Question

1) For each of the following lab execution errors, indicate whether the calculated % of hydration would be too high or too low and briefly state why a.) The crucible was not pre-heated b.) The hydrate and crucible were over heated. c.) The cover was not in place and some of the material was lost by spraying out of the crucible 2.) Iron sulfate exists as a hydrate in the form, FeSO, 7H2O. a.) If 1.000 grams of this compound were heated to dryness, what is the dry weight that would result?

Answer 1

1. a. (low) The crucible need to be preheated as it can contain some adsorbed H2O molecule is pore sizes and temperature difference of crucible before experiment and after experiment, which can result in higher water % percentage after the experiment. Hence crucible need to be preheated first before performing experiment.

b. (High) If the compound or crucible is over heated then it results in decomposition of sample which can result in higher % percentage error (% percentage of hydration is high).

c. (Low or High) As the loss in the final sample can be observable (results in high) or it can be simply lost by escaping (results in low). Hence weight measurements would be wrong.

1. a. FeSO4.7H2O can give FeSO4. Hence molar mass for FeSO4.7H2O = 278.02 g/mol and FeSO4 = 151.91 g/mol

So 1 mol of FeSO4.7H2O (278.02) can give 1 mol of FeSO4 (151.91)

So 1 g of FeSO4.7H2O will give = 151.91/278.02 = 0.546 g of FeSO4.

b. Molar mass of FeO = 71.84

So 1 g of FeSO4.7H2O will give 0.258 g

c. Final mass obtained after heating = 0.655 g

amount need to be formed = 0.546 g

Hence water of hydration present is 0.109 g

So 1 mol of FeSO4.7H20 (278.02) has 7 H2O (126 g). Hence amount of FeSO4.7H2O remain as hydrated is 24%.

Proof = 0.76 g of FeSO4.7H2O will give 0.415 g of FeSO4. and remained 0.24g of FeSO4.7H2O is present in its current form.