Combustion reaction for menthol is as follows; CxHyOz + O₂ ---> xCO₂ + H₂O Mass of CO₂ formed - 28.16 mg Therefore number of moles formed - 28.16/ 44 g/mol = 0.64 mmol Mass of water formed - 11.53 mg number of water moles formed - 11.53 mg/18 g/mol = 0.64 mmol From CO₂, 1 mol of CO₂ - 1 mol of C and 2 mol of O therefore number of C moles - 0.64 mmol O moles - 1.28 mmol from H₂O 1 mol of H₂O - 2 mol of H and 1 mol of O number of H moles - 1.28 mmol O moles - 0.64 mmol
Mass of menthol initially - 10 mg in reactions, the masses of products are equal to the masses of reactants. The excess mass to the products formed is due to O₂ in air Original mass of menthol - 10 mg mass of water and CO₂ - 11.53 mg + 28.16 mg = 39.69 Difference in mass - 39.69 - 10 = 29.69 mg This difference comes from O moles in air - 29.69 mg/ 16 g/mol = 1.8556 mmol then O moles coming from menthol - (1.28 + 0.64) - 1.8556 = 0.064 mmol
In menthol C moles - 0.64 mmol H moles - 1.28 mmol O moles - 0.064 mmol ratios of C:H:O C H O 0.64 1.28 0.064 x1000 x1000 x1000 to get whole numbers 640 1280 64 10 20 1 Simplest ratio of C:H:O is 10:20:1 therefore empirical formula of menthol is C₁₀H₂₀O
The full solution can be found in the image attached. Graham's law was applied to the problem. The rate of diffusion of a gas is inversely proportional to its molar mass or vapour density. Molar mass= 2vapour density
To determine weather the bond is double or triple simply check the electron involved in mutual sharing of an electron if 2 electron takes parts it said to be double or if 3 it said to be triple.
