An organic compound gave the following results C=53.3%, H=15.6 %,N=31.1 %, Molecular weight=45. What is molecular formula of the compound?

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Solution

Given:
The molecular weight of compound=45 u.
Percentage composition is C=53.3 %, H=15.6 % and N=31.1 %
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 Element Atomic mass (u) Atomic ratio Simplest ratio
Carbon (C) 53.3 12 $\frac{53.3}{12} = 4.44$ $\frac{4.44}{2.22} = 2$
Hydrogen (H) 15.6 1 $\frac{15.6}{1} = 15.6$ $\frac{15.6}{2.22} = 7.02$
Nitrogen (N) 31.1 14 $\frac{31.1}{14} = 2.22$ $\frac{2.22}{2.22} = 1$
As the simplest ratio of Carbon is 2, Hydrogen is 7 and Nitrogen is 1, therefore the empirical formula comes out to be C₂H₇N_.
Step-2: Calculate the empirical and molecular mass
Empirical formula weight: Empirical formula weight is obtained by the addition of the atomic weight of the various atoms present in the empirical formula.
The atomic weight of Carbon= 12 u
The atomic weight of Hydrogen= 1 u
The atomic weight of Nitrogen= 14 u
Empirical formula weight= 2(Gram atomic weight of Carbon)+ 7(Gram atomic weight of Hydrogen)+ Gram atomic weight of Nitrogen
Empirical formula weight= 2(12) +7(1) +14= 45 u
Molecular formula weight: Molecular formula weight is obtained by the addition of the atomic weight of the various atoms present in the molecular formula.
Molecular weight =45 (given)
Step-3: Calculate the molecular formula
Molecular formula: Molecular formula is the actual formula of the compound which gives the actual number of atoms present in the molecule of the compound. For example, the molecular formula of glucose is C₆H₁₂O₆.
The relationship between the empirical formula and the molecular formula is given as follows: Molecular formula = n× Empirical formula, where n is an integer such as 1,2,3…..etc.
$n = \frac{M o l e c u l a r w e i g h t o f t h e c o m p o u n d}{E m p i r i c a l f o r m u l a m a s s}$
$n = \frac{45}{45} = 1$
As stated above, the Molecular formula = n× Empirical formula and the empirical formula is C₂H₇N
Therefore, the molecular formula is (C₂H₇N)₁ or C₂H₇N.

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Element	% Of Element	Atomic mass (u)	Atomic ratio	Simplest ratio
Carbon (C)	53.3	12	$\frac{53.3}{12} = 4.44$	$\frac{4.44}{2.22} = 2$
Hydrogen (H)	15.6	1	$\frac{15.6}{1} = 15.6$	$\frac{15.6}{2.22} = 7.02$
Nitrogen (N)	31.1	14	$\frac{31.1}{14} = 2.22$	$\frac{2.22}{2.22} = 1$