Chemistry

An acid of phosphorus has the following percentage composition; Phosphorus = 38.27%; hydrogen = 2.47%; oxygen = 59.26%. Find the empirical formula of the acid and it's molecular formula, given that it's relative molecular mass is 162.

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Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Phosphorus	38.27	31	$\dfrac{38.27 }{31}$ = 1.234	$\dfrac{ 1.234}{1.234 }$ = 1
Hydrogen	2.47	1	$\dfrac{2.47 }{1}$ = 2.47	$\dfrac{2.47 }{1.234 }$ = 2
Oxygen	59.26	16	$\dfrac{59.26}{16}$ = 3.70	$\dfrac{3.70}{ 1.234 }$ = 3

Simplest ratio of whole numbers = P : H : O = 1 : 2 : 3

Hence, empirical formula is H₂PO₃

Empirical formula weight = 31 + 2(1) + 3(16) = 31 + 2 + 48 = 81

Molecular weight = 162

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{162}{81} = 2$

So, molecular formula = 2(H₂PO₃) = H₄P₂O₆

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