Chapter 4, Problem 143QP

(a)

Interpretation Introduction

Interpretation:

Whether ${\text{C}}_3{\text{H}}_4$ and ${\text{C}}_3{\text{H}}_6$ have identical percent compositions of carbon by mass and the compound having a low percentage composition by mass is to be indicated.

Explanation of Solution

${\text{C}}_3{\text{H}}_4$ contains three moles of carbon and four moles of hydrogen. The mass of three moles of carbon is calculated as follows:

$\begin{array}{c}\text{Mass of 3 mol of C}=3\times 12.01\text{ g}\\ =\text{36}\text{.03 g}\end{array}$

The mass of four moles of carbon is calculated as follows:

$\begin{array}{c}\text{Mass of 4 mol of H}=4\times 1.008\text{ g}\\ =4.032\text{ g}\end{array}$

Thus, the total mass of ${\text{C}}_3{\text{H}}_4$ is $40.062\text{ g}$ .

The percentage composition of carbon in ${\text{C}}_3{\text{H}}_4$ is calculated as follows:

$\begin{array}{c}\%\text{C}=\frac{36.03\text{ g}}{40.062\text{ g}}\times 100\\ =89.93\%\end{array}$

${\text{C}}_3{\text{H}}_6$ contains three moles of carbon and six moles of hydrogen. The mass of three moles of carbon is calculated as follows:

$\begin{array}{c}\text{Mass of 3 mol of C}=3\times 12.01\text{ g}\\ =\text{36}\text{.03 g}\end{array}$

The mass of six moles of carbon is calculated as follows:

$\begin{array}{c}\text{Mass of 6 mol of H}=6\times 1.008\text{ g}\\ =6.048\text{ g}\end{array}$

Thus, the total mass of ${\text{C}}_3{\text{H}}_6$ is $42.078\text{ g}$ .

The percentage composition of carbon in ${\text{C}}_3{\text{H}}_6$ is calculated as follows:

$\begin{array}{c}\%\text{C}=\frac{36.03\text{ g}}{42.078\text{ g}}\times 100\\ =85.62\%\end{array}$

Thus, the percentage composition of carbon in ${\text{C}}_3{\text{H}}_6$ is lower.

(b)

Interpretation Introduction

Interpretation:

Whether ${\text{C}}_2{\text{H}}_4$ and ${\text{C}}_3{\text{H}}_4$ have identical percent compositions of carbon by mass and the compound having a low percentage composition by mass is to be indicated.

Explanation of Solution

${\text{C}}_2{\text{H}}_4$ contains two moles of carbon and four moles of hydrogen. The mass of three moles of carbon is calculated as follows:

$\begin{array}{c}\text{Mass of 2 mol of C}=2\times 12.01\text{ g}\\ =\text{24}\text{.02 g}\end{array}$

The mass of four moles of hydrogen is calculated as follows:

$\begin{array}{c}\text{Mass of 4 mol of H}=4\times 1.008\text{ g}\\ =4.032\text{ g}\end{array}$

Thus, the total mass of ${\text{C}}_2{\text{H}}_4$ is $28.052\text{ g}$ .

The percentage composition of carbon in ${\text{C}}_3{\text{H}}_4$ is calculated as follows:

$\begin{array}{c}\%\text{C}=\frac{28.052\text{ g}}{40.062\text{ g}}\times 100\\ =70.02\%\end{array}$

${\text{C}}_3{\text{H}}_4$ contains three moles of carbon and four moles of hydrogen. The mass of three moles of carbon is calculated as follows:

$\begin{array}{c}\text{Mass of 3 mol of C}=3\times 12.01\text{ g}\\ =\text{36}\text{.03 g}\end{array}$

The mass of four moles of carbon is calculated as follows:

$\begin{array}{c}\text{Mass of 4 mol of H}=4\times 1.008\text{ g}\\ =4.032\text{ g}\end{array}$

Thus, the total mass of ${\text{C}}_3{\text{H}}_4$ is $40.062\text{ g}$ .

The percentage composition of carbon in ${\text{C}}_3{\text{H}}_4$ is calculated as follows:

$\begin{array}{c}\%\text{C}=\frac{36.03\text{ g}}{40.062\text{ g}}\times 100\\ =89.93\%\end{array}$

Therefore, ${\text{C}}_2{\text{H}}_4$ has lower percentage composition of carbon.

(c)

Interpretation Introduction

Interpretation:

Whether ${\text{C}}_2{\text{H}}_4$ and ${\text{C}}_4{\text{H}}_2$ have identical percent compositions of carbon by mass and the compound having a low percentage composition by mass is to be indicated.

