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To solve this problem, let's break it into two parts:

### Part 1: Molar Mass Calculation
The first part involves determining the molar mass of the unknown monoprotic organic acid based on the titration data. Here's the process:

1. **Mass of the acid sample** = 0.730 g
2. **Volume of NaOH used** = 51.6 mL = 0.0516 L
3. **Concentration of NaOH solution** = 0.191 M

Since the acid is monoprotic, the reaction between the acid and NaOH is a 1:1 molar ratio:

\[
\text{HA} + \text{NaOH} \rightarrow \text{NaA} + \text{H}_2\text{O}
\]

First, calculate the moles of NaOH used:

\[
\text{Moles of NaOH} = \text{Volume} \times \text{Concentration} = 0.0516 \, \text{L} \times 0.191 \, \text{mol/L} = 0.0098436 \, \text{mol}
\]

Since the acid is monoprotic, the moles of acid (HA) will be the same as the moles of NaOH:

\[
\text{Moles of acid (HA)} = 0.0098436 \, \text{mol}
\]

Now, calculate the molar mass of the acid:

\[
\text{Molar mass of the acid} = \frac{\text{Mass of acid}}{\text{Moles of acid}} = \frac{0.730 \, \text{g}}{0.0098436 \, \text{mol}} \approx 74.15 \, \text{g/mol}
\]

### Part 2: Molecular Formula Calculation
The second part asks for the molecular formula based on the compound's percentage composition: 

- **48.6% C**
- **8.20% H**
- **43.2% O**

We need to determine the empirical formula first.

1. **Assume 100 g of the compound** so the percentages can be treated as grams.
   - Carbon: 48.6 g
   - Hydrogen: 8.20 g
   - Oxygen: 43.2 g

2. **Convert to moles** by dividing by the atomic masses:
   - Carbon: \(\frac{48.6}{12.01} = 4.05 \, \text{mol}\)
   - Hydrogen: \(\frac{8.20}{1.008} = 8.13 \, \text{mol}\)
   - Oxygen: \(\frac{43.2}{16.00} = 2.70 \, \text{mol}\)

3. **Find the simplest whole number ratio**:
   - Divide all by the smallest number of moles (2.70):
     - Carbon: \(\frac{4.05}{2.70} \approx 1.5\)
     - Hydrogen: \(\frac{8.13}{2.70} \approx 3\)
     - Oxygen: \(\frac{2.70}{2.70} = 1\)

4. **Adjust to whole numbers**:
   Multiply everything by 2 to get whole numbers:
   - Carbon: \(1.5 \times 2 = 3\)
   - Hydrogen: \(3 \times 2 = 6\)
   - Oxygen: \(1 \times 2 = 2\)

So, the **empirical formula** is \( \text{C}_3\text{H}_6\text{O}_2 \).

Since the molar mass of the compound is approximately 74 g/mol, compare this with the empirical formula molar mass:

\[
\text{Molar mass of } \text{C}_3\text{H}_6\text{O}_2 = (3 \times 12.01) + (6 \times 1.008) + (2 \times 16.00) = 74.08 \, \text{g/mol}
\]

The empirical formula matches the molar mass, so the molecular formula is also \( \text{C}_3\text{H}_6\text{O}_2 \).

### Final Answers:
- **Molar mass**: 74.15 g/mol
- **Molecular formula**: \( \text{C}_3\text{H}_6\text{O}_2 \)

Would you like further details or have any other questions? Here are some related questions to deepen your understanding:

1. What is the significance of a monoprotic acid in titration?
2. How do different concentrations of titrants affect the volume used in titration?
3. Why is it important to use the empirical formula when determining molecular formulas?
4. How can you confirm that a reaction is completed during titration?
5. What are the potential sources of error in titration experiments?

**Tip**: In titrations, always ensure accurate measurements of the titrant volume and concentration to get precise molar mass calculations.

The identity of an unknown monoprotic organic acid is determined by titration. A 0.730 g sample of the acid is titrated with 0.191 M NaOH. What is the molar mass of the compound if 51.6 mL of the NaOH solution is required to neutralize the sample?
The compound is 48.6% C, 8.20% H, and 43.2% O. What is the molecular formula of the organic acid?

Determine the molar mass and molecular formula of an unknown monoprotic organic acid using titration data and percent composition. This problem involves key stoichiometric concepts and empirical formula determination. Suitable for college-level chemistry students.

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