CHEMISTRY:PRIN.+REACTIONS-OWLV2 ACCESS
CHEMISTRY:PRIN.+REACTIONS-OWLV2 ACCESS
8th Edition
ISBN: 9781305079298
Author: Masterton
Publisher: Cengage Learning
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Chapter 3, Problem 13QAP

Complete the following table for TNT (trinitrotoluene), C7H5(NO2)3.

Chapter 3, Problem 13QAP, Complete the following table for TNT (trinitrotoluene), C7H5(NO2)3.

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The given table should be completed.

Concept introduction:

The number of moles of a substance is related to mass and molar mass as follows:

n=mM

Here, m is mass and M is molar mass of the substance.

Also, according to Avogadro’s law in 1 mol of a substance there are 6.023×1023 units of that substance. Here, 6.023×1023 is known as Avogadro’s number and denoted by symbol NA.

The density of a substance is related to mass and volume as follows:

d=mV

Here, m is mass and V is volume.

Answer to Problem 13QAP

Number of grams Number of moles Number of molecules Number of N atoms
(a)    127.2 g    0.56 mol    3.375×1023    9.0345×1023
(b)    210.1 g    0.9254 mol    5.57×1023    1.67×1024
(c)    4.67×106 g    2.06×104 mol    1.24×1028    3.72×1028
(d)    9.42 g    0.0415 mol    2.5×1022    7.5×1022

Explanation of Solution

The given compound is TNT (trinitrotoluene) with molecular formula C7H5(NO2)3. The molar mass of compound can be calculated as follows:

MC7H5(NO2)3=7MC+5MH+3MN+6MO

The molar mass of carbon, hydrogen, nitrogen and oxygen is 12 g/mol, 1 g/mol, 14 g/mol and 16 g/mol respectively.

Putting the values,

MC7H5(NO2)3=7(12 g/mol)+5(1 g/mol)+3(14 g/mol)+6(16 g/mol)=(84+5+42+96)g/mol=227 g/mol

Step (a)

The mass of TNT is 127.2 g. The number of moles can be calculated as follows:

n=mM

Putting the values,

n=127.2 g227 g/mol=0.56 mol

Since, according to Avogadro’s law in 1 mol of a substance there are 6.023×1023 units of that substance.

Thus, number of molecules in 0.56 mol of TNT will be:

N=0.56 mol(6.023×10231 mol)=3.375×1023

Thus, number of molecules of TNT is 3.375×1023.

Now, the molecular formula of TNT is C7H5(NO2)3 thus, 1 mol TNT contains 3 mol of nitrogen. Thus, in 0.56 mol of TNT, number of moles of N will be:

n=0.5×3=1.5mol

Thus, number of N atoms will be:

NN=1.5 mol(6.023×10231 mol)=9.0345×1023

Therefore, number of N atoms is 9.0345×1023.

Step (b)

The number of moles of TNT is 0.9254 mol and molar mass of TNT is 227 g/mol thus, mass can be calculated as follows:

m=n×M

Putting the values,

m=(0.9254 mol)(227 g/mol)=210.1 g

The number of molecules of TNT can be calculated as follows:

N=(0.9254 mol)(6.023×10231 mol)=5.57×1023

Now, in 1 mol there are 3 nitrogen atoms. Thus, the number of N atoms will be 3 times the number of molecules of TNT.

NN=3×5.57×1023=1.67×1024

Step (c)

The number of molecules of TNT is 1.24×1028. Since, from the molecular formula 1 mol of TNT contains 3 N atoms.

Thus, number of N atoms in 1.24×1028 molecules of TNT will be:

NN=3×1.24×1028=3.72×1028

According to Avogadro’s law, in mol there are 6.023×1023 thus, number of moles from 1.24×1028 number of molecules can be calculated as follows:

n=1.24×1028(1 mol6.023×1023)=2.06×104 mol

Since, molar mass of TNT is 227 g/mol thus, mass can be calculated as follows:

m=n×M=(2.06×104 mol)(227 g/mol)=4.67×106 g

Step (d)

The number of N atoms is 7.5×1022.

Since, the number of N atoms is 3 times the number of TNT molecule. Thus, number of molecules of TNT will be:

N=7.5×10223=2.5×1022

According to Avogadro’s law, in mol there are 6.023×1023 thus, number of moles from 2.5×1022 number of molecules can be calculated as follows:

n=2.5×1022(1 mol6.023×1023)=0.0415 mol

Since, molar mass of TNT is 227 g/mol thus, mass can be calculated as follows:

m=n×M=(0.0415 mol)(227 g/mol)=9.42 g

Conclusion

Therefore, the complete table will be as follows:

Number of grams Number of moles Number of molecules Number of N atoms
(a)    127.2 g    0.56 mol    3.375×1023    9.0345×1023
(b)    210.1 g    0.9254 mol    5.57×1023    1.67×1024
(c)    4.67×106 g    2.06×104 mol    1.24×1028    3.72×1028
(d)    9.42 g    0.0415 mol    2.5×1022    7.5×1022

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Chapter 3 Solutions

CHEMISTRY:PRIN.+REACTIONS-OWLV2 ACCESS

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