Problem 135 - 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 \(\text{C}_5\text{H}_{14}\text{N}_2\).

Step 1: Percent to Mass Conversion

Assume we have 100 g of cadaverine. This means the mass of carbon (C) is 58.77 g, hydrogen (H) is 13.81 g, and nitrogen (N) is 27.40 g. These masses correlate directly to the percentages given.

Step 2: Mass to Moles Conversion

Calculate the moles for each element using the formula \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). - Moles of C: \(\frac{58.77\, \text{g}}{12.01\, \text{g/mol}} \approx 4.89\, \text{moles}\)- Moles of H: \(\frac{13.81\, \text{g}}{1.008\, \text{g/mol}} \approx 13.70\, \text{moles}\)- Moles of N: \(\frac{27.40\, \text{g}}{14.01\, \text{g/mol}} \approx 1.96\, \text{moles}\)

Step 3: Find the Empirical Formula

Divide the moles of each element by the smallest number of moles calculated in step 2 (1.96 moles of N):- C: \(\frac{4.89}{1.96} \approx 2.5\)- H: \(\frac{13.70}{1.96} \approx 7\)- N: \(\frac{1.96}{1.96} = 1\)Since we need whole numbers, multiply each by 2 to avoid fractions: C becomes 5, H becomes 14, and N becomes 2. Thus, the empirical formula is \(\text{C}_5\text{H}_{14}\text{N}_2\).

Step 4: Find the Molecular Formula

Calculate the molar mass of the empirical formula \(\text{C}_5\text{H}_{14}\text{N}_2\):- Molar Mass of \(\text{C}_5\text{H}_{14}\text{N}_2 = (5\times12.01) + (14\times1.008) + (2\times14.01) = 102.2 \, \text{g/mol}\).Since this matches the given molar mass, the empirical formula is also the molecular formula.

Key Concepts

Understanding the Empirical Formula

When analyzing a chemical compound, the first step is often to determine its empirical formula. The empirical formula provides the simplest whole-number ratio of atoms in a compound. For example, if a compound is made up of 58.77% carbon, 13.81% hydrogen, and 27.40% nitrogen, we start by assuming a 100 g sample of this compound. This allows us to directly translate the percentages into grams, simplifying the process of calculation. To find the empirical formula, we need to convert these masses into moles using the molar mass of each element (e.g., for carbon, use 12.01 g/mol). After calculating the moles for each element, we divide all the mole values by the smallest mole result to get the simplest ratio. Sometimes, like in our example, this ratio might not be whole numbers, and a further step of multiplying through by a whole number factor is required to achieve whole number subscripts for the formula:

• C: 4.89 moles / 1.96 = 2.5
• H: 13.70 moles / 1.96 = 7
• N: 1.96 moles / 1.96 = 1

Multiplying each by 2 gives us C: 5, H: 14, and N: 2, resulting in the empirical formula \(\text{C}_5\text{H}_{14}\text{N}_2\). This process helps us establish a foundational understanding of the compound's composition.

Molar Mass Calculation Clarified

To verify or determine a molecular formula, calculating the molar mass of a compound based on its empirical formula is crucial. The molar mass of a compound is simply the sum of the masses of all atoms present in one mole of the compound. For the empirical formula \(\text{C}_5\text{H}_{14}\text{N}_2\), calculate the molar mass as follows: add up the atomic masses (found on the periodic table) of all elements in their respective quantities:- For carbon (C), with 5 atoms, calculate 5 times the atomic mass, 12.01 g/mol.- For hydrogen (H), multiply 14 by its atomic mass, 1.008 g/mol.- For nitrogen (N), using 2 atoms, multiply 2 by 14.01 g/mol.Thus, for \(\text{C}_5\text{H}_{14}\text{N}_2\), the molar mass is \((5 \times 12.01) + (14 \times 1.008) + (2 \times 14.01) = 102.2 \text{ g/mol}\). This calculated molar mass aligns perfectly with the given molar mass of the compound, confirming that the empirical formula is indeed the molecular formula. Molar mass calculation effectively bridges the gap between empirical observations and theoretical predictions.

Chemical Compound Analysis Insights

Chemical compound analysis involves various methods and calculations to deduce the structure and composition of unknown substances. One of the primary steps in this analysis is the determination of the empirical formula, as discussed earlier. By examining the percent composition by mass, we begin to understand the relative amount of each element present in the compound. The analysis further advances into calculating the molar mass to unravel the actual molecular formula when the empirical formula matches the compound's molar mass. It allows scientists and researchers to confirm the composition and structure of the compound they are investigating. This rigorous approach aids in identifying not only the qualitative aspects of a substance but also its quantitative measures. Chemical compound analysis is essential in numerous fields, from forensic science to environmental analysis, where knowing the exact makeup of a substance can provide critical information. With techniques like percent composition analysis, molar mass calculation, and empirical formula determination, chemists can decode many mysteries of unknown substances, giving us insight into their potential uses and effects.