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 empiricalformula unit of the hydrocarbon. (d) Do you think that the hydrocarbon is one of those listed in Appendix C? Explain your answer.

To calculate the moles of CO₂ and H₂O, we use the formula: moles = mass/molar mass Molar mass of CO₂ = (12.01 g/mol of C) + (2 * 16.00 g/mol of O) = 44.01 g/mol Molar mass of H₂O = (2 * 1.01 g/mol of H) + (16.00 g/mol of O) = 18.02 g/mol moles of CO₂ = (21.83 g)/(44.01 g/mol) = 0.496 mol moles of H₂O = (4.47 g)/(18.02 g/mol) = 0.248 mol

From the combustion reaction, it is clear that all the carbon atoms from the hydrocarbon end up in CO₂, and all the hydrogen atoms end up in H₂O. So, moles of C in the hydrocarbon = moles of CO₂ = 0.496 mol And moles of H in the hydrocarbon = 2 * moles of H₂O = 2 * 0.248 mol = 0.496 mol

For the empirical formula, we will find the whole number ratio of C and H atoms. Divide the moles of each element by the smallest value, which is 0.496 mol for both the elements: C ratio: 0.496 / 0.496 = 1 H ratio: 0.496 / 0.496 = 1 The whole number ratio of C to H is 1:1. Therefore, the empirical formula of the hydrocarbon is CH.

Since we calculated the moles of carbon and hydrogen in the hydrocarbon, we can now calculate the total mass. mass of C = moles of C * molar mass of C = 0.496 mol * 12.01 g/mol = 5.96 g mass of H = moles of H * molar mass of H = 0.496 mol * 1.01 g/mol = 0.50 g Total mass of the hydrocarbon = mass of C + mass of H = 5.96 g + 0.50 g = 6.46 g

We will use the given formula: ΔHf° per empirical formula unit = (ΔH of combustion)/ (number of empirical formula units in combustion reaction) ΔHf° per empirical formula unit = (311 kJ) / (0.496 mol) = 627 kJ/mol

When comparing the empirical formula (CH) and ΔHf° (627 kJ/mol) to the values in Appendix C, none of them seem to match directly. However, one of the possibilities includes unsaturated hydrocarbon such as alkynes or alkenes. The student should refer to the specific list in Appendix C for more information, as different books/curriculums may have different listings. To summarize: (a) The mass of the hydrocarbon sample = 6.46 g (b) The empirical formula of the hydrocarbon = CH (c) The value of ΔHf° per empirical formula unit of the hydrocarbon = 627 kJ/mol (d) The hydrocarbon may not be one of those listed in Appendix C but could be one of the unsaturated hydrocarbons. The student should consult the specific list in Appendix C for more details.