Quantity (g) of pure MgSO4 in 2.4 g of MgSO4•7H2O

Question

Jeffery

Chemistry

27 March, 04:32

Quantity (g) of pure MgSO4 in 2.4 g of MgSO4•7H2O

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Saige Torres · Answer 1

1.1724 g.

Explanation:

Firstly we need to calculate the percentage of pure MgSO₄ in (MgSO₄.7H₂O). If we have 1.0 mol of MgSO₄.7H₂O, then we will have the the mass of its molecular mass.

The molecular mass of 1.0 mol (MgSO₄.7H₂O) = 246.4 g/mol.

The molecular mass of pure MgSO₄ = 120.366 g/mol.

The molecular mass of 7 (H₂O) = 7 (18.0 g/mol) = 126.0 g/mol.

∴ The mass % of pure MgSO₄ = [ (mass of pure MgSO₄) / (mass of MgSO₄.7H₂O) ] x 100 = [ (120.366 g/mol) / (246.4 g/mol) ] x 100 = 48.85%.

∴ the quantity of pure MgSO₄ = (mass of MgSO₄.7H₂O) (% of MgSO₄/100) = (2.4 g) (48.85/100) = 1.1724 g.