Question

See page 118 12 Question (2 points) A compound containing only C, H. and O, was extracted from the bark of the sassafras tree. The combustion of 34.5 mg produced 93.7 mg of CO2 and 19.2 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas. 1st attempt

Answer 1

Ans Given data: ombustion of 34.5 mg. Produced Coq = 93.7 mga H90 - 19.2 mg Compound molar mass = 162g/mol organic compound mass = 34.5 mg on = 0.0345 g", Masset H₂o vapour produced from the Combustion = 19.29g = 0.0192gn Mass of co2 produced toom the Combustion = 93.799 20-09379 Yo' 0.0937coy x 1 molco2 x 1 malc uy olg.co2 " I molco2 . 0.03459. 12.01gue 1 mac X100% . Molecular mass of) C022 04.09. Molecular weight of C-72.019. 1 o. 7UX 100 y. = 74.15 % C.. %1H= 0.01929. Hgo X x . 1 mol H₂O - 19.099" Hgo 2 mol 1 mol H₂O X 10689, 4 1molt X 100%. 0.0345 = 0.0622X100% = It 6.22%

3:0 = 100 - (%.c+% ) = (100 -74.15 - 6.29).% 19.63% - loog, of compound Contains 74.159.c, 6.22g. H, & 19.639.0. -) Number of moles of c = 74.159.cx 1molc 12.01guc = 6.17 molcn. - Number of moles of H = 6.29g. Hx - 1 mih 1-0089, H - 6.17 mol Hu . - Number of moles of o=19.639. O x 1 molo 16.0ogo = 1.22 mol Hu -5.05 25 c= 6.17 199 H = .6:17 - 5.05 = 5 . 1:22 0 = 1.22 - 1. -1 -> place these simplest ratio values as the lower subscript to the Concerred clement to get the empirical formula

Cog Hjo - Empirical formula mass of Cg Ho = 81.09g/mol - Molecular formula = nx empirical formula = molecular mass empirical formula mass X empirical formula 162 g/mol 81.099/mon X (Cs Hyg) = = 2X (Eg Hg o) - Go hio O2