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 its molecular formula, given that its relative molecular mass is 162.

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Solution

P = $\frac{38.27}{30.97} = 1.236$
H = $\frac{2.47}{1.00} = 2.47$
O = $\frac{59.26}{16} = 3.704$

Divide through by smallest value
P = 1
H= 2
O = 3

Empirical formula = $H_{2} P O_{3}$

Formula mass = 2+31+48 = 81g/formula

$n = \frac{M o l e c u l a r m a s s}{e m p i r i c a l f o r m u l a m a s s} = \frac{162}{81} = 2$

Molecular formula = $(H_{2} P O_{3}) \times 2 = H_{4} P_{2} O_{6}$

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