Question

A 8.129 g sample of MgSO4 · x H2O was heated until all of the water of hydration is removed. The resulting anhydrous compound weighs 3.967 g.

A. What is the formula of the hydrate?

B. If the label of a reagent bottle containing the hydrate indicates that it has 99.5% purity, how many moles of the hydrate is present in a 70.5 g of the hydrate?

Answer 1

A. Δm= 8.129 – 3.967 = 4.162 g

M(H₂O) = 18 g/mol

n(H₂O) $= \frac{4.162}{18}=0.2312 \;mol$

M(MgSO₄) = 120.36 g/mol

n(MgSO₄) $= \frac{3.967}{120.36} = 0.0329 \; mol$

Proportion:

0.0329 : 0.2312 = 1 : 7.0

MgSO₄x7H₂O is the formula of the hydrate

B. Proportion:

100 % - 70.5 g

99.5% - x g

$x = \frac{99.5 \times 70.5}{100} = 70.14 \;g \\ M(MgSO_4\cdot 7H_2O) = 120.36 + 7 \times 18 =246.36 \;g/mol \\ n = \frac{70.14}{246.36} = 0.284 \;mol$