Question

A compound has a molecular mass of 180 g/mol. It contains 40.8 percent carbon, 5.8 percent hydrogen, and 53.4 percent oxygen. Calculate its empirical and molecular formula, and give its common name.

Answer 1

To determine the empirical and molecular formulas of the compound and its common name, follow these steps:

### Step 1: Determine the moles of each element

1. **Carbon (C)**:
- Percentage: 40.8%
- Moles of Carbon = \(\frac{40.8 \, \text{g}}{12.01 \, \text{g/mol}} \approx 3.40 \, \text{mol}\)

2. **Hydrogen (H)**:
- Percentage: 5.8%
- Moles of Hydrogen = \(\frac{5.8 \, \text{g}}{1.008 \, \text{g/mol}} \approx 5.75 \, \text{mol}\)

3. **Oxygen (O)**:
- Percentage: 53.4%
- Moles of Oxygen = \(\frac{53.4 \, \text{g}}{16.00 \, \text{g/mol}} \approx 3.34 \, \text{mol}\)

### Step 2: Convert moles to the simplest ratio

Next, we divide the number of moles of each element by the smallest number of moles calculated.

- For Carbon:
- \( \frac{3.40}{3.34} \approx 1.02 \)

- For Hydrogen:
- \( \frac{5.75}{3.34} \approx 1.72 \)

- For Oxygen:
- \( \frac{3.34}{3.34} = 1 \)

### Step 3: Simplify these ratios to whole numbers

To express the ratios as whole numbers, we can multiply all the ratios by a common factor. In this case, multiplying by 2 will work well:

- Carbon: \(1.02 \times 2 \approx 2\)
- Hydrogen: \(1.72 \times 2 \approx 3.44 \approx 3.5 \times 2 = 7\)
- Oxygen: \(1 \times 2 = 2\)

### Empirical Formula

The empirical formula is \(C_2H_7O_2\).

### Step 4: Calculate the empirical formula mass

Now, calculate the empirical formula mass:

- Carbon: \(2 \times 12.01 = 24.02 \, \text{g/mol}\)
- Hydrogen: \(7 \times 1.008 = 7.056 \, \text{g/mol}\)
- Oxygen: \(2 \times 16.00 = 32.00 \, \text{g/mol}\)

Total empirical formula mass = \(24.02 + 7.056 + 32.00 = 63.08 \, \text{g/mol}\).

### Step 5: Determine the molecular formula

The molecular formula can be found by dividing the molar mass by the empirical formula mass:

- Molar mass of compound = 180 g/mol
- Empirical formula mass = 63.08 g/mol

Now, divide 180 g/mol by 63.08 g/mol:

\[
\frac{180}{63.08} \approx 2.855 \approx 3
\]

### Step 6: Multiply the empirical formula by this factor

Multiply the empirical formula by 3:

- \(C_2H_7O_2 \Rightarrow (C_2H_7O_2) \times 3 = C_6H_{21}O_6\)

### Molecular Formula

The molecular formula is \(C_6H_{21}O_6\).

### Common Name

This compound is known as **glucose (C₆H₁₂O₆)**. However, note that \(C_6H_{21}O_6\) might suggest a specific structural isomer or derivative of glucose. The more general compound we refer to is glucose (C₆H₁₂O₆).

Thus, the empirical formula is \(C_2H_7O_2\), the molecular formula is \(C_6H_{21}O_6\), and the common name suggested is glucose or a derivative.