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)  C6H8N2  (B)  C7H10N  (C)  C5H6N3  (D)  C4H18N3

Tardigrade Admin · Accepted Answer

Molar mass =108 text g text mol-1 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=8g Number of hydrogen atoms =(8/1)=8 Weight of nitrogen =(3.5/13.5)× 108=28g Number of nitrogen atom =(28/14)=2 Hence, molecular formula =C6H8N2

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 108 g mol−1,C,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

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

$C_{7} H_{10} N$

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

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