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# Solve the following problem: Find the percentage composition of constituent green vitriol crystals (FeSO4.7H2O). Also, find out the mass of iron - Chemistry

Sum

Solve the following problem:

Find the percentage composition of constituent green vitriol crystals (FeSO4.7H2O). Also, find out the mass of iron and the water of crystallisation in 4.54 kg of the crystals. (At. mass : Fe = 56; S = 32; O = 16)

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#### Solution

Given: i. Atomic mass: Fe = 56; S = 32; O = 16
ii. Mass of crystal = 4.54 kg

To find: i. Mass percentage of Fe, S, H, and O
ii. Mass of iron and water of crystallisation in 4.54 kg of crystal

Formula: Percentage (by weight) = "Mass of the element in 1 mole of compound"/"Molar mass of the compound"xx100

i. Molar mass of FeSO4.7H2O = 1 × (56) + 1 × (32) + 14 × (1) + 11 × (16)
= 56 + 32 + 14 + 176
= 278 g mol−1

Percentage of Fe = 56/278xx100 = 20.14%

Percentage of S = 32/278xx100 = 11.51%

Percentage of H = 14/278xx100 = 5.04%

Percentage of O = 176/278xx100 = 63.31%

ii. 278 kg green vitriol = 56 kg iron

∴ 4.54 kg green vitriol = x

∴ x = (56xx4.54)/278 = 0.915 kg

Mass of 7H2O in 278 kg green vitriol = 7 × 18 = 126 kg

∴ 4.54 kg green vitriol = y

∴ y = (126xx4.54)/278 = 2.058 kg

1. Mass percentage of Fe, S, H and O in FeSO4.7H2O are 20.14%, 11.51%, 5.04% and 63.31% respectively.
2. Mass of iron in 4.54 kg green vitriol = 0.915 kg
Mass of water of crystallisation in 4.54 kg green vitriol = 2.058 kg
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#### APPEARS IN

Balbharati Chemistry 11th Standard Maharashtra State Board
Chapter 2 Introduction to Analytical Chemistry
Exercises | Q 4. (F) | Page 24
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