Chapter 3, Problem 154AP

Interpretation Introduction

Interpretation:

The empirical formula of the compound is to be calculated.

Concept introduction:

The number of moles is defined as the ratio of mass to the molar mass:

$n=\frac{m}{M}$

Here, $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass.

Molar mass is calculated by adding the masses of each element multiplied by their number of atoms present (given in subscript). Its S.I. unit is $\text{g}/\text{mol}$.

The mass of compound can be calculated as $m=n\times M$.

Here, $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass.

the Mass of carbon can be calculated as $m_{\text{C}}=n_{{\text{CO}}_2}\times M_{\text{C}}\times n_{\text{C}}$.

The mass of hydrogen can be calculated as $m_{\text{H}}=n_{{\text{H}}_2\text{O}}\times M_{\text{H}}\times n_{\text{H}}$.

Answer to Problem 154AP

Solution: The empirical formula of the first compound is ${\text{PbC}}_{\text{8}}{\text{H}}_{\text{20}}$ and for thesecond compound is ${\text{C}}_5{\text{H}}_{12}\text{O}$

Explanation of Solution

Given information:

For gasoline

$\begin{array}{l}{m}_{{\text{CO}}_2}=55.9\text{ g}\\ m_{{\text{H}}_2\text{O}}=28.61\text{ g}\end{array}$

For methyl tert-butyl ether

$\begin{array}{l}{m}_{{\text{CO}}_2}=30.2\text{ g}\\ m_{{\text{H}}_2\text{O}}=14.8\text{ g}\end{array}$

The molar mass of ${\text{H}}_2\text{O}$ is $18\text{ g}/\text{mol}$

Calculate the number of moles of water as

$n_{{\text{H}}_2\text{O}}=\frac{m_{{\text{H}}_2\text{O}}}{M_{{\text{H}}_2\text{O}}}$

Substitute $28.61\text{ g}$ for $m_{{\text{H}}_2\text{O}}$ and $18\text{ g}/\text{mol}$ for $M_{{\text{H}}_2\text{O}}$ in the equation above

$\begin{array}{c}{n}_{{\text{H}}_2\text{O}}=\frac{28.61\text{ g}}{18\text{ g}/\text{mol}}\\ =1.589\text{ mol}\end{array}$

The molar mass of ${\text{CO}}_2$ is $44\text{ g}/\text{mol}$.

Calculate the number of moles of carbon dioxide as

$n_{{\text{CO}}_2}=\frac{m_{{\text{CO}}_2}}{M_{{\text{CO}}_2}}$

Substitute $55.9\text{ g}$ for $m_{{\text{CO}}_2}$ and $44\text{ g}/\text{mol}$ for $M_{{\text{CO}}_2}$ in the equation above

$\begin{array}{c}{n}_{{\text{CO}}_2}=\frac{55.9\text{ g}}{44\text{ g}/\text{mol}}\\ =1.27\text{ mol}\end{array}$

The molar mass of $\text{C}$ is $12\text{ g}/\text{mol}$.

For carbon in ${\text{CO}}_2$, the number of moles is one.

Calculate the mass of carbon as

$m_{\text{C}}=n_{{\text{CO}}_2}\times M_{\text{C}}\times n_{\text{C}}$

Substitute $1.27\text{ mol}$ for $n_{{\text{CO}}_2}$, $12\text{ g}/\text{mol}$ for $M_C$, and 1 for $n_{\text{C}}$ in the equation above

$\begin{array}{c}{m}_{\text{C}}=1.27\text{ mol}\times 12\text{ g}/\text{mol}\times 1\\ =15.25\text{ g}\end{array}$

The molar mass of $\text{H}$ is $1\text{ g}/\text{mol}$.

For hydrogen in ${\text{H}}_2\text{O}$, the number of moles is two.

Calculate the mass of hydrogen as

$m_{\text{H}}=n_{{\text{H}}_2\text{O}}\times M_{\text{H}}\times n_{\text{H}}$

Substitute $1.589\text{ mol}$ for $n_{{\text{H}}_2\text{O}}$, $1\text{ g}/\text{mol}$ for $M_{\text{H}}$, and 2 for $n_{\text{H}}$ in the equation above

$\begin{array}{c}{m}_{\text{H}}=1.589\text{ mol}\times 1\text{ g}/\text{mol}\times 2\\ =3.178\text{ g}\end{array}$

The mass of the given sample is $51.36\text{ g}$.

