The combustion of \(0.4196 \mathrm{~g}\) of a hydrocarbon releases \(17.55 \mathrm{~kJ}\) of heat. The masses of the products are \(\mathrm{CO}_{2}=1.419 \mathrm{~g}\) and \(\mathrm{H}_{2} \mathrm{O}=0.290 \mathrm{~g}\). (a) What is the empirical formula of the compound? (b) If the approximate molar mass of the compound is \(76 \mathrm{~g}\), calculate its standard enthalpy of formation.

Begin by determining the amounts of hydrogen and carbon present in the combustion products. The mass of carbon can be calculated from the mass of \(\mathrm{CO}_2\) produced, knowing that one mole of \(\mathrm{CO}_2\) contains one mole of carbon. The molar mass of carbon is approximately 12 g/mol and the molar mass of \(\mathrm{CO}_2\) is approximately 44 g/mol. Therefore \((1.419 \mathrm{~g}~\mathrm{CO}_2) \times (1 \mathrm{~mol}~\mathrm{C}/44 \mathrm{~g}~\mathrm{CO}_2) = 0.03225 \mathrm{~mol}~\mathrm{C}\). The mass of hydrogen can be calculated from the mass of \(\mathrm{H}_2 \mathrm{O}\) produced, knowing that one mole of \(\mathrm{H}_2 \mathrm{O}\) contains two moles of hydrogen. The molar mass of hydrogen is approximately 1 g/mol and the molar mass of \(\mathrm{H}_2\mathrm{O}\) is approximately 18 g/mol. Therefore \((0.290 \mathrm{~g}~\mathrm{H}_2 \mathrm{O}) \times (2 \mathrm{~mol}~\mathrm{H}/18 \mathrm{~g}~\mathrm{H}_2 \mathrm{O} ) = 0.03222 \mathrm{~mol}~\mathrm{H}\). The approximate empirical formula is CH (since the ratio of moles of carbon to hydrogen is nearly 1:1).

Next, determine the molecular formula. The empirical formula mass is calculated as approximately 13 g/mol (12 g/mol for C and 1 g/mol for H). Using the given approximate molar mass of 76 g/mol, the empirical formula units per molecule of compound is \(76 \mathrm{~g/mol} / 13 \mathrm{~g/mol} = 5.85\), or approximately 6. Therefore, the molecular formula of the compound is C6H6.

The enthalpy of formation is the heat absorbed or released when one mole of a compound is formed from its elements in their standard states. It's given by \(\Delta H_f = - \frac{q}{n}\), where \(q\) is the heat released and \(n\) is the number of moles of the compound. Here, \(q = 17.55 \mathrm{~kJ}\) (converted to J gives 17550 J) and \(n = 0.4196 \mathrm{~g} / 76 \mathrm{~g/mol} = 0.00552 \mathrm{~mol}\). Therefore, \(\Delta H_f = - \frac{-17550 \mathrm{~J}}{0.00552 \mathrm{~mol}} = -3180000 \mathrm{~J/mol}\) or \(-3180 \mathrm{~kJ/mol}\).