Question

find the percentage composition of constituent green vitriol crystal (FeSO4.7H2O). 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

Answer:

The chemical formula for green vitriol crystal is $FeSO_{4} .7H_{2}O$

so we have to find the molar mass of green vitriol crystal is

                                                                   =55.8+32.1+16*4+7*18

                                                                   =277.9

percentage composition of constituent green vitriol is

                                                             =(7*18)/277.9*100%

                                                             =45.34%

Answer 2

Given:

Atomic masses:

Fe = 56, S = 32, O = 16

Mass of crystal = 4.54 kg

To Find:

The percentage composition of constituents in green vitriol and the mass of iron and the water of crystallization in 4.54 kg of the crystal.

Calculation:

- The molecular weight of Green vitriol = 56 + 32 + 16×4 + 7×18

⇒ M.wt = 278 gm

- Percentage composition of Iron = (56/278) × 100 = 20.144 %

- Percentage composition of Sulphur = (32/278) × 100 = 11.51 %

- Percentage composition of Oxygen = (11 × 16/278) × 100 = 63.31 %

- Percentage composition of Hydrogen = (14 × 1/278) × 100 = 5.04 %

- Mass of iron in 4.54 kg crystal = 4.54 × 20.144/100

⇒ Mass of iron = 0.91454 kg = 914.54 gm

- Mass of water of crystallization = 4.54 × 126/278

⇒ Mass of water of crystallization = 2.0577 kg = 2057.7 gm

- So, the percentage compositions of Fe, S, O and H are 20.144%, 11.51%, 63.31% and 5.04% respectively.

The mass of iron and water of crystallization are 914.54 gm and 2057.7 gm respectively in the given crystal.