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 Chemistry 121 Worksheet 1 (Summer 2014) Answers Data Weight of crucible and lid (g Weight of crucible, lid and hydrate (g) Weight of crucible, lid and anhydrous salt after a) 1st heating (g) b) 2ad heating (g) c) 3rd heating (g) Mass 32.1070 36.8283 35.7347 35.6682 35.6681 Use this information to determine a) the weight of the hydrated salt (CaCh2O) b) the weight of the anhydrous salt (CaCb) Note: the weight after the 2ad and 3rd 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 CaCh 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 CaC2 H2O and you need to solve for x by finding the ratio of number of moles of H20 to the number of moles of CaCl

Explanation / Answer

a) Weight of hydrated salt = Weight of crucible, lid and hydrate - Weight of crucible, lid

=> 36.8283 - 32.1070

=> 4.7213 grams

b) Weight of anhydrous salt = Weight of crucible, lid and anhydrous - Weight of crucible, lid

=> 35.6682 - 32.1070

=> 3.5612 grams

c)

Number of moles of anhydrous salt = Mass/molar mass = 3.5612/110.98 = 0.032088 moles

d)

Mass of water in sample = Mass of hydrated - Mass of anhydrous sample = 4.7213 - 3.5612 = 1.1601 grams

Number of moles of water = Mass/molar mass = 1.1601/18 = 0.06445 moles

e)

The empirical formula of the hydrate will be CaCl2.2H2O

Reason: Since number of moles of hydrate is two times the number of anhydrous salt

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.