Problem 128 - Chemistry The Molecular Science

Question

In chemical research, newly synthesized compounds are often sent to commercial laboratories for analysis that determines the weight percent of \(\mathrm{C}\) and \(\mathrm{H}\) by burning the compound and collecting the evolved \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ( Sec. \(3-9\) ). The molar mass is determined by measuring the osmotic pressure of a solution of the compound. Calculate the empirical and molecular formulas of a compound, \(\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{Cr}\), given this information: (a) The compound contains \(73.94 \% \mathrm{C}\) and \(8.27 \% \mathrm{H} ;\) the remainder is chromium. (b) At \(25^{\circ} \mathrm{C}\), the osmotic pressure of \(5.00 \mathrm{mg}\) of the unknown dissolved in \(100 . \mathrm{mL}\) of chloroform solution is \(3.17 \mathrm{mmHg}\)

Step-by-Step Solution

Answer

Empirical formula: \( \mathrm{C}_{18} \mathrm{H}_{24} \mathrm{Cr} \); Molecular formula: \( \mathrm{C}_{36} \mathrm{H}_{48} \mathrm{Cr}_2 \).\)

Step 1: Calculate Weight Percent of Chromium

Given the compound has a total weight percent of 100%, we start by calculating the weight percent of Chromium (Cr). Since the weight percent of C is 73.94% and H is 8.27%, we find the weight percent of Cr as follows:\[\text{wt. \, \, \, \% Cr} = 100\% - 73.94\% - 8.27\% = 17.79\%\]

Step 2: Convert into Moles (Empirical Formula)

Convert the percentages into moles using the atomic masses of C (12.011 g/mol), H (1.008 g/mol), and Cr (51.996 g/mol):1. Moles of C: \[\frac{73.94}{12.011} = 6.16 \text{ mol C per 100 g compound}\]2. Moles of H: \[\frac{8.27}{1.008} = 8.20 \text{ mol H per 100 g compound}\]3. Moles of Cr: \[\frac{17.79}{51.996} = 0.342 \text{ mol Cr per 100 g compound}\]

Step 3: Determine Empirical Formula

Divide each mole value by the smallest number of moles calculated to convert these mole ratios to the smallest whole numbers.1. Ratio of C: \[\frac{6.16}{0.342} = 18.01 \approx 18\]2. Ratio of H: \[\frac{8.20}{0.342} = 23.98 \approx 24\]3. Ratio of Cr: \[\frac{0.342}{0.342} = 1\]Thus, the empirical formula is \( \mathrm{C}_{18} \mathrm{H}_{24} \mathrm{Cr} \).

Step 4: Calculate Molar Mass from Osmotic Pressure

First, convert the pressure from mmHg to atm: \[3.17 \, \text{mmHg} = \frac{3.17}{760} \, \text{atm} = 0.004171 \, \text{atm}\]Use the osmotic pressure formula to find the molar mass \(M\):\[\Pi = \frac{nRT}{V}\]where \(n = \frac{m}{M}\), \(R = 0.0821 \, L\, atm\, mol^{-1} \, K^{-1}\), and \(T = 273 + 25 = 298 \, K\). Solving for \(M\):\[M = \frac{mRT}{\Pi V} = \frac{5 \times 10^{-3} \, g \times 0.0821 \, \times 298}{0.004171 \, \times 0.1} = 736.56 \, g/mol\]

Step 5: Calculate Molecular Formula

Compare the empirical formula mass with the calculated molar mass.Calculate empirical mass of \( \mathrm{C}_{18} \mathrm{H}_{24} \mathrm{Cr} \):\[18 \times 12.011 + 24 \times 1.008 + 1 \times 51.996 = 306.42 \, g/mol\]Find the ratio:\[\text{Ratio} = \frac{736.56}{306.42} \approx 2.4\]Since we need whole numbers for molecules, we approximate this to 2.Thus, the molecular formula is \( \mathrm{C}_{36} \mathrm{H}_{48} \mathrm{Cr}_2 \).

Key Concepts

Osmotic Pressure

Osmotic pressure is a crucial concept in chemistry, often used to determine the molar mass of a compound. It is the pressure required to prevent the flow of a solvent into a solution through a semipermeable membrane. This pressure is directly related to the concentration of solute particles in the solution.

Here's how you calculate it:

• The osmotic pressure (\( \Pi \)) can be calculated using the formula: \( \Pi = \frac{nRT}{V} \), where \( n \) is the number of moles of solute, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( V \) is the volume of the solution in liters.
• In practical terms, for a given system, you can rearrange this to find the molar mass of the solute by incorporating the mass of the solute, converting it into moles.
• From the osmotic pressure formula, replace \( n = \frac{m}{M} \) to yield \( M = \frac{mRT}{\Pi V} \).

In our example, to find the molar mass given the osmotic pressure and other known variables, you would plug in the values as described in Step 4. This helps link the physical properties of solutions with molecular masses, making it a powerful tool in chemical analysis.

Chemical Analysis

Chemical analysis is the process of identifying the composition of a substance. For organic compounds, analytical techniques often involve combustion analysis, where the compound is burned to convert its elements into known oxides, such as \( CO_2 \) and \( H_2O \).

Here's a typical process used in our scenario:

• The compound is combusted, typically in an excess environment to ensure complete conversion of carbon to \( CO_2 \) and hydrogen to \( H_2O \).
• The quantities of \( CO_2 \) and \( H_2O \) are accurately measured, allowing for the calculation of the amount of carbon and hydrogen in the sample.
• Given the data from such analyses, alongside any metal content like chromium that remains as a residual product, percentage compositions are computed.
• Weight proportions and further conversion of these into mole ratios provide preliminary empirical formulas.

Understanding chemical analysis is pivotal, as it provides the foundational data from which empirical and molecular formulas are derived in the example problem.

Empirical Formula Calculation

Empirical formula calculations involve determining the simplest integer ratios of elements in a compound. These calculations begin with the mass percentages of elements, found through chemical analysis, and proceed by converting these percentages to moles.

Steps to calculate an empirical formula include:

• Start with the percentage composition of each element.
• Convert these percentages into grams (assuming a certain total mass, usually 100 grams).
• Use the atomic masses of the elements to convert these grams to moles for each element.
• Calculate the simplest mole ratio by dividing by the smallest number of moles present.
• For our example, dividing each element's moles by the smallest mole value, ensures finding the lowest whole number ratio, resulting in the empirical formula \( C_{18}H_{24}Cr \).

This method highlights the empirical formula's role as a stepping stone in determining the molecular formula, essential in substantiating compositions derived through different experimental methods like osmotic pressure.