   Chapter 13, Problem 91IL

Chapter
Section
Textbook Problem

An organic compound contains carbon (71.17%), hydrogen (5.12%) with the remainder nitrogen. Dissolving 0.177 g of the compound in 10.0 g of benzene gives a solution with a vapor pressure of 94.16 mm Hg at 25 °C. (The vapor pressure of pure benzene at this temperature is 95.26 mm Hg.) What is the molecular formula for the compound?

Interpretation Introduction

Interpretation Molecular formula of the compound has to be determined.

Concept introduction:

Raoult’s law: In a solution, vapor pressure of solvent is proportional to its mole fraction.

Psolvent=XsolventP0solvent

where,

P0solvent is the vapor pressure of pure solvent.

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture ( nA +  nB +  n...).

Mole fraction of A (χA)= nA nA +  nB +  n...

The number of moles of any substance can be determined using the equation

Numberofmole=GivenmassofthesubstanceMolarmass

Explanation

Molecular formula of the compound is calculated.

Given,

Molar mass of carbon is 12.01g/mol

Molar mass of hydrogen is 1.01g/mol

Molar mass of nitrogen is 14.01g/mol

Vapour pressure of benzene at 250C is Pbenzene=94.16mmHg

Vapour pressure of pure benzene is P0benzene=95.26mmHg

Mass of the compound is 0.177g

Mass of benzene is 10.0g

Molar mass of benzene is 78.11g/mol

The number of moles of any substance can be determined using the equation

Numberofmole=GivenmassofthesubstanceMolarmass

Number of molesofbenzene=10g78.11g/mol=0.128mol

Number of moles of benzene is 0.128mol

Vapour pressure of benzene is calculated by the given equation

Pbenzene=XbenzeneP0benzene

Mole fraction of benzene is,

Xbenzene=PbenzeneP0benzene=94.16mmHg95.26mmHg=0.988

Hence,

Xbenzene=nbenzenenbenzene+ncompound0.988=0.128mol0.128mol+ncompound0.12650.128mol+(0.988)ncompound=0.128molncompound=1.518×103mol

Number of moles of the compound is 1.518×103mol

Hence,

The molar mass of compound is,

Molarmass=0

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