Final answer:
The molecular formula of a compound with an empirical formula of C6H6O and a vapour density of 47 is found to be the same as its empirical formula, as the empirical formula mass is equal to twice the vapour density. Therefore, the molecular formula is C6H6O.
Step-by-step explanation:
The student asked to find the molecular formula of a compound with an empirical formula of C6H6O and a vapour density of 47. To find the molecular formula, one must first calculate the molar mass of the empirical formula.
The atomic masses are approximately 12 amu for C, 1 amu for H, and 16 amu for O, which gives a molar mass of the empirical formula: 6(12) + 6(1) + 1(16) = 94 amu.
Vapour density is defined as half the molar mass of the substance in gaseous state, so the actual molar mass of the compound is twice the vapour density: 47 x 2 = 94 amu.
Here, the empirical formula mass is equal to the actual molar mass calculated from vapour density, which means that the empirical formula is the same as the molecular formula. Hence, the molecular formula of the compound is C6H6O.