Question

Unknown will be dissolved in hydrochloric acid solution. A small excess of (aq) barium cations will be added to precipitate all the sulfate ion as barium sulfate.
Initial mass of 3 samples of metallic sulfate is 0.1204g, 0.1052g, 0.1240g
Mass of barium sulfate precipitate produced from each sulfate sample are is 0.2334g, 0.2041g, 0.2404

Compute mass of sulfate in barium sulfate sample

Compute moles of sulfate in each barium sulfate

Calculate mass of element "M" in sample

Calculate # moles of "M" in each original sample using mole ratio(M to sulfate)

Determine MM of "M"

How I do go about answering these questions? Please show steps. Thank you in advance!

Answer 1

Sample-1: Let the metal be M1 with mlecular mass of W1

Mass of metal sulphate in sample-1 = 0.1204g

Mass of BaSO4 precipitated form sample-1 = 0.2334g

Molecular weight of BaSO4 = 233.4 gmol^-1  

Molar weight of SO₄^2- = 96 gmol^-1

Hence 96 g of SO₄^2- is present in 233.4g BaSO4.

Hence mass of SO₄^2- (sulphate) in barium sulfate sample-1 = (96 g SO₄^2- / 233.4g BaSO4)x(0.2334 g BaSO4)

= 0.096 g SO₄^2-  

Hence 0.096 g SO₄^2- (sulphate) is present in the unknown sulphate sample-1. (answer)

moles of sulphate in BaSO4 sample-1 = mass of SO₄^2- / molar mass of SO₄^2-    = (0.096 g) / (96 gmol^-1)

= 0.001 mol (answer)

Mass of M1 in sulphate sample-1 = (mass of metal sulphate in sample-1) - (mass of  SO₄^2- in sample-1)

= 0.1204 g - 0.096 g = 0.0244 g (answer)

Here the valancy of the metal is not known. Hence we need to apply heat and trail method to consider valancy as 1, 2, 3...

If the valancy is 2,the cation will be (M1)²⁺ and the metal sulphate will be (M1)SO4

Moles of M1 in sulphate sample-1 = moles of SO4^2- = 0.001 mol = (Mass of M1) / molecular mass of M1

=> 0.001mol = 0.0244 g / W1 gmol^-1  

=> W1 = 0.0244 / 0.001 = 24.4 g (Magnesium)

Hence MM of M1 = 24.4 gmol^-1 which is of magnesium. (answer)

sample-2 and 3: We can calculate the above parameters for sample -2 &3 in an similar way we used to calculate for sample-1