Calculate the mass of SO42- present in 2.0 g of BaSO4. Show your work.
The Correct Answer and Explanation is:
To calculate the mass of sulfate ion (SO₄²⁻) present in 2.0 grams of barium sulfate (BaSO₄), we follow these steps:
Step 1: Find the molar mass of BaSO₄
To find this, we add the atomic masses of each element in BaSO₄:
- Barium (Ba) = 137.33 g/mol
- Sulfur (S) = 32.07 g/mol
- Oxygen (O) = 16.00 g/mol × 4 = 64.00 g/mol
Molar mass of BaSO₄ = 137.33 + 32.07 + 64.00 = 233.40 g/mol
Step 2: Determine the fraction of SO₄²⁻ in BaSO₄
The molar mass of SO₄²⁻ is the mass of sulfur plus the mass of four oxygen atoms:
- SO₄²⁻ = 32.07 + 64.00 = 96.07 g/mol
Now, find what fraction of the total molar mass this is:
Fraction of SO₄²⁻ = (96.07 g/mol) / (233.40 g/mol) ≈ 0.4115
Step 3: Calculate the mass of SO₄²⁻ in 2.0 g of BaSO₄
Use the fraction above:
Mass of SO₄²⁻ = 0.4115 × 2.0 g = 0.823 g
✅ Final Answer: The mass of SO₄²⁻ in 2.0 g of BaSO₄ is approximately 0.823 grams.
Explanation
Barium sulfate (BaSO₄) is an ionic compound made of a barium ion (Ba²⁺) and a sulfate ion (SO₄²⁻). When asked to find the mass of the sulfate ion in a sample of BaSO₄, we must first understand the composition and molar mass of the compound. This allows us to determine what portion of the total mass is due to the sulfate group.
The atomic masses used come from the periodic table. Barium is a heavy element, contributing significantly to the total mass of the compound. Sulfur and oxygen, though lighter, still make up a considerable part of the mass due to the presence of four oxygen atoms in SO₄²⁻.
Once the molar mass of BaSO₄ is calculated, we isolate the molar mass of SO₄²⁻ and find its proportion in the whole compound. Multiplying this fraction by the actual mass of the sample (2.0 g) gives the mass of sulfate in the sample.
This method is a direct application of the concept of percent composition in chemistry. It is useful in analytical work and stoichiometry when you need to determine how much of a certain ion or element is present in a given compound. Understanding this calculation also helps in interpreting chemical formulas and performing more advanced quantitative chemical analyses.
