Question

Metal Sulfate Hydrate lab (12 points) 4) An unknown metal sulfate hydrate (MSO XHO) sample with the recorded mass below is dissolved in water, and the sulfate ions from the sample precipitated with Ba2+ ions as BaSO4. A pre-massed filter paper is then used to collect the solid BaSO4. Experimental data was recorded below following same procedure in MSH lab. 0.6939 96.06 g/mol Molar mass of SO42- PARTA Mass of unknown metal sulfate hydrate (MSH) Mass of unknown sample sulfate anhydrous (metal sulfate only) 10.321 g 2nd 0.319 g 233.39 g/mol Molar mass BaSO4 PART B Mass of filter paper heating to constant mass Mass of filter paper + dry BaSO4 sample after drying to a constant mass (precipitates a white solid) 1st 0.234 g 2nd 0.231 g 1st 0.397 g 2nd 0.399 g a) Based on these data, calculate the percent of H20 and the SO42- in the unknown metal sulfate hydrate. b) What was your percent yield of SO42- if your unknown MSH is Na2SO4 10H20?

Answer 1

Q4. (a) mass of metal sulfate + H₂O = 0.693 g

mass of metal sulfate anhydrous = 0.319 g

mass H₂O = (mass of metal sulfate + H₂O) - (mass of metal sulfate anhydrous)

mass H₂O = (0.693 g) - (0.319 g)

mass H₂O = 0.374 g

mass percent H₂O = (mass H₂O / mass of hydrate) * 100

mass percent H₂O = (0.374 g / 0.693 g) * 100

mass percent H₂O = (0.540) * 100

mass percent H₂O = 54.0 %

Mass of BaSO₄ = (mass of dry filter paper + dry BaSO₄) - (mass of filter paper)

Mass of BaSO₄ = (0.399 g) - (0.231 g)

Mass of BaSO₄ = 0.168 g

moles BaSO₄ = (Mass of BaSO₄) / (molar mass BaSO₄)

moles BaSO₄ = (0.168 g) / (233.39 g/mol)

moles BaSO₄ = 7.20 x 10^-4 mol

moles SO₄^2- = moles BaSO₄

moles SO₄^2- = 7.20 x 10^-4 mol

mass SO₄^2- = (moles SO₄^2-) * (molar mass SO₄^2-)

mass SO₄^2- = (7.20 x 10^-4 mol) * (96.06 g/mol)

mass SO₄^2- = 0.0691 g

mass percent SO₄^2- = (mass SO₄^2- / mass of hydrate) * 100

mass percent SO₄^2- = (0.0691 g / 0.693 g) * 100

mass percent SO₄^2- = (0.0998) * 100

mass percent SO₄^2- = 9.98 %

(b) mass Na₂SO₄.10H₂O = 0.693 g

moles Na₂SO₄.10H₂O = (mass Na₂SO₄.10H₂O) / (molar mass Na₂SO₄.10H₂O)

moles Na₂SO₄.10H₂O = (0.693 g) / (322.2 g/mol)

moles Na₂SO₄.10H₂O = 2.15 x 10^-3 mol

moles SO₄^2- = moles Na₂SO₄.10H₂O

moles SO₄^2- = 2.15 x 10^-3 mol

Percent yield of SO₄^2- = (actual moles SO₄^2- / theoretical moles SO₄^2-) * 100

Percent yield of SO₄^2- = (7.20 x 10^-4 mol / 2.15 x 10^-3 mol) * 100

Percent yield of SO₄^2- = (0.335) * 100

Percent yield of SO₄^2- = 33.5 %