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

`C_(2)H_(2)O_(4)`

Explanation

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)`.