Question

give the following chemical formula,molar mass,&% composition of the following problems.Round to the nearest tenths place
1) calcium Hydroxide
2)Iron(III)sulfide

Answer 1

Sure! Let's break down the calculations for both calcium hydroxide and iron(III) sulfide.

### 1) Calcium Hydroxide (Ca(OH)₂)

**Chemical Formula:** Ca(OH)₂

**Molar Mass Calculation:**
- Calcium (Ca): 1 × 40.08 g/mol = 40.08 g/mol
- Oxygen (O): 2 × 16.00 g/mol = 32.00 g/mol
- Hydrogen (H): 2 × 1.01 g/mol = 2.02 g/mol

**Total Molar Mass:**
\[
\text{Molar Mass of Ca(OH)₂} = 40.08 + 32.00 + 2.02 = 74.10 \text{ g/mol}
\]

**% Composition:**
- % Ca = \(\frac{40.08}{74.10} \times 100 \approx 54.1\%\)
- % O = \(\frac{32.00}{74.10} \times 100 \approx 43.2\%\)
- % H = \(\frac{2.02}{74.10} \times 100 \approx 2.7\%\)

### Summary for Calcium Hydroxide
- **Molar Mass:** 74.1 g/mol
- **% Composition:**
- Calcium (Ca): 54.1%
- Oxygen (O): 43.2%
- Hydrogen (H): 2.7%

---

### 2) Iron(III) Sulfide (Fe₂S₃)

**Chemical Formula:** Fe₂S₃

**Molar Mass Calculation:**
- Iron (Fe): 2 × 55.85 g/mol = 111.70 g/mol
- Sulfur (S): 3 × 32.07 g/mol = 96.21 g/mol

**Total Molar Mass:**
\[
\text{Molar Mass of Fe₂S₃} = 111.70 + 96.21 = 207.91 \text{ g/mol}
\]

**% Composition:**
- % Fe = \(\frac{111.70}{207.91} \times 100 \approx 53.7\%\)
- % S = \(\frac{96.21}{207.91} \times 100 \approx 46.3\%\)

### Summary for Iron(III) Sulfide
- **Molar Mass:** 207.9 g/mol
- **% Composition:**
- Iron (Fe): 53.7%
- Sulfur (S): 46.3%

Feel free to ask if you need further assistance!