Mohr's percentage yield

The formula for Mohr's salt is FeSO₄.(NH₄)₂SO₄.6H₂O and this has a molar mass of 392 g mol^-1. We aimed to make 0.05 moles of this salt which has a mass of 0.05 × 392 = 19.6 g - this is the theoretical yield.

We actually made 8.672 g of the salt so the percentage yield was 8.672 ÷ 19.6 × 100 = 44%.

The main reason for such a low yield is that much of the Mohr's salt was lost when we filtered off the solution (which is saturated with Mohr's salt) from the crystals which had formed. We repeated the experiment and allowed the solution to completely crystallize. This produced 17.930 g of crystals which works out as a more impressive 91% yield.

Crystals of Mohr's salt with green solution filtering away

Unreacted iron left behind on the filter paper

You should have noticed some unreacted iron showing that this reaction was incomplete. This means that less than 0.05 moles of iron(II) sulfate was produced so lowering the percentage yield.

There were also some transfer losses with some of the finer crystal remaining on the filter paper.

Magnified traces of Mohr's salt left behind on the filter paper