Question

# Determine the percentage of water of crystallization, iron, Sulphur and oxygen in pure ferrous sulphate $(FeS{{O}_{4}}.7{{H}_{2}}O)$

Hint: Mass percent of any element is calculated by dividing the molar mass of that particular component to total mass of the compound and then multiplying it by 100.
Percent tells the mass of the component per 100ml of solution.

Mass percent of any element is calculated by dividing the molar mass of that particular component to total mass of the compound and then multiplying it by 100.
Mass percent= molar mass of that particular component total mass of compound.
$Mass\text{ }percent=\text{ }\dfrac{molar\text{ }mass\text{ }of\text{ }that\text{ }particular\text{ }component}{total\text{ }mass\text{ }of\text{ }compound}\text{ }\times \text{100}$
Percentage of water of crystallization:
atomic mass of iron, Fe = 56 g/mol
atomic mass of Sulphur, S = 32 g/mol
atomic mass of oxygen, O = 4 g/mol
atomic mass of hydrogen, H = 1 g/mol
total molar mass would be the sum of their respective molar masses multiplied with their respective number of atoms.
now,
molar mass of ferrous sulphate$(FeS{{O}_{4}}.7{{H}_{2}}O)$
\begin{align} & =56\text{ }+\text{ }32\text{ }+\text{ }4\text{ }\times \text{ }16\text{ }+\text{ }7\text{ }\times \text{ }\left\{ \text{ }2\text{ }\times \text{ }1\text{ }+\text{ }16\text{ } \right\} \\ & =\text{ }56\text{ }+\text{ }32\text{ }+\text{ }64\text{ }+\text{ }126 \\ & =\text{ }278\text{ }g/mol \\ \end{align}

Molar mass of 7 molecules of water= $7\text{ }\times \text{ }\left\{ \text{ }2\text{ }\times 1\text{ }+\text{ }16\text{ } \right\}\text{ }=126\text{ }g/mol$
now,
Percentage of water would be molar mass of seven molecules of water divided by total molar mass of compound ferrous sulphate multiplied by 100
$percentage\text{ }of\text{ }water\text{ }in\text{ }FeSO_4.7H_2O\text{ }=\left\{ \dfrac{molar\text{ }mass\text{ }of\text{ }7H_2O}{molar\text{ }mass\text{ }of\text{ }FeSO_4.7H_2O}\text{ } \right\}\times \text{ }100=\text{ }\left\{ \text{ }\dfrac{126}{278} \right\}\text{ }\times \text{ }100=45.32%$
so, the percent of water of crystallization is 45.32%

Now, percentage of iron (Fe) in$(FeS{{O}_{4}}.7{{H}_{2}}O)$
Molar mass of Fe=56g/mol
Number of atoms of Fe=1
Molar mass of compound= 278g/mol
$percentage\text{ }of\text{ Fe }in\text{ }FeS{{O}_{4}}.7H_2O\text{ }=\left\{ \dfrac{molar\text{ }mass\text{ }of\text{ Fe}}{molar\text{ }mass\text{ }of\text{ }FeSO_4.7H_2O}\text{ } \right\}\times \text{ }100=\text{ }\left\{ \text{ }\dfrac{56}{278} \right\}\text{ }\times \text{ }100=20.14%$
So, the percentage of Fe is 20.14%.

Now, percentage of Sulphur(S) in$FeS{{O}_{4}}.7{{H}_{2}}O$
Molar mass of S=32g/mol
Number of atoms of Fe=1
Molar mass of compound= 278g/mol
$percentage\text{ }of\text{ S }in\text{ }FeS{{O}_{4}}.7H_2O\text{ }=\left\{ \dfrac{molar\text{ }mass\text{ }of\text{ S}}{molar\text{ }mass\text{ }of\text{ }FeSO_4.7H_2O}\text{ } \right\}\times \text{ }100=\text{ }\left\{ \text{ }\dfrac{32}{278} \right\}\text{ }\times \text{ }100=11.51%$
So, the percentage of S is 11.51%.

Now, percentage of oxygen (O) in$FeS{{O}_{4}}.7{{H}_{2}}O$
Molar mass of S=16g/mol
Number of atoms of O=4
Molar mass of 4 atoms of O=$4\times 16=64g/mol$
Molar mass of compound= 278g/mol
$percentage\text{ }of\text{ O }in\text{ }FeS{{O}_{4}}.7H_2O\text{ }=\left\{ \dfrac{molar\text{ }mass\text{ }of\text{ O}}{molar\text{ }mass\text{ }of\text{ }FeSO_4.7H_2O}\text{ } \right\}\times \text{ }100=\text{ }\left\{ \text{ }\dfrac{64}{278} \right\}\text{ }\times \text{ }100=23.021%$

So, the percentage of O is 23.021%.

the percentage of water of crystallization, iron, Sulphur and oxygen in pure ferrous sulphate $(FeS{{O}_{4}}.7{{H}_{2}}O)$ is 45.32%,20.14%, 11.51% and 23.021% respectively.

Note: Mass percent of any element is calculated by dividing the molar mass of that particular component to total mass of the compound and then multiplying it by 100.
Mass percent= molar mass of that particular component total mass of compound. $Mass\text{ }percent=\text{ }\dfrac{molar\text{ }mass\text{ }of\text{ }that\text{ }particular\text{ }component}{total\text{ }mass\text{ }of\text{ }compound}\text{ }\times \text{100}$