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Question

An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C= 38.71% and H= 9.67%. What would be the empirical formula of the compound?


A

CHO

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B

CH2O

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C

CH3O

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D

CH4O

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Solution

The correct option is C

CH3O


Given:

Percentage composition is C=38.71 %, H=9.67 %

Percentage composition of Oxygen (O)= 100- (38.71+ 9.67)= 100- 48.38= 51.62 %

Step-1: Calculate the empirical formula

  • Empirical formula: Empirical formula is the simplest formula of the compound which gives the simplest whole-number ratio of the atoms of the various elements present in one molecule of the compound. Example: Molecular formula of glucose is C6H12O6 but the simplest whole-number ratio of atoms in glucose is CH2O. Therefore, CH2O is the empirical formula.
Element % Of ElementAtomic mass (u)Atomic ratioSimplest ratio
Carbon (C)38.711238.7112=3.223.223.22=1
Hydrogen (H)9.6719.671=9.679.673.22=3
Oxygen (O)51. 621651.6216=3.223.223.22=1

As the simplest ratio of Carbon is 1, Hydrogen is 3 and Oxygen is 1, therefore the empirical formula comes out to be C1H3O1 or CH3O.

Hence, option (C) is correct.


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