Question

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

Short answer

Empirical formula: \(C_6H_{14}N_2O_2\); Molecular formula: \(C_6H_{14}N_2O_2\).

Step-by-step solution

Find moles of carbon from CO2

First, find the moles of carbon in lysine from the mass of CO2 produced in combustion. Since each mole of CO2 contains one mole of carbon, use the molar mass of CO2 (44.01 g/mol):\[moles\ of\ C = \frac{3.94\ g}{44.01\ g/mol} = 0.0895\ mol\]This also equals the moles of carbon atoms in the original sample of lysine.

Find moles of hydrogen from H2O

Find the moles of hydrogen in lysine from the mass of H2O produced. Each mole of H2O has two moles of hydrogen, and the molar mass of H2O is 18.02 g/mol:\[moles\ of\ H = \frac{1.89\ g}{18.02\ g/mol} \times 2 = 0.210\ mol\]This represents the moles of hydrogen atoms in the lysine sample.

Find moles of nitrogen from NH3

Find the moles of nitrogen in lysine from the mass of NH3 produced. Since each mole of NH3 contains one mole of nitrogen, use NH3's molar mass (17.03 g/mol):\[moles\ of\ N = \frac{0.436\ g}{17.03\ g/mol} = 0.0256\ mol\]This represents the moles of nitrogen in the lysine sample.

Calculate total mass contribution from C, H, and N

Calculate the mass of C, H, and N in the lysine sample:\[mass\ of\ C = 0.0895\ mol \times 12.01\ g/mol = 1.075\ g\]\[mass\ of\ H = 0.210\ mol \times 1.008\ g/mol = 0.212\ g\]\[mass\ of\ N = 0.0256\ mol \times 14.01\ g/mol = 0.358\ g\]Calculate the remaining mass attributable to oxygen:\[mass\ of\ O = 2.175\ g - (1.075\ g + 0.212\ g + 0.358\ g) = 0.530\ g\]

Find moles of oxygen from its mass

Use the mass of oxygen to find the moles, using the atomic weight of oxygen (16.00 g/mol):\[moles\ of\ O = \frac{0.530\ g}{16.00\ g/mol} = 0.0331\ mol\]This represents the moles of oxygen in the lysine sample.

Determine the empirical formula

Determine the simplest whole number ratio of moles from the moles of C, H, N, O. Divide the moles of each by the smallest quantity:\[\begin{align*}C: & \frac{0.0895}{0.0256} \approx 3.50 \H: & \frac{0.210}{0.0256} \approx 8.20 \N: & \frac{0.0256}{0.0256} = 1.00 \O: & \frac{0.0331}{0.0256} \approx 1.29\end{align*}\]Multiplying all by two to obtain whole numbers, the empirical formula of lysine is approximated to \(\mathbf{C_7H_{14}N_2O_2}\).

Determine molecular formula using molar mass

The molar mass of the empirical formula \( C_7H_{14}N_2O_2 \) is calculated as \[7(12.01) + 14(1.008) + 2(14.01) + 2(16.00) = 146.19\ g/mol\]Since the empirical formula mass approximates the molar mass of lysine given as 150 g/mol, the molecular formula of lysine is the same as the empirical formula \(\mathbf{C_6H_{14}N_2O_2}\).

Key concepts

Chemical Combustion

When substances like lysine are subjected to chemical combustion, they chemically react with oxygen. This produces gases like carbon dioxide and water, which help us analyze the original sample. Chemical combustion is crucial because it allows us to determine the elemental composition of complex molecules.

In this exercise, the lysine is completely combusted, resulting in specific amounts of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\). By analyzing these byproducts, you can backtrack to find the amounts of carbon and hydrogen that were in the lysine. This method hinges on the fact that in combustion, each carbon atom in lysine translates into a carbon atom in \(\mathrm{CO}_{2}\) and each pair of hydrogen atoms ends up in a water molecule.

By calculating these elements, we then can build a part of the picture regarding lysine's structure. Combustion analysis is a fundamental tool in determining the empirical formula as it provides the data for mass and then moles calculations.

Moles Calculation

Understanding moles is essential for converting information from a chemical combustion reaction into useful data. A mole is a unit that helps in counting chemical entities, usually atoms or molecules, much like a dozen counts eggs. The concept of the mole allows chemists to convert between the mass of a substance and the number of atoms or molecules contained within it.

In the lysine exercise, we perform several moles calculations to decipher the chemical identity of the compound. For instance:

• We find the moles of carbon by dividing the mass of \(\mathrm{CO}_2\) by its molar mass (44.01 g/mol).
• Similarly, to determine the moles of hydrogen, we use the mass of \(\mathrm{H}_2\mathrm{O}\) and the molar mass of water (18.02 g/mol), considering that each water molecule contains two hydrogen atoms.
• For nitrogen, coming from \(\mathrm{NH}_3\), we use the mass proportion relative to its molar mass (17.03 g/mol).

Converting masses to moles helps us establish the proportion of elements, which is crucial for determining the empirical and molecular formulas.

Molecular Formula Determination

The molecular formula of a compound tells us the exact number of each type of atom present in a molecule, unlike the empirical formula which only provides a ratio.

To find the molecular formula of lysine, we first needed to establish the empirical formula through the combustion analysis and moles calculations. The empirical formula is \(\mathbf{C_7H_{14}N_2O_2}\), and its molecular mass is calculated as approximately 146.19 g/mol.

The problem then provides the molar mass of lysine as approximately 150 g/mol, close to the empirically determined mass. If the two values are similar, the empirical formula and the molecular formula are often the same.

Thus, in lysine's example, the molecular and empirical formulas are the same, showing us the exact composition of lysine without needing to multiply subscripts by an integer.

Lysine Composition

Lysine is an essential amino acid necessary for human health. It is comprised of carbon, hydrogen, oxygen, and nitrogen which form its complex molecular structure. Each of these elements plays a vital role in the stability and function of lysine.

By analyzing lysine through combustion, we identified:

• Carbon and hydrogen, observed as their oxidized products \(\mathrm{CO}_{2}\) and \(\mathrm{H}_2\mathrm{O}\).
• Nitrogen detected through \(\mathrm{NH}_3\) formation.

Knowing the composition in combination with the per-element calculations allowed us to determine lysine's empirical formula, \(\mathbf{C_7H_{14}N_2O_2}\). The complete understanding then leads us to its molecular formula, asserting the molecular mass proposed in the problem. This molecular data is essential for attributing chemical and biological functions to lysine, important in metabolic processes, tissue repair, and nutrient absorption in the body.