(i) Butyric acid contains only `C, H` and `O.A4.24 mg` sample of butyric acid is completely burned. It gives `8.45 mg` of carbon dioxide `(CO_(2))` and `3.46 mg` of water. What is the mass percentage of each element in butyric acid?
(ii) If the elemental composition of butyric acid is found to be `54.2% C, 9.2% H` and `36.6%O`, determine the empirical formula.
(iii) The molecular mass of butyric acid was determined of experiment to be `88`. What is the molecular formula ?

Explanation

(a) Calculation of mass percentage of different elements
Percentage of carbon can be calculated as :
`CO_(2)=C`
44 mg = 12 mg
44 mg of `CO_(2)` contains C = 12 mg
8.45 mg of `CO_(2)` contiain C `=12/44 xx 8.45` mg
Percentage of C `=("Weight of carbon")/("Weight of compound") xx 100`
`=12/44 xx (8.45)/(4.24) xx 100 = 54.3%`
Percentage of hydrogen can be calculated as:
`H_(2)O =2H`
18 mg = 2 mg
18 mg of `H_(2)O` contain H `=2/18 xx 3.46` mg
Percentage of H `=("Weight of hydrogen")/("Weight of compound") xx 100`
`=2/18 xx (3.46)/(4.24) xx 100 = 9.0%`
The sum of the percentage of C and H =54.3 + 9.0 = 63.3 %
`therefore` Percentage of O =100 - 63.3 = 36.7%
(b) Calculation of molecular formula:

The simplest whole number ratio of atoms is:
C:H:O =2:4:1
`therefore` The empirical formula is `C_(2)H_(4)O`
Empirical formula mass `=2 xx 12 + 4 xx 1+16`
= 44 a.m.u.
Molecular mass = 88 a.m.u.
`n=("Molecular mass")/("Empirical formula mass")`
`n= 88/44 =2`
`therefore` molecular formula of butryic acid `2(C_(2)H_(4)O) = C_(4)H_(8)O_(2)`.