Question

IsepuaraLa Furnace 0 absorber CO; absorber Sample When 5.697 grams of a hydrocarbon, C H, were burned in a combustion analysis apparatus, 18.54 grams of CO2 and 5.694 grams of H2O were produced In a separate experiment, the molar mass of the compound was found to be 54.09 gmol. Determine the empirical formula and the molecular formula of the hydrocarbon. Enter the elements in the order presented in the question. empirical formula- molecular formula

Answer 1

Hydrocarbon (Cx Hy) = xco (9) + 9 H2O ces Amt. of co2= 10.5ug. no. of mole of cos = 18.54 44.0) = 0.4212678 mole TO no. o mole of Tarton - ho. S hole of cos = 0.4212679 no. of mole of H2O= 5.794 13.015 = 0.315 mole no. of mole od H= ax no. of hole Ratio of = 2 x 0.315 = 0.63214 n. of hole of H e no. I mole gc = 0.63214 115 O.VRIR678 no. of mole of t - DUR12 678 no. Bole परा२674 Ratio to , Hi.S = trense empirical Formula = 2H3. empirical ma 55 = R XIR +3X100 = 25.OR4 na Molar mass/ emt. m255 = 54.09 27.024 Molecular Formula Tu H5