Determination of Hydrate Formula:barium chloride hydrate Data: Sample #1 Sample #2Mass of empty crucible and lid: (1) 29.9031 g (1) 29.178 g (2) 29.9028 g (2) 29.1750 gFinal mass of empty crucible and lid: 29.9028 g 29.1750gMass of crucible, lid, and BaCl2•nH2O: 31.3596 g 30.6222 gMass of BaCl2•nH2O (g) 208.3Mass of crucible, lid, and product after: First heating: 31.1450 g 30.4082 g Second heating: 31.1448 g 30.4084 gFinal mass of crucible, lid, and BaCl2 (g): 31.1448 g 30.4084 gMass of anhydrous BaCl2 (g): ???I was told to resubmit this question, not sure if this is the information that was needed

While heating the hydrate BaCl2.nH2O, the water in it escapes as water vapor. The amount of salt…

Answered: Determination of Hydrate Formula:…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Determination of Hydrate Formula:

barium chloride hydrate

Data: Sample #1 Sample #2

Mass of empty crucible and lid: (1) 29.9031 g (1) 29.178 g

(2) 29.9028 g (2) 29.1750 g

Final mass of empty crucible and lid: 29.9028 g 29.1750g
Mass of crucible, lid, and BaCl2•nH2O: 31.3596 g 30.6222 g
Mass of BaCl2•nH2O (g) 208.3
Mass of crucible, lid, and product after:
First heating: 31.1450 g 30.4082 g

Second heating: 31.1448 g 30.4084 g

Final mass of crucible, lid, and BaCl2 (g): 31.1448 g 30.4084 g
Mass of anhydrous BaCl2 (g): ???

I was told to resubmit this question, not sure if this is the information that was needed

Expert Solution

Step 1

While heating the hydrate BaCl₂.nH₂O, the water in it escapes as water vapor. The amount of salt that remains can be used to find out the amount of water that has escaped. The formula of the hydrate is found out as follows.

Sample 1

The amount of BaCl₂.nH₂O used $= Mass of crucible, lid and {BaCl}_{2} . n H_{2} O - Mass of crucible and lid = 31.3596 g - 29.9028 g = 1.4568 g .$

The amount of BaCl₂remained after heating $= Mass of crucible, lid and {BaCl}_{2} after heating - Mass of crucible and lid = 31.1448 g - 29.9028 g = 1.242 g .$

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