Question

The weight loss during the conversion of 1 mole of gypsum to anhydrite is ________. (Atomic weights: Ca = 40.0, S = 32.0, O = 16.0, H = 1.0) (Round off to two decimal places.)

Hint:

Gypsum → Anhydrite involves dehydration: mass loss = 2×(18 g/mol).

Answer 1

Correct Answer: 20.9

Step 1: Chemical equation.
\[ \text{CaSO}_{4}\cdot 2\text{H}_{2}\text{O} \rightarrow \text{CaSO}_{4} + 2\text{H}_{2}\text{O} \] Step 2: Compute molar masses.
Gypsum = 40 + 32 + (4×16) + 2(2×1 + 16) = 172 g/mol.
Anhydrite = 40 + 32 + (4×16) = 136 g/mol.
Step 3: Weight loss.
172 – 136 = 36 g. But since 2 moles of H\textsubscript{2}O are released, total mass lost = 36 g. Step 4: Water per mole.
1 mol of gypsum loses 36 g water. Step 5: Rounding.
Weight loss = 36.00 g/mol.