0.45 g of an organic compound gave 0.44 g of CO_(2) and 0.09 g of H_(2)O. The molecular mass of the compound is 90 amu. Calculate the molecular formula.

0.45 g of an organic compound gave 0.44 g of CO_(2) and 0.09 g of H_(2

1.	0.45 g of an organic compound gave 0.44 g of CO_(2) and 0.09 g of H_(2)O. The molecular mass of the compound is 90 amu. Calculate the molecular formula.
Answer» Solution :Step I. Percentage of elements Percentage of carbon `= (12)/(44) XX ("Mass of "CO_(2))/("Mass of compound")xx100` `=(12)/(44)xx(0.44)/(0.45)xx100=26.67` Percentage of HYDROGEN `=(2)/(18)xx("Mass of" H_(2)O)/("Mass of compound")xx100` `=(2)/(18)xx(0.09)/(0.45)xx100=2.22` Percentage of oxygeb `= 100 - (26.67 + 2.22) = 100-28.89 = 71.11` Step II. Empirical formula of the compound `{:("ELEMENT","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",26.67,12,(26.67)/(12)=2.22,(2.22)/(2.22)=1.0,1),("H",2.22,1,(2.22)/(1)=2.22,(2.22)/(2.22)=1.0,1),("O",71.11,16,(71.11)/(16)=4.44,(4.44)/(2.22)=2.0,2):}` Empirical formula of the compound `= CHO_(2)` Step III. Molecular formula of the compound Empirical formula mass `= 12+1 +2 xx 16 = 45 u` Molecular mass = 90 AMU (Given) `n=("Molecular mass")/("Empirical formula mass")=((90u))/((45u))=2` Molecular formula `= n xx` Empirical formula `= 2 xx CHO_(2)=C_(2)H_(2)O_(4)`.