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

ID: 564193 • Letter: 1

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 1t 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

The initial weight of the complex MgSO4.nH2O: 3.8457 g

The weight of the complex MgSO4.xH2O: 2.4402 g

The initial weight of the complex MgSO4.yH2O: 1.8779 g

A.

The lose of the weight by evaporation of water in second step: 2.4402 -1.8779 = 0.562 g

that equals to (0.562/18) molesof water = 0.031 moles of water

moles MgSO4 fom final step = 1.8779 / 120.42 = 0.0156 moles

so the ratio of water to MgSO4 was 0.031/ 0.0156 = 2:1

That means the formula MgSO4•2H2O

B.

The lose of the weight by evaporation of water in second step: 3.8457 -2.4402 = 1.4055 g

that equals to (1.4055/18) molesof water = 0.078 moles of water

moles MgSO4 fom final step = 1.8779 / 120.42 = 0.0156 moles

so the ratio of water to MgSO4 was 0.078/ 0.0156 = 5:1

That means the formula MgSO4•5H2O

C.

the 3.8457 g sample contained some water. 3.8457 g - 1.8779 = mass water = 1.9678 g

moles water = 1.9678 g / 18.0153 g = 0.109 moles

so the ratio of water to MgSO4 was 0.109 / 0.0155 = 7:1

the formula MgSO4•7H2O