Question

The following data is given to find the formula of a Hydrate:
a. Mass of crucible: 13.56g
b. Mass of crucible & salt CaSO4 .H2O hydrate: 16.05g
c. Mass of crucible & salt CaSO4 anhydrate: 15.07
Find the mass of CaSO4.H2O hydrate, the mass of CaSO4 anhydrate, the mass of water,
the percent of water, the mole of water, the mole of salt (anhydrate), and the ratio of mole
of water to mole of salt (anhydrate).

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