Chapter 2, Problem 2.44AP

Interpretation Introduction

(a)

Interpretation:

The mass of carbon and hydrogen in given compound $X$ is to be calculated.

Concept introduction:

The combustion is the process of heating the compounds in the presence of oxygen. The combustion of alkanes yields carbon dioxide and water. In the process of combustion of alkanes a high amount of heat is evolved. These products are given by all kinds of alkanes generally.

Answer to Problem 2.44AP

The mass percent of carbon and hydrogen in given compound $X$ is $83.86\%$ and $15.98\%$ respectively.

Explanation of Solution

The general balanced reaction for the complete combustion of an alkane is given below as,

${\text{C}}_n{\text{H}}_{2n+2}+\frac{\left(3n+1\right)}{2}{\text{O}}_{\text{2}}\to n{\text{CO}}_{\text{2}}+\left(n+1\right){\text{H}}_{\text{2}}\text{O}$

The carbon and hydrogen are converted to water and carbon dioxide in combustion process. The mass in terms of percentage of carbon and hydrogen can be calculated using the formula as,

$\text{Mass}\%\text{of}\text{C}=\frac{\text{Atomic}\text{mass}\text{of}\text{C}\text{present}\text{in}\text{compound}\times \text{given}\text{mass}\text{of}{\text{CO}}_{\text{2}}}{\text{Molecular}\text{mass}\text{of}{\text{CO}}_{\text{2}}\times \text{given}\text{mass}\text{of}\text{compound}}\times 100$ …(1)

$\text{Mass}\%\text{of}\text{H}=\frac{\text{Atomic}\text{mass}\text{of}\text{H}\text{present}\text{in}\text{compound}\times \text{given}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}}{\text{Molecular}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}\times \text{given}\text{mass}\text{of}\text{compound}}\times 100$…(2)

Substitute the values in equation (1) as,

$\begin{array}{rll}\text{Mass}\%\text{of}\text{C}& =& \frac{\text{Atomic}\text{mass}\text{of}\text{C}\text{present}\text{in}\text{compound}\times \text{given}\text{mass}\text{of}{\text{CO}}_{\text{2}}}{\text{Molecular}\text{mass}\text{of}{\text{CO}}_{\text{2}}\times \text{given}\text{mass}\text{of}\text{compound}}\times 100\\ & =& \frac{\text{12}\text{g}\times \text{24}\text{.60}\text{mg}}{\text{44}\text{g}\times \text{8}\text{.00}\text{mg}}\times 100\\ & =& 83.86\%\end{array}$

Substitute the values in equation (2) as,

$\begin{array}{rll}\text{Mass}\%\text{of}\text{H}& =& \frac{\text{Atomic}\text{mass}\text{of}\text{H}\text{present}\text{in}\text{compound}\times \text{given}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}}{\text{Molecular}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}\times \text{given}\text{mass}\text{of}\text{compound}}\times 100\\ & =& \frac{2\times \text{11}\text{.51}\text{mg}}{\text{18}\times \text{8}\text{.00}\text{mg}}\times 100\\ & =& 15.98\%\end{array}$

Conclusion

The mass percent of carbon and hydrogen in given compound $X$ is $83.86\%$ and $15.98\%$ respectively.

Interpretation Introduction

(b)

Interpretation:

The moles of $\text{H}$ present in $X$ per mole of $\text{C}$ are to be calculated. The formula is to be expressed as ${\text{C}}_{\text{1}}{\text{H}}_x$.

Concept introduction:

The combustion is the process of heating the compounds in the presence of oxygen. The combustion of alkanes yields carbon dioxide and water. In the process of combustion of alkanes a high amount of heat is evolved. These products are given by all kinds of alkanes generally.

Answer to Problem 2.44AP

The number of moles of $\text{H}$ present in $X$ per mole of $\text{C}$ is $2.2$ and the formula can be expressed as ${\text{C}}_{\text{1}}{\text{H}}_{2.2}$.

