Chemistry

A compound has the following percentage composition. Al = 0.2675 g.; P = 0.3505 g.; O = 0.682 g. If the molecular weight of the compound is 122 and it's original weight which on analysis gave the above results 1.30 g. Calculate the molecular formula of the compound. [Al = 27, P = 31, O = 16]

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Original weight = 1.30 g

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Al	$\dfrac{0.2675}{1.30}$ x 100 = 20.57	27	$\dfrac{20.57}{27}$ = 0.7618	$\dfrac{0.7618}{0.7618}$ = 1
P	$\dfrac{0.3505}{1.30}$ x 100 = 26.96	31	$\dfrac{26.96}{31}$ = 0.8696	$\dfrac{0.8696}{0.7618}$ = 1.14 = 1
O	$\dfrac{0.682}{1.30}$ x 100 = 52.46	16	$\dfrac{52.46}{16}$ = 3.278	$\dfrac{3.278}{0.7618}$ = 4.30 = 4

Al : P : O = 1 : 1 : 4

Hence, Simplest ratio of whole numbers = 1 : 1 : 4

Hence, empirical formula is AlPO₄

Empirical formula weight = 27 + 31+ 4(16) = 27 + 31 + 64 = 122

Molecular weight = 122

$\text{n} = \dfrac{\text{Molecular weight}}{\text{empirical formula weight}} \\[0.5em] = \dfrac{122}{122} = 1$

∴ Molecular formula = n[E.F.] = 1[AlPO₄] = AlPO₄

Hence, Molecular formula = AlPO₄

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