Chapter 2, Problem 135GQ

### 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

# The action of bacteria on meat and fish produces a compound called cadaverine. As its name and origin imply, it stinks! (It is also present in bad breath and adds to the odor of urine.) It is 58.77% C, 13.81% H, and 27.40% N. Its molar mass is 102.2 g/mol. Determine the molecular formula of cadaverine.

Interpretation Introduction

Interpretation: The empirical and molecular formulas for the given compound should be determined.

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

Given

Mass percent of carbon, hydrogen and nitrogen in the given compound cadaverine are 58.77â€‰%â€‰,â€‰13.81%â€‰â€‰andâ€‰â€‰27.40â€‰â€‰% respectively.

Mass percent of an element means, 100g of compound contains that mass percent of an element in grams. Therefore, mass of 58.77â€‰gâ€‰ carbon 13.81â€‰g of hydrogen and 27.40â€‰g of nitrogen are present respectively in 100g of cadaverine.

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€‰58.77â€‰â€‰g12.01g/molâ€‰=â€‰â€‰4.893â€‰â€‰mol

The number of moles of hydrogen is,

Â Â Â Â Numberâ€‰ofâ€‰molesâ€‰â€‰=â€‰â€‰â€‰â€‰13.81â€‰â€‰g1â€‰â€‰g/molâ€‰=â€‰â€‰13.81mol

The number of moles of nitrogen is,

Â Â Â Â Numberâ€‰ofâ€‰molesâ€‰â€‰=â€‰â€‰â€‰â€‰27.40â€‰â€‰g14â€‰g/molâ€‰=â€‰â€‰1.957â€‰mol

So, the mole ratio between elements in cadaverine is,

Â Â Â Â C:â€‰â€‰Hâ€‰â€‰:â€‰â€‰N =â€‰â€‰â€‰4

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started