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Butyric acid contains only C, H and O. A...

Butyric acid contains only C, H and O. A 4.24 mg of sample butyric acid is completely burnt. It gives 8.45 mg of carbon dioxide and 3.46 mg of water. What is the mass percentage of each element in butyric acid ? The molecular mass of buryric acid is determined by experiment as 88 amu. What is its molecular formula ?

Text Solution

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Step I : Calculation of mass percentage of different elements
Percentage of carbon. Percentage of carbon can be calculated as follows :
`CO_(2)-=C`
44 parts 12 parts
or 44 mg 12 mg
44 mg of `CO_(2)` contain C = 12 mg
`:. 8.45` g of `CO_(2)` contain `C = (12)/(44)xx8.45 = 2.30` mg
Percentage of `C=("Weight of carbon")/("Weight of compound")xx100=((2.30mg))/((4.24mg))xx100=54.3%`
Percentage of hydrogen. Percentage of hydrogen can be calculated as :
`H_(2)O -= 2H`
18 parts 2 parts
18 mg 2 mg
18 mg of `H_(2)O` contain H = 2mg
`:. 3.46` of `H_(2)O` contain `H= (2)/(18) xx 3.46 = 0.384` mg
Percentage of `H=("Weight of hydrogen")/("Weight of compound")xx100=((0.384mg))/((4.24mg))xx100=9%`
The sum of the percentage of C and H `= 54.3 + 9 = 63.3 %`
As this comes out to be less than 100, the balance must be oxygen.
`:.` Percentage of `O = 100-63.3 = 36.7 %`
Step II. Determination of empirical formula of the compound
`{:("Elemenet","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",54.3,12,(54.3)/(12)=4.52,(4.52)/(2.29)=1.97,2),("H",9.0,1,(9.0)/(1)=9.0,(9.0)/(2.29)=3.93,4),("O",36.7,16,(36.7)/(16)=2.29,(2.29)/(2.29)=1.0,1):}`
The empirical formula of the compound `= C_(2)H_(4)O`
Step III. Determination of molecular formula of the compound
Empirical formula mass `= 2xx 12 + 4 xx1 +16 = 24 + 4+16 = 44` u
Molecular mass = 88 u , `n=("Molecular mass")/("Empirical formula mass")=(88)/(44)=2`
`:.` Molecular formula `= n xx` Empirical formula `= 2xx C_(2)H_(4)O=C_(4)H_(8)O_(2)`
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