Explanation of Solution

The moles of $\text{H}$ present in $X$ per mole of $\text{C}$ can be calculated by finding out the atomic ratio and simple ratio. The atomic ratio can be calculated by taking the ratio of percent composition of element to its atomic weight. The percent composition for carbon and hydrogen calculated in part (a) is $83.86\%$ and $15.98\%$ respectively. The atomic mass of carbon and hydrogen is $12\text{g}$ and $1\text{g}$ respectively. Thus, the atomic ratio is calculated as,

$\text{Atomic}\text{ratio}\text{of}\text{element}=\frac{\text{Percent composition of}\text{element}}{\text{Atomic}\text{mass}\text{of}\text{element}}$ … (3)

Substitute the values in equation (3) for carbon as,

$\begin{array}{rll}\text{Atomic}\text{ratio}\text{of}\text{C}& =& \frac{\text{Percent composition of}\text{C}}{\text{Atomic}\text{mass}\text{of}\text{C}}\\ & =& \frac{83.86}{12}\\ & =& 6.98\end{array}$

Substitute the values in equation (3) for hydrogen as,

$\begin{array}{rll}\text{Atomic}\text{ratio}\text{of}\text{H}& =& \frac{\text{Percent composition of}\text{H}}{\text{Atomic}\text{mass}\text{of}\text{H}}\\ & =& \frac{15.98}{1}\\ & =& 15.98\end{array}$

The simplest ratio is the ratio of atomic ratio of carbon and hydrogen by atomic ratio of carbon. It is calculated to get the number of moles of element in the compound. The simplest ratio for carbon is as follows.

$\begin{array}{rll}\text{Simplest}\text{ratio}\text{of}\text{C}& =& \frac{\text{Atomic}\text{ratio}\text{of}\text{C}}{\text{Atomic}\text{ratio}\text{of}\text{C}}\\ & =& \frac{6.98}{6.98}\\ & =& 1\end{array}$

The simplest ratio for hydrogen is as follows.

$\begin{array}{rll}\text{Simplest}\text{ratio}\text{of}\text{H}& =& \frac{\text{Atomic}\text{ratio}\text{of}\text{H}}{\text{Atomic}\text{ratio}\text{of}\text{C}}\\ & =& \frac{15.98}{6.98}\\ & =& 2.2\end{array}$

Thus, the number of moles of $\text{H}$ present in $X$ per mole of $\text{C}$ is $2.2$ and the formula can be expressed as ${\text{C}}_{\text{1}}{\text{H}}_{2.2}$.

Conclusion

The number of moles of $\text{H}$ present in $X$ per mole of $\text{C}$ is $2.2$ and the formula can be expressed as ${\text{C}}_{\text{1}}{\text{H}}_{2.2}$.

Interpretation Introduction

(c)

Interpretation:

The formula obtained in (b) is to be multiplied by successive integers until the amount of $\text{H}$ becomes an integer.

Concept introduction:

The combustion is the process of heating the compounds in the presence of oxygen. The combustion of alkanes yields carbon dioxide and water. In the process of combustion of alkanes a high amount of heat is evolved. These products are given by all kind of alkanes generally.

Answer to Problem 2.44AP

The formula obtained in (b) is multiplied by successive integers and the formula for compound $X$ obtained is ${\text{C}}_5{\text{H}}_{11}$.

Explanation of Solution

The formula obtained in (b) is ${\text{C}}_{\text{1}}{\text{H}}_{2.2}$. The formulae obtained on multiplying the number of moles in ${\text{C}}_{\text{1}}{\text{H}}_{2.2}$ by $2$, $3$, $4$ and $5$ are ${\text{C}}_2{\text{H}}_{4.4}$, ${\text{C}}_3{\text{H}}_{6.6}$, ${\text{C}}_4{\text{H}}_{8.8}$ and ${\text{C}}_5{\text{H}}_{11}$ respectively. In the above formulae, the formula obtained by multiplying the number of moles in ${\text{C}}_{\text{1}}{\text{H}}_{2.2}$ by $5$ represents the number of moles of hydrogen as an integer. Thus, the number of moles of carbon and hydrogen is $5$ and $11$ respectively in the compound $X$.

Conclusion

The formula obtained in (b) is multiplied by successive integers and the formula obtained for compound $X$ is ${\text{C}}_5{\text{H}}_{11}$.