Chapter 3, Problem 145AP

Lysine, an essential amino acid in the human body, contains $\text{C, H, O, and N}$. In one experiment, the complete combustion of 2.175 g of lysine gave 3.94 g CO, and 1.89 g ${\text{H}}_{\text{2}}\text{O}$. In a separate experiment. 1.873 g of lysine gave 0.436 g ${\text{NH}}_{\text{3}}$. (a) Calculate the empirical formula of lysine. (b) The approximate molar mass of lysine is 150 g. What is the molecular formula of the compound?

Interpretation Introduction

Interpretation:

The empirical and molecular formula of lysine is to be determined with given mass of lysine, mass of carbon dioxide, and mass of water.

Concept introduction:

Empirical formula is the simplest formula of any compound when written in the least possible whole number without altering the relative number of atoms.

The molecular mass of any compound is the sum of the atomic masses of every element that is present in the molecule of the compound. It is denoted by M. The unit of molecular mass isamu.

When any hydrocarbon has completely reacted with oxygen, then the product formed contains ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$. In these products, the carbon and hydrogen come entirely from the hydrocarbon. The oxygen may come from the hydrocarbon if it contains any oxygen atom, but it primarily comes from the added oxygen.

The percent composition of an element by its mass in a compound can be calculated as

$\text{Percent by mass of an element}=\frac{n\times \text{ mass of an element}}{\text{molecular mass of the compound}}\times 100\%$

Here, n represents the number of atoms in the given molecule or compound.

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.

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 molecular formula can be calculated as

$\text{Molecular formula}=n\times \text{Empirical formula}$

Answer to Problem 145AP

Solution:

(a) ${\text{C}}_3{\text{H}}_7\text{NO}$

(b) ${\text{C}}_6{\text{H}}_{14}{\text{N}}_2{\text{O}}_2$

Explanation of Solution

a) The empirical formula of lysine

$2.175\text{ g}$ lysine gave $\text{3}\text{.94 g}$ of ${\text{CO}}_2$ and $1.89\text{ g}$ of ${\text{H}}_2\text{O}$.

$\text{1}\text{.873 g}$ lysine gave $\text{0}\text{.436 g}$ of ${\text{NH}}_{\text{3}}$.

The molar mass of carbon dioxide is $12.01\text{ g}/\text{mol}$.

Calculate the mass of carbon dioxide as

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

Substitute $\frac{1\times \text{3}\text{.94 g}}{\text{44}\text{.01 g}/\text{mol}}$ for $n_{{\text{CO}}_2}$ and $12.01\text{ g}/\text{mol}$ for $M_{{\text{CO}}_2}$ in the above equation

$\begin{array}{c}{m}_{{\text{CO}}_2}=\frac{1\times \text{3}\text{.94 g}}{\text{44}\text{.01 g}/\text{mol}}\times 12.01\text{ g}/\text{mol}\\ =\text{1}\text{.075 g}\end{array}$

The molar mass of hydrogen is $1.008\text{ g}/\text{mol}$.

Calculate the mass of water as

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

Substitute $\frac{2\times 1.89\text{ g}}{\text{18}\text{.02 g}/\text{mol}}$ for $n_{{\text{H}}_2\text{O}}$ and $1.008\text{ g}/\text{mol}$ for $M_{{\text{H}}_2\text{O}}$ in the above equation

$\begin{array}{c}{m}_{{\text{H}}_2\text{O}}=\frac{2\times 1.89\text{ g}}{\text{18}\text{.02 g}/\text{mol}}\times 1.008\text{ g}/\text{mol}\\ =0.2114\text{ g}\end{array}$

The molar mass of nitrogen is $14.01\text{ g}/\text{mol}$.

Calculate the mass of ammonia as

$m_{{\text{NH}}_3}=n_{{\text{NH}}_3}\times M_{{\text{NH}}_3}$

Substitute $\frac{1\times \text{0}\text{.436 g}}{\text{17}\text{.03 g}/\text{mol}}$ for $n_{{\text{NH}}_3}$ and $14.01\text{ g}/\text{mol}$ for $M_{{\text{NH}}_3}$ in the above equation

$\begin{array}{c}{m}_{{\text{NH}}_{\text{3}}}=\frac{1\times \text{0}\text{.436 g}}{\text{17}\text{.03 g}/\text{mol}}\times 14.01\text{ g}/\text{mol}\\ =0.3587\text{ g}\end{array}$

Calculate the mass percent of carbon as

$\begin{array}{c}\text{Mass percentage of C = }\frac{\text{1}\text{.075 g }}{\text{2}\text{.175 g}}\text{ × 100\%}\\ \text{= 49}\text{.43\% C}\end{array}$

