Molecular Formula from Elemental Analysis and Molecular Mass Determination An un

Question

Molecular Formula from Elemental Analysis and Molecular Mass Determination An unidentified covalent molecular compound contains only carbon, hydrogen, and oxygen. When 9.20 mg of this compound is burned, 22.47 mg of CO2 and 3.07 mg of H2O are produced The freezing point of camphor is lowered by 26.4°C when 2.319 g of the compound is dissolved in 13.00 g of camphor (Kf= 40.0°C kg/mol) What is the molecular formula of the unidentified compound? Choose appropriate coefficients in the molecular formula belovw CHO

Explanation / Answer

Given the mass of CO2 = 22.47 mg or 0.02247 g CO2

Mass of H2O = 3.07 mg or 0.00307 g H2O

0.02247 g CO2 / 44.01 g/mol = 0.000510 moles CO2
0.000510 moles CO2 contains 0.000510 moles C atoms

0.000510 moles x 12.011 g/mole = 0.00613 C atoms (6.13 mg)

0.00307 g H2O / 18.0152 g/mol = 0.00017 mole H2O
0.00017 mole H2O contains 0.00085 mol H atoms
0.00085 mole H x 1.0079 g/mol = 0.000857 g H atoms

9.20 mg sample – 6.13 mg C - 0.857 mg H = 2.043 mg O atoms
0.002043 g O / 15.9994 g/mol = 0.0001278 mol O atoms

C: 0.510 mmol / 0.1278 = 3.99 ?4
H: 0.85 mmol / 0.1278 = 6.7?7
O: 0.1278 mmol / 0.1278 = 1

C4H7O (molar mass = 71.089.0 g/mol) = empirical formula

26.4 deg K = 40.0 deg K / m x molality
molality = 0.66
moles / 0.01300 kg = 0.86
moles = 0.00858
2.319 g / 0.00858 moles = 270.2 g/mol
270.2 / 71.0 = 4 times the empirical formula Molecular formula from elemental analysis and molecular mass determination

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