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CHEMISTRY 121 1 Department of Chemistry University of British Columbia Problem Set 6 Answers Molecular Identification Problems 1. Heating calcium carbonate yields a solid compound A (molecular weight = 56.1 g/mol) and a gas B . 0.5 mole of compound A reacts with 0.5 mole of water to give 0.5 mole of compound C . Reacting 0.5 mole of compound B with 0.5 mole of C yields 0.5 mole of compound D and 0.5 mole of water. What are A , B , C and D ? Answer: A: CaO B: CO 2 C: Ca(OH) 2 D: CaCO 3 Compound A has a molecular weight of 56.1. It can be assumed to contain Ca since the only other product, a gas, is unlikely to contain this element. Calcium carbonate contains only C and O in addition to Ca, so we may deduce that compound A is CaO, which has the correct molecular weight. Compound B is the only other product of the decomposition of CaCO 3 , thus must be CO 2 . Reaction of A (CaO) with H 2 O in a 1:1 mole ratio gives compound C which must contain Ca, H and O in a 1:2:2 ratio. We may therefore deduce that this is Ca(OH) 2 . Reaction of B (CO 2 ) with C (Ca(OH) 2 ) in a 1:1 ratio produces D and one equivalent of H 2 O. The elemental composition of D is 1 Ca: 3 O: 1 C, thus it is CaCO 3 . 2. A sample of sulfur dioxide is treated with water to give a solution of compound E . When 0.1 mol of compound E dissolved in water is treated with 0.2 mol of NaOH, 0.1 mol of compound F (molecular weight = 126 g/mol) is formed. Boiling this solution of F in the presence of 0.0125 mol of elemental sulfur (S 8 ) gives compound G as the sole product. (a) Identify E , F , and G . SO 2 + H 2 O H 2 SO 3 (compound E ) H 2 SO 3 + 2 NaOH Na 2 SO 3 (compound F ) + 2 H 2 O Na 2 SO 3 + 1/8 S 8 Na 2 S 2 O 3 (compound G )
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CHEMISTRY 121 2 Department of Chemistry University of British Columbia
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This note was uploaded on 11/03/2009 for the course ECON 210 taught by Professor James during the Spring '09 term at The University of British Columbia.

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