exactly 9.00g of a hydrate of CoCl2 was heated strongly, then allowed to cool. t
ID: 1054336 • Letter: E
Question
exactly 9.00g of a hydrate of CoCl2 was heated strongly, then allowed to cool. the mass of the anhydrate remaining was found to be 5.89g a. calculate the mass of H2O lost from the hydrate. b). calculate % of water in the hydrate c. determine the formula of the CoCl2 hyrate. c) determine the formula of the CoCl2 hydrate. (kindly show your calculations).
2. when 4.86g of magnesium reacted with oxygen the magnesium oxide product has a mass of 8.06g. calculate the simplest or empirical formula of the magnesium oxide.
3. An oxide of gold thermally decomposes. A 0.205g sample of the oxide is heated forming 0.183g of gold metal. a) what is the empirical formula of the gold oxide.
Explanation / Answer
Hydrate of CoCl2 = 9g
anhydrate CoCl2 = 5.89g
mass of H2O = 9-5.89 = 3.11g
b. % of water in the hydrate = 3.11*100/9 = 34.55%
c. no of moles of H2O = 3.11/18 = 0.172
no of moles of anhydrate coCl2 = 5.89/130 = 0.045 moles
CoCl2 : H2O
0.045 : 0.172
0.045/0.045 : 0.172/0.045
1: 4
formula is CoCl2.4H2O
2. mass of Mg = 4.86g
mass of magnesium oxide = 8.06g
mass of oxgen = 8.06-4.86 = 3.2g
no of moles of Mg = 4.86/24 = 0.2025 moles
no of moles of O = 3.2/16 = 0.2 moles
Mg : O
0.2: 0.2
0.2/0.2 : 0.2/0.2
1:1
empirical formula = MgO
3. mass of gold oxide = 0.205g
mass of gold = 0.183g
mass of oxygen = 0.205-0.183 = 0.022g
no of moles of Au = 0.183/197 = 0.00093moles
no of moles of oxygen = 0.022/16 = 0.001375 moles
Au : O
0.00093 : 0.001375
0.00093/0.00093 : 0.001375/0.00093
1 :1.5
1*2 : 1.5*2
2 : 3
Empirical formula = Au2O3
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