To determine the empirical formula of a molecule with given percentages of its constituent elements, we need to follow a series of steps that include converting mass percentages to moles and then finding the simplest whole number ratio of the elements. Here's the detailed process:
1. Convert mass percentages to moles:
- The atomic mass of Carbon (C) is approximately 12.01 g/mol.
- The atomic mass of Hydrogen (H) is approximately 1.008 g/mol.
- The atomic mass of Chlorine (Cl) is approximately 35.45 g/mol.
Using the mass percentage as a direct mass in grams (since we assume 100 grams of the compound for simplicity):
- Moles of Carbon, [tex]\(C\)[/tex]:
[tex]\[
\text{moles of } C = \frac{24 \text{ grams}}{12.01 \text{ g/mol}} = 1.998 \approx 2
\][/tex]
- Moles of Hydrogen, [tex]\(H\)[/tex]:
[tex]\[
\text{moles of } H = \frac{6 \text{ grams}}{1.008 \text{ g/mol}} = 5.952 \approx 6
\][/tex]
- Moles of Chlorine, [tex]\(Cl\)[/tex]:
[tex]\[
\text{moles of } Cl = \frac{70 \text{ grams}}{35.45 \text{ g/mol}} = 1.975 \approx 2
\][/tex]
2. Determine the simplest whole number mole ratio:
- Calculate and simplify the mole ratios by dividing each by the smallest number of moles calculated:
- Smallest number of moles here is approximately 2.
[tex]\[
\text{Ratio of } C: \frac{1.998}{1.998} = 1
\][/tex]
[tex]\[
\text{Ratio of } H: \frac{5.952}{1.998} \approx 3
\][/tex]
[tex]\[
\text{Ratio of } Cl: \frac{1.975}{1.998} \approx 1
\][/tex]
3. Write the empirical formula based on the simplest ratio:
- From the mole ratios, we determine the formula to be [tex]\(CH_3Cl\)[/tex].
Our answer is:
[tex]\[ \boxed{CH_3Cl} \][/tex]
Thus, the empirical formula of the molecule is given by option:
[tex]\[ \text{B. } CH_3Cl \][/tex]
