Mohr's salt with a yellow-brown crust: recrystallise or discard?
A student prepares Mohr's salt by dissolving 2.80 g of FeSO₄·7H₂O (M = 278 g mol⁻¹) in 20 cm³ of water with 2 cm³ of dilute sulphuric acid, warming gently, then adding 1.40 g of ammonium sulphate (M = 132 g mol⁻¹). The solution is concentrated and cooled, and 3.10 g of pale-green crystals is collected. Some crystals carry a yellow-brown crust and the filtrate is pale yellow. In the procedure the student used tap water rather than distilled water, heated the mixture to near boiling to complete dissolution, and dried the crystals in an oven at 110 °C. The teacher describes the product as 'Mohr's salt with a defect' and asks for the theoretical yield, the percentage yield, and a decision on whether to recrystallise or discard the batch.
1. Calculate the theoretical yield of Mohr's salt, identify the limiting reagent, and calculate the percentage yield obtained.
Read the worked answer
Moles of FeSO₄·7H₂O = 2.80/278 = 0.01007 mol. Moles of (NH₄)₂SO₄ = 1.40/132 = 0.01061 mol.
The salt is FeSO₄·(NH₄)₂SO₄·6H₂O, formed in a 1:1 ratio, so the iron(II) sulphate is limiting: 0.01007 mol of product at most.
Molar mass of the product: FeSO₄ (152) + (NH₄)₂SO₄ (132) + 6H₂O (108) = 392 g mol⁻¹. Alternatively, 278 − 7(18) = 152 for the anhydrous FeSO₄, then add 132 and 108, which gives the same 392 g mol⁻¹.
Theoretical yield = 0.01007 × 392 = 3.95 g.
Percentage yield = 3.10/3.95 × 100 = 78.5 %.
So the mass is respectable; the problem is purity, not quantity. Note also that the product contains 56/392 = 14.3 % iron by mass, which is the figure an assay would check.
2. Interpret the yellow-brown crust and the pale-yellow filtrate. Decide whether the batch should be discarded or recrystallised, giving your reasoning.
Read the worked answer
The crust is not a Mohr's salt colour; Fe²⁺ salts are pale green. Yellow-brown indicates iron(III), formed when dissolved oxygen oxidises Fe²⁺: 4Fe²⁺ + O₂ + 4H⁺ → 4Fe³⁺ + 2H₂O. The student's conditions all favour this: heating to near boiling, a solution not sufficiently acidic, and tap water containing dissolved oxygen and chloride.
Discarding the whole batch would be an overreaction. The colour is a surface crust and the bulk crystals remain pale green, so most of the iron is still Fe(II). Mohr's salt is very soluble in water and much less soluble in cold, slightly acidified water, so recrystallisation separates the product from most of the Fe³⁺ and from tap-water salts, which stay in the mother liquor.
The judgement is therefore to recrystallise, then check purity: a good sample should be pale green with no brown tinge, dissolve to a clear solution, and give an Fe²⁺ assay close to 14.3 %. The sample should be discarded only if the entire mass is brown or if a permanganate titration shows Fe²⁺ well below the expected value.
3. Recommend two procedural changes and predict, with reasons, their effect on yield and purity.
Read the worked answer
First, replace tap water with freshly boiled and cooled distilled water, and increase the dilute sulphuric acid to about 3 cm³. Boiling expels dissolved oxygen, which removes the oxidant, and the extra acid suppresses hydrolysis of Fe³⁺ and shifts the Fe²⁺/Fe³⁺ equilibrium back towards Fe²⁺. Purity should improve markedly; yield should also rise slightly, because less iron is lost as basic ferric sulphate that stays in the mother liquor.
Second, dissolve at room temperature or at most around 50 °C with stirring, and dry the crystals between filter papers or in a desiccator rather than at 110 °C. Heating accelerates air oxidation and can drive off water of crystallisation, so cooling the procedure protects both the Fe²⁺ content and the six water molecules that define the double salt. Overall yield may fall slightly, since less material is forced out of solution, but the product will be closer to the theoretical composition.
Finally, wash the crystals with a little ice-cold water rather than warm water, to remove mother liquor without redissolving a significant fraction of the product.