Question

2.516 g of a compound containing carbon, hydrogen and oxygen (CXHYOZ) is subjected to combustion analysis. The results show that 3.082 g of CO2 and 2.705 g of H2O were produced.
   (i). What is the empirical formula for the compound?
   (ii). If the molecular weight of the compound is 160.2 g/mol, what is the molecular formula of the       compound?

Answer 1

(1) No of moles of (0g = Mass = 3.082 g Molar Mass 44.olgimal = 6.07001 male - No of moles H₂O = 2.705g = 0.1502 mol - 18.015 g/mol h 1 male (o contains y male c - No of 'moles of ( in given combound = 0.07001 mol As f mole Ho contains 2 mole H o No of moles of o in given compound=2X0.1501 mol " =0.300 4 mol ☆ Mass of ( in given compounda No of moles x Malar mass = 0.07061 mol x 12.0lgimol = 0.8408 g Mass of Hin-given compound = 0.3004 mol X1.0078 mal * = 0. 3027 g 8 Mass of o in given compound= Total mass of compound - -(Mass of Ht Mass of E 2.5169 – 10.30219+0.8908)g 3 1.37259 & No of moles of 0= 1.37259/15.999 g/mol = 0.085787 mal Molar Slatio of C: H: 0 = 0.07001: 0.3004: 0.085787 Dinding by 0.07001 and converting to whole no. (: H: 0 = 1: 04: 1 - Empirical formula = (Hyo

(1) Molar Mass of CH40 = 12.01+ (4x 1.01)+(16.00) glmul = 32.05 g/mol . No of Chuo units in complete Molecule = Molecular weight of - Compound - Molar Mass of Chuo unit = 160.2 g/mol = 5. 32-05 g/mol . Malecular formula = (CH4O)4 = (5 Hgo DE