A certain compound composed of carbon hydrogen and fluorine has a percent composition of 48% carbon 6.44% hydrogen and 45.56% fluorine. Determine the empirical formula

To determine the empirical formula of a compound based on its percent composition, we need to:

1. Convert the percentage composition to grams (assuming a 100 g sample).
2. Convert grams to moles for each element.
3. Find the mole ratio of the elements.
4. Determine the simplest whole number ratio and write the empirical formula.

### Step 1: Assume a 100 g Sample
- Carbon (C): 48.00 g
- Hydrogen (H): 6.44 g
- Fluorine (F): 45.56 g

### Step 2: Convert Grams to Moles
Using the atomic masses:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.008 g/mol
- Fluorine (F): 19.00 g/mol

\[
\text{Moles of Carbon} = \frac{48.00 \, \text{g}}{12.01 \, \text{g/mol}} = 3.998 \, \text{mol}
\]

\[
\text{Moles of Hydrogen} = \frac{6.44 \, \text{g}}{1.008 \, \text{g/mol}} = 6.388 \, \text{mol}
\]

\[
\text{Moles of Fluorine} = \frac{45.56 \, \text{g}}{19.00 \, \text{g/mol}} = 2.398 \, \text{mol}
\]

### Step 3: Find the Mole Ratios
To find the simplest mole ratio, divide each of the moles by the smallest value (2.398 mol in this case):

\[
\text{Ratio of Carbon} = \frac{3.998}{2.398} \approx 1.67
\]

\[
\text{Ratio of Hydrogen} = \frac{6.388}{2.398} \approx 2.66
\]

\[
\text{Ratio of Fluorine} = \frac{2.398}{2.398} = 1
\]

### Step 4: Convert to the Simplest Whole Number Ratios
The ratios are approximately \(1.67 : 2.66 : 1\). To convert them to whole numbers:

- Multiply all by 3 to avoid fractions:

\[
1.67 \times 3 \approx 5, \quad 2.66 \times 3 \approx 8, \quad 1 \times 3 = 3
\]

### Conclusion: Empirical Formula
The empirical formula is **C\(_5\)H\(_8\)F\(_3\)**.

Would you like more details or have any questions?

Here are some related questions that might interest you:

1. How do you determine the molecular formula if given the molar mass?
2. What is the difference between empirical and molecular formulas?
3. How can empirical formulas be used in combustion analysis?
4. What are the steps to convert percent composition to a molecular formula?
5. How does the law of definite proportions relate to empirical formulas?

**Tip:** Remember that the empirical formula represents the simplest whole number ratio of atoms in a compound, not necessarily the actual number of atoms in a molecule.

A certain compound composed of carbon, hydrogen, and fluorine has a percent composition of 48% carbon, 6.44% hydrogen, and 45.56% fluorine. Determine the empirical formula.

Learn how to determine the empirical formula for a compound composed of 48% carbon, 6.44% hydrogen, and 45.56% fluorine. This problem covers key chemistry concepts, including stoichiometry, percent composition, and the law of definite proportions. Suitable for high school chemistry students.

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