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 \(65.2-\mathrm{mg}\) sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 119 ), giving \(35.6 \mathrm{~mL} \mathrm{~N}_{2}\) at 740 . torr and \(25^{\circ} \mathrm{C}\). iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{~mL} / \mathrm{min}\). The effusion rate of argon gas, under identical conditions, is \(26.4 \mathrm{~mL} / \mathrm{min}\). What is the molecular formula of the compound?

Short answer

The molecular formula of the unknown organic compound containing carbon, hydrogen, and nitrogen is \(C_{2} H_{12} N_{4}\).

Step-by-step solution

Calculate the mass percentage of C, H, and N in the compound.

First, use the weights of CO2 and H2O obtained from the combustion: 1. The mass of C in \(CO_2\): \( \frac{12}{44} \times 33.5 \mathrm{mg} = 9.136 \mathrm{mg} \) 2. The mass of H in \(H_2O\): \( \frac{2}{18} \times 41.1 \mathrm{mg} = 4.567 \mathrm{mg} \) Now, using the sample mass of \(35.0 \mathrm{mg}\), subtract the mass of C and H to find the mass of N: 3. Mass of N: \(35.0 - 9.136 - 4.567 = 21.297 \mathrm{mg} \) Now, calculate the mass percentages of C, H, and N: 4. Mass percentage of C: \( \frac{9.136}{35.0} \times 100 = 26.1 \% \) 5. Mass percentage of H: \( \frac{4.567}{35.0} \times 100 = 13.0 \% \) 6. Mass percentage of N: \( \frac{21.297}{35.0} \times 100 = 60.8 \% \)

Determine the empirical formula.

To find the empirical formula, first, calculate moles of C, H, and N in 100 mg of the compound: 1. Moles of C: \( \frac{26.1 \mathrm{mg}}{12.01\ \mathrm{mg\ / mmol}} = 2.169 \mathrm{mmol} \) 2. Moles of H: \( \frac{13.0 \mathrm{mg}}{1.008\ \mathrm{mg\ / mmol}} = 12.896 \mathrm{mmol} \) 3. Moles of N: \( \frac{60.8 \mathrm{mg}}{14.01\ \mathrm{mg\ / mmol}} = 4.341 \mathrm{mmol} \) Now, divide all mole values by the smallest value (C moles) to obtain the empirical formula: \( \text{C} : \frac{2.169}{2.169} = 1 \\ \text{H} : \frac{12.896}{2.169} = 5.95 \approx 6 \\ \text{N} : \frac{4.341}{2.169} = 2.00 \approx 2 \) Thus, the empirical formula is \(C H_{6} N_{2}\).

Calculate the molecular weight of the compound using Graham's law.

We are given the effusion rates of the compound and argon gas. Using Graham's law of effusion: \( \frac{r_\text{compound}}{r_\text{Ar}} = \sqrt{\frac{M_\text{Ar}}{M_\text{compound}}} \\ \frac{24.6}{26.4} = \sqrt{\frac{39.95}{M_\text{compound}}} \) Now, square both sides and solve for the molecular weight of the compound: \( M_\text{compound} = \frac{39.95 \times 24.6^2}{26.4^2} = 52.6 \)

Calculate the molecular formula of the compound.

Now, determine the ratio between the molecular weight of the compound and the empirical formula weight: \( \text{Empirical Formula Weight} = 12.01 + 6 \times 1.008 + 2 \times 14.01 = 30.08 \\ \text{Multiplication factor} = \frac{52.6}{30.08} = 1.748 \approx 2 \) Now, multiply the empirical formula by the multiplication factor to obtain the molecular formula: Molecular Formula: \(2 \times C H_{6} N_{2} = C_{2} H_{12} N_{4}\) So, the molecular formula of the unknown compound is \(C_{2} H_{12} N_{4}\).

Key concepts

Combustion Analysis

Combustion analysis is a laboratory technique used to determine the elemental composition of a compound, especially organic compounds containing carbon (C), hydrogen (H), and nitrogen (N). In this technique, the compound is completely combusted and the resulting combustion products, typically carbon dioxide (CO extsubscript{2}) and water (H extsubscript{2}O), are measured.

In the exercise, 35.0 mg of an organic compound was burned, producing 33.5 mg of CO extsubscript{2} and 41.1 mg of H extsubscript{2}O. To find the mass of carbon in the compound, use the conversion factor derived from the molar mass of carbon dioxide: \( \frac{12}{44} \) for carbon in CO extsubscript{2}. Thus, the calculation is \( \frac{12}{44} \times 33.5 \mathrm{mg} = 9.136 \mathrm{mg} \) of carbon. Similarly, to find hydrogen, use \( \frac{2}{18} \) from the molar mass of water: \( \frac{2}{18} \times 41.1 \mathrm{mg} = 4.567 \mathrm{mg} \) of hydrogen.

Subtract these masses from the total to find the nitrogen's mass. The combustion analysis also allows you to deduce the empirical formula by converting mass percentages into moles and determining the nearest whole number ratio among the elements.

Dumas Method

The Dumas method is an analytical chemistry technique used to determine the nitrogen content in a sample. This method involves converting organic nitrogen into molecular nitrogen (N extsubscript{2}) by heating the sample with an excess of a strong base, typically copper oxide. The resulting nitrogen gas is then collected and measured.

In the provided exercise, a sample weighing 65.2 mg yielded 35.6 mL of nitrogen gas under specific conditions of pressure (740 torr) and temperature (25°C). Using the ideal gas law, we can relate the volume of nitrogen collected to the amount in moles, and hence back-calculate it to its mass.

The Dumas method is especially useful for determining the empirical formula of compounds containing nitrogen. Once you find the nitrogen's contribution in terms of moles, it is easier to understand how nitrogen integrates into the compound's total composition.

Graham's Law of Effusion

Graham's Law of Effusion provides insight into the rate at which gases travel through a small opening. It states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This property allows for the comparison of effusion rates between different gases to determine unknown molecular weights.

In the problem, the effusion rate for the unknown compound was measured alongside argon under the same conditions. By applying Graham's Law, where \( \frac{r_\text{compound}}{r_\text{Ar}} = \sqrt{\frac{M_\text{Ar}}{M_\text{compound}}} \), you can solve for the molecular weight of the unknown compound: \( M_\text{compound} = \frac{39.95 \times 24.6^2}{26.4^2} = 52.6 \).

This information is pivotal because, once the molecular weight is computed, it is used alongside the empirical formula weight to determine if further scaling is needed for the molecular formula, establishing the molecular identity of the compound.