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 it's molecular formula?

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(i)

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	14.4	12	$\dfrac{14.4}{12}$ = 1.2	$\dfrac{1.2}{1.2}$ = 1
Hydrogen	1.2	1	$\dfrac{1.2}{1}$ = 1.2	$\dfrac{1.2}{1.2}$ = 1
chlorine	84.5	35.5	$\dfrac{84.5}{35.5}$ = 2.38	$\dfrac{2.38}{1.2}$ = 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

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{168}{84} = 2$

Molecular formula = n[E.F.] = 2[CHCl₂] = C₂H₂Cl₄

∴ Molecular formula = C₂H₂Cl₄

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