2. During a Chem 121 experiment, a student obtained the following data after hea

Question

2. During a Chem 121 experiment, a student obtained the following data after heating a sample of Calcium Chloride hydrate. Data Weight of crucible and lid (g) Weight of crucible, lid and hydrate (g) Weight of crucible, lid and anhydrous salt after a) 1 heating (g) b) 2nd heating (g) c) 3"d heating (g) 32.1070 36.8283 5.7347 35.6682 35.6681 Use this information to determine: a) the weight of the hydrated salt (CaCl.xHO) b) the weight of the anhydrous salt (CaCl) Note: the weight after the 2nd and 3 heating are close enough in value to assume that constant weight has been reached and all the water molecules have been removed c) the weight of water removed d) the number of moles of the anhydrous salt Note: the molar mass of CaCl, is 110.98 g/mol e) the number of moles of water f) the empirical formula of the hydrate Note: the formula of the hydrate is CaCl.xH.O and you need to solve for x by finding the ratio of number of moles of H O to the number of moles of CaCl

Explanation / Answer

Ans. #a. Weight of hydrated salt =

(Weight of crucible & lid plus hydrated salt) - (Weight of crucible & lid)

= 36.8283 g – 32.1070 g

= 4.7213 g

#b. Weight of anhydrous salt =

Weight after 3rd drying - (Weight of crucible & lid)

= 35.6681 g – 32.1070 g

= 3.5611 g

#c. Weight of water removed =

(Weight of crucible & lid plus hydrated salt) – Weight after 3rd drying

= 36.8283 g – 35.6681 g

= 1.1602 g

#d. Moles of anhydrous salt = Mass of anhydrous salt / Molar mass

                        = 3.5611 g / (110.98 g/ mol)

                        = 0.03209 mol

#e. Moles of water lost = 1.1602 g / (18.02 g/ mol) = 0.06439 mol

#f. Moles of H2O / Moles of anhydrous salt

= 0.06439 mol / 0.03209 mol

= 2.0 (whole number)      

That is, there is 2 mole of water per mol of anhydrous salt.

So, x = 2.

# Empirical formula of the hydrate = CaCl2.2H2O

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