(i) A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. (H = 1; C = 12; Cl = 35.5)
(ii) The relative molecular mass of this compound is 168, so what is its molecular formula?

Question

(i) A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. (H = 1; C = 12; Cl = 35.5)
(ii) The relative molecular mass of this compound is 168, so what is its molecular formula?

KnowledgeBoat · Accepted Answer

(i) Element | % composition | At. wt. | Relative no. of atoms| Simplest ratio --------|---------------|---------|----------------------|---- Carbon | 14.4 | 12 | = 1.2 | = 1 Hydrogen | 1.2 | 1 | = 1.2 | = 1 chlorine | 84.5 | 35.5 | = 2.38 | = 1.98 = 2 Simplest ratio of whole numbers = C : H : Cl = 1 : 1 : 2 Hence, empirical formula is CHCl₂ Empirical formula weight = 12 + 1 + 2(35.5) = 84 g Relative molecular mass = 168 Molecular formula = n[E.F.] = 2[CHCl₂] = C₂H₂Cl₄ Molecular formula = C₂H₂Cl₄

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