When burning some organic compound weighing 3.9 g, the relative vapor density of which for hydrogen is 39

When burning some organic compound weighing 3.9 g, the relative vapor density of which for hydrogen is 39, carbon monoxide weighing 13.2 g and water 2.7 g were formed. Derive the molecular formula of the substance.

Let the formula given by the condition be CxHyOz, then we compose the equation of the combustion reaction
1CxHyOz + 7.5O2 = 6CO2 + 3H2O

M (CxHyOz) = 39 * 2 = 78
n (CxHyOz) = 3.9 / 78 = 0.05 mol / 0.05 = 1 mol
n (CO2) = 13.2 / 44 = 0.3 mol / 0.05 = 6 mol
n (H2O) = 2.7 / 18 = 0.15 mol / 0.05 = 3 mol
Let’s check M (C6H6) = 78 g / mol
the answer is C6H6 benzene