Calculate the molar mass of Fe2(SO4)3. Then calculate the mass of iron(III) sulfate of 4.05×10^23 formula units. Answer: 269g Fe2(SO4)3
The Correct Answer and Explanation is:
To calculate the molar mass of Fe₂(SO₄)₃, we need to determine the mass contribution of each element in the formula.
Step 1: Calculate Molar Mass
The formula Fe₂(SO₄)₃ contains:
- 2 atoms of Fe (iron)
- 3 sulfate ions (SO₄)
Each sulfate ion contains:
- 1 sulfur (S)
- 4 oxygen (O)
Now calculate the total atoms:
- Iron (Fe): 2 atoms × 55.85 g/mol = 111.70 g/mol
- Sulfur (S): 3 atoms × 32.07 g/mol = 96.21 g/mol
- Oxygen (O): 12 atoms × 16.00 g/mol = 192.00 g/mol
Total molar mass = 111.70 + 96.21 + 192.00 = 399.91 g/mol
Step 2: Use Avogadro’s Number
Avogadro’s number is 6.022 × 10²³ formula units/mol. The problem gives 4.05 × 10²³ formula units. We need to find how many moles this represents.Moles of Fe₂(SO₄)₃=4.05×10236.022×1023=0.6726 mol\text{Moles of Fe₂(SO₄)₃} = \frac{4.05 × 10^{23}}{6.022 × 10^{23}} = 0.6726 \text{ mol}Moles of Fe₂(SO₄)₃=6.022×10234.05×1023=0.6726 mol
Step 3: Calculate the Mass
Now multiply the number of moles by the molar mass:Mass=0.6726 mol×399.91 g/mol=268.9 g\text{Mass} = 0.6726 \text{ mol} × 399.91 \text{ g/mol} = 268.9 \text{ g}Mass=0.6726 mol×399.91 g/mol=268.9 g
Rounded to three significant figures, the mass is:269 g Fe₂(SO₄)₃\boxed{269 \text{ g Fe₂(SO₄)₃}}269 g Fe₂(SO₄)₃
Explanation
Iron(III) sulfate has a complex formula that requires careful accounting of each atom in the compound. We begin by breaking down the molar mass using atomic masses from the periodic table. After summing the contributions of all atoms, we use Avogadro’s number to relate formula units to moles. Finally, multiplying the moles by the molar mass gives the total mass of the compound in grams. The precision of three significant figures matches the input value, which ensures consistency. This approach is fundamental in stoichiometry for converting between particles and measurable mass in chemistry.
