Problem 125 - Chemistry: The Central Science in SI Units, Global Edition

Question

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.

Step-by-Step Solution

Answer

(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.

Step 1: Calculating the moles of CO₂ and H₂O produced

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

Step 2: Calculating the moles of C and H in the hydrocarbon

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

Step 3: Calculating the empirical formula of the hydrocarbon

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.

Step 4: Calculating the mass of the hydrocarbon

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

Step 5: Calculating the value of ΔHf° per empirical formula unit of the hydrocarbon

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

Step 6: Deciding whether the hydrocarbon is one of those listed in Appendix C

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.

Key Concepts

Empirical Formula Calculation

The empirical formula of a compound represents the simplest whole-number ratio of the atoms present. To find it, we first determine how many moles of each element are in the compound. In the context of hydrocarbon combustion analysis, after the complete combustion of a hydrocarbon sample, the elements' moles can be inferred from the products formed. When a hydrocarbon combusts, all carbon atoms convert into carbon dioxide and all hydrogen atoms form water.

In our exercise, the number of moles of carbon ( amage of the compound)) is determined using the moles of CO₂ produced, which we found to be 0.496 mol. The moles of hydrogen are derived from the water produced. Since each water molecule contains two hydrogen atoms, we calculate the moles of hydrogen as 2 times the moles of H₂O, resulting in 0.496 mol as well.

Next, we find the ratio of carbon to hydrogen by dividing each element's mole quantity by the smallest number among them, which in this case is 0.496. This results in a 1:1 ratio for C:H, leading to the empirical formula CH. It reflects the proportional representation in the simplest form of atoms in the hydrocarbon.

Molar Mass Concept

The concept of molar mass is vital in stoichiometry and identifying chemical formulas. Molar mass is the weight of one mole of a given substance expressed in grams per mole. Each element's molar mass comes from its atomic weight on the periodic table. For example, the molar mass of carbon is approximately 12.01 g/mol, and for hydrogen, it's about 1.01 g/mol.

In the combustion analysis context, knowing how to use molar mass allows us to convert between grams and moles, which is essential when calculating how much of each element was involved in forming the combustion products. From our example, we calculated the moles of CO₂ and H₂O using their molar masses, 44.01 g/mol for CO₂ and 18.02 g/mol for H₂O. This conversion is essential for determining the moles of elements involved, which then helps in calculating the empirical formula.

Understanding molar mass is crucial for computing other quantities like the mass of the hydrocarbon before combustion occurred. By multiplying the moles by their respective molar masses, we find the substance's actual weight. This process culminates in revealing the mass of the original hydrocarbon sample as calculated in the exercise.

Heat of Formation

The heat of formation, denoted as \(\Delta H_{f}^{\circ}\), represents the change in enthalpy when one mole of a compound is formed from its elements under standard conditions. It is crucial in understanding reaction energetics, particularly in combustion processes where hydrocarbons release energy.

In our problem, we focus on the \(\Delta H_{f}^{\circ}\) per empirical formula unit of the hydrocarbon. The total energy change from combustion was provided as 311 kJ. This value aids in finding energy release per mole of the empirical formula. By dividing the total energy by the moles of empirical formula units (0.496 mol, calculated earlier), we find \(\Delta H_{f}^{\circ}\) to be 627 kJ/mol.

This calculated energy reflects on whether the hydrocarbon might be listed in a reference appendix, such as Appendix C from a chemistry text. These references include typical heats of formation for known substances, helping identify unknowns based on their energetics. While our example did not match one exactly, such a heat of formation suggests potential matches with unsaturated hydrocarbons, such as alkenes or alkynes, which should be further clarified by cross-referencing with specific databases.