Chapter 13, Problem 13.55QP

Interpretation Introduction

Interpretation:

The molecular formula and molar mass of compound Pheromones with given mass and percent composition values should be calculated.

Concept Introduction:

Molecular Formula: It is the simplest representation of composition of chemical substance where each and every type of atom is written with the total number of the atom present in the substance.

Molar mass: It is obtained by dividing the mass of substance with the amount of substance and the S.I. unit of molar mass is $\text{kg/mol}$. It is numerically equal to the molecular weight since molecular weight is sum of all individual atom weight present in molecule which is represented in $amu$.

$\text{Molar}\text{mass}\left(\text{g/mol}\right)=\frac{\text{Given}\text{mass}\text{of}\text{substance}\left(\text{g}\right)}{\text{Moles}\text{of}\text{substance}\left(mol\right)}$

Freezing point is the temperature at which liquid turns into solid.

The Freezing point depression ${\text{(ΔT}}_{\text{f}}\text{)}$ is indicated as follows,

${\text{ΔT}}_{\text{f}}{\text{= K}}_{\text{f}}\text{m}$

Where,

${\text{ΔT}}_{\text{f}}-$ Change in freezing point

${\text{K}}_{\text{f}}-$ Molal freezing point constant

$m-$ Molality of the solution