Problem 91 - Chemistry and Chemical Reactivity

Question

An organic compound contains carbon \((71.17 \%)\) hydrogen \((5.12 \%)\) with the remainder nitrogen. Dissolving 0.177 g of the compound in \(10.0 \mathrm{g}\) of benzene gives a solution with a vapor pressure of \(94.16 \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C} .\) (The vapor pressure of pure benzene at this temperature is \(95.26 \mathrm{mm}\) Hg. What is the molecular formula for the compound?

Step-by-Step Solution

Answer

The molecular formula is C\(_7\)H\(_6\)N\(_2\).

Step 1: Determine Percent Composition of Nitrogen

The compound contains carbon and hydrogen percentages, so we calculate the percentage of nitrogen. Since the total percentage is 100%, the percentage of nitrogen is given by: \[ 100 ext{%} - (71.17 ext{%} + 5.12 ext{%}) = 23.71 ext{%} \] Thus, the compound is composed of 71.17% carbon, 5.12% hydrogen, and 23.71% nitrogen.

Step 2: Find Empirical Formula

Assume a 100 g sample of the compound for easy calculation. Then, convert the percentages to grams: - Carbon: 71.17 g- Hydrogen: 5.12 g- Nitrogen: 23.71 gConvert grams to moles by dividing by atomic masses:- Moles of C: \( \frac{71.17}{12.01} = 5.93 \text{ moles} \)- Moles of H: \( \frac{5.12}{1.008} = 5.08 \text{ moles} \)- Moles of N: \( \frac{23.71}{14.01} = 1.69 \text{ moles} \)Divide by the smallest number of moles to obtain the simplest ratio:- C: \( \frac{5.93}{1.69} = 3.51 \approx 3.5 \)- H: \( \frac{5.08}{1.69} = 3.01 \approx 3 \)- N: \( \frac{1.69}{1.69} = 1 \)Thus, the empirical formula is approximately C\(_{3.5}\)H\(_{3}\)N.

Step 3: Adjust Empirical Formula

Since empirical formulas must have whole numbers, multiply each subscript by 2 (to make the subscript for carbon a whole number):- Carbon: C\(_{7}\)- Hydrogen: H\(_{6}\)- Nitrogen: N\(_{2}\)The adjusted empirical formula is C\(_7\)H\(_6\)N\(_2\).

Step 4: Calculate Mole Fraction of Benzene

Use the vapor pressure difference to find the mole fraction of benzene, using Raoult's Law:\[ P_{solvent} = P^0_{solvent} \times X_{solvent} \]where \(P_{solvent}\) is 94.16 mm Hg and \(P^0_{solvent}\) is 95.26 mm Hg.Solve for \(X_{benzene}\):\[ X_{benzene} = \frac{94.16}{95.26} \approx 0.988 \].This means the mole fraction of benzene in the solution is 0.988.

Step 5: Calculate Moles of Compound

First, find moles of benzene: \( \frac{10.0 \text{ g}}{78.11 \text{ g/mol}} = 0.128 \text{ mol} \).Use the mole fraction to solve for the total moles in solution:\[ X_{benzene} = \frac{0.128}{0.128 + n_{compound}} = 0.988 \]Solve for \(n_{compound}\):\( n_{compound} = \frac{0.128 - (0.988 \times 0.128)}{0.988} \approx 0.0016 \text{ mol} \).

Step 6: Find Molar Mass of Compound

Given 0.177 g of compound and 0.0016 mol, calculate the molar mass:\[ \text{Molar mass} = \frac{0.177}{0.0016} \approx 110.63 \text{ g/mol} \].

Step 7: Determine Molecular Formula

The empirical formula mass of C\(_7\)H\(_6\)N\(_2\) is approximately 110.72 g/mol, which matches the calculated molar mass of 110.63 g/mol. Hence, the molecular formula is the same as the empirical formula: C\(_7\)H\(_6\)N\(_2\).

Key Concepts

Percent Composition in Chemistry

Percent composition in chemistry helps us understand the make-up of compounds by revealing the percentage of each element relative to the entire formulation. It's crucial in determining empirical formulas and guiding molecular mass calculations.

In the context of the given exercise, the compound consists of carbon, hydrogen, and nitrogen. To find the percent composition, we first subtract the known percentages from 100%, calculating the missing percentage for nitrogen. This step is essential to fully profile the compound's elemental structure.

Here's how it's calculated:

• The percent carbon is 71.17%.
• The percent hydrogen is 5.12%.
• The percent nitrogen is determined by subtracting the given percentages of carbon and hydrogen from 100%, which yields 23.71% nitrogen.

Understanding percent composition allows chemists to break down complex mixtures into individual elemental contributions efficiently.

Empirical Formula Calculation

Empirical formulas represent the simplest whole-number ratio of atoms in a compound, providing a foundation for calculating the more complex molecular formulas.

In the exercise solution, we proceed to determine the empirical formula of the compound using percent composition. Assuming a 100 g sample simplifies the conversion from percentages to grams, making it easier to ultimately convert to moles:

• The carbon content is 71.17 g, hydrogen is 5.12 g, and nitrogen is 23.71 g.

Next, each mass is divided by its respective atomic mass:

• Moles of Carbon: \( \frac{71.17}{12.01} = 5.93 \) moles
• Moles of Hydrogen: \( \frac{5.12}{1.008} = 5.08 \) moles
• Moles of Nitrogen: \( \frac{23.71}{14.01} = 1.69 \) moles

To find the empirical formula, we divide by the smallest mole value (1.69), and adjust to the nearest whole number. Thus, the empirical formula becomes \( \text{C}_3.5\text{H}_3\text{N} \). By multiplying the coefficients by 2, we adjust to whole numbers, resulting in \( \text{C}_7\text{H}_6\text{N}_2 \). This method accurately simplifies the elemental ratio, guiding the determination of the compound's molecular structure.

Raoult's Law in Chemistry

Raoult's Law in chemistry describes how the vapor pressure of a solution differs from that of a pure solvent. It's instrumental in calculating mole fractions and is essential in mixture analysis.

In the original exercise, Raoult's Law is a pivotal tool. It allows us to relate the vapor pressure of the benzene solution to the mole fraction present. Here's a breakdown of its application:

• The true vapor pressure of the solution is noted as 94.16 mm Hg.
• The pure benzene's vapor pressure is known: 95.26 mm Hg.

The mole fraction \( (X_{benzene}) \) is calculated using the formula:\[ X_{benzene} = \frac{P_{solvent}}{P^0_{solvent}} = \frac{94.16}{95.26} \approx 0.988 \].

This mole fraction tells us the presence of benzene in the solution, indicating most of the solution is benzene, with a small fraction of solute. By understanding Raoult's Law, we can explore how a solute influences solvent behaviors, highlighting its importance in chemistry for mixture and solution behavior analysis.