🎯 Objective
To determine the number of water molecules (x) of crystallization in Mohr's salt [(NH₄)₂Fe(SO₄)₂·xH₂O] by gravimetric analysis (heating to constant weight).
📖 Principle / Theory
Mohr's salt is a double salt with formula (NH₄)₂Fe(SO₄)₂·xH₂O (x = 6). When heated at 100–110°C, the water of crystallization is driven off and the anhydrous salt is obtained.
x = (loss in weight × M_anhydrous) / (weight of anhydrous salt × 18)
where M_anhydrous = 284 g/mol (for Mohr's salt without water).
🧰 Apparatus Required
Crucible, crucible tongs, china dish, desiccator, analytical balance, oven/burner.
🧪 Chemicals Required
Mohr's salt [(NH₄)₂Fe(SO₄)₂·xH₂O] crystals, silica gel (for desiccator).
⚗️ Procedure
- Clean and dry a crucible. Heat to red hot, cool in desiccator, and weigh (W₁).
- Weigh accurately about 3–4 g of Mohr's salt crystals in the crucible (W₂).
- Heat the crucible in an oven at 110°C for 30 minutes. Cool in desiccator and weigh (W₃).
- Repeat heating and weighing until constant weight is obtained (W₃ constant).
- Calculate: Loss in weight = (W₂ − W₁) − (W₃ − W₁) = mass of water lost.
- Calculate x = (loss in weight / 18) × (284 / mass of anhydrous salt).
📊 Observations & Calculations
ℹ️
Weight of empty crucible (W₁) = ___g. Weight of crucible + Mohr's salt (W₂) = ___g. Weight after drying (W₃) = ___g. x = ?
| Measurement | Heating 1 | Heating 2 | Heating 3 (constant) |
|---|---|---|---|
| Weight of empty crucible, W₁ (g) | ______ | ______ | ______ |
| Weight of crucible + Mohr's salt, W₂ (g) | ______ | ______ | ______ |
| Weight of crucible + anhydrous salt, W₃ (g) | ______ | ______ | ______ |
| Weight of Mohr's salt taken (W₂ − W₁) (g) | ______ | ______ | ______ |
| Loss in weight = water driven off (g) | ______ | ______ | ______ |
| Weight of anhydrous salt (W₃ − W₁) (g) | ______ | ______ | ______ |
| x = moles water / moles anhydrous salt | ______ | ||
Calculation:
Moles of water lost = Loss in weight / 18
Moles of anhydrous Mohr's salt = Weight of anhydrous salt / 284
x = Moles of water / Moles of anhydrous salt ≈ ______ (Theoretical: 6)
Moles of water lost = Loss in weight / 18
Moles of anhydrous Mohr's salt = Weight of anhydrous salt / 284
x = Moles of water / Moles of anhydrous salt ≈ ______ (Theoretical: 6)
✅ Result
The number of water molecules of crystallization in Mohr's salt = x ≈ ______. Theoretical value = 6.
⚠️ Precautions
- Heat the crucible and cool in desiccator before each weighing.
- Do not overheat — above 110°C the salt begins to decompose.
- Ensure constant weight is achieved (two consecutive weighings within 0.01 g).
- Handle the crucible with tongs only after cooling.
❓ Viva-Voce Questions
1. What is water of crystallization? Give three examples of hydrated salts.
Refer to your lab manual, textbook (Rattan or Vogel), and lecture notes. Discuss with your batch partners and prepare for the viva-voce examination.
2. What is the formula of Mohr's salt? Why is it called a double salt?
Refer to your lab manual, textbook (Rattan or Vogel), and lecture notes. Discuss with your batch partners and prepare for the viva-voce examination.
3. Why is Mohr's salt preferred over FeSO₄ as a primary standard?
Refer to your lab manual, textbook (Rattan or Vogel), and lecture notes. Discuss with your batch partners and prepare for the viva-voce examination.
4. What is a desiccator? Why is it used in gravimetric analysis?
Refer to your lab manual, textbook (Rattan or Vogel), and lecture notes. Discuss with your batch partners and prepare for the viva-voce examination.
5. What is the principle of gravimetric analysis?
Refer to your lab manual, textbook (Rattan or Vogel), and lecture notes. Discuss with your batch partners and prepare for the viva-voce examination.
6. Distinguish between 'water of crystallization' and 'water of constitution'.
Refer to your lab manual, textbook (Rattan or Vogel), and lecture notes. Discuss with your batch partners and prepare for the viva-voce examination.
📚 References
- Vogel's Textbook of Quantitative Chemical Analysis
- Day, R.A. – Quantitative Analytical Chemistry