Question

An unidentified covalent molecular compound contains only carbon, hydrogen, and oxygen. When 7.40 mg of this compound is burned, 21.98 mg of CO₂ and 3.00 mg of H₂O are produced. The freezing point of camphor is lowered by 10.4

Answer 1

The mass of CO2 produced by burning the given mass of the covalent molecular compound = 21.98 mg = 0.02198 g

i.e. The no. of moles of CO2 produced = 0.02198 g/(44 g/mol) = 5*10^-4 mol = no. of moles of Carbon

Therefore, the mass of Carbon in the required compound = 5*10^-4 mol * 12 g/mol = 6 mg

The mass of H2O produced by burning the given mass of the covalent molecular compound = 3 mg = 0.003 g

i.e. The no. of moles of H2O produced = 0.003 g/(18 g/mol) = 1.667*10^-4 mol

Therefore, the no. of moles of Hydrogen in the required compound = 2*1.667*10^-4 mol = 3.334*10^-4 mol

Therefore, the mass of Hydrogen in the required compound = 3.334*10^-4 mol * 1 g/mol = 0.33 mg

Therefore, the mass of Oxygen in the required compound = 7.4 - (6 + 0.33) = 1.07 mg

Therefore, the no. of moles of Oxygen in the required compound = 0.00107 g/(16 g/mol) = 0.669*10^-4 mol

Therefore, the no. of atoms of C = 5*10^-4/0.669*10^-4 ~ 7.5

The no. of atoms of H = 3.334*10^-4/0.669*10^-4 ~ 5

And the no. of atoms of oxygen = 0.669*10^-4/0.669*10^-4 = 1

Hence, the molecular formula of the compound = C_7.5H₅O₁ = C₁₅H₁₀O₂