(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?

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 = \(0.015 L \times 0.1008 M = 0.001512 \mathrm{~moles}\)

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

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 = \( \frac{mass}{moles} = \frac{0.2053 \mathrm{~g}}{0.001512 \mathrm{~moles}} = 135.7 \mathrm{~g/mol}\)

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 = \( \frac{23.5 \mathrm{~g}} {16.00 \mathrm{~g/mol}} = 1.469 \mathrm{~moles}\)

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

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