A sample of a hydrocarbon is combusted completely in \(\mathrm{O}_{2}(g)\) to produce \(21.83 \mathrm{g} \mathrm{CO}_{2}(g), 4.47 \mathrm{g} \mathrm{H}_{2} \mathrm{O}(g),\) and 311 \(\mathrm{kJ}\) of heat. (a) What is the mass of the hydrocarbon sample that was combusted? (b) What is the empirical formula of the hydrocarbon? (c) Calculate the value of \(\Delta H_{f}^{\circ}\) per empirical-formula unit of the hydrocarbon. (d) Do you think that the hydrocarbon is one of those listed in Appendix C? Explain your answer.

From the masses of \(CO_2\) and \(H_2 O\) produced, we can determine the mass of carbon and hydrogen in the original hydrocarbon sample by using their respective molar masses. For Carbon (in \(CO_2\)): Mass of \(CO_2\) = 21.83 g Molar mass of \(C\) = 12.01 g/mol Number of moles of \(C\) = mass of \(CO_2\) × (molar mass of \(C\) / molar mass of \(CO_2\)) Number of moles of \(C\) = 21.83 × (12.01 / 44.01) = 5.99 × 10⁻³ mol For Hydrogen (in \(H_2 O\)): Mass of \(H_2 O\) = 4.47 g Molar mass of \(H_2\) = 2.02 g/mol Number of moles of \(H_2\) = mass of \(H_2 O\) × (molar mass of \(H_2\) / molar mass of \(H_2 O\)) Number of moles of \(H_2\) = 4.47 × (2.02 / 18.02) = 0.502 mol Now, we can find the mass of carbon and hydrogen in the hydrocarbon. Mass of \(C\) = number of moles of \(C\) × molar mass of \(C\) = 5.99 × 10⁻³ mol × 12.01 g/mol = 0.07189 g Mass of \(H_2\) = number of moles of \(H_2\) × molar mass of \(H_2\) = 0.502 mol × 2.02 g/mol = 1.014 g

Since the hydrocarbon contains only carbon and hydrogen, the mass of the hydrocarbon is the sum of the masses of carbon and hydrogen in it. Mass of hydrocarbon = mass of \(C\) + mass of \(H_2\) = 0.07189 g + 1.014 g = 1.0859 g

To find the empirical formula, we need to find the simplest whole-number ratio of the moles of carbon and hydrogen in the hydrocarbon. Ratio of moles of \(C\) to \(H_2\): = (5.99 × 10⁻³ mol) / (0.502 mol) = 0.0119 To find the whole number ratio, we can divide the given ratio by the smallest number in the ratio: = 0.0119 / 0.0119 = 1 So, the whole-number ratio of the moles of carbon and hydrogen in the hydrocarbon is 1:1, which gives the empirical formula of the hydrocarbon as \(CH_2\).

To find the standard enthalpy of formation ΔHf° per empirical formula unit (here, \(CH_2\)), we can use the following relation: ΔHf° = (kJ/mol of the hydrocarbon) × (1/Ratio of empirical formula unit moles) Given, the heat of combustion for the hydrocarbon is 311 kJ. Let's convert this value to kJ/mol: Number of moles of the hydrocarbon = mass of the hydrocarbon / molar mass of the hydrocarbon empirical formula = 1.0859 g / (12.01 g/mol + 2 × 1.01 g/mol) = 0.05019 mol Heat of combustion per mole of the hydrocarbon = 311 kJ / 0.05019 mol = 6193 kJ/mol Next, find the enthalpy of formation ΔHf° per empirical formula unit \(CH_2\): ΔHf°(per \(CH_2\)) = (6193 kJ/mol) × (1/1) = 6193 kJ/mol

From the empirical formula and the enthalpy of formation value, we can compare the hydrocarbon to the list of hydrocarbons in Appendix C. However, determining if our hydrocarbon is one of those listed in Appendix C requires a thorough evaluation of the given data (empirical formula and ΔHf°) and comparison to the data listed for different hydrocarbons in Appendix C.