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?

Step by step solution

01

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: \( \frac{21.297}{35.0} \times 100 = 60.8 \% \)

02

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 \(C H_{6} N_{2}\).

03

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: \( M_\text{compound} = \frac{39.95 \times 24.6^2}{26.4^2} = 52.6 \)

04

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 \(C_{2} H_{12} N_{4}\).

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