Calculate the mass of lead as

$51.36\text{ g}=\text{mass C}+\text{mass H}+\text{mass}\mathrm{Pb}$

Substitute $3.178\text{ g}$ for $\text{mass H}$ and $15.25\text{ g}$ for $\text{mass C}$ in the equation above

$\begin{array}{c}51.36\text{ g} = 15.25\text{ g}+3.201\text{ g}+\text{mass Pb}\\ =18.451\text{ g}+\text{mass Pb}\end{array}$

Rearrange the above equation for $\text{mass Pb}$ as

$\begin{array}{l}\text{Mass Pb}=51.36\text{ g}-18.451\text{ g}\\ \text{Mass Pb} = 32.91\text{ g}\end{array}$

Now, the number of moles of each element is to be calculated as

The molar mass of $\text{C}$ is $12\text{ g}/\text{mol}$.

Calculate the number of moles of carbon as

$n_{\text{C}}=\frac{m_{\text{C}}}{M_{\text{C}}}$

Substitute $15.25\text{ g}$ for $m_{\text{C}}$ and $12\text{ g}/\text{mol}$ for $M_{\text{C}}$ in the equation above

$\begin{array}{c}{n}_{\text{C}}=\frac{15.25\text{ g}}{12\text{ g}/\text{mol}}\\ =1.27\text{ mol}\end{array}$

The molar mass of hydrogen is $1\text{ g}/\text{mol}$.

Calculate the mass of hydrogen as

$n_{\text{H}}=\frac{m_{\text{H}}}{M_{\text{H}}}$

Substitute $3.178\text{ g}$ for $m_{\text{H}}$ and $1\text{ g}/\text{mol}$ for $M_{\text{H}}$ in the equation above

$\begin{array}{c}{n}_{\text{H}}=\frac{3.178\text{ g}}{1\text{ g}/\text{mol}}\\ =3.17\text{ mol}\end{array}$

The molar mass of lead is $207.2\text{ g}/\text{mol}$.

Calculate the mass of lead as

$n_{\text{Pb}}=\frac{m_{\text{Pb}}}{M_{\text{Pb}}}$

Substitute $32.91\text{ g}$ for $m_{\text{Pb}}$ and $207.2\text{ g}/\text{mol}$ for $M_{\text{Pb}}$ in the equation above

$\begin{array}{c}{n}_{\text{Pb}}=\frac{32.91\text{ g}}{207.2\text{ g}/\text{mol}}\\ =0.1588\text{ mol}\end{array}$

So, the molecular formula of the compound can be written as

${\text{Pb}}_{0.1588}{\text{C}}_{1.270}{\text{H}}_{3.176}$

As the empirical formula are always has whole number multiples, on simplifying the number of moles of each element, it is calculated as

For H

$n_{\text{H}}=\frac{3.176}{0.1588}\approx 20$

For C

$n_{\text{C}}=\frac{1.270}{0.1588}\approx 8$

For Pb

$n_{\text{Pb}}=\frac{0.1588}{0.1588}=1$

So, the empirical formula of the compound is ${\text{PbC}}_{\text{8}}{\text{H}}_{\text{20}}$.

Now, for tert-butyl ether:

The molar mass of ${\text{H}}_2\text{O}$ is $18\text{ g}/\text{mol}$

Calculate the number of moles of water as

$n_{{\text{H}}_2\text{O}}=\frac{m_{{\text{H}}_2\text{O}}}{M_{{\text{H}}_2\text{O}}}$

Substitute $\text{14}\text{.8 g}$ for $m_{{\text{H}}_2\text{O}}$ and $18\text{ g}/\text{mol}$ for $M_{{\text{H}}_2\text{O}}$ in the equation above

$\begin{array}{c}{n}_{{\text{H}}_2\text{O}}=\frac{14.8\text{ g}}{18\text{ g}/\text{mol}}\\ =0.822\text{ mol}\end{array}$

The molar mass of ${\text{CO}}_2$ is $44\text{ g}/\text{mol}$.

Calculate the number of moles of carbon dioxide as

$n_{{\text{CO}}_2}=\frac{m_{{\text{CO}}_2}}{M_{{\text{CO}}_2}}$

Substitute $\text{30}\text{.2 g}$ for $m_{{\text{CO}}_2}$ and $44\text{ g}/\text{mol}$ for $M_{{\text{CO}}_2}$ in the equation above

$\begin{array}{c}{n}_{{\text{CO}}_2}=\frac{30.2\text{ g}}{44\text{ g}/\text{mol}}\\ =0.686\text{ mol}\end{array}$

The molar mass of $\text{C}$ is $12\text{ g}/\text{mol}$.

For carbon in ${\text{CO}}_2$, the number of moles is one.

