Problem 130 - Chemistry An Atoms First Approach

Question

An organic compound containing only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\) yields the following data. i. Complete combustion of \(35.0 \mathrm{mg}\) of the compound produced \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\) ii. A 65.2 -mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129 ), giving \(35.6 \mathrm{mL}\) of dry \(\mathrm{N}_{2}\) at \(740 .\) torr and \(25^{\circ} \mathrm{C}\) iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{mUmin.}\) The effusion rate of argon gas, under identical conditions, is \(26.4 \mathrm{mL} / \mathrm{min.}\) What is the molecular formula of the compound?

Step-by-Step Solution

Answer

The molecular formula of the organic compound is \(CH_{6}N\).

Step 1: Determine Empirical Formula

Using the combustion data, we will find the moles of carbon and hydrogen in the compound: Moles of Carbon: Moles of carbon in \(\mathrm{CO}_{2}\) = \(\frac{33.5 \mathrm{mg}}{44.01 \mathrm{mg/mol}}\) = \(0.761 \, \mathrm{moles}\) Moles of Hydrogen: Moles of hydrogen in \(\mathrm{H}_{2} \mathrm{O}\) = 2 * \(\frac{41.1 \mathrm{mg}}{18.015 \mathrm{mg/mol}}\) = \(4.56 \, \mathrm{moles}\) The ratio of moles of Carbon to Hydrogen is \(C:H \approx 1:6\). So, the empirical formula is \(\mathrm{CH}_{6}N\).

Step 2: Calculate Mass of Nitrogen in the Compound

Using the Dumas method data, we find the mass of nitrogen: n = \(\frac{PV}{RT}\) \(P = 740 \mathrm{torr} * \frac{101.325 \mathrm{Pa}}{760 \mathrm{torr}} = 100,000 \mathrm{Pa}\) \(V = 35.6 \mathrm{mL} * \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.0356 \mathrm{L}\) \(R = 8.314 \mathrm{J \, mol^{-1} \, K^{-1}} * \frac{1 \mathrm{L \, atm}}{0.0821 \mathrm{J \, K^{-1} \, mol^{-1}}}=0.0821 \mathrm{L \, atm \, mol^{-1} \, K^{-1}}\) \(T = 25^{\circ}\mathrm{C} + 273.15 = 298.15 \mathrm{K}\) Moles of Nitrogen = \(n = \frac{PV}{RT} = \frac{100,000 \mathrm{Pa} * 0.0356 \mathrm{L}}{0.0821 \mathrm{L \, atm \, mol^{-1} \, K^{-1}} * 298.15 \mathrm{K}} = 1.53 \, \mathrm{moles}\) Mass of Nitrogen = \(1.53 \, \mathrm{moles} * 14.007 \mathrm{mg/mol} = 21.4 \mathrm{mg}\)

Step 3: Calculate Molar Mass of the Compound

Using the effusion rate comparison: Effusion rate of \(\mathrm{C} \mathrm{H}_{6} \mathrm{N}\) = \(0.246 \, \mathrm{L} / \mathrm{min}\) Effusion rate of \(\mathrm{Ar}\) = \(0.264 \, \mathrm{L} / \mathrm{min}\) Ratio = \(\frac{\text{Effusion rate of } \mathrm{C} \mathrm{H}_{6} \mathrm{N}}{\text{Effusion rate of } \mathrm{Ar}} = \frac{0.246 \, \mathrm{L} / \mathrm{min}}{0.264 \, \mathrm{L} / \mathrm{min}} = 0.932\) Using Graham's law of effusion: \(\frac{\sqrt{M_{\mathrm{CH}_{6} \mathrm{N}}}}{\sqrt{M_{\mathrm{Ar}}}} = 0.932\) Solve for \(M_{\mathrm{CH}_{6} \mathrm{N}}\): \(M_{\mathrm{CH}_{6} \mathrm{N}} = (0.932)^{2} * M_{\mathrm{Ar}} = (0.932)^{2} * 39.95 \mathrm{g/mol} = 34.43 \, \mathrm{g/mol}\)

Step 4: Determine the Molecular Formula

Using the empirical formula and the molar mass of the compound: Empirical formula mass of \(\mathrm{CH}_{6}\mathrm{N}\) = \(1.01 \mathrm{g/mol} + 6 * 1.008 \mathrm{g/mol} + 14.01 \mathrm{g/mol} = 28.05 \mathrm{g/mol}\) Molecular formula factor = \(\frac{34.43 \, \mathrm{g/mol}}{28.05 \, \mathrm{g/mol}} \approx 1.23 \approx 1\) Since the molecular formula factor is close to 1, the molecular formula of the compound is the same as the empirical formula. The molecular formula of the organic compound is \(\boxed{\mathrm{CH}_{6} \mathrm{N}}\).

Key Concepts

Combustion Analysis

Combustion analysis is a chemical engineering technique used to determine the elemental composition of a compound by fully burning it and analyzing the products. For a compound containing carbon (C), hydrogen (H), and nitrogen (N), a known mass is combusted, producing carbon dioxide (CO2) and water (H2O), which are then used to calculate the amounts of C and H.

By measuring the mass of CO2 and H2O produced, the moles of carbon and hydrogen are determined. Since each molecule of CO2 contains one carbon atom and each molecule of H2O contains two hydrogen atoms, the ratio of their moles provides the stoichiometric coefficients for C and H in the empirical formula. However, this method doesn't directly provide the quantity of nitrogen and thus other techniques, like the Dumas method, are needed to determine the nitrogen content.

Dumas Method

The Dumas method is a classical procedure for determining the molar mass of a volatile substance. In this method, the substance is vaporized and its volume, temperature, and pressure are measured. Applying the ideal gas law (\( PV = nRT \)), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature, the molar mass can be deduced.

Calculating Nitrogen Content with the Dumas Method

In the Dumas method for nitrogen analysis, the nitrogen gas produced from a sample is collected and its volume recorded at a known temperature and pressure. The volume of nitrogen gas is then used to calculate the amount of nitrogen present in the original compound, providing the mass of nitrogen that contributes to the complete molar mass of the unknown compound.

Graham's Law of Effusion

Graham's law of effusion describes the rate at which a gas will effuse through a tiny orifice or permeable membrane into a vacuum. It states that the rate of effusion is inversely proportional to the square root of the gas's molar mass. Mathematically, it can be expressed as \( \text{Rate}_1 / \text{Rate}_2 = \sqrt{M_2 / M_1} \), where \( Rate_1 \) and \( Rate_2 \) are the effusion rates of two gases, and \( M_1 \) and \( M_2 \) are their respective molar masses.

By comparing the effusion rate of a gas of unknown molar mass to that of a reference gas with a known molar mass, the molar mass of the unknown gas can be determined. This application of Graham's law is crucial in our exercise, which involves comparing the effusion rates of the unknown organic compound and argon gas to calculate the molar mass of the organic compound.

Empirical Formula Calculation

The empirical formula of a compound represents the simplest whole number ratio of the elements within it. To calculate the empirical formula, the mass or percentage composition of each element must be known or determined through experimental methods. Moles of each element are computed from the mass, and then the simplest whole-number mole ratio is established.

In the context of our problem, the empirical formula was calculated using the moles of carbon, hydrogen, and nitrogen derived from combustion analysis and the Dumas method. By comparing the moles of these elements, we achieve the simplest stoichiometric ratio, which corresponds to the empirical formula. However, note that the empirical formula may not reflect the actual number of atoms in the molecule if the molecular mass is an integer multiple of the empirical formula mass; the latter scenario necessitates additional steps to identify the exact molecular formula.