Question

A 1.543 gram sample containing sulfate ion was treated with barium chloride reagent, and 0.2243 grams of barium sulfate was isolated. calculate the percentage of sulfate ion in the sample.

Answer 1

Answer: The mass percent of sulfate ion in the sample is 5.98 %

Explanation:

We are given:

Mass of barium sulfate = 0.2243 g

We know that:

Molar mass of barium sulfate = 233.4 g/mol

Molar mass of sulfate = 96.06 g/mol

To calculate the mass of sulfate ion present in given amount of barium sulfate, we use unitary method:

In 233.4 g of barium sulfate, the mass of sulfate ion present is 96.06 g

So, in 0.2243 g of barium sulfate, the mass of sulfate ion present will be = $\frac{96.06}{233.4}\times 0.2243=0.0923g$

To calculate mass of sulfate ion present in the sample, we use the equation:

$\text{Mass percent of sulfate ion}=\frac{\text{Mass of sulfate ion}}{\text{Mass of sample}}\times 100$

Mass of sulfate ion = 0.0923 g

Mass of sample = 1.543 g

Putting values in above equation, we get:

$\text{Mass percent of sulfate ion}=\frac{0.0923g}{1.543g}\times 100\\\\\text{Mass percent of sulfate ion}=5.98\%$

Hence, the mass percent of sulfate ion in the sample is 5.98 %

Answer 2

Chemical reaction: SO₄²⁻ + Ba²⁺ → BaSO₄.
m(sample) = 1,543 g.
m(BaSO₄) = 0,2243 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,2243 g ÷ 233,4 g/mol. 
n(BaSO₄) = 0,00096 mol.
n(BaSO₄) = n(SO₄²⁻).
ω(SO₄²⁻) = m(SO₄²⁻) ÷ m(sample).
ω(SO₄²⁻) = 0,00096 mol · 96 g/mol ÷ 1,543 g.
ω(SO₄²⁻) = 0,059 = 5,9%.