Question

An organic substance containing carbon, hydrogen and oxygen gave the following percentage composition. $ C=40.687%;\text{ }H=5.085%\text{ }and\text{ }O=54.228% $ What is the empirical formula of the compound?

• A. $ C{{H}_{3}}O $
• B. $ {{C}_{2}}{{H}_{2}}O $
• C. $ C{{H}_{4}}O $
• D. $ {{C}_{2}}{{H}_{3}}{{O}_{2}} $

Answer 1

Answer: (D)

Symbol
Percentage
Relative no. of moles
Simplest molar ration
Simplest whole no. molar ratio
$ C $
$ 40.687 $
$ \frac{40.687}{12}=3.390 $
$ \frac{3.390}{3.389}=1 $
2
$ H $
$ 5.085 $
$ \frac{5.085}{1}=5.085 $
$ \frac{5.085}{3.389}=1.5 $
3
$ O $
$ 54.228 $
$ \frac{54.228}{16}=3.389 $
$ \frac{3.389}{3.389}=1 $
2
Hence, the epitomical formula of the compound is $ {{C}_{2}}{{H}_{3}}{{O}_{2}} $ .