Question

Compound \(\mathrm{X}\) contains only the elements \(\mathrm{C}_{2} \mathrm{H}, \mathrm{O}\), and \(\mathrm{S} . \mathrm{A}\) \(5.00 \mathrm{~g}\) sample undergoes complete combustion to give \(4.83 \mathrm{~g}\) of \(\mathrm{CO}_{2}, 1.48 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\), and a certain amount of \(\mathrm{SO}_{2}\) that is further oxidized to \(\mathrm{SO}_{3}\) and dissolved in water to form sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) On titration of the \(\mathrm{H}_{2} \mathrm{SO}_{4}, 109.8 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) is needed for complete reaction. (Both \(\mathrm{H}\) atoms in sulfuric acid are acidic and react with \(\mathrm{NaOH}\).) (a) What is the empirical formula of \(\mathrm{X} ?\) (b) When \(5.00 \mathrm{~g}\) of \(\mathrm{X}\) is titrated with \(\mathrm{NaOH}\), it is found that \(\mathrm{X}\) has two acidic hydrogens that react with \(\mathrm{NaOH}\) and that \(54.9 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) is required to completely neutralize the sample. What is the molecular formula of \(\mathrm{X} ?\)

Short answer

(a) The empirical formula is C2H3S. (b) The molecular formula is C6H9S3.

Step-by-step solution

Determine Moles of Carbon from CO2

Firstly, convert the mass of \( \mathrm{CO}_2 \) obtained from combustion to moles of carbon. Since the molar mass of \( \mathrm{CO}_2 \) is approximately \( 44.01 \text{ g/mol} \), the number of moles of \( \mathrm{CO}_2 \) is \( 4.83 \text{ g} / 44.01 \text{ g/mol} \approx 0.1097 \text{ moles}. \) Each mole of \( \mathrm{CO}_2 \) contains one mole of carbon, so the moles of carbon are also \( 0.1097 \text{ moles}. \)

Determine Moles of Hydrogen from H2O

Next, convert the mass of \( \mathrm{H}_2 \mathrm{O} \) to moles of hydrogen. The molar mass of \( \mathrm{H}_2 \mathrm{O} \) is approximately \( 18.02 \text{ g/mol} \). So, moles of \( \mathrm{H}_2 \mathrm{O} = 1.48 \text{ g} / 18.02 \text{ g/mol} \approx 0.0821 \text{ moles}. \) Since each mole of \( \mathrm{H}_2 \mathrm{O} \) contains two moles of hydrogen, the moles of hydrogen are \( 2 \times 0.0821 = 0.1642 \text{ moles}. \)

Determine Moles of Sulfur from NaOH Titration

The volume of NaOH solution used to neutralize sulfuric acid can give the moles of sulfur. The reaction is \( \mathrm{H}_2 \mathrm{SO}_4 + 2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{SO}_4 + 2 \mathrm{H}_2 \mathrm{O} \). Since \( 109.8 \text{ mL} \) of \( 1.00 \text{ M} \) NaOH is used, the moles of \( \mathrm{NaOH} \) is \( 0.1098 \text{ moles} \). Therefore, the moles of \( \mathrm{H}_2 \mathrm{SO}_4 \) and hence \( \mathrm{SO}_4 \) are \( 0.1098 / 2 = 0.0549 \text{ moles}. \) Each mole of \( \mathrm{SO}_4 \) results from one mole of sulfur, so there are \( 0.0549 \text{ moles} \) of sulfur.

Calculate Empirical Formula

Calculate the ratio of moles in the compound. Use the previously determined moles: \( \mathrm{C}: 0.1097 \), \( \mathrm{H}: 0.1642 \), \( \mathrm{S}: 0.0549 \). Dividing each by the smallest value (0.0549), we get: \( \mathrm{C} = 2, \mathrm{H} = 3, \mathrm{S} = 1 \). The empirical formula is \( \mathrm{C}_2 \mathrm{H}_3 \mathrm{S} \).

Determine Moles of Acidic Hydrogens in X

From the additional titration with NaOH, \( 54.9 \text{ mL} \) of \( 1.00 \text{ M} \) NaOH neutralized the acidic hydrogens in \( \mathrm{X} \). So, the moles of \( \mathrm{NaOH} = 0.0549. \) Since \( \mathrm{X} \) has two acidic hydrogens per molecule, this implies \( 0.02745 \text{ moles of } \mathrm{X} \) were in the sample.

Calculate Molecular Weight and Molecular Formula

Given \( 5.00 \text{ g} \) of \( \mathrm{X} \) contains \( 0.02745 \text{ moles} \), calculate the molecular weight: \( 5.00 \text{ g} / 0.02745 \approx 182.1 \text{ g/mol}. \) The empirical formula mass of \( \mathrm{C}_2 \mathrm{H}_3 \mathrm{S} \) is approximately \( 59.11 \text{ g/mol}. \) Comparing: \( 182.1 / 59.11 \approx 3.08 \), which suggests the molecular formula is \( \mathrm{C}_6 \mathrm{H}_9 \mathrm{S}_3 \).

Key concepts

Combustion Analysis

Combustion analysis is a powerful method to identify the proportions of elements in an unknown compound. When a substance is combusted, it reacts with oxygen to form products like carbon dioxide (CO extsubscript{2}), water (H extsubscript{2}O), and possibly sulfur dioxide (SO extsubscript{2}). By measuring the mass of these products, you can deduce the amount of each element in the original compound. For example, in the given exercise, the compound X is combusted, resulting in the formation of CO extsubscript{2} and H extsubscript{2}O.

• CO extsubscript{2} helps us determine the amount of carbon present.
• H extsubscript{2}O indicates the amount of hydrogen.
• Additional titration allows us to find the sulfur by transforming SO extsubscript{2} into H extsubscript{2}SO extsubscript{4}.

By knowing the molar masses, you can convert the mass of combustion products back into moles of the original elements. This precise method forms the basis of empirical formula determination in the exercise.

Titration

Titration is a laboratory method used to determine the concentration of an acidic or basic solution, revealing more insights about the sample's composition. In this process, a solution of known concentration, known as the titrant, is gradually added to a test solution until a certain reaction completion is indicated, usually by a color change. In the context of the exercise, titration is performed using NaOH to find the moles of sulfuric acid resulting from the combustion of sulfur in compound X. Sulfuric acid contains two acidic hydrogens that react with NaOH, which is crucial for determining the empirical and molecular formulas.

• The equation for the reaction is: H extsubscript{2}SO extsubscript{4} + 2 NaOH → Na extsubscript{2}SO extsubscript{4} + 2 H extsubscript{2}O.
• From the volume of NaOH used, you can calculate the moles of sulfuric acid, and hence, the moles of sulfur in the original compound.

This reaction allows us to trace back to the original sulfur content in the compound X, essential for molecular formula calculations.

Molecular Formula Determination

Determining the molecular formula of a compound involves understanding its elemental makeup and combining this knowledge with its molar mass. The molecular formula indicates the actual number of atoms of each element in a molecule, unlike the empirical formula which gives the simplest ratio. In the exercise, after finding the empirical formula (C extsubscript{2}H extsubscript{3}S) using combustion data, the molar mass is calculated from an additional titration step.

• The molar mass is found by dividing the sample's mass by the moles calculated from titration.
• The empirical formula mass of C extsubscript{2}H extsubscript{3}S is then used to find the multiple, which relates the empirical to the molecular formula.

By multiplying the empirical formula by this factor, you get the molecular formula, which in this case is C extsubscript{6}H extsubscript{9}S extsubscript{3}. This approach solidifies the exact composition of the compound and aligns empirical data with theoretical molecular structures.