2. The formula of a compound of Tin (Sn) and Oxygen (O) was determined using the

Question

2. The formula of a compound of Tin (Sn) and Oxygen (O) was determined using the following data. A sampl of Sn was weighed into a crucible. The crucible was covered and heated to allow the Sn and O to react. The crucible was then cooled and weighed. The following data were collected. (Molecular mass of Sn is g/mol and of O is 16.0 g/mol Mass of Sn reacted: 1.897 g Mass of product: 2.146 g. (a) Calculate mass of O reacted. (b) Calculate moles of O reacted. (c) Calculate moles of Sn reacted. (d) Calculate the empirical formula of the compound.

Explanation / Answer

Ans. #a. Calculated mass of O reacted = Final mass after reaction – Initial mass of Sn

                                                            = 2.146 g – 1.897 g

                                                            = 0.249 g

#b. Moles of O-reacted = Mass of O reacted / MW

                                                = 0.249 g / (16.0 g/ mol)

                                                = 0.0155625 mol

#c. Moles of Sn-reacted = 1.897 g / (118.71 g/ mol) = 0.015980 mol

#d. Molar ratio of reactants = Moles of Sn / Moles of O reacted

                                                = 0.015980 mol / 0.0155625 mol

                                                = 1.03

                                                = 1.0 (nearest whole number)

The molar ratio of reactants 1.0 indicates that there is 1 atom O-atom every 1 Sn atom.

Therefore,

            Empirical formula of the compound = SnO

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