Question

Vanillin is the compound containing carbon, hydrogen, and oxygen that gives vanilla beans their distinctive flavor. The combustion of 30.4 mg of vanillin produces 70.4 mg of CO2 and 14.4 mg of H2O. The mass spectrum of vanillin shows a molecular-ion peak at 152 amu. Use this information to determine the molecular formula of vanillin

Answer 1

70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C 14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O molar mass of C=12 g/mole molar mass of H=1 g/mole 0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C 0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O molar mass of O=16 g/mole 9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602 C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602 C2.66H2.66O1 is the empirical formula; to obtain whole numbers multiply by 3 3[C2.66H2.66O1] = C8H8O3 above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu The empirical formula weight and the molecular formula weight are the same . Molecular formula is C8H8O3.