Question

Question #a7489

Answer 1

Percentage Composition of constituent elements in a compound

`C->74.0%`

`H->7.4%`

`N->8.6%`

`O->10.0%`

Considering their atomic masses as

`C->12" g/mol"`

`H->1" g/mol"`

`N->14" g/mol"`

`O->16" g/mol"`

Now dividing each percentage composition by the respective atomic mass and then taking the ratio we get the ratio of number of atoms of constituent elements in the compound as

`C:H:N:O=74/12:7.4/1:8.6/14:10/16`

`=6.16:7.4:0.614:0.625`

`=6.16/0.614:7.4/0.614:0.614/0.614:0.625/0.614`

`=10:12:1:1`

Hence the empirical formula is

`C_10H_12NO`