Problem 131 - Chemistry and Chemical Reactivity

Question

The action of bacteria on meat and fish produces a compound called cadaverine. As its name and origin imply, it stinks! (It is also present in bad breath and adds to the odor of urine.) It is \(58.77 \%\) C, \(13.81 \%\) H and \(27.40 \%\) N. Its molar mass is \(102.2 \mathrm{g} / \mathrm{mol}\) Determine the molecular formula of cadaverine.

Step-by-Step Solution

Answer

The molecular formula of cadaverine is \(C_5H_{14}N_2\).

Step 1: Assume 100 grams of Compound

To find the empirical formula of cadaverine, we start by assuming we have 100 grams of it. This makes it easy to work with percentages directly as grams: 58.77 grams of carbon, 13.81 grams of hydrogen, and 27.40 grams of nitrogen.

Step 2: Convert Grams to Moles

Convert the mass of each element to moles using their respective atomic masses: \(C: 58.77 \text{ g} / 12.01 \text{ g/mol} = 4.895 \text{ mol}\), \(H: 13.81 \text{ g} / 1.008 \text{ g/mol} = 13.70 \text{ mol}\), \(N: 27.40 \text{ g} / 14.01 \text{ g/mol} = 1.956 \text{ mol}\).

Step 3: Determine the Simplest Whole Number Ratio

Divide all mole values by the smallest number of moles calculated, 1.956 mol of nitrogen in this case, to find the simplest ratio: \(C: 4.895/1.956 = 2.5\), \(H: 13.70/1.956 = 7\), \(N: 1.956/1.956 = 1\).

Step 4: Convert to Whole Numbers

To convert the ratio to whole numbers, multiply each number by 2 because 2.5 is the smallest number requiring scaling: \(C_5:H_{14}:N_2\).

Step 5: Calculate Empirical Formula Molar Mass

Calculate the molar mass of the empirical formula to ensure it fits the given molar mass: \((5\times12.01) + (14\times1.008) + (2\times14.01) = 102.2 \text{ g/mol}\). This matches the provided molar mass.

Step 6: Write the Molecular Formula

Since the empirical formula molar mass matches the given molar mass, the empirical formula \(C_5H_{14}N_2\) is the molecular formula for cadaverine.

Key Concepts

Understanding Empirical Formulas

The empirical formula of a compound is a simple way to express the smallest whole-number ratio of the elements. It provides a basic snapshot of the proportion of each type of atom that makes up the substance. For cadaverine, the given percentages of carbon (C), hydrogen (H), and nitrogen (N) allow us to calculate its empirical formula. Here's how:

• We start by assuming a total mass of 100 grams for simplicity. This turns percentage values into direct mass values in grams: 58.77g of carbon, 13.81g of hydrogen, and 27.40g of nitrogen.
• We then convert these masses to moles using each element's atomic mass from the periodic table. For carbon, hydrogen, and nitrogen, the moles are calculated as follows: \( C: \frac{58.77}{12.01} = 4.895 \), \( H: \frac{13.81}{1.008} = 13.70 \), \( N: \frac{27.40}{14.01} = 1.956 \).
• The smallest number of moles calculated is for nitrogen. We divide all mole quantities by this smallest value to get a ratio. However, these quantities are not whole numbers, so we multiply them to convert to whole numbers, giving us the empirical formula \( C_5H_{14}N_2 \).

It's essential to understand that this approach gives you the simplest recipe, so to speak, of the molecules, not the absolute numbers of atoms.

Molar Mass Calculation of Compounds

Calculating the molar mass of a compound is crucial in determining its molecular formula. The molar mass is simply the sum of masses of all atoms in a molecule. Once we have the empirical formula, we can find the mass by adding the atomic masses of the constituent atoms according to their proportions:

• For our empirical formula \( C_5H_{14}N_2 \), we calculate its molar mass by multiplying the atomic mass of each element by the number of atoms of that element in the compound and summing them up: \( 5 \times 12.01 \) (for carbon) + \( 14 \times 1.008 \) (for hydrogen) + \( 2 \times 14.01 \) (for nitrogen).
• This results in a total empirical formula mass of \( 102.2 \, \text{g/mol} \).

This calculation matches the provided molar mass for the compound, confirming that our empirical formula \( C_5H_{14}N_2 \) is indeed the molecular formula for cadaverine. Understanding how to add these atomic masses and relate them to molecular structure is vital for chemists.

Basic Stoichiometry in Determining Formulas

Stoichiometry forms the bridge between quantitative measurements and chemical formulations in reactions. It involves using balanced equations to calculate quantities of reactants or products. Although in this exercise we aren't balancing a reaction, similar principles of proportion and conversion apply when deducing molecular formulas.Here's how stoichiometry helps in finding formulas:

• First, we convert mass percentages into moles, relying on atomic masses, which are stoichiometric principles in action.
• Then, by determining the smallest whole number mole ratios, we effectively "balance" our imaginary equation for the compound's composition.
• Multiplying any non-integers in our ratio to achieve whole numbers is akin to balancing stoichiometric coefficients in reactions.

Ultimately, the outcome of stoichiometric methods allows us to determine precise formulas like \( C_5H_{14}N_2 \) for cadaverine, which correctly portrays its molecular structure based on empirical data.