Explanation of Solution

${\text{C}}_2{\text{H}}_4$ contains two moles of carbon and four moles of hydrogen. The mass of three moles of carbon is calculated as follows:

$\begin{array}{c}\text{Mass of 2 mol of C}=2\times 12.01\text{ g}\\ =\text{24}\text{.02 g}\end{array}$

The mass of four moles of hydrogen is calculated as follows:

$\begin{array}{c}\text{Mass of 4 mol of H}=4\times 1.008\text{ g}\\ =4.032\text{ g}\end{array}$

Thus, the total mass of ${\text{C}}_2{\text{H}}_4$ is $28.052\text{ g}$ .

The percentage composition of carbon in ${\text{C}}_2{\text{H}}_4$ is calculated as follows:

$\begin{array}{c}\%\text{C}=\frac{28.052\text{ g}}{40.062\text{ g}}\times 100\\ =70.02\%\end{array}$

${\text{C}}_4{\text{H}}_2$ contains four moles of carbon and two moles of hydrogen. The mass of four moles of carbon is calculated as follows:

$\begin{array}{c}\text{Mass of 4 mol of C}=4\times 12.01\text{ g}\\ =\text{48}\text{.04 g}\end{array}$

The mass of four moles of carbon is calculated as follows:

$\begin{array}{c}\text{Mass of 2 mol of H}=2\times 1.008\text{ g}\\ =2.016\text{ g}\end{array}$

Thus, the total mass of ${\text{C}}_4{\text{H}}_2$ is $50.056\text{ g}$ .

The percentage composition of carbon in ${\text{C}}_4{\text{H}}_2$ is calculated as follows:

$\begin{array}{c}\%\text{C}=\frac{48.04\text{ g}}{50.056\text{ g}}\times 100\\ =95.97\%\end{array}$

Therefore, ${\text{C}}_2{\text{H}}_4$ has lower percentage composition of carbon.

(d)

Interpretation Introduction

Interpretation:

Whether ${\text{C}}_2{\text{H}}_4$ and ${\text{C}}_3{\text{H}}_6$ have identical percent compositions of carbon by mass and the compound having a low percentage composition by mass is to be indicated.

Explanation of Solution

The empirical formula of ${\text{C}}_2{\text{H}}_4$ and ${\text{C}}_3{\text{H}}_6$ is the same, that is ${\text{CH}}_2$ . So, the percentage composition of ${\text{C}}_2{\text{H}}_4$ and ${\text{C}}_3{\text{H}}_6$ is identical.

(e)

Interpretation Introduction

Interpretation:

Whether ${\text{C}}_4{\text{H}}_8$ and ${\text{C}}_3{\text{H}}_8$ have identical percent compositions of carbon by mass and the compound having a low percentage composition by mass is to be indicated.

Explanation of Solution

${\text{C}}_4{\text{H}}_8$ contains four moles of carbon and eight moles of hydrogen. The mass of four moles of carbon is calculated as follows:

$\begin{array}{c}\text{Mass of 4 mol of C}=4\text{ mol}\times \text{12}\text{.01 g}\\ =\text{48}\text{.04 g}\end{array}$

The mass of eight moles of hydrogen is calculated as follows:

$\begin{array}{c}\text{Mass of 8 mol of H}=8\text{ mol}\times \text{1}\text{.008 g}\\ =\text{8}\text{.064 g}\end{array}$

Thus, the total mass of ${\text{C}}_4{\text{H}}_8$ is $56.104\text{ g}$ .

The percentage composition of carbon in ${\text{C}}_4{\text{H}}_8$ is calculated as follows:

$\begin{array}{c}\%\text{C}=\frac{48.04\text{ g}}{56.104\text{ g}}\times 100\\ =85.62\%\end{array}$

${\text{C}}_3{\text{H}}_8$ contains three moles of carbon and eight moles of hydrogen. The mass of three moles of carbon is calculated as follows:

$\begin{array}{c}\text{Mass of 3 mol of C}=3\text{ mol}\times \text{12}\text{.01 g}\\ =\text{36}\text{.03 g}\end{array}$

The mass of eight moles of hydrogen is calculated as follows:

$\begin{array}{c}\text{Mass of 8 mol of H}=8\text{ mol}\times \text{1}\text{.008 g}\\ =\text{8}\text{.064 g}\end{array}$

Thus, the total mass of ${\text{C}}_3{\text{H}}_8$ is $44.094\text{ g}$ .

The percentage composition of carbon in ${\text{C}}_4{\text{H}}_8$ is calculated as follows:

$\begin{array}{c}\%\text{C}=\frac{36.03\text{ g}}{44.094\text{ g}}\times 100\\ =81.71\%\end{array}$

Therefore, ${\text{C}}_3{\text{H}}_8$ has lower percentage composition by mass.