A compound contains 38.8 % C, 16 % H and 45.2 % N. The empirical formula of the compound will be:

Question

A compound contains 38.8 % C, 16 % H and 45.2 % N. The empirical formula of the compound will be: (A)  CH5N  (B)  C2H5N  (C)  CHN  (D)  CH10N2

Tardigrade Admin · Accepted Answer

Element % Relative no. of atoms Simple ratio C 38.8 38.8/12=3.2 3.2/3.2=1 H 16.0 16/1=16.0 16.0/3.2=5 N 45.2 45.2/14=3.2 3.2/3.2=1 ∴ Empirical formula = CH 5 N

Q. A compound contains $38.8% C, 16%$ H and $45.2% N$ . The empirical formula of the compound will be:

$C H_{5} N$

$C_{2} H_{5} N$

$C H N$

$C H_{10} N_{2}$

Element % Relative no. of atoms Simple ratio

C 38.8 38.8/12=3.2 3.2/3.2=1

H 16.0 16/1=16.0 16.0/3.2=5

N 45.2 45.2/14=3.2 3.2/3.2=1

$∴$ Empirical formula $= C H_{5} N$

Element	%	Relative no. of atoms	Simple ratio
C	38.8	38.8/12=3.2	3.2/3.2=1
H	16.0	16/1=16.0	16.0/3.2=5
N	45.2	45.2/14=3.2	3.2/3.2=1

Q. A compound contains 38.8%C,16% H and 45.2%N. The empirical formula of the compound will be:

CH5​N

C2​H5​N

CHN

CH10​N2​