Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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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](https://content.bartleby.com/qna-images/question/0e109430-7b34-4d38-8cab-38f4ccc5aed5/f983fd1d-428f-4c95-8744-5cf3224719c3/21z23u_processed.jpeg)
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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