Question

( % ) of ( c=x, n=y quad & quad N=2 )
( frac{x}{4}=6, x+4=1: 52, x=6 y )
- (ii i
from
(i) 4
[
begin{array}{c}
6 y+4=1 cdot 5 z
7 y=1 cdot 5 z
14 y=3 z
frac{y}{y}=frac{14}{3}
end{array}
]
( frac{x}{y}: frac{y}{7} ; frac{2}{7}=frac{5}{1}: 1 ; frac{14}{7} )
[
Rightarrow 18: 3: 14
]
mole ratio ( n ) o ( C ). ( mathrm{H} ); N ( frac{18}{12}: frac{3}{1}: frac{14}{14} )
( 1 cdot 5: 3: 1 )
Simplest orido: ( sqrt{3}=6 quad therefore 2 )
Empirical bormana ig ( angle 3 mathrm{H}_{6} mathrm{N}_{2} )
rorss of ( mathrm{c}_{3} mathrm{H}_{6} mathrm{N}_{2}=70 )
given mol. Wh ( >107 )
( therefore ) least value in 0 in ill be 2
therefore least mol wit will be
[
70 times 2=190
]

# In an organic compound of molar mass greater than 100 containing only C. H and N. the percentage of C is 6 times the percentage of H while the sum of the percentages of Cand His 1.5 times the percentage of N. What is the least molar mass?

Solution