Problem 106 - Chemistry The Central Science In Si Units

Question

(a) By titration, \(15.0 \mathrm{~mL}\) of \(0.1008 \mathrm{M}\) sodium hydroxide is needed to neutralize a \(0.2053-\mathrm{g}\) 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 \(5.89 \% \mathrm{H}, 70.6 \% \mathrm{C},\) and \(23.5 \% \mathrm{O}\) by mass. What is its molecular formula?

Step-by-Step Solution

Answer

The acid's molar mass is 135.81 g/mol, and its molecular formula is \(\text{C}_8\text{H}_8\text{O}_2\).

Step 1: Calculate Moles of NaOH

The first step is to calculate the moles of sodium hydroxide (NaOH) used in the titration. Use the formula:\[ ext{Moles of NaOH} = ext{Volume (L)} \times ext{Molarity (mol/L)} \]The volume of NaOH is given in milliliters, so it must be converted to liters:\[15.0 ext{ mL} = 0.0150 ext{ L}\]Thus, the moles of NaOH are:\[ ext{Moles of NaOH} = 0.0150 ext{ L} \times 0.1008 ext{ mol/L} = 0.001512 ext{ mol}\]

Step 2: Determine Moles of Acid

Since the acid is monoprotic, it reacts with NaOH at a one-to-one molar ratio:\[ ext{Moles of acid} = ext{Moles of NaOH} = 0.001512 ext{ mol}\]Thus, the number of moles of acid is equal to the number of moles of NaOH used.

Step 3: Calculate Molar Mass of the Acid

Use the formula for molar mass, which is the mass of the acid sample divided by the number of moles:\[ ext{Molar mass of acid} = \frac{0.2053 ext{ g}}{0.001512 ext{ mol}} = 135.81 ext{ g/mol}\]This is the molar mass of the acid.

Step 4: Determine Empirical Formula by Mass Percent

Convert the mass percentages to grams (assuming a 100 g total mass for convenience):- H: 5.89 g- C: 70.6 g- O: 23.5 gConvert these masses to moles:- Moles of H: \(\frac{5.89}{1.01} = 5.83\)- Moles of C: \(\frac{70.6}{12.01} = 5.88\)- Moles of O: \(\frac{23.5}{16.00} = 1.47\)Find the simplest ratio by dividing each by the smallest number of moles (1.47):\[\text{H: } \frac{5.83}{1.47} \approx 3.97 \approx 4\]\[\text{C: } \frac{5.88}{1.47} \approx 4.00 \approx 4\]\[\text{O: } \frac{1.47}{1.47} = 1\]Thus, the empirical formula is \(\text{C}_4\text{H}_4\text{O}\).

Step 5: Determine Molecular Formula

Calculate the empirical formula mass:\[ ext{Empirical formula mass} = (4 \times 12.01) + (4 \times 1.01) + (16.00) = 80.08 ext{ g/mol}\]Determine the ratio of the molar mass to the empirical formula mass:\[ n = \frac{135.81}{80.08} \approx 1.70\]Since this must be a whole number, the nearest whole number is 2. Thus, multiply the empirical formula by 2 to find the molecular formula:\[ ext{Molecular formula} = \text{C}_8\text{H}_8\text{O}_2\]

Step 6: Conclusion

The molar mass of the acid is 135.81 g/mol, and its molecular formula is \(\text{C}_8\text{H}_8\text{O}_2\).

Key Concepts

Titration

Titration is a common laboratory technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration. It involves gradually adding a titrant from a burette to the analyte until the reaction reaches an equivalence point, where stoichiometrically equivalent quantities of the titrant and analyte have reacted. The point at which the titration is complete is often indicated by a color change due to an indicator or a dramatic change in the measured pH.

In the provided exercise, titration was used to find the moles of a weak acid. We knew how much sodium hydroxide (NaOH), a solution of known molarity, we needed to completely neutralize the acid. When calculating, it is crucial to convert volumes from milliliters to liters, as it ensures consistency with molar units. Here, the titration data helped us deduce that the moles of acid equaled the moles of NaOH because the acid is monoprotic, meaning it donates one proton when reacted with a base. The accurate determination of moles through titration is essential, as it feeds directly into calculating more complex properties of the acid.

Empirical Formula

The empirical formula of a compound expresses the simplest whole-number ratio of the atoms of each element present. It can be determined using the mass percent of each element, which represents how much of each element's mass is present per 100 grams of compound. This percentage can be directly converted into grams for computation convenience, assuming a hypothetical 100 g sample.

To illustrate, in the step-by-step solution, we used the elemental analysis to determine the empirical formula of the acid. By converting the percentage of each atom (hydrogen, carbon, and oxygen) to grams, and then to moles, we could then calculate the simplest ratio by dividing each mole value by the smallest calculated value. This process gives us the empirical formula. In the exercise, after simplifying the ratios, we arrived at the empirical formula \(\text{C}_4\text{H}_4\text{O}\), which helps in better understanding the basic composition of the substance.

Molecular Formula

The molecular formula represents the actual number of atoms of each element in a molecule of the compound. It can be a multiple of the empirical formula. Determining the molecular formula involves finding out how many times the empirical formula's molar mass fits into the compound's actual molar mass. The molecular formula is often more informative than the empirical formula, as it reflects the compound's true molar mass and structure.

In our exercise, once the molar mass of the compound (determined from titration data) and the empirical formula mass were known, the next step was finding their ratio. By dividing the molar mass by the empirical formula's mass and rounding it to the nearest whole number, we determine how many "units" of the empirical formula must be combined to form the molecular structure. For the given exercise, this involves taking \(\text{C}_4\text{H}_4\text{O}\) twice, leading to the molecular formula \(\text{C}_8\text{H}_8\text{O}_2\). This approach gives a clearer picture of the molecule's structure and composition.