In an organic compound of molar mass 108 g mol -1 C , H and N atoms are present in 9: 1: 3.5 by weight. Molecular formula can be :

Question

In an organic compound of molar mass 108 g mol -1 C , H and N atoms are present in 9: 1: 3.5 by weight. Molecular formula can be : (A) C 6 H 8 N 2 (B) C 7 H 10 N (C) C 5 H 6 N 3 (D) C 4 H 18 N 3

Tardigrade Admin · Accepted Answer

Molar mass 108 g Total part by weight =9+1+3.5=13.5 Weight of carbon =(9/13.5) × 108=72 g Number of carbon atoms =(72/12)=6 Weight of hydrogen =(1/13.5) × 108=8 g Number of hydrogen atoms =(8/1)=8 Weight of nitrogen =(3.5/13.5) × 108=28 g Number of nitrogen atom =(28/14)=2 Hence, molecular formula = C 6 H 8 N 2

Q. In an organic compound of molar mass $108 g$ $m o l^{- 1} C, H$ and $N$ atoms are present in 9: 1: 3.5 by weight. Molecular formula can be :

$C_{6} H_{8} N_{2}$

$C_{7} H_{10} N$

$C_{5} H_{6} N_{3}$

$C_{4} H_{18} N_{3}$

Q. In an organic compound of molar mass 108g mol−1C,H and N atoms are present in 9: 1: 3.5 by weight. Molecular formula can be :

C6​H8​N2​

C7​H10​N

C5​H6​N3​

C4​H18​N3​

Q. In an organic compound of molar mass $108 g$ $m o l^{- 1} C, H$ and $N$ atoms are present in 9: 1: 3.5 by weight. Molecular formula can be :

$C_{6} H_{8} N_{2}$

$C_{7} H_{10} N$

$C_{5} H_{6} N_{3}$

$C_{4} H_{18} N_{3}$