Chapter 7, Problem 7.5VP

Interpretation Introduction

To find:

Which two hydrocarbons with the molecular structures shown in Figure P7.5 have the same percent composition?

Answer to Problem 7.5VP

Solution:

Molecules (b) $C_2{H}_4$ and (d) $C_6{H}_{12}$ have same percent composition.

Explanation of Solution

1) Formula:

Percent composition is the percent by mass of an element in a compound:

$Percent composition of element= \frac{Mass of element}{Molecular mass}*100$

2) Calculations:

In the given Figure P7.5 black sphere represents $C$ atoms and white sphere represents $H$ atoms.

a) Molecule (a) contains six $C$ atoms and six $H$ atoms i.e. molecular formula is $C_6{H}_6$

$Mass of 6 atoms C=6 atoms C\times \frac{12.01 amu}{1 atom C}=72.06 amu$

$Mass of 6 atoms H=6 atoms H\times H= 6.048 amu$

$Molecular mass of C_6{H}_6=72.06 amu+6.048 amu= 78.108 amu.$

$Percent composition of C= \frac{Mass of C}{Molecular mass }*100$

$= \frac{72.06}{78.108}*100=92.26\%$

$Percent composition of H= \frac{Mass of H}{Molecular mass }*100$

$= \frac{6.048}{78.108}*100=7.74\%$

b) Molecule (b) contains two $C$ atoms and four $H$ atoms i.e. molecular formula is $C_2{H}_4$

$Mass of 2 atoms C=2 atoms C\times \frac{12.01 amu}{1 atom C}=24.02 amu$

$Mass of 4 atoms H=4 atoms H\times \frac{1.008 amu}{1 atom H}= 4.032 amu$

$Molecular mass of C_2{H}_4=24.02 amu+4.032 amu=28.052 amu.$

$Percent composition of C= \frac{24.02}{28.052}*100=85.63\%$

$Percent composition of H= \frac{4.032 }{28.052}*100=14.37\%$

c) Molecule (c) contains three $C$ atoms and eight $H$ atoms i.e. molecular formula is $C_3{H}_8$

$Mass of 3 atoms C=3 atoms C\times \frac{12.01 amu}{1 atom C}=36.03 amu$

$Mass of 8 atoms H=8 atoms H\times \frac{1.008 amu}{1 atom H}=8.064 amu$

$Molecular mass of C_3{H}_8=8.064 amu+36.03 amu=44.094 amu.$

$Percent composition of C= \frac{36.03 }{44.094}*100=\%$

$Percent composition of H= \frac{8.064}{44.094}*100=18.29\%$

d) Molecule (d) contains six $C$ atoms and twelve $H$ atoms i.e. molecular formula is $C_6{H}_{12}$

$Mass of 6 atoms C=6 atoms C\times \frac{12.01 amu}{1 atom C}=72.06 amu$

$Mass of 12 atoms H=12 atoms H\times \frac{1.008 amu}{1 atom H}= 12.096 amu$

$Molecular mass of C_6{H}_{12}=72.06 amu+12.096 amu= 84.156 amu.$

$Percent composition of C= \frac{72.06}{84.156}*100=85.63\%$

$Percent composition of H= \frac{12.096}{84.156}*100=14.37\%$

Therefore, molecule (b) and (d) have same percent composition.

Conclusion:

First, we calculate the mass of the number of atoms of C and H, and then calculate the percent composition of C and H using molecular mass of compound.