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
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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