Calculate the mass of carbon as

$m_{\text{C}}=n_{{\text{CO}}_2}\times M_{\text{C}}\times n_{\text{C}}$

Substitute $\text{0}\text{.686 mol}$ for $n_{{\text{CO}}_2}$, $12\text{ g}/\text{mol}$ for $M_C$, and 1 for $n_{\text{C}}$ in the equation above

$\begin{array}{c}{m}_{\text{C}}=0.686\text{ mol}\times 12\text{ g}/\text{mol}\times 1\\ =8.24\text{ g}\end{array}$

The molar mass of $\text{H}$ is $1\text{ g}/\text{mol}$.

For hydrogen in ${\text{H}}_2\text{O}$, the number of moles is two.

Calculate the mass of hydrogen as

$m_{\text{H}}=n_{{\text{H}}_2\text{O}}\times M_{\text{H}}\times n_{\text{H}}$

Substitute $\text{0}\text{.822 mol}$ for $n_{{\text{H}}_2\text{O}}$, $1\text{ g}/\text{mol}$ for $M_{\text{H}}$, and 2 for $n_{\text{H}}$ in the equation above

$\begin{array}{c}{m}_{\text{H}}=0.822\text{ mol}\times 1\text{ g}/\text{mol}\times 2\\ =1.6\text{ g}\end{array}$

The mass of the given sample is $12.1\text{ g}$.

Calculate the mass of oxygen as

$\text{mass C}+\text{mass H}+\text{mass O}=12.1\text{ g}$

Substitute $1.6\text{ g}$ for $\text{mass H}$ and $8.24\text{ g}$ for $\text{mass H}$ in the equation above

$\begin{array}{c}12.1\text{ g}= 8.24\text{ g}+1.6\text{ g}+\text{mass O}\\ =9.9\text{ g}+\text{mass O}\end{array}$

Rearrange the above equation for $\text{mass O}$ as

$\begin{array}{c}\text{mass O}=12.1\text{ g}-9.9\text{ g}\\ \text{mass O} = 2.2\text{ g}\end{array}$

Now, the number of moles of each element is to be calculated as

The molar mass of carbon is $12\text{ g}/\text{mol}$.

Calculate the mass of carbon as

$n_{\text{C}}=\frac{m_{\text{C}}}{M_{\text{C}}}$

Substitute $\text{8}\text{.24 g}$ for $m_{\text{C}}$ and $12\text{ g}/\text{mol}$ for $M_{\text{C}}$ in the above equation

$\begin{array}{c}{n}_{\text{C}}=\frac{\text{8}\text{.24 g}}{12\text{ g}/\text{mol}}\\ =0.686\text{ mol}\end{array}$

The molar mass of hydrogen is $1\text{ g}/\text{mol}$.

Calculate the mass of hydrogen as

$n_{\text{H}}=\frac{m_{\text{H}}}{M_{\text{H}}}$

Substitute $\text{1}\text{.6 g}$ for $m_{\text{H}}$ and $1\text{ g}/\text{mol}$ for $M_{\text{H}}$ in the equation above

$\begin{array}{c}{n}_{\text{H}}=\frac{1.6\text{ g}}{1\text{ g}/\text{mol}}\\ =1.6\text{ mol}\end{array}$

The molar mass of oxygen is $16\text{ g}/\text{mol}$.

Calculate the mass of oxygen as

$n_{\text{O}}=\frac{m_{\text{O}}}{M_{\text{O}}}$

Substitute $\text{2}\text{.2 g}$ for $m_{\text{O}}$ and $\text{16}\text{.00 g}/\text{mol}$ for $M_{\text{O}}$ in the equation above

$\begin{array}{c}{n}_{\text{O}}=\frac{2.2\text{ g}}{16\text{ g}/\text{mol}}\\ =0.14\text{ mol}\end{array}$

So, the molecular formula of the compound is ${\text{C}}_{0.686}{\text{H}}_{1.65}{\text{O}}_{0.14}$.

As the empirical formula are always whole number multiples, on simplifying the number of moles of each element, it is calculated as

For H

$\begin{array}{c}{n}_{\text{H}}=\frac{1.65}{0.14}\\ \approx 12\end{array}$

For C

$\begin{array}{c}{n}_{\text{C}}=\frac{0.686}{0.14}\\ \approx 5\end{array}$

For O

$\begin{array}{c}{n}_{\text{Pb}}=\frac{0.14}{0.14}\\ =1\end{array}$

So, the empirical formula of the compound is ${\text{C}}_5{\text{H}}_{12}\text{O}$.

Conclusion

The empirical formula of the first compound is ${\text{PbC}}_{\text{8}}{\text{H}}_{\text{20}}$ and for the second compound is ${\text{C}}_5{\text{H}}_{12}\text{O}$.