Question

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

Answer 1

To determine the empirical and molecular formulas, we first need to convert the percentages of each element into moles, and then find the simplest whole number ratio of the elements.

### Step 1: Convert percentages to grams

Assume we have 100 grams of the compound:
- Carbon (C): 40.8 g
- Hydrogen (H): 5.8 g
- Oxygen (O): 53.4 g

### Step 2: Convert grams to moles

Molar masses:
- Carbon (C): ~12.01 g/mol
- Hydrogen (H): ~1.008 g/mol
- Oxygen (O): ~16.00 g/mol

Calculating moles:
- Moles of Carbon: \( \frac{40.8 \, \text{g}}{12.01 \, \text{g/mol}} \approx 3.40 \, \text{mol} \)
- Moles of Hydrogen: \( \frac{5.8 \, \text{g}}{1.008 \, \text{g/mol}} \approx 5.75 \, \text{mol} \)
- Moles of Oxygen: \( \frac{53.4 \, \text{g}}{16.00 \, \text{g/mol}} \approx 3.34 \, \text{mol} \)

### Step 3: Find the simplest ratio

Now, we take the smallest number of moles (which is approximately 3.34 for Oxygen) and divide all mole amounts by it:

- Carbon: \( \frac{3.40}{3.34} \approx 1.02 \)
- Hydrogen: \( \frac{5.75}{3.34} \approx 1.72 \)
- Oxygen: \( \frac{3.34}{3.34} \approx 1.00 \)

To avoid fractions, we can approximate these ratios. Since \( 1.02 \) is very close to \( 1 \) and \( 1.72 \) suggests close to \( 1.5 \), we can multiply all ratios by \( 2 \) to find whole numbers:

- Carbon: \( 1.02 \times 2 \approx 2 \)
- Hydrogen: \( 1.72 \times 2 \approx 3.44 \) (rounds to \( 3 \))
- Oxygen: \( 1.00 \times 2 \approx 2 \)

Thus, the empirical formula is approximately \( C_2H_4O_2 \).

### Step 4: Calculate molecular formula

The molar mass of the empirical formula \( C_2H_4O_2 \):
- \( 2 \times 12.01 + 4 \times 1.008 + 2 \times 16.00 = 24.02 + 4.032 + 32.00 \approx 60.05 \, \text{g/mol} \)

Now, to find the molecular formula:
- Given molecular mass = \( 180 \, \text{g/mol} \)
- Divide by empirical formula mass: \( \frac{180}{60.05} \approx 3 \)

Thus, the molecular formula is \( C_{2 \times 3}H_{4 \times 3}O_{2 \times 3} = C_6H_{12}O_6 \).

### Common Name

The compound \( C_6H_{12}O_6 \) is commonly known as **Glucose**.

### Summary
- **Empirical Formula:** \( C_2H_4O_2 \)
- **Molecular Formula:** \( C_6H_{12}O_6 \)
- **Common Name:** Glucose