Problem 106 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) By titration, \(15.0 \mathrm{~mL}\) of \(0.1008 \mathrm{Msodium}\) hydroxide is needed to neutralizea \(0.2053-\mathrm{g}\) sample of a weak acid. What is the molar mass of the acid if it is monoprotic? (b) An elemental analysis of the acid indicates that it is composed of \(5.899 \mathrm{H}, 70.6 \% \mathrm{C},\) and \(23.5 \% \mathrm{O}\) by mass. What is its molecular formula?

Step-by-Step Solution

Answer

The molar mass of the weak monoprotic acid is 135.7 g/mol, and its molecular formula is C8H8O2.

Step 1: Calculate moles of sodium hydroxide

First, we need to find the moles of sodium hydroxide (NaOH) used in the titration using the given volume and molarity. Moles of NaOH = Volume of NaOH × Molarity of NaOH Moles of NaOH = \(15.0 mL \times 0.1008 M\) Note: 1 mL = 0.001 L, so we will convert mL to L Moles of NaOH = \(0.015 L \times 0.1008 M = 0.001512 \mathrm{~moles}\)

Step 2: Calculate moles of the weak acid

Since we are given that the weak acid is monoprotic (meaning it donates one proton per molecule), the stoichiometric ratio between the weak acid and sodium hydroxide is 1:1. Therefore, the moles of the weak acid are equal to the moles of sodium hydroxide. Moles of weak acid = Moles of NaOH = 0.001512 moles

Step 3: Calculate molar mass of the weak acid

Now, we can find the molar mass of the weak acid by dividing its mass by the moles we found earlier. Molar mass of weak acid = \( \frac{mass}{moles} = \frac{0.2053 \mathrm{~g}}{0.001512 \mathrm{~moles}} = 135.7 \mathrm{~g/mol}\)

Step 4: Find the empirical formula of the weak acid

To find the empirical formula, let's assume we have 100 grams of the weak acid. We can calculate the grams of each element from the given percentages and then convert them to moles. Mass of H = \(5.899 \% \times 100 \mathrm{~g} = 5.899 \mathrm{~g}\) Mass of C = \(70.6 \% \times 100 \mathrm{~g} = 70.6 \mathrm{~g}\) Mass of O = \(23.5 \% \times 100 \mathrm{~g} = 23.5 \mathrm{~g}\) Now, convert these masses to moles using the molar masses of H, C, and O. Moles of H = \( \frac{5.899 \mathrm{~g}} {1.01 \mathrm{~g/mol}} = 5.836 \mathrm{~moles}\) Moles of C = \( \frac{70.6 \mathrm{~g}} {12.01 \mathrm{~g/mol}} = 5.879 \mathrm{~moles}\) Moles of O = \( \frac{23.5 \mathrm{~g}} {16.00 \mathrm{~g/mol}} = 1.469 \mathrm{~moles}\)

Step 5: Determine the ratio of the elements in the empirical formula

Dividing each number of moles by the smallest number of moles will give us the mole ratio of the elements in the empirical formula. Mole ratio of H:C:O = \(\frac{5.836}{1.469} : \frac{5.879}{1.469} : \frac{1.469}{1.469} = 3.97 : 4.0 : 1\) Rounding off these ratios, we get: Empirical Formula = C4H4O1, which can be written as C4H4O

Step 6: Calculate the molecular formula

Now, we can find the molecular formula by comparing the molar mass of the empirical formula and the molar mass of the weak acid we calculated earlier. Empirical formula molar mass = (4 × 12.01) + (4 × 1.01) + (1 × 16.00) = 60.09 g/mol Next, we calculate the ratio between the molar mass of the weak acid and the empirical formula molar mass: n = \( \frac{135.7 \mathrm{~g/mol}}{60.09 \mathrm{~g/mol}} = 2.26 \approx 2\) Since n is approximately 2, the molecular formula is twice the empirical formula: Molecular Formula = 2 × (C4H4O) = C8H8O2

Key Concepts

Titration

Titration is a laboratory method used to determine the concentration of an unknown solution by reacting it with a solution of known concentration. This process involves the gradual addition of the titrant, the known solution, to the analyte, the unknown solution, until the reaction reaches its endpoint. The endpoint is often indicated by a color change of an indicator that is added to the analyte.

In our exercise, sodium hydroxide (NaOH) is used as the titrant to neutralize a weak acid, which means we are looking at an acid-base titration. By measuring the volume of NaOH required to neutralize the acid, we can calculate the moles of NaOH used in the reaction. This is done using the formula: \[\text{Moles of NaOH} = \text{Volume of NaOH} \times \text{Molarity of NaOH}\] Here, both reactants participate in a direct stoichiometric relationship, especially since we're dealing with a monoprotic acid.

Monoprotic Acid

A monoprotic acid is an acid that can donate only one proton or hydrogen ion ( H^+ ) per molecule in a chemical reaction. This means for every molecule of acid, only one hydrogen ion participates in the reaction.

In the context of the titration problem, knowing that the weak acid is monoprotic simplifies the calculations because the stoichiometry is 1:1. This means the moles of the acid are equal to the moles of NaOH since one molecule of the acid reacts with one molecule of NaOH.

For example, if we calculated that there are 0.001512 moles of NaOH, there would be an identical amount of moles of the monoprotic acid present. This relationship is very helpful in determining quantities like the acid's molar mass. By dividing the mass of the acid sample by the number of moles, you get the molar mass of the acid.

Empirical Formula

An empirical formula of a compound gives the simplest whole-number ratio of the atoms in the compound. To derive it, you need the percentage composition by mass of each element in the compound, as given in the exercise for hydrogen, carbon, and oxygen.

Here's the step-by-step process to find the empirical formula:

• Assume a 100 g sample of the compound; percentages can then be directly converted into grams.
• Convert the mass of each element to moles using the respective atomic masses.
• Determine the simplest ratio by dividing each mole value by the smallest number of moles calculated.
• Adjust to the smallest whole-number ratio if necessary, resulting in the empirical formula.

In our example, H, C, and O have the mole ratios of about 4:4:1, giving the empirical formula C\(_4\)H\(_4\)O.

Molecular Formula

The molecular formula of a compound represents the actual number of atoms of each element in a molecule of the compound. It can be determined by comparing the compound's molar mass to the molar mass of its empirical formula.

In the given exercise, the molar mass of the weak acid is already known, and we have calculated the molar mass of its empirical formula. Here's how you find the molecular formula:

• Divide the compound's molar mass by the empirical formula's molar mass to find a factor, which we can denote as \(n\).
• If \(n\) is approximately a whole number, multiply the subscripts in the empirical formula by \(n\) to get the molecular formula.

Based on the exercise, \(n\) is about 2. Thus, the molecular formula is C\(_8\)H\(_8\)O\(_2\), indicating twice the number of atoms as in the empirical formula.