Question

A 0.649-g sample containing only K_2SO_4 and (NH_4)2SO_4 was dissolved in water and treated with Ba (NO_3)_2 to precipitate all SO^2-_4 as BaSO_4. Find the weight percent of K_2SO_4 in the sample if 0.977 g of precipitate was formed.

Answer 1

Given,

Mass of K₂SO₄ + Mass of (NH₄)₂SO₄ = 0.649 g

Mass of precipitate(BaSO₄) formed = 0.977 g

Assume,

Mass of K₂SO₄ = "x"

Thus,

Mass of (NH₄)₂SO₄ = (0.649 - x)

Now, Calculating the number of moles of precipitate(BaSO₄) from the given mass,

= 0.977 g BaSO₄ x ( 1 mol / 233.38 g)

= 0.004186 mol BaSO₄

Now, the separate balanced reactions for K₂SO₄ and (NH₄)₂SO₄ are,

K₂SO₄(aq) + Ba(NO₃)₂$\rightarrow$ BaSO₄(s) + 2KNO₃(aq)

Similarly,

(NH₄)₂SO₄(aq) + Ba(NO₃)₂$\rightarrow$ BaSO₄(s) + 2NH₄NO₃(aq)

Now, the moles of K₂SO₄ and (NH₄)₂SO₄ are,

Moles of K₂SO₄ = (x / 174.25)

Similarly,

Moles of (NH₄)₂SO₄ = [(0.649-x) / 132.14]

Now, using the moles of  K₂SO₄ and moles of (NH₄)₂SO₄, calculating the number of moles of precipitate forming from the balanced chemical reaction,

= (x / 174.25) mol K₂SO₄ x ( 1 mol BaSO₄ /1 mol K₂SO₄)

= (x / 174.25) mol BaSO₄

Similarly,

= [(0.649-x) / 132.14] mol (NH₄)₂SO₄ x ( 1 mol BaSO₄ /1 mol (NH₄)₂SO₄)

= [(0.649-x) / 132.14] mol BaSO₄

Now,

Moles of BaSO₄ formed from K₂SO₄ + Moles of BaSO₄ formed from (NH₄)₂SO₄ = 0.004186 mol BaSO₄

(x / 174.25) + [(0.649-x) / 132.14] = 0.004186

(132..14x + 113.09 - 174.25x)/ 23025.4 = 0.004186

(132..14x + 113.09 - 174.25x) = 96.39

16.6969 = 42.11x

x = 0.3965

Thus, Mass of K₂SO₄ in the sample = 0.3965 g

Now, We know, the weight percent formula,

Weight percent of K₂SO₄ = ( Weight of K₂SO₄ / Total weight of sample) x 100

Weight percent of K₂SO₄ = (0.3965 g / 0.649 g) x 100

Weight percent of K₂SO₄ = 61.1 %