448 mL of a hydrocarbon, having C = 87.8%, H = 12.19%, weigh 1.64 g at S.T.P Determine the molecular formula of the compound.

Explanation

To determine the molecular formula of the hydrocarbon given the data, we can follow these steps:

### Step 1: Calculate the Molar Mass of the Hydrocarbon
We know that at Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters (or 22,400 mL).

Given:
- Volume of hydrocarbon = 448 mL
- Mass of hydrocarbon = 1.64 g

Using the formula for molar mass:
\[ \text{Molar Mass} = \frac{\text{Mass}}{\text{Volume}} \times 22.4 \text{ L} \]
Convert 448 mL to liters:
\[ 448 \text{ mL} = 0.448 \text{ L} \]
Now, calculate the molar mass:
\[ \text{Molar Mass} = \frac{1.64 \text{ g}}{0.448 \text{ L}} \times 22.4 \text{ L} = \frac{1.64 \times 22.4}{0.448} = 82 \text{ g/mol} \]

### Step 2: Determine the Empirical Formula
Given the percentages of carbon and hydrogen:
- C = 87.8%
- H = 12.19%

Convert these percentages to grams (assuming 100 g of the compound):
- Mass of C = 87.8 g
- Mass of H = 12.19 g

Now, convert grams to moles:
\[ \text{Moles of C} = \frac{87.8 \text{ g}}{12 \text{ g/mol}} = 7.3167 \text{ mol} \]
\[ \text{Moles of H} = \frac{12.19 \text{ g}}{1 \text{ g/mol}} = 12.19 \text{ mol} \]

### Step 3: Find the Simplest Ratio
To find the simplest ratio, divide the number of moles of each element by the smallest number of moles:
\[ \text{Ratio of C} = \frac{7.3167}{7.3167} = 1 \]
\[ \text{Ratio of H} = \frac{12.19}{7.3167} \approx 1.66 \]

### Step 4: Convert to Whole Numbers
To convert the ratio to whole numbers, multiply both by 3:
- C: \(1 \times 3 = 3\)
- H: \(1.66 \times 3 \approx 5\)

Thus, the empirical formula is:
\[ \text{Empirical Formula} = C_3H_5 \]

### Step 5: Calculate the Empirical Formula Mass
Calculate the mass of the empirical formula:
\[ \text{Mass of } C_3H_5 = (3 \times 12) + (5 \times 1) = 36 + 5 = 41 \text{ g/mol} \]

### Step 6: Determine the Molecular Formula
Now, we can find \(n\):
\[ n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}} = \frac{82 \text{ g/mol}}{41 \text{ g/mol}} = 2 \]

Finally, the molecular formula is:
\[ \text{Molecular Formula} = n \times \text{Empirical Formula} = 2 \times C_3H_5 = C_6H_{10} \]

### Final Answer
The molecular formula of the compound is:
\[ \boxed{C_6H_{10}} \]