Question

Calcium sulfate is known to form hydrates (ionic compounds with water that have well-defined ratios of the ionic formula unit to molecules of water, written CaSO nH,O). The water content of the hydrate can be determined by gravimetric analysis. How many molecules of water are present for every formula unit of Caso4 in an unknown calcium sulfate hydrate if a complete drying of a sample to anhydrous CaSO, caused its mass to decrease from 8.609 g to 6.807 g? for the formula mass of Caso . Use 1.36.13 paor the formula mass of H,0 136.14 . Use 18.02for the formula mass of H2O The answer should be a whole number. Provide your answer below:

Answer 1

Answer:

The formula of hydrate is $CaSO_4.2H_2O$.

Explanation:

Mass of hydrate of calcium sulfate =  8.609 g

Mass of  of anhydrous calcium sulfate =  6.807 g

Moles of anhydrous calcium sulfate =  $\frac{6.807 g}{136.14 g/mol}=0.05 mol$

Mass of water present in hydrate of calcium sulfate :

= 8.609 g - 6.807 g = 1.802 g

Moles of water = $\frac{1.802 g}{18.02 g/mol}=0.1 mol$

$CaSO_4.xH_2O\rightarrow CaSO_4+xH_2O$

According to reaction 1 mole of anhydrous calcium sulfate is obtained from 1 mole of hydrate of calcium sulfate.

Then from 0.05 mole of anhydrous calcium sulfate will be obtained from:

$\frac{1}{1}\times 0.05 mol=0.05 mol$ hydrate of calcium sulfate.

As we can see that above, 0.05 mole of hydrated calcium sulfate gives 0.1 moles of water.

Then 1 mole of hydrated calcium sulfate will give:

$\frac{0.1 mol}{0.05}=2 mol$

So, the reaction becomes:

$CaSO_4.2H_2O\rightarrow CaSO_4+2H_2O$

The formula of hydrate is $CaSO_4.2H_2O$.