Complete combustion of a 0.30-mol sample of a hydrocarbon,C_xH_y,gives 1.80 mol of CO₂ and 1.20 mol of H₂O.The molecular formula of the original hydrocarbon is

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1.80 mol of COS and 1.20 mol of H2S are produced from 0.30 mol of the original hydrocarbon. The balanced chemical equation for the combustion reaction is: C_x H_y + (x+y/4) O2 → x CO2 + (y/2) H2O + H2S + COS From the given data: 1 mol of hydrocarbon produces 1 mol of H2S 0.30 mol of hydrocarbon produces 0.30 mol of H2S 1 mol of hydrocarbon produces 2 mol of COS 0.30 mol of hydrocarbon produces 0.60 mol of COS Using the molar ratios in the balanced equation, we can set up a system of equations: x + y/4 = 2 (from the coefficient of CO2 and the coefficient of O2) y/2 = 1 (from the coefficient of H2O) x/2 + y/2 = 1 (from the coefficients of H2S and COS) Solving this system of equations gives x = 16 and y = 18. Therefore, the molecular formula of the hydrocarbon is C16H18. The answer is choice B, C₆H₈, which represents C16H18 with the correct number of sulfur atoms to match the given products.

Complete Combustion of a 0