On complete combustion in air, a sample of an unknown hydrocarbon yielded 176 grams of CO2 and 108 grams of H2O with no additional products. Which one of the following might be the formula of the hydrocarbon? (Relative atomic masses: H=1, C =12, O =16)

a) CH₂b) CH₃c) CH₄d) C₂H₄ e) C₂H₈

The equation for the combustion of the hydrocarbon is: C_xH_y+ (A + 0.5B)O₂ --> A.CO₂+ B.H₂OMoles = mass/A_rMoles of CO₂= 176/44 = 4 = AMoles of H₂O = 108/18 = 6 = BSubstitute the moles into the equation: C_xH_y+ 7O₂ --> 4CO₂ + 6H₂OTo balance the equation, there must be 4 carbon atoms and 12 hydrogen atoms on the left side. The formula of the hydrocarbon must either contain 4 carbon and 12 hydrogen atoms, or contain a number of carbon and hydrogen atoms that are factors of 4 and 12, respectively.Therefore, the options for the formula of the hydrocarbon are: C₄H₁₂, C₂H₆or CH₃ Out of those, CH₃is the only option given, so the answer is B.