Chemistry
(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?
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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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