Anorganiccompoundcontains C = 16.27%, H = 0.677%, CI = 72.2% and O=10.8%. Its molecular mass is 147.5 amu. Find its molecular formula.

Anorganiccompoundcontains C = 16.27%, H = 0.677%, CI = 72.2% and O=10.

1.	Anorganiccompoundcontains C = 16.27%, H = 0.677%, CI = 72.2% and O=10.8%. Its molecular mass is 147.5 amu. Find its molecular formula.
Answer» Solution :Determination of EMPIRICAL formula : Percentage of oxygen `= 100 - (16.27 + 0.677 + 72.2) = 10.8` `therefore` The empirical formula of the given compound as CALCULATED in the table given ahead is `C_2HCl_3O`. CALCULATION of molecular formula Empirical formula mass `=(12.01 xx 2) + 1.008 + (35.45 xx 3) + 16.0 = 147.4` Molecular mass (given) = 147.5 `therefore n=("Molecular mass")/("Empirical formula mass") = 147.5/147.4 =1` `therefore` Molecular formula `=1 xx` Empirical formula `= 1 xx C_(2)HCl_(3)O` `=C_(2)HCl_(3)O` Hence, the molecular formula of the given compound is `C_(2)HCl_(3)O`.