Answer to Question #310695 in General Chemistry for candyfloss

Question

Anhydrous copper(II) sulfate is difficult to dry completely. What mass of copper(II) sulfate would remain after removing 90% of the water from 250. g of CuSO₄×5H₂O? 

Answer 1

M(CuSO₄×5H₂O) = 250 g/mole;

M(CuSO₄) = 160 g/mole;

X(CuSO₄) = M(CuSO₄)/M(CuSO₄×5H₂O) = 160/250 = 0.64 = 64 %;

X(H₂O) = 1 - X(CuSO₄) = 1 - 0.64 = 0.36 = 36 %

m(CuSO₄) = m_substance * X(CuSO₄) = 250 * 0.64 = 160 g;

m(H₂O) = m_substance * X(H₂O) = 250 * 0.36 = 90 g;

Find the water content after removing 90% of the water from 250 g of CuSO₄×5H₂O:

m(H₂O)_remain = m(H₂O) * 100%-90%/100% = 90 * 0.1 = 9 g;

Find the mass of copper(II) sulfate would remain after removing 90% of the water from 250 g of CuSO₄×5H₂O:

m(substance)_remain = m(CuSO₄) + m(H₂O)_remain = 160 + 9 = 169 g.

Answer: 9 g.