Use percent composition to determine an empirical formula. A compound is found to contain S, O, and CI with the following percent composition: S: 26.96% O: 13.45% CI: 59.59% What is the empirical formula of this compound? Enter the elements in the order S, O, CI.

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**Determining an Empirical Formula Using Percent Composition**

To determine the empirical formula of a compound using its percent composition, follow these steps. Here, we are working with a compound composed of sulfur (S), oxygen (O), and chlorine (Cl) with the following percent compositions:

- Sulfur (S): 26.96%
- Oxygen (O): 13.45%
- Chlorine (Cl): 59.59%

**Objective:**
Find the empirical formula of this compound.

**Instructions:**
1. **Convert Percentages to Masses:** Assume a 100 gram sample of the compound. This makes the percentages equal to the masses in grams. Therefore:
   - Sulfur (S): 26.96 grams
   - Oxygen (O): 13.45 grams
   - Chlorine (Cl): 59.59 grams

2. **Convert Masses to Moles:** Use the molar masses of each element to convert the masses to moles.
   - Sulfur (S): The molar mass of sulfur is approximately 32.07 g/mol.
   - Oxygen (O): The molar mass of oxygen is approximately 16.00 g/mol.
   - Chlorine (Cl): The molar mass of chlorine is approximately 35.45 g/mol.

     \[
     \text{Moles of S} = \frac{26.96 \text{ g}}{32.07 \text{ g/mol}} = 0.840 \text{ mol}
     \]
     \[
     \text{Moles of O} = \frac{13.45 \text{ g}}{16.00 \text{ g/mol}} = 0.841 \text{ mol}
     \]
     \[
     \text{Moles of Cl} = \frac{59.59 \text{ g}}{35.45 \text{ g/mol}} = 1.682 \text{ mol}
     \]

3. **Divide by the Smallest Number of Moles:** 
   Determine the smallest number of moles calculated and divide each of the mole values by this number:
   \[
   \text{Moles of S} = \frac{0.840}{0.840} = 1
   \]
   \[
   \text{Moles of O} = \frac{0.841}{0.840} = 1
Transcribed Image Text:**Determining an Empirical Formula Using Percent Composition** To determine the empirical formula of a compound using its percent composition, follow these steps. Here, we are working with a compound composed of sulfur (S), oxygen (O), and chlorine (Cl) with the following percent compositions: - Sulfur (S): 26.96% - Oxygen (O): 13.45% - Chlorine (Cl): 59.59% **Objective:** Find the empirical formula of this compound. **Instructions:** 1. **Convert Percentages to Masses:** Assume a 100 gram sample of the compound. This makes the percentages equal to the masses in grams. Therefore: - Sulfur (S): 26.96 grams - Oxygen (O): 13.45 grams - Chlorine (Cl): 59.59 grams 2. **Convert Masses to Moles:** Use the molar masses of each element to convert the masses to moles. - Sulfur (S): The molar mass of sulfur is approximately 32.07 g/mol. - Oxygen (O): The molar mass of oxygen is approximately 16.00 g/mol. - Chlorine (Cl): The molar mass of chlorine is approximately 35.45 g/mol. \[ \text{Moles of S} = \frac{26.96 \text{ g}}{32.07 \text{ g/mol}} = 0.840 \text{ mol} \] \[ \text{Moles of O} = \frac{13.45 \text{ g}}{16.00 \text{ g/mol}} = 0.841 \text{ mol} \] \[ \text{Moles of Cl} = \frac{59.59 \text{ g}}{35.45 \text{ g/mol}} = 1.682 \text{ mol} \] 3. **Divide by the Smallest Number of Moles:** Determine the smallest number of moles calculated and divide each of the mole values by this number: \[ \text{Moles of S} = \frac{0.840}{0.840} = 1 \] \[ \text{Moles of O} = \frac{0.841}{0.840} = 1
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