If an organic compound contain 92.3 % C and 7.7 % H, than number of sp3,sp2 and sp hybridized carbon atoms in all possible structures of compound respectively are (molecular mass = 52 g/mol)

Question

If an organic compound contain 92.3 % C and 7.7 % H, than number of sp3,sp2 and sp hybridized carbon atoms in all possible structures of compound respectively are (molecular mass = 52 g/mol) (A) 1, 2, 5 (B) 0, 4, 4 (C) 0, 8, 4 (D) None of these

Tardigrade Admin · Accepted Answer

Let amount of compound =100 g No. of moles of C =(92.3/12)=7.69=7.7 No. of moles of H =(7.7/1)=7.7 Empirical formula = CH Empirical formula mass =12+1=13 g / mol Molecular mass =52 g / mol n=(52/13)=4 ∴ Molecular foumula = Empirical formula × 4= C 4 H 4 Possible structures

Q. If an organic compound contain $92.3% C$ and $7.7% H$ , than number of $s p^{3}, s p^{2}$ and $s p$ hybridized carbon atoms in all possible structures of compound respectively are (molecular mass $= 52 g / m o l)$

$1, 2, 5$

$0, 4, 4$

$0, 8, 4$

None of these

Q. If an organic compound contain 92.3%C and 7.7%H, than number of sp3,sp2 and sp hybridized carbon atoms in all possible structures of compound respectively are (molecular mass =52g/mol)

1,2,5

0,4,4

0,8,4

None of these

Let amount of compound =100g No. of moles of C=1292.3​=7.69=7.7 No. of moles of H=17.7​=7.7 Empirical formula =CH Empirical formula mass =12+1=13g/mol Molecular mass =52g/mol n=1352​=4 ∴ Molecular foumula = Empirical formula ×4=C4​H4​ Possible structures

Q. If an organic compound contain $92.3% C$ and $7.7% H$ , than number of $s p^{3}, s p^{2}$ and $s p$ hybridized carbon atoms in all possible structures of compound respectively are (molecular mass $= 52 g / m o l)$

$1, 2, 5$

$0, 4, 4$

$0, 8, 4$