   Chapter 2, Problem 92PS

Chapter
Section
Textbook Problem

Nicotine, a poisonous compound found in tobacco leaves, is 74.0% C, 8.65% H, and 17.35% N. Its molar mass is 162 g/mol. What are the empirical and molecular formulas of nicotine?

Interpretation Introduction

Interpretation:

The empirical formula and molecular formula of the nicotine is to be determined if the molar mass is 162g/mol.

Concept introduction:

• Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.
• Equation for finding Molecular formula from the empirical formula,

MolarmassEmpiricalformula mass × Empirical formula

• Equation for number moles from mass and molar mass,

Numberofmoles=MassingramsMolarmass

• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
Explanation

Mass percent of C, H and N in nicotine is given as 74%, 8.65% and 17.35%.

Mass percent of an element means, 100g of compound contains that mass percent of an element in grams. Therefore, mass of C, H and N are 74g, 8.65g and 17.35g respectively in 100g of nicotine.

Equation for number moles from mass and molar mass is,

Numberofmoles=MassingramsMolarmass

Therefore,

The number of moles of carbon is,

Numberofmoles=74g12.01g/mol=6.16mol

The number of moles of hydrogen is,

Numberofmoles=8.65g1.008g/mol=8.58mol

The number of moles of nitrogen is,

Numberofmoles=17

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