Question

0.265g of an organic compound produced on evaporation 102cm cube of vapour at 373K and 775mmHg. Percentage composition of the constituent elements are 92.24% C and 7.76% H. Find the molecular mass and molecular formula of the composition.​

Answer 1

Using PV=nRT

• since
• 
  `n. \: no \: of \: moles= (mass)/(molar \: mass)`
• PV= nRT

P=775mmhg/760mmhg

P=1.01atm

so,n=PV/RT

n= (1.01*0.102)/(0.0821*373)

n= 0.00339mole

n =mass/ molecular mass

molecular mass =0.265/0.00339

molecular mass= 78.17

empirical formular = C. H

C= 92.24. 92.24/12= 7.688

H= 7.76. 7.76/1=7.76

therefore Empirical formula is CH

(CH)n=78.16

(12+1)n=78.14

n=78.14/13

n= 6.01

therefore, molecular formula is

C6H6

Answer 2

No results found for 0.265g of an organic compound produced on evaporation 102cm cube of vapour at 373K and 775mmHg. Percentage composition of the constituent elements are 92.24% C and 7.76% H. Find the molecular mass and molecular formula of the composition..

Results for 0.265g of an organic compound produced on evaporation 102cm cube of vapour at 373K and 775mmHg Percentage composition of the constituent elements are 92.24 C and 7.76 H Find the molecular mass and molecular formula of the composition