An organic compound has a composition …

CBSE, JEE, NEET, CUET

Question Bank, Mock Tests, Exam Papers

NCERT Solutions, Sample Papers, Notes, Videos

An organic compound has a composition of compound as carbon is equals to40.6 8% and hydrogen is equals to 5.085 and oxygen is equals to 50 4.22 percent the vapour density of a compound is 59 calculate the molecular formula of a compound

Posted by Tript Preet???? 6 years, 5 months ago

CBSE > Class 11 > Chemistry

1 answers

Sia ? 6 years, 5 months ago

Given Percentage of C= 40.687% , Percentage of H = 5.085%.

Therefore, Percentage of O = 100 - (40.687+5.085) = 54.228%.
Step I: To calculate the empirical formula of the compound.

<th scope="col">Element</th> <th scope="col">Symbol</th> <th scope="col">Percentage of element</th> <th scope="col">Moles of the element

Carbon	C	40.687	{tex}\frac { 40.687 } { 12 } = 3.390{/tex}	{tex}\frac { 3.390 } { 3.389 } = 1{/tex}	2
Hydrogen	H	5.085	{tex}\frac { 5.0885 } { 1 } = 5.085{/tex}	{tex}\frac { 5.085 } { 3.389 } = 1.5{/tex}	3
Oxygen	O	16	{tex}\frac { 54.228 } { 16 } = 3.389{/tex}	{tex}\frac { 3.389 } { 3.389 } = 1{/tex}	2

Since, ration of C : H : O = 2 :3 :2.

{tex}\therefore{/tex} An empirical formula is C₂H₃O₂.
Step II: The empirical formula of the compound = C₂H₃O₂.
{tex}\therefore{/tex} Empirical formula mass = 2 {tex}\times{/tex} C +3 {tex}\times{/tex} H + 2{tex}\times{/tex} O = {tex}( 2 \times 12 ) + ( 3 \times 1 ) + ( 2 \times 16 ) = 59{/tex}
Step III: To calculate the molecular mass of the salt
The vapour density of the compound = 59 (Given)
Using the relation between vapour density and molecular mass.

Therefore, Molecular mass of compound = 2 {tex}\times{/tex} vapour density of compound = 2 {tex}\times{/tex} 59 = 118
Step IV: The value of n = {tex}\frac { \text { molecular mass } } { \text { empirical formula mass } } = \frac { 118 } { 59 } = 2{/tex}
Step V: Calculation of the molecular formula of the salt,
Molecular formula = n {tex}\times{/tex} empirical formula = {tex}2 \times \mathrm { C } _ { 2 } \mathrm { H } _ { 3 } \mathrm { O } _ { 2 } = \mathrm { C } _ { 4 } \mathrm { H } _ { 6 } \mathrm { O } _ { 4 }{/tex}
Thus, the molecular formula is C₄H₆O₄.