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0.279 g of the hydrate CoCl2•nH2O is heated to produce 0.197 g of anhydrous CoCl2 (MM = 129.83 g/mol).  We want to use this to determine n, how many waters of hydration were in the original sample, be the end of the questions that follow. How many grams of water were in the original 0.279 g of hydrate? How many grams of CoCl2 were present on the original 0.279 g of hydrate?  So ... we now have (as a result of the previous two questions) what we need to know to determine the empirical formula of CoCl2•nH2O. We can look at CoCl2•nH2O  in this way (CoCl2)x•nH2O.  Recalling that x and n are whole number molar ratios ... use the results of the last two problems to determine the correct formula.

0.279 g of the hydrate CoCl2•nH2O is heated to produce 0.197 g of anhydrous CoCl2 (MM = 129.83 g/mol).  We want to use this to determine n, how many waters of hydration were in the original sample, be the end of the questions that follow. How many grams of water were in the original 0.279 g of hydrate? How many grams of CoCl2 were present on the original 0.279 g of hydrate?  So ... we now have (as a result of the previous two questions) what we need to know to determine the empirical formula of CoCl2•nH2O. We can look at CoCl2•nH2O  in this way (CoCl2)x•nH2O.  Recalling that x and n are whole number molar ratios ... use the results of the last two problems to determine the correct formula.

Chemistry
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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0.279 g of the hydrate CoCl2•nH2O is heated to produce 0.197 g of anhydrous CoCl2 (MM = 129.83 g/mol).  We want to use this to determine n, how many waters of hydration were in the original sample, be the end of the questions that follow.

How many grams of water were in the original 0.279 g of hydrate?

How many grams of CoCl2 were present on the original 0.279 g of hydrate? 

So ... we now have (as a result of the previous two questions) what we need to know to determine the empirical formula of CoCl2•nH2O.

We can look at CoCl2•nH2O  in this way (CoCl2)x•nH2O.  Recalling that x and n are whole number molar ratios ... use the results of the last two problems to determine the correct formula.

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