Question

The following data is given to find the formula of a Hydrate:

Answer 1

Answer:

$m_{CaSO_4.H_2O}=2.49\ g$

$m_{CaSO_4}=1.51\ g$

$m_{water}=0.98\ g$

$\%\ of\ water=39.36\ \%$

$moles_{water}= 0.0544\ mol$

$moles_{CaSO_4}= 0.0111\ mol$

$CaSO_4:H_2O$ = 5 : 1

Explanation:

Given :

$m_{crucible}=13.56\ g$

$m_{crucible}+m_{CaSO_4.H_2O}=16.05\ g$

$m_{crucible}+m_{CaSO_4}=15.07\ g$

Mass of salt hydrate:

$m_{crucible}=13.56\ g$

$m_{crucible}+m_{CaSO_4.H_2O}=16.05\ g$

$m_{CaSO_4.H_2O}=16.05-m_{crucible}\ g=16.05-13.56\ g=2.49\ g$

Mass of salt anhydrous:

$m_{crucible}=13.56\ g$

$m_{crucible}+m_{CaSO_4}=15.07\ g$

$m_{CaSO_4}=15.07-m_{crucible}\ g=15.07-13.56\ g=1.51\ g$

Mass of water:

$m_{water}=m_{CaSO_4.H_2O}-m_{CaSO_4}=2.49-1.51\ g=0.98\ g$

$m_{water}=2.49-1.51\ g=0.98\ g$

Percentage of water:

$\%\ of\ water=\frac{Mass_{water}}{Total\ mass\ of\ hydrated\ salt}\times 100$

$\%\ of\ water=\frac{0.98}{2.49}\times 100$

$\%\ of\ water=39.36\ \%$

Moles of water:

Mass of water = 0.98 g

Molar mass of $H_2O$ = 18 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles are:

$moles= \frac{0.98\ g}{18\ g/mol}$

$moles_{water}= 0.0544\ mol$

Moles of anhydrate salt:

Amount = 1.51 g

Molar mass of $CaSO_4$ = 136.14 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles are:

$moles= \frac{1.51\ g}{136.14\ g/mol}$

$moles_{CaSO_4}= 0.0111\ mol$

The simplest ration of the two are:

$CaSO_4:H_2O$ = 0.0111 : 0.0544 = 5 : 1