Problem 137 - Chemistry An Atoms First Approach

Question

A compound contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\). It is \(58.51 \%\) C and \(7.37 \%\) H by mass. Helium effuses through a porous frit \(3.20\) times as fast as the compound does. Determine the empirical and molecular formulas of this compound.

Step-by-Step Solution

Answer

The empirical formula of the compound is \(\mathrm{C_2H_3N}\), and the molecular formula is also \(\mathrm{C_2H_3N}\) as the molecular formula ratio is approximately 1.

Step 1: Calculate the mole ratio of C, H, and N in the compound

First, we need to find the mole ratio of C, H, and N in the compound. We will assume that we have 100 g of the compound, so the mass percentages can be used as masses in grams. Let's calculate the number of moles of each element using their atomic masses: Moles of C = (58.51 g C) / (12.01 g/mol) = 4.87 mol C Moles of H = (7.37 g H) / (1.01 g/mol) = 7.30 mol H Since the compound contains only C, H, and N (and we know the mass percentage of C and H), we can calculate the mass percentage of N and then determine the moles of N: Mass % of N = 100% - (58.51% + 7.37%) = 34.12 % Moles of N = (34.12 g N) / (14.01 g/mol) = 2.44 mol N

Step 2: Convert the mole ratios to the simplest whole numbers

Now, let's convert the mole ratios calculated above to the simplest whole numbers. We will divide each mole value by the smallest mole value: Moles of C = 4.87 mol / 2.44 ≈ 2 Moles of H = 7.30 mol / 2.44 ≈ 3 Moles of N = 2.44 mol / 2.44 ≈ 1 The simplest whole number ratio for this compound is 2:3:1. So, the empirical formula is \(\mathrm{C_2H_3N}\).

Step 3: Use Graham's Law to find the molar mass of the compound

Graham's Law of Effusion states that the rate of effusion of two different gases at the same temperature and pressure is inversely proportional to the square root of their molar masses: \( \frac{ Rate_{1} } { Rate_{2} } = \sqrt{ \frac{MM_{2}}{MM_{1}} }\) In this problem, helium (He) effuses 3.20 times as fast as the compound: \( \frac{Rate_{He}}{Rate_{Compound}} = \sqrt{ \frac{MM_{Compound}}{MM_{He}} }\) We know the molar mass of helium (He) is 4.00 g/mol: \( 3.20 = \sqrt{ \frac{MM_{Compound}}{4.00} }\) Now, let's solve for the molar mass of the compound: \(MM_{Compound} = 4.00 \times 3.20^{2} \) \(MM_{Compound} ≈ 40.96\, g/mol \)

Step 4: Calculate the molecular formula of the compound

To find the molecular formula, we need to determine the ratio of the molar mass of the compound to the molar mass of the empirical formula: Molecular Formula Ratio = \( \frac{Molar\,Mass\,of\,Compound}{Molar\,Mass\,of\,Empirical\,Formula} \) Molar mass of empirical formula (\(\mathrm{C_2H_3N}\)) = (2 × 12.01) + (3 × 1.01) + (1 × 14.01) = 30.05 g/mol Molecular Formula Ratio = \( \frac{40.96}{30.05} ≈ 1.364 \) Since the molecular formula must consist of whole-number multiples of the empirical formula, we can round this ratio to the nearest whole number: Molecular Formula Ratio ≈ 1 Therefore, the molecular formula of the compound is the same as the empirical formula, which is \(\mathrm{C_2H_3N}\).

Key Concepts

Moles Calculation

Understanding the concept of moles is crucial in chemistry as it acts as the bridge between the atomic world and the mass we observe in the laboratory. A mole represents Avogadro's number, which is approximately \(6.022 \times 10^{23}\) of whatever entities – atoms, molecules, ions, etc. In this exercise, we used moles to calculate the proportion of each element in the compound by mass. For example, by taking the given mass percentages and assuming a 100 g sample, we found the mass of carbon and hydrogen in the compound. From there, we could determine the moles for each element by dividing by their respective atomic masses:

• Carbon (\(12.01\) g/mol)
• Hydrogen (\(1.01\) g/mol)
• Nitrogen (\(14.01\) g/mol)

Calculating the moles helps us determine the empirical formula by converting mass percentages into a simple whole number ratio.

Graham's Law of Effusion

Graham's Law of Effusion provides a way to compare the rate of effusion between two gases. This law is particularly useful when you have a gaseous unknown, as the rate at which it leaks can help you deduce its molar mass. According to the law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Using Graham’s Law:\[ \frac{ Rate_{He} } { Rate_{Compound} } = \sqrt{ \frac{MM_{Compound}}{MM_{He}} } \]The problem tells us that helium effuses 3.20 times faster than our compound, meaning: \[3.20 = \sqrt{ \frac{MM_{Compound}}{4.00} } \]After rearranging and solving this equation, we can determine the molar mass of the unknown compound. This method is particularly effective for determining the properties of gases without needing extensive physical measurement.

Molecular Formula

A molecular formula reveals the actual number of each type of atom present in a molecule of the compound, making it crucial for understanding molecular structure. Once we've calculated the empirical formula, discovering the molecular formula requires determining the compound's molar mass. We compute the molecular formula ratio using:\[ \text{Molecular Formula Ratio} = \frac{\text{Molar Mass of Compound}}{\text{Molar Mass of Empirical Formula}} \]Using the empirical formula \(\text{C}_2\text{H}_3\text{N}\), its molar mass is calculated as:

• Carbon: \((2 \times 12.01)\)
• Hydrogen: \((3 \times 1.01)\)
• Nitrogen: \((1 \times 14.01)\)

Adding these yields a molar mass of \(30.05\) g/mol. With the compound's molar mass obtained from Graham's Law (\(40.96\) g/mol), we calculate:\[ \frac{40.96}{30.05} \approx 1.364 \]The ratio rounds practically to 1, indicating the molecular formula largely mirrors the empirical formula: \(\text{C}_2\text{H}_3\text{N}\). This check ensures the empirical formula perfectly aligns with the actual structure of the molecule.

Stoichiometry

Stoichiometry is the part of chemistry which uses the quantitative relationships between the reactants and products in chemical reactions. In this context, stoichiometry helps us balance equations and predict the amounts of substances consumed and produced. Even though the exercise doesn't involve chemical reactions, understanding stoichiometry allows us to convert between moles and mass, offering a deeper comprehension of the composition of compounds. Through stoichiometric relationships:

• Empirical formulas are derived from percentage composition of the elements in the compound.
• They establish the simplest whole-number atom ratios.
• Molecular formulas are then scaled versions of empirical formulas based on molar mass comparisons.

Stoichiometry thus underpins these calculations, ensuring that mass and atomic ratios are consistent across moles, helping us connect the theoretical with experimental data.