656a CO2 and 0.4039 H2O upon combustion. If the MW= 45 + 5, calculate the molecular formula. (Answer: C2H60) 3. 0.768g of a compound containing C, H and possibly O, yields 2.376g CO2 and 0.792g H2O upon combustion. Calculate the molecular formula if the MW=380 + 10. (Answer: C27H440) 4. A 0.157g of a compound containing C, H, N and possibly O, yields 0.213g CO2 and 0.0310g H2O upon combustion. Another sample of the same compound weighing 0.1039 yields 0.023g NH3 in a nitrogen analysis. Calculate the empirical and molecular formulae given the MW= 450 +20 (Answer: C7H5O6N3: C14H10012N6) 5. A 0.411g sample of a compound yields 1.055g CO2 and 0.413g water upon combustion. Given the MW=310 + 10 calculate the molecular formula (Answer: C20H3804 10

Answer 1

ven mass= 0.3428 o compound) + Co, (g) + H₂O 0.656g 0.403g motor mass of co=44, in which ing of t is present. In qag of co mass of cis 12 g 12 0.6569 12 x 0.656 44 -0.1789 .. Mass of c in Compound ñ 0.179 g similarly molar mass of H₂O=18 In 18g of Ho, mass of t present = 2 g 0.003 2 x 403 g = 0.0 447 = 0.045g X 100 f age of c = mass of c total mass of compound 0-342 -52'34% 0.129_*100 = 52'34%

age of H = 0.045 x 100 = 134.167 0.342 total amount = 100% 7-age of Oxygen = 100-52.34 - 13:16 - - 100 -6515 = 34.5% Cal" for molecular formule C Ho ! 34.5 g mass 52.34 12. en 100g no of atom 52.34 13.16 13:16 34.5 12 13:15 = 245 ~ 4.35 deyde by 2.15 ñ 2.02 =6 ) Empirical & B stolectar formule is H₂O . Aus & Empirical 46 Empion Constecatar mass=2x12 + 6 + 16 . Molecular mas = CH GO same as 2760 ginen molecular formula given mars

to (Compound river = 0.7688 co + H₂O 2.3768 0.792g mass of c in coa = mass of c in compound mass of ca 12 4 X2 * 2:376 = 0.6487 similare of H = 2 x 0.792 =0.0889 nass of oxygen total mass of Compound - (mass of ct masof H) = 0.768 -0.648 -0.088), = 0.768 -0.736 = 0.0 32% I formula calh of Empirical H hay 0.6484 0.088 o.032 atom [ 0:554 0-088 0.002 simplest ratio ř 27 44

". Empirical formula = Cath ago Empirical mass= (27X12)+44 +16 - 324 + 60 = 384 في | لعل لعلعلمي - ". Molecular formule= Empirical formula Note :- for calculation of Empirical formula, we can consider any amount of compound because empirical formula is independent of exact mass of compound!
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