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]

Original weight = 1.30 g

Element % composition At. wt. Relative no. of atoms Simplest ratio
Al (0.2675/1.30) × 100 = 20.57 27 20.57/27 = 0.7618 0.7618/0.7618 = 1
P (0.3505/1.30) × 100 = 26.96 31 26.96/31 = 0.8696 0.8696/0.7618 = 1.14 = 1
O (0.682/1.30) × 100 = 52.46 16 52.46/16 = 3.278 3.278/0.7618 = 4.30 = 4

Al : P : O = 1 : 1 : 4

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

Hence, the empirical formula is AlPO4

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

= 27 + 31 + 64 = 122

Molecular weight = 122

n = Molecular\ weight\over Emperical\ Formula\ Weight

= 122\over 122

= 1

∴ Molecular formula = (AlPO4)1 = AlPO4

Hence, Molecular formula = AlPO4

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