Question

HOMEWORK 0.0239 sample of Cr.Cl, undergoes a process where all chlorine is converted to 1.6929 of AgCl. Identify the empirical formula of Cr.CI, (Answer: CrCl) 2. 0.342g of a compound yield 0.656 CO, and 0.4039 H,O upon combustion. If the MW= 45 + 5, calculate the molecular formula. (Answer: C.H.0) 3. 0.768g of a compound containing C, H and possibly O, yields 2.376g CO, and 0.792g H2O upon combustion. Calculate the molecular formula if the MW= 380 + 10. (Answer: C2H4.0) 4. A 0.157g of a compound containing C, H, N and possibly o. yields 0.213g CO, and 0.0310g H2O upon combustion. Another sample of the same compound weighing 0.103g yields 0.023g NH, in a nitrogen analysis. Calculate the empirical and molecular formulae given the MW= 450 +20 (Answer: C.H.O.N.C.H1,02N.) 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: C20H30.

Answer 1

10 molar mass of agce - 143.32ginal nolar mass of clog = 35.45 gimol 1 unit of Agce weighs = 143.32 contains 35.45 g ce 143.32 g Ace 35.45 g ce fore 1. 692 g of Agce, mass of ce = 0.4189 Total mass of sample = 0.632g mass of cu= 0.632-0.418 = 0.2149 moles of ce= 0.418 = 1.18x 10 moles 35.45 51.9.9 moles of Ce = 0.214 = 4.17 x 10 roles Сk : 4. 18 X 10 -3 : 1.18 x 10-2 1 : 3 in Empirical formula (Gillz]