Question

An organic compound on analysis gave the following results : `C = 54.5% , O = 36.4 %, H = 9.1%` . The Empirical formula of the compound is
A. `C_(2)H_(4)`
B. `C_(3)H_(4)O`
C. `C_(3)H_(4)O`
D. `C_(4)H_(8)O`

Answer 1

Correct Answer - B
`{:("Element"," No. of moles"," Simple Ratio"),(C=54.5,54.5//12=4.54,=2(2.2)),(H=9.1,9.1//1=9.1,=4(4.5)),(O=36.4,36.4//16=2.27,=1(1)):}`
Hence `C_(2)H_(4)O`.