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Question

A 4.24 mg sample of butyric acid is completely burnt.It gives 8.45 mg of carbon dioxide and 3.46 mg of H2O what is mass percent of each element in butyric acid ? 

Elaborated answer please


Solution

To calculate the mass percentage of C.
CO subscript 2 space identical to space straight C 44 mg space space identical to space 12 mg space 44 mg space CO subscript 2 space end subscript contains space straight C space equals space 12 mg therefore space 8.45 space CO subscript 2 space would space contain space straight C space equals space 12 over 44 cross times 8.45 equals 2.304 m g percent sign space o f space C space equals space fraction numerator 2.304 over denominator 4.24 end fraction cross times 100 equals 54.30 percent sign P e r c e n t a g e space o f space H H subscript 2 O space identical to space 2 H 18 mg space space identical to space 2 mg space therefore space 3.46 space H subscript 2 O space would space contain space H space equals space 2 over 18 cross times 3.46 equals 0.384 m g percent sign space o f space H space equals space fraction numerator 0.384 over denominator 4.24 end fraction cross times 100 equals 9.06 percent sign P e r c e n t a g e space o f space O percent sign space o f space H space equals space 100 minus left parenthesis percent sign space o f thin space C plus percent sign o f space H right parenthesis equals 100 minus left parenthesis 54.30 plus 9.06 right parenthesis equals 36.54 percent sign

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