Question

Find the percentage composition of constituent green vitriol crystal ($FeS{{O}_{4}}.7{{H}_{2}}O$). Also, find out the mass of iron and the water of crystallization in 4.54 kg of the crystal. (At. mass: Fe = 56, S = 32, O = 16)​

Answer 1

Hint: A chemical substance that comprises sulfates of metals like copper or iron are generally known as vitriols and are originally found as crystals.
To solve this question, we first need to find out the molar mass of green vitriol crystal of iron (II) sulfate heptahydrate ($FeS{{O}_{4}}.7{{H}_{2}}O$).

Complete answer:
It is given to us that the atomic mass of Fe = 56, the atomic mass of S = 32, the atomic mass of O = 16, the atomic mass of H = 1.
So, the molar mass of $FeS{{O}_{4}}.7{{H}_{2}}O$ will be
\[\begin{align}
  & {{M}_{FeS{{O}_{4}}.7{{H}_{2}}O}}={{M}_{Fe}}+{{M}_{S}}+(4\times {{M}_{O}})+(7\times [(2\times {{M}_{H}})+{{M}_{O}}]) \\
 & {{M}_{FeS{{O}_{4}}.7{{H}_{2}}O}}=56+32+(4\times 16)+(7\times [(2\times 1)+16]) \\
 & {{M}_{FeS{{O}_{4}}.7{{H}_{2}}O}}=278\text{ g/mol} \\
\end{align}\]
Now, the formula to calculate the percentage composition of a constituent in a compound is given by:
\[\text{percentage composition = }\dfrac{\text{mass of element in 1 mole of compound}}{\text{molar mass of compound}}\times 100\]
So, the percentage compositions of elements Fe, S, O, and H are as follows
\[\begin{align}
  & (Fe)=\dfrac{56}{278}\times 100=20.14\% \\
 & (S)=\dfrac{32}{278}\times 100=11.51\% \\
 & (O)=\dfrac{11\times 16}{278}\times 100=63.31\% \\
 & (H)=\dfrac{14\times 1}{278}\times 100=5.04\% \\
\end{align}\]
Now, we know that the molar mass of green vitriol crystal is 278 g/mol.
i.e., 278 g of green vitriol crystal contains 56 g of iron.
So, 4.54 kg of the crystal will contain
\[\begin{align}
  & {{m}_{Fe}}=\dfrac{56g}{278g}\times 4.54\text{ }kg \\
 & {{m}_{fe}}=0.915\text{ }kg \\
\end{align}\]
Similarly, 278 g of green vitriol crystal contains 126 g of water of crystallization since there are 7 molecules of water of crystallization in green vitriol crystal.
So, 4.54 kg of the crystal will contain
\[\begin{align}
  & {{m}_{7{{H}_{2}}O}}=\dfrac{126g}{278g}\times 4.54\text{ kg} \\
 & {{\text{m}}_{7{{H}_{2}}O}}=2.058\text{ kg} \\
\end{align}\]
So, percentage compositions of constituents Fe = 20.14%, S = 11.51%, O = 63.31%, and H = 5.04%.
Also, 4.54 kg of the crystal will contain 0.915 kg of iron and 2.058 kg of water of crystallization.

Note:
It should be noted that the color of the vitriol can be used to identify the mineral present in the substance.
Hydrated copper (II) sulfate ($CuS{{O}_{4}}.5{{H}_{2}}O$) forms blue vitriol.
Hydrated iron (II) sulfate ($FeS{{O}_{4}}.7{{H}_{2}}O$) forms green vitriol.
Hydrated cobalt (II) sulfate ($CoS{{O}_{4}}.7{{H}_{2}}O$) forms red vitriol.
Hydrated zinc (II) sulfate ($ZnS{{O}_{4}}.7{{H}_{2}}O$) forms white vitriol.
A mixture of heptahydrate sulfates of Cu, Fe, Co, Mg, Mn, and Ni form black vitriol.