Question

25 grams of Al2(SO4)3. How many moles of aluminum sulfate would
you have? How many molecules of aluminum sulfate would be present?

Answer 1

To find the number of moles and molecules of aluminum sulfate (Al₂(SO₄)₃) in 25 grams, we need to follow these steps:

1. **Calculate the molar mass of Al₂(SO₄)₃**:
- Aluminum (Al): 26.98 g/mol
- Sulfur (S): 32.07 g/mol
- Oxygen (O): 16.00 g/mol

The formula for aluminum sulfate is Al₂(SO₄)₃, which contains:
- 2 Aluminum atoms
- 3 Sulfate groups (SO₄), each containing 1 Sulfur and 4 Oxygen atoms.

Therefore, the molar mass is calculated as follows:
\[
\text{Molar mass} = 2 \times (26.98 \, \text{g/mol}) + 3 \times (32.07 \, \text{g/mol}) + 12 \times (16.00 \, \text{g/mol})
\]
\[
= 2 \times 26.98 + 3 \times 32.07 + 12 \times 16.00
\]
\[
= 53.96 + 96.21 + 192.00
\]
\[
= 342.17 \, \text{g/mol}
\]

2. **Calculate the number of moles**:
Using the formula:
\[
\text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
\]
\[
\text{Number of moles} = \frac{25 \, \text{g}}{342.17 \, \text{g/mol}} \approx 0.073 \, \text{moles}
\]

3. **Calculate the number of molecules**:
To find the number of molecules, we use Avogadro's number (6.022 × 10²³ molecules/mol):
\[
\text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's number}
\]
\[
\text{Number of molecules} \approx 0.073 \, \text{moles} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 4.40 \times 10^{22} \, \text{molecules}
\]

**Summary**:
- Moles of Al₂(SO₄)₃ in 25 grams = **0.073 moles**
- Molecules of Al₂(SO₄)₃ in 25 grams ≈ **4.40 × 10²² molecules**.