Question

A 2.794-g sample of a mixture of anhydrous copper(II) sulfate, CuSO4, and copper(II) sulfate pentahydrate, CuSO4∙5H2O,was analyzed by heating todrive off the water in the hydrate.CuSO4∙5H2O(s)→CuSO4(s)+ 5 H2O(g)If the mass after heating was 2.578 g, determine the percentage of the hydrate in the original sample

Answer 1

Molar mass of CuSO₄*5H₂O = 249.685g/mol

Weight loss of salt = 2.794g- 2.578g = 0.216g

Moles of water released = 0.216g/(18g/mol) = 0.012 moles

CuSO₄*5H₂O CuSO₄(s) + 5H₂O(g)

Moles of hydrated salt = moles of water ÷ 5 = 0.012 ÷ 5

= 0.0024moles

Mass of hydrated salt = number of moles × molar mass

= 0.0024mole × (249.685g/mol) = 0.599244g

% of hydrated salt in sample = (0.599244g)/(2.794g) × 100

= 21.45 % (answer)