Chapter 2, Problem 92PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
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,Â HÂ andÂ 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,Â HÂ andÂ N are 74g,Â 8.65gÂ andÂ 17.35g respectively in 100g of nicotine.

Equation for number moles from mass and molar mass is,

â€‚Â Numberâ€‰ofâ€‰molesâ€‰â€‰=â€‰â€‰Massâ€‰inâ€‰gramsâ€‰Molarâ€‰mass

Therefore,

The number of moles of carbon is,

Â Â Â Â Numberâ€‰ofâ€‰molesâ€‰â€‰=â€‰â€‰74â€‰g12.01g/molâ€‰=â€‰6.16â€‰mol

The number of moles of hydrogen is,

Â Â Â Â Numberâ€‰ofâ€‰molesâ€‰â€‰=â€‰â€‰8.65â€‰g1.008g/molâ€‰=â€‰8.58â€‰mol

The number of moles of nitrogen is,

Â Â Â Â Numberâ€‰ofâ€‰molesâ€‰â€‰=â€‰â€‰17

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