Question

give the chemical formula molar mass and % of the following problem iron(III)sulfide

Answer 1

Iron(III) sulfide is a compound composed of iron (Fe) and sulfur (S) with the formula \( \text{Fe}_2\text{S}_3 \).

### Chemical Formula:
- **Chemical Formula**: \( \text{Fe}_2\text{S}_3 \)

### Molar Mass Calculation:
1. **Atomic Masses** (from the periodic table):
- Iron (Fe): approximately 55.85 g/mol
- Sulfur (S): approximately 32.07 g/mol

2. **Calculate Molar Mass**:
- For \( Fe_2 \): \( 2 \times 55.85 \, \text{g/mol} = 111.70 \, \text{g/mol} \)
- For \( S_3 \): \( 3 \times 32.07 \, \text{g/mol} = 96.21 \, \text{g/mol} \)

Adding these together gives:
\[
\text{Molar Mass of } \text{Fe}_2\text{S}_3 = 111.70 \, \text{g/mol} + 96.21 \, \text{g/mol} = 207.91 \, \text{g/mol}
\]

### Percentage Composition:
To find the percent composition of each element in \( \text{Fe}_2\text{S}_3 \):

1. **Percent of Iron (Fe)**:
\[
\% \text{Fe} = \left( \frac{111.70 \, \text{g/mol}}{207.91 \, \text{g/mol}} \right) \times 100 \approx 53.67\%
\]

2. **Percent of Sulfur (S)**:
\[
\% \text{S} = \left( \frac{96.21 \, \text{g/mol}}{207.91 \, \text{g/mol}} \right) \times 100 \approx 46.33\%
\]

### Summary:
- **Chemical Formula**: \( \text{Fe}_2\text{S}_3 \)
- **Molar Mass**: \( 207.91 \, \text{g/mol} \)
- **Percent Composition**:
- Iron (Fe): \( \approx 53.67\% \)
- Sulfur (S): \( \approx 46.33\% \)