Calculate the mass percent of hydrogen as

$\begin{array}{c}\text{Mass percentage of H = }\frac{\text{0}\text{.2114 g}}{\text{2}\text{.175 g}}\text{×100\%}\\ \text{= 9}\text{.720\% H}\end{array}$

Calculate the mass percent of nitrogen as

$\begin{array}{c}\text{Mass percentage of N = }\frac{\text{0}\text{.3587 g}}{\text{1}\text{.873 g}}\text{×100\%}\\ \text{= 19}\text{.15\% N}\end{array}$

Calculate the mass percent of oxygen as

$\begin{array}{c}\text{Mass percentage of O} = 100\% - (49.43\%+9.720\%+19.15\%)\\ = 21.70\%\text{ O}\end{array}$

Now, calculate the number of moles by assuming the sample to be 100 g.

The molar mass of carbon is $12.01\text{ g}/\text{mol}$.

Calculate the number of moles of carbon as

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

Substitute $\text{49}\text{.93 g}$ for $m_{\text{C}}$ and $\text{12}\text{.01 g}/\text{mol}$ for $M_{\text{C}}$ in the above equation

$\begin{array}{c}{n}_{\text{C}}=\frac{49.93\text{ g}}{12.01\text{ g}/\text{mol}}\\ =4.116\text{ mol}\end{array}$

The molar mass of hydrogen is $1.008\text{ g}/\text{mol}$.

Calculate the number of moles of hydrogen as

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

Substitute $\text{9}\text{.720 g}$ for $m_{\text{H}}$ and $\text{1}\text{.008 g}/\text{mol}$ for $M_{\text{H}}$ in the above equation

$\begin{array}{c}{n}_{\text{H}}=\frac{\text{9}\text{.720 g}}{1.008\text{ g}/\text{mol}}\\ =9.643\text{ mol}\end{array}$

The molar mass of nitrogen is $\text{14}\text{.01 g}/\text{mol}$.

Calculate the number of moles of nitrogen as

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

Substitute $\text{19}\text{.15 g}$ for $m_{\text{N}}$ and $\text{14}\text{.01 g}/\text{mol}$ for $M_{\text{N}}$ in the above equation

$\begin{array}{c}{n}_{\text{N}}=\frac{19.15\text{ g}}{14.01\text{ g}/\text{mol}}\\ =1.367\text{ mol}\end{array}$

The molar mass of oxygen is $16.00\text{ g}/\text{mol}$.

Calculate the number of moles of oxygen as

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

Substitute $21.70\text{ g}$ for $m_{\text{O}}$ and $16.00\text{ g}/\text{mol}$ for $M_{\text{O}}$ in the above equation

$\begin{array}{c}{n}_{\text{O}}=\frac{21.70\text{ g}}{16.00\text{ g}/\text{mol}}\\ =1.367\text{ mol}\end{array}$

Calculate the ratio of carbon, hydrogen, nitrogen, and oxygen as

$\frac{4.116\text{ mol}}{1.356\text{ mol}}:\frac{\text{9}\text{.643 mol}}{\text{1}\text{.356 mol}}\text{: }\frac{\text{1}\text{.367 mol}}{\text{1}\text{.356 mol}}\text{: }\frac{\text{1}\text{.356 mol}}{\text{1}\text{.356 mol}}=3.00:7.00:1.00:1.00$

Hence, the empirical formula of lysine is ${\text{C}}_3{\text{H}}_7\text{NO}$.

b)The molecular formula of the compound

The molecular mass of lysine is 150 g.

Calculate the empirical formula mass of lysine as

$\begin{array}{c}\text{Empirical formula mass }=\text{3}\times \text{12+7}\times \text{1+14+16}\\ =\text{62}\text{g}\end{array}$

Now, the whole number multiple $\left(n\right)$ can be calculated as

$n=\frac{\text{Molecular mass}}{\text{Empirical formula mass}}$

Substitute $150\text{ g}$ for molecular mass and $\text{62}\text{g}$ for empirical formula mass in the above equation

$\begin{array}{c}n=\frac{150\text{ g}}{\text{62}\text{g}}\\ \approx 2\end{array}$

Calculate the molecular formula as

$\text{Molecular formula}=n\times \text{Empirical formula}$

Substitute 2 for $n$ and ${\text{C}}_3{\text{H}}_7\text{NO}$ for empirical formula in the above equation

$\begin{array}{c}\text{Molecular formula}=\text{2}\times {\text{C}}_3{\text{H}}_7\text{NO}\\ ={\text{C}}_6{\text{H}}_{14}{\text{N}}_2{\text{O}}_2\end{array}$

Hence, themolecular formula of lysine is ${\text{C}}_6{\text{H}}_{14}{\text{N}}_2{\text{O}}_2$.