Question

A compound of carbon, hydrogen and nitrogen contaians three element in the ratio 9:1:3.5 calculate molecular formula of the compound vepour density is 54​

Answer 1

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In a certain compound, the ratio of C:H:N=9:1:3.5

Molar ratio of C:H:N =

$12</p><p>9</p><p> </p><p> : </p><p>1</p><p>1</p><p> </p><p> : </p><p>14</p><p>3.5</p><p> </p><p> =0.75:1:10.25$

Integer ratio =

$0.25</p><p>0.75</p><p> </p><p> : </p><p>0.25</p><p>1</p><p> </p><p> : </p><p>0.25</p><p>0.25</p><p> </p><p> =3:4:1</p><p>$

Therefore, the empirical formula is

$C </p><p>3</p><p> </p><p> H </p><p>4</p><p> </p><p> N.$

Empirical weight =12×3+1×4+14=54 g.

Given the molecular weight is 108 g.

$</p><p></p><p>n= </p><p>Mw/EFw</p><p> </p><p> = </p><p>108/54 </p><p> </p><p> =2</p><p>$

Molecular formula =

$C6H8N2</p><p></p><p></p><p>$