1. A student carried out a dehydration experiment using MgSO4. nH20. The student

Question

1. A student carried out a dehydration experiment using MgSO4. nH20. The student placed an evaporating dish which contained 3.8457 g of MgSO4. nH20 in an oven at 110 °C. At this temperature, some but not all of the water in this hydrate was lost. When a constant mass was achieved, the salt weighed 2.4402 g. The student then placed the evaporating dish and contents back in the oven at 160 o°c. The residue came to a constant mass of 1.8779 g. Stronger heating produced no further loss in mass so we can assume all the water has evaporated. Assume that this final residue is the completely anhydrous compound (see 1st page for definition). The water is lost in molecular steps and the formulas must be written with integers and not fractions a. What is the formula of the 1.8779 g sample? b. What is the formula of the 2.4402 g sample? c. What is the formula of the 3.8457 g sample?

Explanation / Answer

(a)

Formula of 1.8779 mass is MgSO4

(b)

Mass loss =3.8457 g - 2.4402g = 1.4055 g

This is due to water. Moles of water = 1.4055g/18g/mol

= 0.078 mol

0.1093 mol corresponds to 7 molecules of H2O (see Part C).

So, 0.078 mol corresponds to 7 *0.078/0.1093 = 4.99~5 molecules

Formula of 2.4402 g mass = MgSO4.2H2O

(c)

mass loss = 3.8457 g-1.8779 g

= 1.9678 g

This mass corresponds to H2O.

moles of H2O = 1.9678 g/18g/mol

= 0.1093 mol

Mass of anhydrous MgSO4 = 1.8779g

Moles of anhydrous MgSO4 = 1.8779g/120.37g/mol

= 0.0156 mol

So, 0.0156 mol MgSO4 contains 0.1093 moles water.

moles of water in 1 mol MgSO4 = 7.007 ~7

Formula of 3.8457g sample = MgSO4.7H2O

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