Question 21(b)(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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Accepted Answer

(i)Element% compositionAt. wt.Relative no. of atomsSimplest ratioCarbon14.41214.412\dfrac{14.4}{12}1214.4​ = 1.21.21.2\dfrac{1.2}{1.2}1.21.2​ = 1Hydrogen1.211.21\dfrac{1.2}{1}11.2​ = 1.21.21.2\dfrac{1.2}{1.2}1.21.2​ = 1chlorine84.535.584.535.5\dfrac{84.5}{35.5}35.584.5​ = 2.382.381.2\dfrac{2.38}{1.2}1.22.38​ = 1.98 = 2Simplest ratio of whole numbers = C : H : Cl = 1 : 1 : 2Hence, empirical formula is CHCl2Empirical formula weight = 12 + 1 + 2(35.5) = 84 gRelative molecular mass = 168n=Molecular weightEmpirical formula weight=16884=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{168}{84} = 2n=Empirical formula weightMolecular weight​=84168​=2Molecular formula = n[E.F.] = 2[CHCl2] = C2H2Cl4∴ Molecular formula = C2H2Cl4

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

Mole Concept & Stoichiometry Miscellaneous Exercises — Question 22

Question 21(b)