Answer to Question #189026 in General Chemistry for Eduardo Velazquez

Question

Beryllium sulfate is a hydrated compound whose formula can be written BeSO4 • xH2O, where z is the number of moles of H2O per mole of BeSO4. When a 3.284 g sample of this hydrate was heated to 130°C for 10 minutes, all of the water of hydration was lost, leaving 1.948 g of anhydrous beryllium sulfate. Calculate the value of x, and write the formula and name of the hydrate.

Answer 1

Δm = 3.284 – 1.948 = 1.336 g

M(H₂O) = 18 g/mol

n(H₂O) $= \frac{1.336}{18} = 0.07422 \; mol$

M(BeSO₄) = 105.07 g/mol

n(BeSO₄) $= \frac{1.948}{105.07} = 0.01854 \;mol$

0.01854 : 0.07422 = 1: 4

x = 4

BeSO₄ • 4H₂O Beryllium sulfate tetrahydrate